Proving $ \sum_{n=1}^{ \infty } f_n(1) $ converges when $ \sum_{n=1}^{ \infty } f_n(x) $ converges uniformly

Question

I have a question which I tried solving for a few hours.
My only solid direction so far is trying using Cauchy.

Let $(f_n(x))$ be a series of continuous functions in $[0,1]$.
Prove; If $ \sum_{n=1}^{ \infty } f_n(x) $ converges uniformly at $[0,1)$, then $ \sum_{n=1}^{ \infty } f_n(1) $ converges.

Thank you in advance, been trying a lot solving this. Much appreciated.

Answer 1

If $f(x)=\sum_{n=1}^\infty f_n(x)$ converges uniformly on $(0,1)$, then $\forall \epsilon>0$ there exists $N$ s.t. $$\left|\sum_{n=N}^\infty f_n(x) \right| < \epsilon$$ for all $x\in (0,1)$. Furthermore $\sum_{n=N}^\infty f_n(x)$ is uniformly continuous on $(0,1)$ (and can therefore be continued to $[0,1]$) since (by uniform continuity of $f_n(x)$ on $(0,1)$ and uniform convergence of $f(x)$) $\forall \epsilon'>0$ there is $N'>N$ and $\delta(N')=\delta'>0$, s.t. $\forall x,x' \in (0,1)$ with $|x-x'|<\delta'$ $$\left|\sum_{n=N}^\infty f_n(x)-f_n(x') \right| \leq \sum_{n=N}^{N'-1} \underbrace{\left| f_n(x)-f_n(x') \right|}_{<\frac{\epsilon'}{N'-N}} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x)\right|}_{<\epsilon'} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x') \right|}_{<\epsilon'}<3\epsilon' \, .$$

For $x=1-\delta$ with $\delta >0$ this gives $$\left|\sum_{n=N}^\infty f_n(1) \right| \leq \left|\sum_{n=N}^\infty f_n(1) - f_n(1-\delta) \right| + \left|\sum_{n=N}^\infty f_n(1-\delta) \right|<2\epsilon$$ The second term is less than $\epsilon$ by definition of uniform convergence, while the first term can be made less than $\epsilon$ for small enough $\delta$ due to the continuity of $\left|\sum_{n=N}^\infty f_n(x) \right|$ on $[0,1]$.

Answer 2

Maybe not the most elegant, but a detailed way to get it by "splitting $\varepsilon$."

Let $S_n\colon[0,1]\to\mathbb{R}$ be the function defined by $S_n(x) = \sum_{k=1}^n f_k(x)$, for $n\geq 1$ and $x\in[0,1]$. Since the $f_n$'s are continuous on a compact set, they are uniformly continuous, so is $S_n$. Moreover, by assumption the sequence $(S_n)_n$ converges uniformly on $(0,1)$, to a limit which we will call $S\colon(0,1)\to\mathbb{R}$.

Note that $S$ is then uniformly continuous (can you see why?), and thus can be extended on $[0,1]$ to a continuous function. I.e., we get that $S$ is continuous on $[0,1]$.

• Fix $\varepsilon > 0$. By uniform convergence, there exists $n_\varepsilon\geq 1$ such that, for every $n\geq n_\varepsilon$, $\sup_{x\in(0,1)} |S_n(x)-S(x)| \leq \varepsilon/3$.

• Moreover, since $S_{n_\varepsilon}$ is continuous on $[0,1]$, there exists $\delta_\varepsilon>0$ such that, for all $x\in(1-\delta_\varepsilon,1)$, $|S_{n_\varepsilon}(1)-S_{n_\varepsilon}(x)| \leq \varepsilon/3$. (Using continuity at $1$).

• Moreover, since $S$ is also continuous on $[0,1]$, there exists $\delta'_\varepsilon>0$ such that, for all $x\in(1-\delta'_\varepsilon,1)$, $|S(1)-S(x)| \leq \varepsilon/3$. (Using continuity at $1$).

By the triangle inequality, combining the two inequalities for $S_{n_\varepsilon}$, we get that for all $x\in(1-\min(\delta_\varepsilon,\delta'_\varepsilon),1)$, $$ |S_{n_\varepsilon}(1)-S(1)| \leq \varepsilon $$ Since $\varepsilon>0$ was arbitrary, this shows that the sequence $(S_n(1))_n$ converges, which is what we wanted.