If series is not uniformly convergent, can we still integrate term by term? The 2019 Stack Overflow Developer Survey Results Are Inproving that some series converges uniformlyprove that the series is not uniformly convergentShow a Fourier series converges uniformlyseries of functions and term by term differentiationpolynomial of the term of convergent series convergesCondition for term-by-term differentiation of a non-convergent seriesCan series converge even if the general term does not have limit?If sequence of function does not converges uniformly, so doesn't the series?Power series uniform convergence: equivalent statements or not?Term by term integration of $int_0^1 sum_n=0^inftya_nx^n g(x)dx$

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If series is not uniformly convergent, can we still integrate term by term?



The 2019 Stack Overflow Developer Survey Results Are Inproving that some series converges uniformlyprove that the series is not uniformly convergentShow a Fourier series converges uniformlyseries of functions and term by term differentiationpolynomial of the term of convergent series convergesCondition for term-by-term differentiation of a non-convergent seriesCan series converge even if the general term does not have limit?If sequence of function does not converges uniformly, so doesn't the series?Power series uniform convergence: equivalent statements or not?Term by term integration of $int_0^1 sum_n=0^inftya_nx^n g(x)dx$










3












$begingroup$


We know that if $sum a_nx^n$ converges uniformly, then we can integrate term by term. So this is just a sufficient condition, right? Does there exists a series not converging uniformly and still we can perform term by term integration? I am looking for some general results in this direction.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
    $endgroup$
    – Conrad
    Mar 23 at 17:20











  • $begingroup$
    This is helpful. I will look further into this stuff(and Fourier series). Thanks.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:22






  • 1




    $begingroup$
    The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
    $endgroup$
    – KCd
    Mar 25 at 2:35
















3












$begingroup$


We know that if $sum a_nx^n$ converges uniformly, then we can integrate term by term. So this is just a sufficient condition, right? Does there exists a series not converging uniformly and still we can perform term by term integration? I am looking for some general results in this direction.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
    $endgroup$
    – Conrad
    Mar 23 at 17:20











  • $begingroup$
    This is helpful. I will look further into this stuff(and Fourier series). Thanks.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:22






  • 1




    $begingroup$
    The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
    $endgroup$
    – KCd
    Mar 25 at 2:35














3












3








3


1



$begingroup$


We know that if $sum a_nx^n$ converges uniformly, then we can integrate term by term. So this is just a sufficient condition, right? Does there exists a series not converging uniformly and still we can perform term by term integration? I am looking for some general results in this direction.










share|cite|improve this question









$endgroup$




We know that if $sum a_nx^n$ converges uniformly, then we can integrate term by term. So this is just a sufficient condition, right? Does there exists a series not converging uniformly and still we can perform term by term integration? I am looking for some general results in this direction.







real-analysis calculus sequences-and-series functional-analysis analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 23 at 16:55









ProblemBookProblemBook

816




816







  • 1




    $begingroup$
    This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
    $endgroup$
    – Conrad
    Mar 23 at 17:20











  • $begingroup$
    This is helpful. I will look further into this stuff(and Fourier series). Thanks.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:22






  • 1




    $begingroup$
    The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
    $endgroup$
    – KCd
    Mar 25 at 2:35













  • 1




    $begingroup$
    This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
    $endgroup$
    – Conrad
    Mar 23 at 17:20











  • $begingroup$
    This is helpful. I will look further into this stuff(and Fourier series). Thanks.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:22






  • 1




    $begingroup$
    The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
    $endgroup$
    – KCd
    Mar 25 at 2:35








1




1




$begingroup$
This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
$endgroup$
– Conrad
Mar 23 at 17:20





$begingroup$
This is a complicated question in general but Fourier series can be integrated term by term regardless of uniformity of convergence or boundness in the sense that if $c_0$ is the free term of a Fourier series of a lebesgue integrable periodic function $fin L^1([0,2pi])$ and $F$ is any indefinite integral of $f$, $F(x)-c_0x$ has as Fourier series the term by term integral of the Fourier series of $f-c_0$ up to a constant and the convergence is uniform for the periodic $F-c_0x$ regardless of how the Fourier series of $f$ behaves (could even be divergent ae)
$endgroup$
– Conrad
Mar 23 at 17:20













$begingroup$
This is helpful. I will look further into this stuff(and Fourier series). Thanks.
$endgroup$
– ProblemBook
Mar 23 at 17:22




$begingroup$
This is helpful. I will look further into this stuff(and Fourier series). Thanks.
$endgroup$
– ProblemBook
Mar 23 at 17:22




1




1




$begingroup$
The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
$endgroup$
– KCd
Mar 25 at 2:35





$begingroup$
The main issue is not termwise integration of power series, but integration with uniform vs. pointwise limits: if $f_n rightarrow f$ on a set $A$, does $int_A f_n(x), dx rightarrow int_A f(x), dx$? The last limit is just numbers but the first is functions: pointwise or uniform or something else? Here measure theory is important, as it gives practical conditions to justify commuting integration and pointwise limits. See the Dominated Convergence Theorem!
$endgroup$
– KCd
Mar 25 at 2:35











2 Answers
2






active

oldest

votes


















2












$begingroup$

With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.



