Equivalence condition of Absolute ContinuityWhy doesn't induction extend to infinity? (re: Fourier series)Why is the Cantor function not absolutely continuous?On Lipschitz condition and absolute continuityContinuous function with finitely many discontinuities is Riemann IntegralHow to prove function $f(x,y)=frac1xy$ is not uniformly continuous?Question About the Proof of Absolute Continuity of the Total Variation FunctionProving that $lim_kto inftyfracf(x_k)-f(y_k)x_k-y_k=f'(c)$ where $lim_kto inftyx_k=lim_kto inftyy_k=c$ with $x_k<c<y_k$Is $sqrtx, xin [0,1]$ absolutely continuous?Prove a multilinear map is differentiableHow to prove $f(x,y)=sin(xy)$ is uniformly continuous in $mathbbR^2$?Absolute continuity of increasing functions on an intervalWhy doesn't separate continuity imply continuity?
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Equivalence condition of Absolute Continuity
Why doesn't induction extend to infinity? (re: Fourier series)Why is the Cantor function not absolutely continuous?On Lipschitz condition and absolute continuityContinuous function with finitely many discontinuities is Riemann IntegralHow to prove function $f(x,y)=frac1xy$ is not uniformly continuous?Question About the Proof of Absolute Continuity of the Total Variation FunctionProving that $lim_kto inftyfracf(x_k)-f(y_k)x_k-y_k=f'(c)$ where $lim_kto inftyx_k=lim_kto inftyy_k=c$ with $x_k<c<y_k$Is $sqrtx, xin [0,1]$ absolutely continuous?Prove a multilinear map is differentiableHow to prove $f(x,y)=sin(xy)$ is uniformly continuous in $mathbbR^2$?Absolute continuity of increasing functions on an intervalWhy doesn't separate continuity imply continuity?
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Let me restate my point. The intuition behind this construction is straightforward. Let $f$ be a continuous increasing function. Take out pieces of the graph of $f$ which is correspond to a collection of finite disjoint subintervals, then glue them together to make a new function $g$. If we claim every $g$ must be continuous, then by the definition of continuity $forall deltaexistsepsilon$ such that “total length of the intervals $<epsilon$” implies “total variation of $f$ on these intervals<$delta$” which means $f$ is absolute continuous. This is not so sensible, so I hypotheze that $g$ is not necessarily continuous even if $f$ is.
$f:Itomathbb R$ is continuous. Take finite number of subintervals $[x_1,y_1]cup [x_2,y_2],...,cup[x_n,y_n]=mathcal Ksubseteq I$, where $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$.
Let's consider the the part of graph of $f$ constrained in $mathcal K$. Rigorous definition follows.
Define function $g:[x_1,x_1+K]tomathbb R$ such that, if $xin[x_1,y_1],$ then $g(x)=f(x)$, if $xin[y_1,y_1+y_2-x_2]$, then $g(x)=f(x-y_1+x_2)+c_1$, where $c_2=f(y_1)-f(x_2)$. We thus define $x_1'=x_1$, $y_1'=y_1$, $x_2'=y_1$, $y_2'=x_2'+y_2-x_2$, and, recursively, $x_k'=y_k-1', y_k'=x_k'+y_k-x_k$.
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
We call $g$ a clipping of $f$ at $mathcal K$:
$g=clip(f,mathcal K)$. Intuitively, $g(x)$ is exactly equivalent to $f(x)$ on each interval up to the equivalent class of affine transformation.
Consider the following condition:
Condition 1: $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Is this condition a necessary and sufficient condition for the absolute continuity of $f$ on $I$? Can it be extended to multidimensional functions?
Is the morphism "clipping" well studied?
This is showing that $g$ may not continuous even if $f$ is continuous. $g$ is not continuous at $x_1$ means: $existsdeltaforallepsilonexists y$ such that $|x_1-y|<epsilon$ but $|f(x_1)-f(y)|=delta$
Let $f$ the Cantor function. Let $n$ be a natural number. At the $n$-th stage of the construction of the Cantor set, a disjoint collection $[x_k, y_k]_1<k<2^n $of $2^n$ subintervals of $I=[0, 1]$ have been constructed that cover the Cantor set, each of which has length $(1/3)^n$. The Cantor-Lebesgue function is constant on each of the intervals that comprise the complement in $[0, 1]$ of this collection of intervals.
We have $sum_kleq 2^n(y_k-x_k)=K<(2/3)^n$ while $sum_kleq 2^n(f(y_k)-f(x_k))=1$
$forall epsilon>0$ $exists n>0$ such that $K<epsilon$.
By definition, $K=y_k'-x_k$, and $sum_kleq 2^n(f(y_k)-f(x_k))=g(y'_k)-g(x_k)$.
That is: $existsdelta=1forallepsilonexists y=y_k'$ such that $|x_1-y'_k|<epsilon$ but $|g(x_1)-g(y_k')|=delta$. $g$ is not continuous.
It is well known that, in the definition of absolute continuity, the word "finite" can be replaced by "countably infinite":
A function $f: I to mathbbR$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies
$$ sum_k |y_k - x_k| < delta$$
then
$$sum_k |f(y_k) - f(x_k)| < epsilon$$
We know that the square wave can be written as the infinite sum of forms of sine functions: $lim_ntoinftysum_i=1^n h_i(x)$. Each $h_i$ is continuous but the limit is not. Let $H_n=sum_i=1^n h_i(x)$. $H_n$ should be continuous by induction. If $H_n$ is indeed continuous at $x=0$ then $forall deltaexistsepsilonforall yinmathbb R$ such that $|y-x|<epsilon$ implies $|f(y)-f(x)|<delta$. However, $forall epsilon>0$ $exists n>0$ such that $|y-x|<epsilon$ and $f(y)-f(0)=1$, contradition!
