Let $f : [-1,1] \to \mathbb{R}$ be a function such that $f$ and $f'$ are continuous in $[-1,1]$. Show that the series $\sum_{n=1}^{\infty}f(\dfrac{x}{n^2})$ converges uniformly if and only if $f(0)= 0$. Hint: use the mean value theorem.
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If $\sum_{n=1}^{\infty}f(\dfrac{x}{n^2})$ converges uniformly on $[-1,1]$. Then, $\sum_{n=1}^{\infty}f(0)$ converges,so we must have $f(0)=0$. Next, if $f(0)=0$.
Now,We have $f'$ is continuous on $[-1,1]$ hence bounded say, $|f'(x)|<M....(1)$
Then by MVT and using $(1)$ we have, $|f(\dfrac{x}{n^2})|<\frac{M}{n^2}$
So, by Weierstsras's M test, the sum converges uniformly on $[-1,1]$.
"Only if" part: Assume the series converges uniformly. Then for an arbitrary $\epsilon >0$, there exists $n$ large enough such that $|f(\frac{x}{n^2})|<\epsilon,\ \forall x\in [-1,1]$. This means $f([\frac{-1}{n^2},\frac{1}{n^2}])\subset (-\epsilon,\epsilon)$, which implies $f(0)=0$.
"If" part: Assume that $f(0)=0$. Then $f(x)=f(x)-f(0)=xf'(c)$ for some $c$ lies between $0$ and $x$. For $\epsilon >0$, there is $\delta>0$ such that $|f'(x)-f'(0)|<\epsilon$ for all $x\in (-\delta,\delta)$. Hence $|f'(x)|<|f'(0)|+\epsilon,\ \forall x\in (-\delta,\delta)$. There exists $N$ large enough such that $\forall n>N$, $\frac{x}{n^2}\in (-\delta,\delta)$ so $$|f(\frac{x}{n^2})|=|\frac{x}{n^2}||f'(c)|<\frac{1}{n^2}(|f'(0)|+\epsilon),\ \forall n>N$$ It follows that $$\sum\limits_{n>N}|f(\frac{x}{n^2})|<(|f'(0)|+\epsilon)(\sum\limits_{n>N}\frac{1}{n^2}),\ \forall x\in [-1,1]$$ The right-hand side is converges, so the series converges uniformly.