The only place we can't use uniform convergence? The boundary.



And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots$ gives us the alternating harmonic series $ln(2)=1-frac12+frac13-frac14+cdots$.



Relating to the term-by-term integration your interested in, that logarithm series is the integral of $frac11+x=1-x+x^2-x^3+cdots$, and we then have
$$int_0^1 1-x+x^2-x^3+cdots,dx = left.x-fracx^22+fracx^33-fracx^44+cdotsright|_x=1 = ln(2)$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:12



















1












$begingroup$

In response to a comment by the proposer to another A:



For $nin Bbb Z^+$ let $g_n(x)=0$ for $xnot in [2^-n-1,2^-n]$ and let $g_n$ be integrable on $[2^-n-1,2^-n]$ such that $sup ge 1.$ And such that $lim_nto inftyint_0^1g_n(t)dt=0.$



Let $g_0(x)=0$ for all $x.$



For $nge 0$ let $f_n=g_n-g_n+1 .$



Then $sum_j=0^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $sum_j=0^nint_0^1f_j(t)dt=-int_0^1 g_n(t)dt$ converges to $0.$



For example, for $nin Bbb Z^+:$ Let $a_n=(2^-n-1+2^-n)/2.$ Let $g_n$ be linear on $[2^-n-1,a_n]$ and on $[a_n,2^-n]$ with $g_n(2^-n-1)=0=g_n(2^-n)$ and $g_n(a_n)=n.$



You can easily produce variations on this theme.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you. This is a nice example.
    $endgroup$
    – ProblemBook
    Mar 27 at 2:28











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.



The only place we can't use uniform convergence? The boundary.



And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots$ gives us the alternating harmonic series $ln(2)=1-frac12+frac13-frac14+cdots$.



Relating to the term-by-term integration your interested in, that logarithm series is the integral of $frac11+x=1-x+x^2-x^3+cdots$, and we then have
$$int_0^1 1-x+x^2-x^3+cdots,dx = left.x-fracx^22+fracx^33-fracx^44+cdotsright|_x=1 = ln(2)$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:12
















2












$begingroup$

With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.



The only place we can't use uniform convergence? The boundary.



And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots$ gives us the alternating harmonic series $ln(2)=1-frac12+frac13-frac14+cdots$.



Relating to the term-by-term integration your interested in, that logarithm series is the integral of $frac11+x=1-x+x^2-x^3+cdots$, and we then have
$$int_0^1 1-x+x^2-x^3+cdots,dx = left.x-fracx^22+fracx^33-fracx^44+cdotsright|_x=1 = ln(2)$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:12














2












2








2





$begingroup$

With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.



The only place we can't use uniform convergence? The boundary.



And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots$ gives us the alternating harmonic series $ln(2)=1-frac12+frac13-frac14+cdots$.



Relating to the term-by-term integration your interested in, that logarithm series is the integral of $frac11+x=1-x+x^2-x^3+cdots$, and we then have
$$int_0^1 1-x+x^2-x^3+cdots,dx = left.x-fracx^22+fracx^33-fracx^44+cdotsright|_x=1 = ln(2)$$






share|cite|improve this answer









$endgroup$



With a power series like that, there's a radius of convergence $r$ - and for any $s<r$, the series converges uniformly for $|x|<s$. With that, we can use the uniform convergence to integrate term by term with the same radius of convergence.



The only place we can't use uniform convergence? The boundary.



And on that boundary... Abel's theorem. If the series converges at a point on the boundary, then it's the limit of the values inside, approaching head-on. For example, this theorem applied to the series $ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots$ gives us the alternating harmonic series $ln(2)=1-frac12+frac13-frac14+cdots$.



Relating to the term-by-term integration your interested in, that logarithm series is the integral of $frac11+x=1-x+x^2-x^3+cdots$, and we then have
$$int_0^1 1-x+x^2-x^3+cdots,dx = left.x-fracx^22+fracx^33-fracx^44+cdotsright|_x=1 = ln(2)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 23 at 17:07









jmerryjmerry

17k11633




17k11633







  • 1




    $begingroup$
    So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:12













  • 1




    $begingroup$
    So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
    $endgroup$
    – ProblemBook
    Mar 23 at 17:12








1




1




$begingroup$
So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
$endgroup$
– ProblemBook
Mar 23 at 17:12





$begingroup$
So I have to look beyond the power series. Can I get some example of a non-uniformly convergent series which is not a power series where term by term integration is performed? I am going to accept your answer after some time. I am expecting new comments.
$endgroup$
– ProblemBook
Mar 23 at 17:12