To Ramiro:
Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_n'$ such that $|y_n'-x_1|<epsilon$ implies $|g(y_n')-g(x_1)|<delta$; we know $K=sum_ileq n |y_i-x_i|,$ and $ sum_ileq n |f(y_i)-f(x_i)|=|g(y_n')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum_ileq n |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_n')-g_mathcal K(x_1)|<delta$,
which implies $sum_ileq n |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!
I am not sure that a single $g$ is required.
real-analysis calculus analysis continuity absolute-continuity
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show 4 more comments
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Let me restate my point. The intuition behind this construction is straightforward. Let $f$ be a continuous increasing function. Take out pieces of the graph of $f$ which is correspond to a collection of finite disjoint subintervals, then glue them together to make a new function $g$. If we claim every $g$ must be continuous, then by the definition of continuity $forall deltaexistsepsilon$ such that “total length of the intervals $<epsilon$” implies “total variation of $f$ on these intervals<$delta$” which means $f$ is absolute continuous. This is not so sensible, so I hypotheze that $g$ is not necessarily continuous even if $f$ is.
$f:Itomathbb R$ is continuous. Take finite number of subintervals $[x_1,y_1]cup [x_2,y_2],...,cup[x_n,y_n]=mathcal Ksubseteq I$, where $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$.
Let's consider the the part of graph of $f$ constrained in $mathcal K$. Rigorous definition follows.
Define function $g:[x_1,x_1+K]tomathbb R$ such that, if $xin[x_1,y_1],$ then $g(x)=f(x)$, if $xin[y_1,y_1+y_2-x_2]$, then $g(x)=f(x-y_1+x_2)+c_1$, where $c_2=f(y_1)-f(x_2)$. We thus define $x_1'=x_1$, $y_1'=y_1$, $x_2'=y_1$, $y_2'=x_2'+y_2-x_2$, and, recursively, $x_k'=y_k-1', y_k'=x_k'+y_k-x_k$.
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
We call $g$ a clipping of $f$ at $mathcal K$:
$g=clip(f,mathcal K)$. Intuitively, $g(x)$ is exactly equivalent to $f(x)$ on each interval up to the equivalent class of affine transformation.
Consider the following condition:
Condition 1: $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Is this condition a necessary and sufficient condition for the absolute continuity of $f$ on $I$? Can it be extended to multidimensional functions?
Is the morphism "clipping" well studied?
This is showing that $g$ may not continuous even if $f$ is continuous. $g$ is not continuous at $x_1$ means: $existsdeltaforallepsilonexists y$ such that $|x_1-y|<epsilon$ but $|f(x_1)-f(y)|=delta$
Let $f$ the Cantor function. Let $n$ be a natural number. At the $n$-th stage of the construction of the Cantor set, a disjoint collection $[x_k, y_k]_1<k<2^n $of $2^n$ subintervals of $I=[0, 1]$ have been constructed that cover the Cantor set, each of which has length $(1/3)^n$. The Cantor-Lebesgue function is constant on each of the intervals that comprise the complement in $[0, 1]$ of this collection of intervals.
We have $sum_kleq 2^n(y_k-x_k)=K<(2/3)^n$ while $sum_kleq 2^n(f(y_k)-f(x_k))=1$
$forall epsilon>0$ $exists n>0$ such that $K<epsilon$.
By definition, $K=y_k'-x_k$, and $sum_kleq 2^n(f(y_k)-f(x_k))=g(y'_k)-g(x_k)$.
That is: $existsdelta=1forallepsilonexists y=y_k'$ such that $|x_1-y'_k|<epsilon$ but $|g(x_1)-g(y_k')|=delta$. $g$ is not continuous.
It is well known that, in the definition of absolute continuity, the word "finite" can be replaced by "countably infinite":
A function $f: I to mathbbR$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies
$$ sum_k |y_k - x_k| < delta$$
then
$$sum_k |f(y_k) - f(x_k)| < epsilon$$
We know that the square wave can be written as the infinite sum of forms of sine functions: $lim_ntoinftysum_i=1^n h_i(x)$. Each $h_i$ is continuous but the limit is not. Let $H_n=sum_i=1^n h_i(x)$. $H_n$ should be continuous by induction. If $H_n$ is indeed continuous at $x=0$ then $forall deltaexistsepsilonforall yinmathbb R$ such that $|y-x|<epsilon$ implies $|f(y)-f(x)|<delta$. However, $forall epsilon>0$ $exists n>0$ such that $|y-x|<epsilon$ and $f(y)-f(0)=1$, contradition!
To Ramiro:
Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_n'$ such that $|y_n'-x_1|<epsilon$ implies $|g(y_n')-g(x_1)|<delta$; we know $K=sum_ileq n |y_i-x_i|,$ and $ sum_ileq n |f(y_i)-f(x_i)|=|g(y_n')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum_ileq n |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_n')-g_mathcal K(x_1)|<delta$,
which implies $sum_ileq n |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!
I am not sure that a single $g$ is required.
real-analysis calculus analysis continuity absolute-continuity
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1
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I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
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– Ramiro
Mar 22 at 11:40
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@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
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– High GPA
Mar 22 at 19:11
1
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If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
1
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By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
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– Ramiro
Mar 22 at 22:43
1
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Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
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– Ramiro
Mar 22 at 23:20
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show 4 more comments
$begingroup$
Let me restate my point. The intuition behind this construction is straightforward. Let $f$ be a continuous increasing function. Take out pieces of the graph of $f$ which is correspond to a collection of finite disjoint subintervals, then glue them together to make a new function $g$. If we claim every $g$ must be continuous, then by the definition of continuity $forall deltaexistsepsilon$ such that “total length of the intervals $<epsilon$” implies “total variation of $f$ on these intervals<$delta$” which means $f$ is absolute continuous. This is not so sensible, so I hypotheze that $g$ is not necessarily continuous even if $f$ is.
