$forallvarepsilon > 0,exists a >0 : |f(x)|,le, a|f|_2 + varepsilon|f'|_2$ for $fin H^1(0,1)$ The 2019 Stack Overflow Developer Survey Results Are InExample of an Hilbert space operatornorm of differential operator on $P^n[0,1]$Prove that $ell^p$ and $C[0,1]$ are infinite dimensional$L^1 ([0,1])$, bouned linear functional, absolute continuous functionShowing the compactness of a limit operator.Why do we need the extra condition of being 'Fredholm of index zero' when showing that an operator has a bounded inverse?Is the functional $E: L^2[0,1] to mathbbR$, $E(f) = |f|_2^2$, continuous?Is there any diagonalisable operator that is not compact?Why delta function is a tempered distribution?Existence of some continuous functions when a distribution is given
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$forallvarepsilon > 0,exists a >0 : |f(x)|,le, a|f|_2 + varepsilon|f'|_2$ for $fin H^1(0,1)$
The 2019 Stack Overflow Developer Survey Results Are InExample of an Hilbert space operatornorm of differential operator on $P^n[0,1]$Prove that $ell^p$ and $C[0,1]$ are infinite dimensional$L^1 ([0,1])$, bouned linear functional, absolute continuous functionShowing the compactness of a limit operator.Why do we need the extra condition of being 'Fredholm of index zero' when showing that an operator has a bounded inverse?Is the functional $E: L^2[0,1] to mathbbR$, $E(f) = |f|_2^2$, continuous?Is there any diagonalisable operator that is not compact?Why delta function is a tempered distribution?Existence of some continuous functions when a distribution is given
$begingroup$
I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(mathbb R)$).
So, $delta_x : H^1(0,1)tomathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $fin H^1(0,1)$.
As $D$-compact operators have $D$-bound zero, for any $varepsilon>0$, I should find an $a > 0$ such that
$$
|f(x)|,le,a|f|_2 + varepsilon|f'|_2,quad fin H^1(0,1).
$$
But I can't find such $a$ for given $varepsilon$. Can anyone help me out?
functional-analysis inequality sobolev-spaces distribution-theory dirac-delta
edited Mar 23 at 7:37
Andrews
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
$endgroup$
add a comment |
$begingroup$
I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(mathbb R)$).
So, $delta_x : H^1(0,1)tomathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $fin H^1(0,1)$.
As $D$-compact operators have $D$-bound zero, for any $varepsilon>0$, I should find an $a > 0$ such that
$$
|f(x)|,le,a|f|_2 + varepsilon|f'|_2,quad fin H^1(0,1).
$$
But I can't find such $a$ for given $varepsilon$. Can anyone help me out?
functional-analysis inequality sobolev-spaces distribution-theory dirac-delta
edited Mar 23 at 7:37
Andrews
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
$endgroup$
$begingroup$
What is $H^1(0,1)$?
$endgroup$
– user56834
Mar 23 at 6:14
$begingroup$
@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
$endgroup$
– amsmath
Mar 23 at 6:20
add a comment |
$begingroup$
I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(mathbb R)$).
So, $delta_x : H^1(0,1)tomathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $fin H^1(0,1)$.
As $D$-compact operators have $D$-bound zero, for any $varepsilon>0$, I should find an $a > 0$ such that
$$
|f(x)|,le,a|f|_2 + varepsilon|f'|_2,quad fin H^1(0,1).
$$
But I can't find such $a$ for given $varepsilon$. Can anyone help me out?
functional-analysis inequality sobolev-spaces distribution-theory dirac-delta
edited Mar 23 at 7:37
Andrews
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
$endgroup$
I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(mathbb R)$).
So, $delta_x : H^1(0,1)tomathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $fin H^1(0,1)$.
As $D$-compact operators have $D$-bound zero, for any $varepsilon>0$, I should find an $a > 0$ such that
$$
|f(x)|,le,a|f|_2 + varepsilon|f'|_2,quad fin H^1(0,1).
$$
But I can't find such $a$ for given $varepsilon$. Can anyone help me out?
functional-analysis inequality sobolev-spaces distribution-theory dirac-delta
functional-analysis inequality sobolev-spaces distribution-theory dirac-delta
edited Mar 23 at 7:37
Andrews
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
edited Mar 23 at 7:37
Andrews
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
edited Mar 23 at 7:37
Andrews
1,2812423
edited Mar 23 at 7:37
Andrews
1,2812423
edited Mar 23 at 7:37
Andrews
1,2812423
1,2812423
asked Mar 23 at 4:59
amsmathamsmath
3,287421
asked Mar 23 at 4:59
amsmathamsmath
3,287421
asked Mar 23 at 4:59
amsmathamsmath
3,287421
3,287421
$begingroup$
What is $H^1(0,1)$?
