For $m, d\ge 1$, let $f_n: K\to \mathbb{R}^m$ be a continuous function for every $n\ge 0$, where $K\subset \mathbb{R}^d$ is compact. Show that if $\sum_{n=0}^\infty \|f_n\|_K <\infty$ then $f(x):=\sum_{n=0}^\infty f_n(x) <\infty$ is also a continuous function $f:K\to\mathbb{R}^m$.
Proof:
$\infty>\sum_{n=0}^\infty \|f_n\|_K =\sum_{n=0}^\infty \max\{\|f_n(x)\|:x\in K\} \ge \>\sum_{n=0}^\infty \|f_n(x)\|$, thus $\sum_{n=0}^\infty f_n(x)$ converges for any $x\in K$, which implies that $f(x)$ exists. This also implies that $\forall \varepsilon > 0, \exists N>0$, such that $k>N\implies \|\sum_{n=0}^k f_n(x) – f(x)\|<\varepsilon/2$.
Since $f_n(x)$ is continuous for every $x\in K$, then $\forall \varepsilon >0, \exists \delta > 0$ such that, for $x_0\in K$, $\|x-x_0\|<\delta\implies \|f_n(x)-f_n(x_0)\| < \varepsilon$.
Now, let $k>N$, then $\left\| \sum_{n=0}^k f_n(x)-\sum_{n=0}^k f_n(x_0)\right\|
=\left\| \sum_{n=0}^k f_n(x)+f(x_0)-f(x_0)-\sum_{n=0}^k f_n(x_0)\right\|\le$
$\left\| \sum_{n=0}^k f_n(x)-f(x_0)\right\|+\left\|\sum_{n=0}^k f_n(x_0)-f(x_0)\right\|$,
and this is where I get stuck since I do not know what to do with the $\left\| \sum_{n=0}^k f_n(x)-f(x_0)\right\|$ term, so some help would be appreciated. Is there something that I do not know about compact sets which can be applied here?
Update: I've just found out that sequences of continuous functions are uniformly convergent on compact sets, so I would change my last inequality as follows:
$\left\| \sum_{n=0}^k f_n(x)-\sum_{n=0}^k f_n(x_0)\right\|
=\left\| \sum_{n=0}^k f_n(x)+f(x)-f(x)-\sum_{n=0}^k f_n(x_0)\right\|\le$
$\left\| \sum_{n=0}^k f_n(x)-f(x)\right\|+\left\|\sum_{n=0}^k f_n(x_0)-f(x)\right\|
\le \varepsilon/2+\sup\limits_{x\in K}\left\|\sum_{n=0}^k f_n(x)-f(x)\right\|<\varepsilon/2+\varepsilon/2=\varepsilon$
Do you think this is correct now?
Best Answer
Since $\sum_{n=0}^\infty \|f_n\|_K < \infty$ for any $\epsilon > 0$ there exists $M \in \mathbb{N}$ such that
$$\sum_{n=M+1}^\infty \|f_n\|_K < \frac{\epsilon}{3},$$
and for every $x \in K$ we have
$$\begin{align}\left\|f(x) - \sum_{n=0}^M f_n(x)\right\| &= \left\|\sum_{n=M+1}^\infty f_n(x)\right\| \\ &\leqslant \sum_{n=M+1}^\infty \|f_n(x)\| \\ &\leqslant \sum_{n=M+1}^\infty \|f_n\|_K \\ &< \frac{\epsilon}{3}\end{align}.$$
Hence, if $x,x_0 \in K$ we have
$$\left\|f(x) - f(x_0) \right\| \leqslant \left\|f(x) - \sum_{n=0}^M f_n(x)\right\| + \left\|f(x_0) - \sum_{n=0}^M f_n(x_0)\right\| + \left\|\sum_{n=0}^M f_n(x) - \sum_{n=0}^M f_n(x_0)\right\| \\ < \frac{2\epsilon}{3} + \left\|\sum_{n=0}^Mf_n(x) - \sum_{n=0}^M f_n(x_0)\right\|. $$
Since $f_n$ is continuous on $K$ for every $n$, the finite sum $\sum_{n=0}^M f_n$ (where $M$ is fixed) is continuous at $x_0 \in K$. There exists $\delta > 0$ (which may depend on $x_0$ and $M$) such that if $\|x - x_0\| < \delta$ then
$$\left\|\sum_{n=0}^Mf_n(x) - \sum_{n=0}^M f_n(x_0)\right\| < \frac{\epsilon}{3},$$
and, consequently,
$$\left\|f(x) - f(x_0) \right\| < \epsilon.$$
Therefore $f$ is continuous at each point $x_0 \in K$. Note that compactness of $K$ was not needed to prove continuity. If, however, $K$ is compact then we have automatically uniform continuity of $f$ on $K.$
Regarding your update -- a sequence of continuous functions on a compact set need not be uniformly convergent. A counterexample is $f_n(x) = x^n$ where each $f_n$ is continuous on the compact interval $[0,1]$ but the sequence is not uniformly convergent.