$f:Itomathbb R$ is continuous. Take finite number of subintervals $[x_1,y_1]cup [x_2,y_2],...,cup[x_n,y_n]=mathcal Ksubseteq I$, where $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$.
Let's consider the the part of graph of $f$ constrained in $mathcal K$. Rigorous definition follows.
Define function $g:[x_1,x_1+K]tomathbb R$ such that, if $xin[x_1,y_1],$ then $g(x)=f(x)$, if $xin[y_1,y_1+y_2-x_2]$, then $g(x)=f(x-y_1+x_2)+c_1$, where $c_2=f(y_1)-f(x_2)$. We thus define $x_1'=x_1$, $y_1'=y_1$, $x_2'=y_1$, $y_2'=x_2'+y_2-x_2$, and, recursively, $x_k'=y_k-1', y_k'=x_k'+y_k-x_k$.
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
We call $g$ a clipping of $f$ at $mathcal K$:
$g=clip(f,mathcal K)$. Intuitively, $g(x)$ is exactly equivalent to $f(x)$ on each interval up to the equivalent class of affine transformation.
Consider the following condition:
Condition 1: $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Is this condition a necessary and sufficient condition for the absolute continuity of $f$ on $I$? Can it be extended to multidimensional functions?
Is the morphism "clipping" well studied?
This is showing that $g$ may not continuous even if $f$ is continuous. $g$ is not continuous at $x_1$ means: $existsdeltaforallepsilonexists y$ such that $|x_1-y|<epsilon$ but $|f(x_1)-f(y)|=delta$
Let $f$ the Cantor function. Let $n$ be a natural number. At the $n$-th stage of the construction of the Cantor set, a disjoint collection $[x_k, y_k]_1<k<2^n $of $2^n$ subintervals of $I=[0, 1]$ have been constructed that cover the Cantor set, each of which has length $(1/3)^n$. The Cantor-Lebesgue function is constant on each of the intervals that comprise the complement in $[0, 1]$ of this collection of intervals.
We have $sum_kleq 2^n(y_k-x_k)=K<(2/3)^n$ while $sum_kleq 2^n(f(y_k)-f(x_k))=1$
$forall epsilon>0$ $exists n>0$ such that $K<epsilon$.
By definition, $K=y_k'-x_k$, and $sum_kleq 2^n(f(y_k)-f(x_k))=g(y'_k)-g(x_k)$.
That is: $existsdelta=1forallepsilonexists y=y_k'$ such that $|x_1-y'_k|<epsilon$ but $|g(x_1)-g(y_k')|=delta$. $g$ is not continuous.
It is well known that, in the definition of absolute continuity, the word "finite" can be replaced by "countably infinite":
A function $f: I to mathbbR$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies
$$ sum_k |y_k - x_k| < delta$$
then
$$sum_k |f(y_k) - f(x_k)| < epsilon$$
We know that the square wave can be written as the infinite sum of forms of sine functions: $lim_ntoinftysum_i=1^n h_i(x)$. Each $h_i$ is continuous but the limit is not. Let $H_n=sum_i=1^n h_i(x)$. $H_n$ should be continuous by induction. If $H_n$ is indeed continuous at $x=0$ then $forall deltaexistsepsilonforall yinmathbb R$ such that $|y-x|<epsilon$ implies $|f(y)-f(x)|<delta$. However, $forall epsilon>0$ $exists n>0$ such that $|y-x|<epsilon$ and $f(y)-f(0)=1$, contradition!
To Ramiro:
Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_n'$ such that $|y_n'-x_1|<epsilon$ implies $|g(y_n')-g(x_1)|<delta$; we know $K=sum_ileq n |y_i-x_i|,$ and $ sum_ileq n |f(y_i)-f(x_i)|=|g(y_n')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum_ileq n |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_n')-g_mathcal K(x_1)|<delta$,
which implies $sum_ileq n |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!
I am not sure that a single $g$ is required.
real-analysis calculus analysis continuity absolute-continuity
$endgroup$
Let me restate my point. The intuition behind this construction is straightforward. Let $f$ be a continuous increasing function. Take out pieces of the graph of $f$ which is correspond to a collection of finite disjoint subintervals, then glue them together to make a new function $g$. If we claim every $g$ must be continuous, then by the definition of continuity $forall deltaexistsepsilon$ such that “total length of the intervals $<epsilon$” implies “total variation of $f$ on these intervals<$delta$” which means $f$ is absolute continuous. This is not so sensible, so I hypotheze that $g$ is not necessarily continuous even if $f$ is.
$f:Itomathbb R$ is continuous. Take finite number of subintervals $[x_1,y_1]cup [x_2,y_2],...,cup[x_n,y_n]=mathcal Ksubseteq I$, where $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$.
Let's consider the the part of graph of $f$ constrained in $mathcal K$. Rigorous definition follows.
Define function $g:[x_1,x_1+K]tomathbb R$ such that, if $xin[x_1,y_1],$ then $g(x)=f(x)$, if $xin[y_1,y_1+y_2-x_2]$, then $g(x)=f(x-y_1+x_2)+c_1$, where $c_2=f(y_1)-f(x_2)$. We thus define $x_1'=x_1$, $y_1'=y_1$, $x_2'=y_1$, $y_2'=x_2'+y_2-x_2$, and, recursively, $x_k'=y_k-1', y_k'=x_k'+y_k-x_k$.
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
We call $g$ a clipping of $f$ at $mathcal K$:
$g=clip(f,mathcal K)$. Intuitively, $g(x)$ is exactly equivalent to $f(x)$ on each interval up to the equivalent class of affine transformation.
