Find, with proof, a compact set $K$ and a Banach space $W$ such that $C(K,W)$ equipped with the sup norm (i.e. $\|f\|_\infty = \sup\{\|f(x)\|_W : x \in K\}$) the set of continuous functions from $K$ to $W$ is not a Banach space.
Using a similar proof to the proof that $C([0,1]) = C([0,1], \mathbb{R})$ is a Banach space, it can be shown that $C([a,b],\mathbb{R})$ for any $a<b\in\mathbb{R}$ is a Banach space. However, I'm unsure how to find the given set $K$ and Banach space $W$. I think it might have something to do with uniform continuity of functions not "behaving" as well in certain Banach spaces as in $\mathbb{R}$.
Best Answer
The condition $K$ compact and $W$ Banach always makes $C(K,W)$ Banach.
The fact that $K$ is compact guarantees that $\|\cdot\|_\infty$ is a norm.
Let $\{f_n\}\subset C(K,W)$ be a Cauchy sequence. For each $k\in K$, we have $$\|f_n(k)-f_m(k)\|=\|(f_n-f_m)(k)\|\leq\|f_n-f_m\|_\infty,$$ so $\{f_n(k)\}$ is Cauchy in $W$. As $W$ is Banach, there exists $f(k)=\lim_nf_n(k)\in W$. Now fix $\varepsilon>0$. Take $n_0$ such that $\|f_n-f_m\|<\varepsilon$ when $n,m>n_0$. For $n,m>n_0$, $$ \|f_n(k)-f(k)\|\leq\|f_n(k)-f_{m}(k)\|+\|f_{n_0}(k)-f(k)\|\leq \varepsilon +\|f_{m}(k)-f(k)\|. $$ As this holds for any $m$, we obtain $\|f_n(k)-f(k)|<\varepsilon$ for all $n>n_0$, and so the convergence is uniform. This makes $f$ continuous: indeed, fix $\varepsilon>0$. Then there exists $n$ such that $\|f_n-f\|<\varepsilon/3$. Let $\delta>0$ such that $\|f_n(k)-f_n(h)\|<\varepsilon/3$ when $\|k-h\|<\delta$; then $$ \|f(k)-f(h)\|\leq\|f(k)-f_n(k)\|+\|f_n(k)-f_n(h)\|+\|f_n(h)-f(h)\|\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon $$ and $f$ is continuous. So $C(K,W)$ is Banach.