$endgroup$
– user56834
Mar 23 at 6:14
$begingroup$
@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
$endgroup$
– amsmath
Mar 23 at 6:20
add a comment |
$begingroup$
What is $H^1(0,1)$?
$endgroup$
– user56834
Mar 23 at 6:14
$begingroup$
@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
$endgroup$
– amsmath
Mar 23 at 6:20
$begingroup$
What is $H^1(0,1)$?
$endgroup$
– user56834
Mar 23 at 6:14
$begingroup$
What is $H^1(0,1)$?
$endgroup$
– user56834
Mar 23 at 6:14
$begingroup$
@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
$endgroup$
– amsmath
Mar 23 at 6:20
$begingroup$
@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
$endgroup$
– amsmath
Mar 23 at 6:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A somehow clumsy argument:
By approximation assume $f$ is $C^1$. First assume $xleq frac 12$. If $|f(y)|>frac 12 |f(x)$| for all $yin [x, x+frac 14 epsilon^2]$, then
$$int_0^1f^2(y)dygeq int_x^x+frac 14 epsilon^2f(y)^2 dygeq frac 14epsilon^2frac 14 f(x)^2; $$
thus in this case $|f(x)|leq 4epsilon^-1|f|_2$.
If for some $yin [x, x+frac 14 epsilon^2]$ we have $|f(y)|leq frac 12 |f(x)|$, then
$$
f(x)^2leq 4[f(x)-f(y)]^2=4Big(int_x^y -f'(t) dtBig)^2 leq 4int_x^y|f'(t)|^2dt cdot |y-x|
leq epsilon^2int_0^1|f'(t)|^2dt,
$$
i.e. $|f(x)|leq epsilon |f'|_2$.
If $x>frac 12$ we can argue similarly by considering $yin [x-frac 14 epsilon^2, x]$.
Overall we see that one can take $a=4 epsilon^-1$.
answered Mar 23 at 6:24
Yu DingYu Ding
7187
$endgroup$
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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votes
active
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votes
$begingroup$
A somehow clumsy argument:
By approximation assume $f$ is $C^1$. First assume $xleq frac 12$. If $|f(y)|>frac 12 |f(x)$| for all $yin [x, x+frac 14 epsilon^2]$, then
$$int_0^1f^2(y)dygeq int_x^x+frac 14 epsilon^2f(y)^2 dygeq frac 14epsilon^2frac 14 f(x)^2; $$
thus in this case $|f(x)|leq 4epsilon^-1|f|_2$.
If for some $yin [x, x+frac 14 epsilon^2]$ we have $|f(y)|leq frac 12 |f(x)|$, then
$$
f(x)^2leq 4[f(x)-f(y)]^2=4Big(int_x^y -f'(t) dtBig)^2 leq 4int_x^y|f'(t)|^2dt cdot |y-x|
leq epsilon^2int_0^1|f'(t)|^2dt,
$$
i.e. $|f(x)|leq epsilon |f'|_2$.
If $x>frac 12$ we can argue similarly by considering $yin [x-frac 14 epsilon^2, x]$.
Overall we see that one can take $a=4 epsilon^-1$.
answered Mar 23 at 6:24
Yu DingYu Ding
7187
$endgroup$
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
add a comment |
$begingroup$
A somehow clumsy argument:
By approximation assume $f$ is $C^1$. First assume $xleq frac 12$. If $|f(y)|>frac 12 |f(x)$| for all $yin [x, x+frac 14 epsilon^2]$, then
$$int_0^1f^2(y)dygeq int_x^x+frac 14 epsilon^2f(y)^2 dygeq frac 14epsilon^2frac 14 f(x)^2; $$
thus in this case $|f(x)|leq 4epsilon^-1|f|_2$.
If for some $yin [x, x+frac 14 epsilon^2]$ we have $|f(y)|leq frac 12 |f(x)|$, then
$$
f(x)^2leq 4[f(x)-f(y)]^2=4Big(int_x^y -f'(t) dtBig)^2 leq 4int_x^y|f'(t)|^2dt cdot |y-x|
leq epsilon^2int_0^1|f'(t)|^2dt,
$$
i.e. $|f(x)|leq epsilon |f'|_2$.