Consider the following condition:
Condition 1: $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Is this condition a necessary and sufficient condition for the absolute continuity of $f$ on $I$? Can it be extended to multidimensional functions?
Is the morphism "clipping" well studied?
This is showing that $g$ may not continuous even if $f$ is continuous. $g$ is not continuous at $x_1$ means: $existsdeltaforallepsilonexists y$ such that $|x_1-y|<epsilon$ but $|f(x_1)-f(y)|=delta$
Let $f$ the Cantor function. Let $n$ be a natural number. At the $n$-th stage of the construction of the Cantor set, a disjoint collection $[x_k, y_k]_1<k<2^n $of $2^n$ subintervals of $I=[0, 1]$ have been constructed that cover the Cantor set, each of which has length $(1/3)^n$. The Cantor-Lebesgue function is constant on each of the intervals that comprise the complement in $[0, 1]$ of this collection of intervals.
We have $sum_kleq 2^n(y_k-x_k)=K<(2/3)^n$ while $sum_kleq 2^n(f(y_k)-f(x_k))=1$
$forall epsilon>0$ $exists n>0$ such that $K<epsilon$.
By definition, $K=y_k'-x_k$, and $sum_kleq 2^n(f(y_k)-f(x_k))=g(y'_k)-g(x_k)$.
That is: $existsdelta=1forallepsilonexists y=y_k'$ such that $|x_1-y'_k|<epsilon$ but $|g(x_1)-g(y_k')|=delta$. $g$ is not continuous.
It is well known that, in the definition of absolute continuity, the word "finite" can be replaced by "countably infinite":
A function $f: I to mathbbR$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies
$$ sum_k |y_k - x_k| < delta$$
then
$$sum_k |f(y_k) - f(x_k)| < epsilon$$
We know that the square wave can be written as the infinite sum of forms of sine functions: $lim_ntoinftysum_i=1^n h_i(x)$. Each $h_i$ is continuous but the limit is not. Let $H_n=sum_i=1^n h_i(x)$. $H_n$ should be continuous by induction. If $H_n$ is indeed continuous at $x=0$ then $forall deltaexistsepsilonforall yinmathbb R$ such that $|y-x|<epsilon$ implies $|f(y)-f(x)|<delta$. However, $forall epsilon>0$ $exists n>0$ such that $|y-x|<epsilon$ and $f(y)-f(0)=1$, contradition!
To Ramiro:
Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_n'$ such that $|y_n'-x_1|<epsilon$ implies $|g(y_n')-g(x_1)|<delta$; we know $K=sum_ileq n |y_i-x_i|,$ and $ sum_ileq n |f(y_i)-f(x_i)|=|g(y_n')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum_ileq n |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_n')-g_mathcal K(x_1)|<delta$,
which implies $sum_ileq n |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!
I am not sure that a single $g$ is required.
real-analysis calculus analysis continuity absolute-continuity
real-analysis calculus analysis continuity absolute-continuity
edited Mar 25 at 5:20
High GPA
asked Mar 22 at 1:46
High GPAHigh GPA
1,014421
1,014421
1
$begingroup$
I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
$endgroup$
– Ramiro
Mar 22 at 11:40
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@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
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– High GPA
Mar 22 at 19:11
1
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If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
1
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By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
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– Ramiro
Mar 22 at 22:43
1
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Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
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– Ramiro
Mar 22 at 23:20
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show 4 more comments
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I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
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– Ramiro
Mar 22 at 11:40
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@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
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– High GPA
Mar 22 at 19:11
1
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If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
1
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By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
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– Ramiro
Mar 22 at 22:43
1
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Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
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– Ramiro
Mar 22 at 23:20
1
1
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I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
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– Ramiro
Mar 22 at 11:40
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I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
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– Ramiro
Mar 22 at 11:40
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@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
$endgroup$
– High GPA
Mar 22 at 19:11
$begingroup$
@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
$endgroup$
– High GPA
Mar 22 at 19:11
1
1
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If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
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If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
1
1
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By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
$endgroup$
– Ramiro
Mar 22 at 22:43
$begingroup$
By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
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– Ramiro
Mar 22 at 22:43
1
1
$begingroup$
Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
$endgroup$
– Ramiro
Mar 22 at 23:20
$begingroup$
Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
$endgroup$
– Ramiro
Mar 22 at 23:20
|
show 4 more comments
1 Answer
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Suppose that $f:Itomathbb R$ is continuous, where $I$ is an interval. Take any finite set $mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]subseteq I$, where $ x_1 < y_1leqslant x_2<y_2leqslant,...,<x_n leqslant y_n $ and $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$. Then $clip(f,mathcal K)$ is continuous.
Proof: Let $g=clip(f,mathcal K)$. We will prove it by induction that for any $kin 1, ... n$, $g$ is continuous on $[x_1, y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that $g$ is continuous on $[x_1,y_k']$, where $1leqslant k<n$. Then we have that
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k+1']$$
But $ y_k+1' = x_k+1'+y_k+1-x_k+1= y_k'+y_k+1-x_k+1 $ so we have
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k'+y_k+1-x_k+1]$$
Since, for all $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$. Thus $g$ is continuous on $[y_k',y_k'+y_k+1-x_k+1]$ and we have that $y_k'$ is
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+x_k+1)+ g(y_k')-f(x_k+1)=g(y_k')=g|_[x_1,y_k'](y_k')$$
So $g$ is continuous on $[x_1,y_k+1']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$.
Conclusion: If $f$ is continuous, then $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Remark 1:
The general expression to define $g$ is
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$,
which is equivalent to:
If $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$.
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be:
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes:
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_[x_1,y_k'](y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So, it is a necessary condition for the absolute continuity of $f$ on $I$.
