Suppose $\lim_{h\to 0}f_n(h) = 0$ such that $g(h)=\sum_{n=1}^{\infty}f_n(h)$ converges for any $h$. Can we tell that $\lim_{h\to0}g(h) = 0$ ? i.e can we change the order of limit and sum here ? If not what is needed to make it happen ?

I don't know how to use DCT/BCT here.

Is it true if $f_n(h) = \int_{E_n} |f(x+h) -f(x)| dx$ where $E_n$ is an interval of length $[nh,(n+1)h]$ where $f$ is bounded and integrable.

  • $\begingroup$ I can't understand why you are asking to 'change the order of integration'! $\endgroup$ – Gaurav Tiwari Mar 1 '14 at 12:04
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    $\begingroup$ Let $f_n$ be the continuous function whose graph consists of the straight line segments connecting $(0,0)$, $(1/n,1)$, and $(2/n,0)$. I think $(f_n)$ will give you a counterexample. $\endgroup$ – David Mitra Mar 1 '14 at 12:18

No: Consider $f_n(h)=h^{n-1}$ for $h\in(-1,1),n\in\mathbb N$. Then $f_n(h)\to0$ for $h\to0$ and any $n$, but $$ g(h)=\sum_{n\in\mathbb N}f_n(h)=\sum_{n\in\mathbb N}h^{n-1}=\frac{1}{1-h}, $$ which does not tend to 0 as $h\to0$. To interchange limits you need that the series converges uniformly

EDIT: This example doesnt work (see comment). But how about this: Since $\mathbb Q\backslash\{0\}$ is countable, there is a bijective mapping $a:\mathbb N\to\mathbb Q\backslash\{0\}$. Now define $f_n(h)=0$ if $h\neq a(n)$ and $f_n(h)=1$ if $h=a(n)$. Then each $f_n$ is continuous at $0$, since $f_n$ is 0 everywhere except at $a(n)\neq0$. Since $a$ is bijective we have that $g(h)=1$ if $h\in\mathbb Q\backslash\{0\}$ and $g(h)=0$ otherwise. In particular, $g(1/n)=1$ for any $n\in\mathbb N$.

  • $\begingroup$ $f_1$ does not have limit $0$ at $x=0$. Also, your sum does converge uniformly on a closed subinterval of $(-1,1)$. $\endgroup$ – David Mitra Mar 1 '14 at 12:15
  • $\begingroup$ Yeah youre right. Your example seems a lot better :) $\endgroup$ – sranthrop Mar 1 '14 at 12:30
  • $\begingroup$ @sranthrop: I have added something. $\endgroup$ – aaaaaa Mar 1 '14 at 13:03

$$\lim_{h\to0}g(h) = \lim_{h \to 0} \sum_{n=1}^{\infty} f_n(h)=\sum_{n=1}^{\infty} \left({\lim_{h \to 0} f_n(h)}\right) = 0 $$ as $$\sum_{n=1}^{\infty} \left({\lim_{h \to 0} f_n(h)}\right) = \sum_{n=1}^{\infty} 0 = 0 $$

  • $\begingroup$ this is not true in general! $\endgroup$ – sranthrop Mar 1 '14 at 12:06
  • $\begingroup$ $\sum_{n=1}^\infty0=\lim_{m\to\infty}\sum_{n=1}^m0=\lim_{m\to\infty}0=0...$ $\endgroup$ – sranthrop Mar 1 '14 at 12:11
  • $\begingroup$ What about, $\sum_{n=1}^{\infty} 0 = 0 \times \infty$ ? $\endgroup$ – Gaurav Tiwari Mar 1 '14 at 12:16
  • $\begingroup$ this doesnt make sense since the left hand side is defined as the limit of its partial sums which is 0. $\endgroup$ – sranthrop Mar 1 '14 at 12:21
  • $\begingroup$ If it's so, then my first answer appears correct. $\endgroup$ – Gaurav Tiwari Mar 1 '14 at 12:31

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