Let $(V_1, \lVert \cdot \rVert_{V_1})$ be a normed vector space and $(V_1, \lVert \cdot \rVert_{V_2})$ be a complete normed vector space. Let $C_b(V_1, V_2)$ be the set of all continuous bounded functions $f:V_1\to V_2$. Prove that $\left(C_b(V_1, V_2),\lVert\cdot\rVert_\infty\right)$ is complete.
I'm a bit puzzled here. Let $\{x_n\}\subseteq V_1$ be a Cauchy sequence. Since $V_1$ is not necessarily complete, $\{x_n\}$ may not converge in $V_1$. Now let $\{f_m\}\subseteq C_b(V_1,V_2)$ be a sequence of functions. Then is it true that $f_m(\lim\limits_{n\to \infty} x_n)$ is generally undefined if $V_1$ is not complete?
Also, we know that if $f_m$ converges to some function $f$, then $\lim\limits_{m\to \infty} \lVert f_m - f_k \rVert = 0$ implies that $f_m$ converges uniformly. Thus $f \in C(V_1,V_2)$. All we need to prove in this case is that $f$ is also bounded. Is this correct?
My approach to proof:
Let $\{f_m\}\subseteq C_b(V_1, V_2)$ be a Cauchy sequence of functions, then $f_m$ converges uniformly, which implies that $f_m(x)$ converges pointwise for all $x\in V_1$. So, $f_m(x)\subseteq V_2$. Let $S\subset V_1$ be a compact set and restrict $f_m$ to $f_m:S\to V_2$. Then, for all $x\in S$, $f_m$ maps $x$ to some $y\in K\subset V_2$, where $K$ is a compact set. Since $V_2$ is complete, $K$ is complete. Thus, for all $x\in S$, $f_m(x)$ converges to $f(x) \in K$, so $f(x)$ is bounded. Since $\{f_m\}$ converges uniformly, $f$ must be bounded, hence $f\in C_b(V_1, V_2)$.
Now I just realized that I have an error in my proof, since $f_m$ is bounded on $V_1$, regardless of whether or not $f_m$ maps compact sets to compact sets. So, what if we suppose that $f:V_1\to H\subset V_2$ and extend $H$ to $closure(H)$, so $closure(H)$ is compact? Then $closure(H)$ is complete, so $f_m$ must converge either in $H$ or on its boundary.
Please let me know if this is correct.