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1 is true for any continuous function, independently if the function is absolutely continuous or not.
Finally, note that when "clipping" $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.
Remark 4: (about the Cantor function)
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work.
If $f$ is the Cantor function, each "clipping" is continuous.
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".
Remark 5:
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$sum |y_i-x_i|<epsilon$ implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$.
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".
Remark 6:
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_k')-g_mathcal K(x_1)|<delta$,
which implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.
- First let us assume that $f$ is a monotonous function (like Cantor function). Then $g_mathcal K$ is also a monotonous function.
Let us go step by step. The function $g$ before the "SO" is already a $g_mathcal K'$ for some $mathcal K'$.
By its continuity, given $delta> 0$ there is an $epsilon>0$ (which depends on $delta$ and on $g_mathcal K'$) such that,
for any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$, we have
$$|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$$
where $y_k''$ corresponds to the "leftmost" point of $mathcal K$ (not of $mathcal K'$).
We also have that:
$$ sum_(x_i,y_i)inmathcal K' |f(y_i)-f(x_i)|=|g_mathcal K'(y_k')-g_mathcal K'(x_1)|$$
and
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k'')-g_mathcal K(x_1)|$$
but none of those two equations combine with $|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$.
The flaw in your argument is a result that, by writing just $g$, you think of different $g_mathcal K$ functions as being just one and the same function. The same applies $y_k'$, which actually depends on which $mathcal K$ we are considering.
- If $f$ is not supposed to a monotonous function, then even
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k')-g_mathcal K(x_1)|$$
it may not be true for all $mathcal K$, because of the absolute values used in the summation.
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Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
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– High GPA
Mar 24 at 4:43
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Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
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– Ramiro
Mar 24 at 12:38
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In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
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– High GPA
Mar 25 at 0:29
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Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
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– High GPA
Mar 25 at 0:37
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@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
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– Ramiro
Mar 25 at 0:47
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Suppose that $f:Itomathbb R$ is continuous, where $I$ is an interval. Take any finite set $mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]subseteq I$, where $ x_1 < y_1leqslant x_2<y_2leqslant,...,<x_n leqslant y_n $ and $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$. Then $clip(f,mathcal K)$ is continuous.
Proof: Let $g=clip(f,mathcal K)$. We will prove it by induction that for any $kin 1, ... n$, $g$ is continuous on $[x_1, y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that $g$ is continuous on $[x_1,y_k']$, where $1leqslant k<n$. Then we have that
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k+1']$$
But $ y_k+1' = x_k+1'+y_k+1-x_k+1= y_k'+y_k+1-x_k+1 $ so we have
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k'+y_k+1-x_k+1]$$
Since, for all $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$. Thus $g$ is continuous on $[y_k',y_k'+y_k+1-x_k+1]$ and we have that $y_k'$ is
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+x_k+1)+ g(y_k')-f(x_k+1)=g(y_k')=g|_[x_1,y_k'](y_k')$$
So $g$ is continuous on $[x_1,y_k+1']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$.
Conclusion: If $f$ is continuous, then $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Remark 1:
The general expression to define $g$ is
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$,
which is equivalent to:
If $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$.
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be:
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes:
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_[x_1,y_k'](y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So, it is a necessary condition for the absolute continuity of $f$ on $I$.
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1 is true for any continuous function, independently if the function is absolutely continuous or not.
Finally, note that when "clipping" $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.
Remark 4: (about the Cantor function)
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work.
If $f$ is the Cantor function, each "clipping" is continuous.
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".
Remark 5:
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$sum |y_i-x_i|<epsilon$ implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$.
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".
Remark 6:
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_k')-g_mathcal K(x_1)|<delta$,
which implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.
- First let us assume that $f$ is a monotonous function (like Cantor function). Then $g_mathcal K$ is also a monotonous function.
Let us go step by step. The function $g$ before the "SO" is already a $g_mathcal K'$ for some $mathcal K'$.
By its continuity, given $delta> 0$ there is an $epsilon>0$ (which depends on $delta$ and on $g_mathcal K'$) such that,
for any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$, we have
$$|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$$
where $y_k''$ corresponds to the "leftmost" point of $mathcal K$ (not of $mathcal K'$).
We also have that:
$$ sum_(x_i,y_i)inmathcal K' |f(y_i)-f(x_i)|=|g_mathcal K'(y_k')-g_mathcal K'(x_1)|$$
and
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k'')-g_mathcal K(x_1)|$$
but none of those two equations combine with $|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$.
The flaw in your argument is a result that, by writing just $g$, you think of different $g_mathcal K$ functions as being just one and the same function. The same applies $y_k'$, which actually depends on which $mathcal K$ we are considering.
- If $f$ is not supposed to a monotonous function, then even
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k')-g_mathcal K(x_1)|$$
it may not be true for all $mathcal K$, because of the absolute values used in the summation.
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Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
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– High GPA
Mar 24 at 4:43
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Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
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– Ramiro
Mar 24 at 12:38
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In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
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– High GPA
Mar 25 at 0:29
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Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
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– High GPA
Mar 25 at 0:37
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@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
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– Ramiro
Mar 25 at 0:47
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show 13 more comments
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Suppose that $f:Itomathbb R$ is continuous, where $I$ is an interval. Take any finite set $mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]subseteq I$, where $ x_1 < y_1leqslant x_2<y_2leqslant,...,<x_n leqslant y_n $ and $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$. Then $clip(f,mathcal K)$ is continuous.
