Pointwise convergence of series of functions implies uniform convergence?

Question

In my Real Analysis class, a question came up, but I can't come up with a proof or disproof. Here's the conjecture:

If $\sum f_n \rightarrow f$ pointwise on $(a, b) \subseteq \mathbb{R}$ then $\sum f_n \rightarrow f$ converges uniformly on $[c, d] \subseteq (a, b)$.

I'm convinced that divergence may occur at the endpoints and having $[c, d]$ gives us a better interval to work with by getting rid of the bad stuff. To be more specific, since $\sum f_n \rightarrow f$ pointwise, the number series $\sum f_n(x)$ converges $\forall x \in (a, b) \implies f_n(x) \rightarrow 0$. But, we can't come up with a number sequence $(x_n)$ where each $x_n \in [c, d]$ such that $f_n(x_n) \nrightarrow 0$ since we got rid of our endpoints.

Is my conjecture valid? If so, how would I prove or disprove my conjecture?

Answer 1

A counterexample is the series with partial sums

$$S_n(x) = xe^{-x^2}+\sum_{k=1}^{n-1} \left[(k+1)x e^{-(k+1)^2 x^2} - kxe^{-k^2x^2}\right] = nxe^{-n^2x^2},$$

where $S_n(x) \to S(x) =0$ as $n \to \infty$ pointwise for all $x \in \mathbb{R}$. However, convergence is not uniform on $[0,1]$ since $|S_n(1/n) - S(1/n)| = e^{-1} \not \to 0$.

Answer 2

Let $a=-1, b=1, c=-\frac 1 2, d=\frac 1 2$. Let $g_n(x)=n|x|$ for $|x| \leq \frac 1 n$ and $1$ for $|x| > \frac 1 n$. Let $g(x)=1$ for $x \neq 0$ and $g(0)=0$. Then $g_n \to g$ pointwise in $(a,b)$ but not uniformly in $[c,d]$. Now take $f_1=g_1, f_2=g_2-g_1,f_3=g_3-g_2,...$.

Answer 3

Since the counterexample is already given (by Kavi), I'd like to comment about pointwise and uniform convergence.

First, remember that a series is just a sequence (of partial sums) and a sequence can be written as a series as Kavi's answer shows. So, we'll talk only about sequences.

Pointwise convergence means: for each individual $x$, the sequence $f_n(x)$ converges. This means that once you've picked $x$ you can tell if the sequence $f_n(x)$ is close to its limit. Different $x$'s may converge at different rates (I mean their respective sequences $(f_n(x))_n$). You cannot know if the function sequence $(f_n)_n$ itself is close to the (pointwise) limiting function. That's what the notion of uniform limit is for...

Uniform convergence means: Now you can tell if the whole function sequence $(f_n)_n$ is close to the (pointwise) limiting function. By this I mean comparing their graphs. From some $N\in\mathbb{N}$ on, the graph of $f_n$ is really close to the graph of its limit for $n\geq N$.