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Let $f_n$ be a sequence of functions ($[a,b] \to \mathbb{R}$) converging uniformely to zero. If it's also the case that the series $\sum f_n$ is pointwise convergent, can I conclude that $\sum f_n$ is uniformely convergent? I think the answer should be YES but I'm confused on how to prove it. I didn't find any theorem or proposition stating this in my notes or my book, maybe its too obvious. What I did find in my book is a similar result but with the special case of power series, namely, if $f_n(x)=a_nx^n$ then $\sum f_n$ converges uniformely in any compact subset of the interval of (pointwise) convergence of the series.

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    Clarification: is the criterion that the sequence $\{f_n\}$ converges uniformity to $0$ or that for every $i$, $f_i$ converges uniformity to zero? The former makes the claim false but I think the later makes the claim true. – 2017-02-02
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    @StellaBiderman the sequence $\{f_n\}$ converges uniformely to zero, but doesn't it imply that for each $i$ $f_i$ converges to zero? I don't understand what you mean, if the former makes the claim false how can the later make it true? I'm quite confused, can you provide some example or counterexample? Thanks – 2017-02-02
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    Sorry, I slightly misread the problem. I understand now. I thought there was a sequence of evaluation points for some reason. My second case isn't meaningful in this context. – 2017-02-02

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Counterexample:

$$[a,b] := [0,1]$$ $$f_n(x) := \left\{\begin{matrix}1/n&\text{ if }0

Then $$\sup_{x\in[0,1]}\left|f_n(x)\right|=1/n\to_{n\to\infty}0\text,$$

hence $(f_n)$ converge uniformly to $0$. Also, for every $x\in[0,1]$, the sum $\sum_{n=1}^\infty f_n(x)$ is finite, thus $(\sum f_n)$ converges pointwise.

Let $f(x):=\sum_{n=1}^\infty f_n(x)$ be the pointwise limit. Then suppose the convergence were uniform, so by definition we would have

$$\sup_{x\in[0,1]}\left|f(x)-\sum_{n=1}^k f_n(x)\right|\to_{k\to\infty} 0\text.$$

Hence for any sequence $(x_k)$ in $[0,1]$:

$$\left|f(x_k)-\sum_{n=1}^k f_n(x_k)\right|\le\sup_{x\in[0,1]}\left|f(x)-\sum_{n=1}^k f_n(x)\right|\to_{k\to\infty} 0\text.$$

Let $x_k:=1/k^2$. Then

$$\left|f(x_k)-\sum_{n=1}^k f_n(x_k)\right| = \sum_{n=k+1}^\infty f_n(1/k^2) = \sum_{n=k+1}^{k^2} 1/n\ge \sum_{n=k+1}^{k^2} 1/k^2 = \frac{k^2-k}{k^2}\to_{k\to\infty}1\text.$$

Contradiction.

$f_n$ continuous

Another counterexample is

$$f_n(x) := \left\{\begin{matrix}\sqrt{x}&\text{ if }x\le 1/n\\0&\text{ if }x \ge 1/(n-1)\\ (1-n)\sqrt{n}\left(x + \frac1{1-n}\right)&\text{ otherwise}\end{matrix}\right.$$

where the last line is obtained by linear interpolation on $[1/n, 1/(n-1)]$.

     ↑ 
1/√n +     .
     |   ◜  \
     | ◜     \
     |◜       \
     •-----+---+•••••••+-->
    0    1/n  1/(n-1)  1

Note that

  • the $f_n$ are continuous
  • $\sup_{x\in[0,1]}\left|f_n(x)\right|=1/\sqrt n$
  • for all $t\in[0,1]$ we have $n\ge t+1\implies f_n(1/t) = 0$

From there, you can easily use the same technique as above to show that $\sum f_n$ does not converge uniformly.