Proof: Let $g=clip(f,mathcal K)$. We will prove it by induction that for any $kin 1, ... n$, $g$ is continuous on $[x_1, y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that $g$ is continuous on $[x_1,y_k']$, where $1leqslant k<n$. Then we have that
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k+1']$$
But $ y_k+1' = x_k+1'+y_k+1-x_k+1= y_k'+y_k+1-x_k+1 $ so we have
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k'+y_k+1-x_k+1]$$
Since, for all $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$. Thus $g$ is continuous on $[y_k',y_k'+y_k+1-x_k+1]$ and we have that $y_k'$ is
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+x_k+1)+ g(y_k')-f(x_k+1)=g(y_k')=g|_[x_1,y_k'](y_k')$$
So $g$ is continuous on $[x_1,y_k+1']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$.
Conclusion: If $f$ is continuous, then $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Remark 1:
The general expression to define $g$ is
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$,
which is equivalent to:
If $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$.
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be:
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes:
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_[x_1,y_k'](y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So, it is a necessary condition for the absolute continuity of $f$ on $I$.
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1 is true for any continuous function, independently if the function is absolutely continuous or not.
Finally, note that when "clipping" $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.
Remark 4: (about the Cantor function)
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work.
If $f$ is the Cantor function, each "clipping" is continuous.
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".
Remark 5:
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$sum |y_i-x_i|<epsilon$ implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$.
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".
Remark 6:
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_k')-g_mathcal K(x_1)|<delta$,
which implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.
- First let us assume that $f$ is a monotonous function (like Cantor function). Then $g_mathcal K$ is also a monotonous function.
Let us go step by step. The function $g$ before the "SO" is already a $g_mathcal K'$ for some $mathcal K'$.
By its continuity, given $delta> 0$ there is an $epsilon>0$ (which depends on $delta$ and on $g_mathcal K'$) such that,
for any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$, we have
$$|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$$
where $y_k''$ corresponds to the "leftmost" point of $mathcal K$ (not of $mathcal K'$).
We also have that:
$$ sum_(x_i,y_i)inmathcal K' |f(y_i)-f(x_i)|=|g_mathcal K'(y_k')-g_mathcal K'(x_1)|$$
and
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k'')-g_mathcal K(x_1)|$$
but none of those two equations combine with $|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$.
The flaw in your argument is a result that, by writing just $g$, you think of different $g_mathcal K$ functions as being just one and the same function. The same applies $y_k'$, which actually depends on which $mathcal K$ we are considering.
- If $f$ is not supposed to a monotonous function, then even
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k')-g_mathcal K(x_1)|$$
it may not be true for all $mathcal K$, because of the absolute values used in the summation.
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Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
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– High GPA
Mar 24 at 4:43
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Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
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– Ramiro
Mar 24 at 12:38
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In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
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– High GPA
Mar 25 at 0:29
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Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
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– High GPA
Mar 25 at 0:37
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@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
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– Ramiro
Mar 25 at 0:47
|
show 13 more comments
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Suppose that $f:Itomathbb R$ is continuous, where $I$ is an interval. Take any finite set $mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]subseteq I$, where $ x_1 < y_1leqslant x_2<y_2leqslant,...,<x_n leqslant y_n $ and $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$. Then $clip(f,mathcal K)$ is continuous.
Proof: Let $g=clip(f,mathcal K)$. We will prove it by induction that for any $kin 1, ... n$, $g$ is continuous on $[x_1, y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that $g$ is continuous on $[x_1,y_k']$, where $1leqslant k<n$. Then we have that
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k+1']$$
But $ y_k+1' = x_k+1'+y_k+1-x_k+1= y_k'+y_k+1-x_k+1 $ so we have
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k'+y_k+1-x_k+1]$$
Since, for all $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$. Thus $g$ is continuous on $[y_k',y_k'+y_k+1-x_k+1]$ and we have that $y_k'$ is
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+x_k+1)+ g(y_k')-f(x_k+1)=g(y_k')=g|_[x_1,y_k'](y_k')$$
So $g$ is continuous on $[x_1,y_k+1']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$.
Conclusion: If $f$ is continuous, then $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Remark 1:
The general expression to define $g$ is
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$,
which is equivalent to:
If $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$.
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be:
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes:
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_[x_1,y_k'](y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So, it is a necessary condition for the absolute continuity of $f$ on $I$.
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1 is true for any continuous function, independently if the function is absolutely continuous or not.
Finally, note that when "clipping" $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.
Remark 4: (about the Cantor function)
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work.
If $f$ is the Cantor function, each "clipping" is continuous.
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".
Remark 5:
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$sum |y_i-x_i|<epsilon$ implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$.
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".
Remark 6:
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_k')-g_mathcal K(x_1)|<delta$,
which implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.
- First let us assume that $f$ is a monotonous function (like Cantor function). Then $g_mathcal K$ is also a monotonous function.
Let us go step by step. The function $g$ before the "SO" is already a $g_mathcal K'$ for some $mathcal K'$.
By its continuity, given $delta> 0$ there is an $epsilon>0$ (which depends on $delta$ and on $g_mathcal K'$) such that,
for any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$, we have
$$|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$$
where $y_k''$ corresponds to the "leftmost" point of $mathcal K$ (not of $mathcal K'$).
We also have that:
$$ sum_(x_i,y_i)inmathcal K' |f(y_i)-f(x_i)|=|g_mathcal K'(y_k')-g_mathcal K'(x_1)|$$
and
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k'')-g_mathcal K(x_1)|$$
but none of those two equations combine with $|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$.
The flaw in your argument is a result that, by writing just $g$, you think of different $g_mathcal K$ functions as being just one and the same function. The same applies $y_k'$, which actually depends on which $mathcal K$ we are considering.
- If $f$ is not supposed to a monotonous function, then even
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k')-g_mathcal K(x_1)|$$
it may not be true for all $mathcal K$, because of the absolute values used in the summation.