If $x>frac 12$ we can argue similarly by considering $yin [x-frac 14 epsilon^2, x]$.
Overall we see that one can take $a=4 epsilon^-1$.
answered Mar 23 at 6:24
Yu DingYu Ding
7187
$endgroup$
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
add a comment |
$begingroup$
A somehow clumsy argument:
By approximation assume $f$ is $C^1$. First assume $xleq frac 12$. If $|f(y)|>frac 12 |f(x)$| for all $yin [x, x+frac 14 epsilon^2]$, then
$$int_0^1f^2(y)dygeq int_x^x+frac 14 epsilon^2f(y)^2 dygeq frac 14epsilon^2frac 14 f(x)^2; $$
thus in this case $|f(x)|leq 4epsilon^-1|f|_2$.
If for some $yin [x, x+frac 14 epsilon^2]$ we have $|f(y)|leq frac 12 |f(x)|$, then
$$
f(x)^2leq 4[f(x)-f(y)]^2=4Big(int_x^y -f'(t) dtBig)^2 leq 4int_x^y|f'(t)|^2dt cdot |y-x|
leq epsilon^2int_0^1|f'(t)|^2dt,
$$
i.e. $|f(x)|leq epsilon |f'|_2$.
If $x>frac 12$ we can argue similarly by considering $yin [x-frac 14 epsilon^2, x]$.
Overall we see that one can take $a=4 epsilon^-1$.
answered Mar 23 at 6:24
Yu DingYu Ding
7187
$endgroup$
A somehow clumsy argument:
By approximation assume $f$ is $C^1$. First assume $xleq frac 12$. If $|f(y)|>frac 12 |f(x)$| for all $yin [x, x+frac 14 epsilon^2]$, then
$$int_0^1f^2(y)dygeq int_x^x+frac 14 epsilon^2f(y)^2 dygeq frac 14epsilon^2frac 14 f(x)^2; $$
thus in this case $|f(x)|leq 4epsilon^-1|f|_2$.
If for some $yin [x, x+frac 14 epsilon^2]$ we have $|f(y)|leq frac 12 |f(x)|$, then
$$
f(x)^2leq 4[f(x)-f(y)]^2=4Big(int_x^y -f'(t) dtBig)^2 leq 4int_x^y|f'(t)|^2dt cdot |y-x|
leq epsilon^2int_0^1|f'(t)|^2dt,
$$
i.e. $|f(x)|leq epsilon |f'|_2$.
If $x>frac 12$ we can argue similarly by considering $yin [x-frac 14 epsilon^2, x]$.
Overall we see that one can take $a=4 epsilon^-1$.
answered Mar 23 at 6:24
Yu DingYu Ding
7187
answered Mar 23 at 6:24
Yu DingYu Ding
7187
answered Mar 23 at 6:24
Yu DingYu Ding
7187
answered Mar 23 at 6:24
Yu DingYu Ding
7187
7187
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
add a comment |
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Thank you very much!!! :o) Actually, I don't see any point where you need that $fin C^1$.
$endgroup$
– amsmath
Mar 23 at 15:12
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Before we proved this inequality, we don't know if we can define $f(x)$ (i.e. one tries to make sense of the value of an$L^2$ function at a particular point), as well as the use of fundamental theorem of calculus. So one starts with smooth functions then get the general case by approximation.
$endgroup$
– Yu Ding
Mar 23 at 15:59
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Yu, please note that $fin H^1(0,1)$ implies that $f$ has an absolutely continuous representative. And for these you have the generalized (Lebesgue-) fundamental theorem of calculus. So, everything is nice and well settled.
$endgroup$
– amsmath
Mar 23 at 16:42
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
Well, you have to quote that result, which in my view is not completely "trivial", because it (continuous representative) is not true for higher dimensions...
$endgroup$
– Yu Ding
Mar 23 at 17:24
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
$begingroup$
It is neither trivial that $C^1$ is dense in $H^1$. ;-)
$endgroup$
– amsmath
Mar 23 at 17:43
add a comment |
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What is $H^1(0,1)$?
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– user56834
Mar 23 at 6:14
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@user56834 It's the usual Sobolev space over $(0,1)$ based on $L^2(0,1)$.
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– amsmath
Mar 23 at 6:20