1












$begingroup$

In response to a comment by the proposer to another A:



For $nin Bbb Z^+$ let $g_n(x)=0$ for $xnot in [2^-n-1,2^-n]$ and let $g_n$ be integrable on $[2^-n-1,2^-n]$ such that $sup ge 1.$ And such that $lim_nto inftyint_0^1g_n(t)dt=0.$



Let $g_0(x)=0$ for all $x.$



For $nge 0$ let $f_n=g_n-g_n+1 .$



Then $sum_j=0^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $sum_j=0^nint_0^1f_j(t)dt=-int_0^1 g_n(t)dt$ converges to $0.$



For example, for $nin Bbb Z^+:$ Let $a_n=(2^-n-1+2^-n)/2.$ Let $g_n$ be linear on $[2^-n-1,a_n]$ and on $[a_n,2^-n]$ with $g_n(2^-n-1)=0=g_n(2^-n)$ and $g_n(a_n)=n.$



You can easily produce variations on this theme.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you. This is a nice example.
    $endgroup$
    – ProblemBook
    Mar 27 at 2:28















1












$begingroup$

In response to a comment by the proposer to another A:



For $nin Bbb Z^+$ let $g_n(x)=0$ for $xnot in [2^-n-1,2^-n]$ and let $g_n$ be integrable on $[2^-n-1,2^-n]$ such that $sup ge 1.$ And such that $lim_nto inftyint_0^1g_n(t)dt=0.$



Let $g_0(x)=0$ for all $x.$



For $nge 0$ let $f_n=g_n-g_n+1 .$



Then $sum_j=0^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $sum_j=0^nint_0^1f_j(t)dt=-int_0^1 g_n(t)dt$ converges to $0.$



For example, for $nin Bbb Z^+:$ Let $a_n=(2^-n-1+2^-n)/2.$ Let $g_n$ be linear on $[2^-n-1,a_n]$ and on $[a_n,2^-n]$ with $g_n(2^-n-1)=0=g_n(2^-n)$ and $g_n(a_n)=n.$



You can easily produce variations on this theme.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you. This is a nice example.
    $endgroup$
    – ProblemBook
    Mar 27 at 2:28













1












1








1





$begingroup$

In response to a comment by the proposer to another A:



For $nin Bbb Z^+$ let $g_n(x)=0$ for $xnot in [2^-n-1,2^-n]$ and let $g_n$ be integrable on $[2^-n-1,2^-n]$ such that $sup ge 1.$ And such that $lim_nto inftyint_0^1g_n(t)dt=0.$



Let $g_0(x)=0$ for all $x.$



For $nge 0$ let $f_n=g_n-g_n+1 .$



Then $sum_j=0^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $sum_j=0^nint_0^1f_j(t)dt=-int_0^1 g_n(t)dt$ converges to $0.$



For example, for $nin Bbb Z^+:$ Let $a_n=(2^-n-1+2^-n)/2.$ Let $g_n$ be linear on $[2^-n-1,a_n]$ and on $[a_n,2^-n]$ with $g_n(2^-n-1)=0=g_n(2^-n)$ and $g_n(a_n)=n.$



You can easily produce variations on this theme.






share|cite|improve this answer









$endgroup$



In response to a comment by the proposer to another A:



For $nin Bbb Z^+$ let $g_n(x)=0$ for $xnot in [2^-n-1,2^-n]$ and let $g_n$ be integrable on $[2^-n-1,2^-n]$ such that $sup ge 1.$ And such that $lim_nto inftyint_0^1g_n(t)dt=0.$



Let $g_0(x)=0$ for all $x.$



For $nge 0$ let $f_n=g_n-g_n+1 .$



Then $sum_j=0^nf_j=g_0-g_n=-g_n$ converges point-wise, but not uniformly, to $0$. And $sum_j=0^nint_0^1f_j(t)dt=-int_0^1 g_n(t)dt$ converges to $0.$



For example, for $nin Bbb Z^+:$ Let $a_n=(2^-n-1+2^-n)/2.$ Let $g_n$ be linear on $[2^-n-1,a_n]$ and on $[a_n,2^-n]$ with $g_n(2^-n-1)=0=g_n(2^-n)$ and $g_n(a_n)=n.$



You can easily produce variations on this theme.







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answered Mar 25 at 1:06









DanielWainfleetDanielWainfleet

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  • $begingroup$
    Thank you. This is a nice example.
    $endgroup$
    – ProblemBook
    Mar 27 at 2:28
















  • $begingroup$
    Thank you. This is a nice example.
    $endgroup$
    – ProblemBook
    Mar 27 at 2:28















$begingroup$
Thank you. This is a nice example.
$endgroup$
– ProblemBook
Mar 27 at 2:28




$begingroup$
Thank you. This is a nice example.
$endgroup$
– ProblemBook
Mar 27 at 2:28

















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