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Suppose that $f:Itomathbb R$ is continuous, where $I$ is an interval. Take any finite set $mathcal K$ of subintervals $[x_1,y_1], [x_2,y_2],...,[x_n,y_n]subseteq I$, where $ x_1 < y_1leqslant x_2<y_2leqslant,...,<x_n leqslant y_n $ and $sum_kmu([x_k,y_k])=K$, i.e. $|mathcal K|=K$. Then $clip(f,mathcal K)$ is continuous.
Proof: Let $g=clip(f,mathcal K)$. We will prove it by induction that for any $kin 1, ... n$, $g$ is continuous on $[x_1, y_k']$.
For $k=1$, we have that, $g=f$ on $[x_1,y_1]$ and $y_1'=y_1$. So $g$ is continuous on $[x_1,y_1']$.
Now suppose that we know that $g$ is continuous on $[x_1,y_k']$, where $1leqslant k<n$. Then we have that
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k+1']$$
But $ y_k+1' = x_k+1'+y_k+1-x_k+1= y_k'+y_k+1-x_k+1 $ so we have
$$ [x_1,y_k+1'] = [x_1,y_k'] cup [y_k', y_k'+y_k+1-x_k+1]$$
Since, for all $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$. Thus $g$ is continuous on $[y_k',y_k'+y_k+1-x_k+1]$ and we have that $y_k'$ is
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+x_k+1)+ g(y_k')-f(x_k+1)=g(y_k')=g|_[x_1,y_k'](y_k')$$
So $g$ is continuous on $[x_1,y_k+1']$. By induction, $g$ is continuous on $[x_1,y_n']$. and, since $y_n'=x_1+K$, we have that $g$ is continuous on $[x_1,x_1+K]$.
Conclusion: If $f$ is continuous, then $forallmathcal Ksubseteq I$, $g=clip(f,mathcal K)$ is continuous.
Remark 1:
The general expression to define $g$ is
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$,
which is equivalent to:
If $xin[y_k',y_k'+y_k+1-x_k+1]$, then $g(x)=f(x-y_k'+x_k+1)+c_k$, and $c_k=g(y_k')-f(x_k+1)$.
Remark 2:
if for some reason you really want the general expreesion to define $g$ to be:
If $(x+y_k')in[y_k',y_k'+y_k+1-x_k+1]$, then $g(x+y_k')=f(x+y_k)+c_k$, and $c_k=g(y_k')-f(y_k)$.
the proof above remains valid with minor changes:
$$g|_[y_k',y_k'+y_k+1-x_k+1](y_k')=f(y_k'-y_k'+y_k)+ g(y_k')-f(y_k)=g(y_k')=g|_[x_1,y_k'](y_k')$$
Remark 3 (answers to your questions):
As the proof above shows, your condition 1 is a necessary condition for the continuity of $f$ on $I$. So, it is a necessary condition for the absolute continuity of $f$ on $I$.
Also, as a consequence of the proof above, your condition 1 is NOT a sufficient condition for the absolute continuity of $f$ on $I$. In fact, condition 1 is true for any continuous function, independently if the function is absolutely continuous or not.
Finally, note that when "clipping" $f$, a key mechanism was that two "glued" compact intervals have just one point in common, and so, we could adjust the pieces of $f$ by simply adding a constant. In higher dimensions, this mechanism does not work. For instance, in general, two "glued" rectangles will have an edge in common, not just one point.
Remark 4: (about the Cantor function)
In "clipping" as you define, the number of intervals is always finite, so the proof I post (based on FINITE induction) always work.
If $f$ is the Cantor function, each "clipping" is continuous.
To apply the argument you added about the Cantor function, you have to use limits and have an infinite countable sequence of intervals. So the argument you posted is not a counter-example to the proof above. It does not apply to "clipping" as you define it.
If it was this kind of "argument" that you want to capture in the "clipping" definition, then you need to change your definition of "clipping".
Remark 5:
You wrote: "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO
$sum |y_i-x_i|<epsilon$ implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous!"
In order for this argument to be correct, you need to define a single function $g$ such that for any finite sequence of sub-intervals $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$.
However, according definition of $g$, for each finite sequence of sub-intervals, we have a different continuous "$g$".
Remark 6:
You wrote "Let assume $f$ is increasing. Your answer seems imply that, if $f$ is continuous, then $g$ is continuous, then $forall deltaexistsepsilonforall y_k'$ such that $|y_k'-x_1|<epsilon$ implies $|g(y_k')-g(x_1)|<delta$; we know $K=sum |y_i-x_i|,$ and $ sum_ileq k |f(y_i)-f(x_i)|=|g(y_k')-g(x_1)|$, SO:
Take any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$,
this implies $K<epsilon$,
which implies $|g_mathcal K(y_k')-g_mathcal K(x_1)|<delta$,
which implies $sum |f(y_i)-f(x_i)|<delta$. Then $f$ is absolute continuous! "
No. This argument does not work. Let me explain.
- First let us assume that $f$ is a monotonous function (like Cantor function). Then $g_mathcal K$ is also a monotonous function.
Let us go step by step. The function $g$ before the "SO" is already a $g_mathcal K'$ for some $mathcal K'$.
By its continuity, given $delta> 0$ there is an $epsilon>0$ (which depends on $delta$ and on $g_mathcal K'$) such that,
for any one collection of disjoint interval $mathcal K$ such that $sum |y_i-x_i|<epsilon$, we have
$$|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$$
where $y_k''$ corresponds to the "leftmost" point of $mathcal K$ (not of $mathcal K'$).
We also have that:
$$ sum_(x_i,y_i)inmathcal K' |f(y_i)-f(x_i)|=|g_mathcal K'(y_k')-g_mathcal K'(x_1)|$$
and
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k'')-g_mathcal K(x_1)|$$
but none of those two equations combine with $|g_mathcal K'(y_k'')-g_mathcal K'(x_1)|<delta$.
The flaw in your argument is a result that, by writing just $g$, you think of different $g_mathcal K$ functions as being just one and the same function. The same applies $y_k'$, which actually depends on which $mathcal K$ we are considering.
- If $f$ is not supposed to a monotonous function, then even
$$ sum_(x_i,y_i)inmathcal K |f(y_i)-f(x_i)|=|g_mathcal K(y_k')-g_mathcal K(x_1)|$$
it may not be true for all $mathcal K$, because of the absolute values used in the summation.
edited Mar 25 at 16:23
answered Mar 23 at 5:16
RamiroRamiro
7,35421535
7,35421535
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Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
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– High GPA
Mar 24 at 4:43
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Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
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– Ramiro
Mar 24 at 12:38
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In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
$endgroup$
– High GPA
Mar 25 at 0:29
$begingroup$
Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
$endgroup$
– High GPA
Mar 25 at 0:37
$begingroup$
@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
$endgroup$
– Ramiro
Mar 25 at 0:47
|
show 13 more comments
$begingroup$
Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
$endgroup$
– High GPA
Mar 24 at 4:43
$begingroup$
Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
$endgroup$
– Ramiro
Mar 24 at 12:38
$begingroup$
In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
$endgroup$
– High GPA
Mar 25 at 0:29
$begingroup$
Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
$endgroup$
– High GPA
Mar 25 at 0:37
$begingroup$
@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
$endgroup$
– Ramiro
Mar 25 at 0:47
$begingroup$
Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
$endgroup$
– High GPA
Mar 24 at 4:43
$begingroup$
Thanks for writing, upvoted. I updated in the main question the counterexample that $g$ is not always continuous if $f$ is continuous. Note that the method of induction does not always work when the repetition goes to infinity (math.stackexchange.com/questions/98093/…); finite and countably infinite are somewhat equivalent in the context of continuity.
$endgroup$
– High GPA
Mar 24 at 4:43
$begingroup$
Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
$endgroup$
– Ramiro
Mar 24 at 12:38
$begingroup$
Induction does not apply to Fourier series continuity, not because the "repetition goes to infinity", but because we consider the pointwise limit of a sequence of functions. Finite and countably infinite are NOT in general equivalent in the the context of continuity exactly because, in this context, when we have countably infinite we need to apply some kind of limit, which may not preserve some properties. For instance, the pointwise limit of a sequence of continuous functions may not be continuous.
$endgroup$
– Ramiro
Mar 24 at 12:38
$begingroup$
In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
$endgroup$
– High GPA
Mar 25 at 0:29
$begingroup$
In the cantor example, my $n$ is always finite. No matter how you choose an $epsilon$, I can always find a finite $n$ such that $K<epsilon$ and $|g(y_k')-g(x_1)|=1$; that is, $g$ is not continuous.
$endgroup$
– High GPA
Mar 25 at 0:29
$begingroup$
Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
$endgroup$
– High GPA
Mar 25 at 0:37
$begingroup$
Despite our disagreement, you answer is still very, very helpful and constructive to me. I do appreciate it. Can we at least agree on that, if I change the word "finite" into "countable" on the first line, $g$ is not necessarily continuous when $f$ is continuous?
$endgroup$
– High GPA
Mar 25 at 0:37
$begingroup$
@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
$endgroup$
– Ramiro
Mar 25 at 0:47
$begingroup$
@HighGPA In your Cantor example, for each $epsilon >0$, you can find a finite $n$ such that $K < epsilon $ and **the correspinding** "g_epsilon" satisfy $|g_epsilon (y_k')-g_epsilon (x_1)|=1$. The functions $g_epsilon $ are not a single function and each one of them is continous.
$endgroup$
– Ramiro
Mar 25 at 0:47
|
show 13 more comments
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I assume that $I$ is an interval. Then, since $f$ is continuous, the way you defined $clip(f,mathcal K)$ makes it continuous, no matter if $f$ is absolute continuous or not.
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– Ramiro
Mar 22 at 11:40
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@Ramiro If $f$ is Cantor function, it is possible to find a set of subintervals that $K$ is arbitrarily small but $sum_i(f(y_i)-f(x_i))>1$; thus $g$ is not continuous?
$endgroup$
– High GPA
Mar 22 at 19:11
1
$begingroup$
If I understand your idea, you are defining $g:[x_1,x_1+K]tomathbb R$ by pieces (compact intervals). In the interior of each interval $g$ is continuous (because its an affine transformation of a piece of $f$). In the endpoint of between any two intervals, you add constants to $f$ to ensure that the piece on the right will coincide to the piece on the left, so $g$ is also continuous at those points. So $g$ is continuous in all points in $[x_1,x_1+K]$.
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– Ramiro
Mar 22 at 22:28
1
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By the way, I think there is a small typo when your wrote: "if $xin[y_k,y_k+y_k+1-x_k+1]$, then $g(x)=f(x-y_k+x_k+1)+c_k+1$, where $c_k+1=f(y_k)-f(x_k+1)$". The general expression for the intervals (from the second on) is NOT $[y_k,y_k+y_k+1-x_k+1]$. The second interval is $[y_1,y_1+y_2-x_2]$, but the third one is not $[y_2,y_2+y_3-x_3]$. The third one should be $[y_1+y_2-x_2,y_1+y_2-x_2+y_3-x_3]$. The third interval should "start" where the second one "ends".
$endgroup$
– Ramiro
Mar 22 at 22:43
1
$begingroup$
Ok. Now the intervals are fine. If I understand your idea, for each interval there is also a $c_n $ to be added to $f $, like the $c_2$. Right?
$endgroup$
– Ramiro
Mar 22 at 23:20