I'm trying to prove the following claim in my lecture notes on ODEs.
Theorem:
Let $f: [a,b] \to G$ where $G \subset \mathbb{R}^{d}$ is closed and define
$\| f \|_{\infty}=sup_{t}\lvert f(t) \rvert=max_{t}\lvert f(t) \rvert$ where $\vert f(t) \rvert$ denotes a norm on $\mathbb{R}^{d}$.
Then $C([a, b],G)$ with norm $\| f \|_{\infty}$ is a Banach space.
The author notes that it's not necessary to specify which norm we mean since all norms on $\mathbb{R}^{d}$ are equivalent.
I've first looked at a proof of the theorem for $d=1$. A proof can be found here.
The idea is to show that the space of all real-valued bounded functions on $X$, call it $B(X)$, is complete and if $X$ is compact, then the space of continuous functions on $X$ is a closed subset of $B(X)$.
I think this proof can be generalized to real vector-valued functions. So consider the space of bounded real vector-valued functions $B(X,\mathbb{R}^{d})$ and define
$\| f \|_{\infty,vector}=\sup \limits_{t}\| f(t) \|_{2}$ where $\| f(t) \|_{2}$ denotes the Euclidean norm on $\mathbb{R}^{d}$.
Proof: Let $f_{n}$ be a Cauchy sequence in $B(X,\mathbb{R}^{d})$. Given $\epsilon>0$, there exists $N>0$ such that $n,m \geq N \implies \| f_{n} - f_{m} \|_{\infty,vector}<\epsilon$. Hence for all $x \in X$, $\| f_{n}(x) - f_{m}(x) \|_{2} \leq \| f_{n} - f_{m} \|_{\infty,vector}<\epsilon$.
This implies that for each $x \in X$, the sequence of real vectors $\{f_{n}(x)\}$ is a Cauchy sequence. Since $\mathbb{R}^{d}$ is complete, $\{f_{n}(x)\}$ is convergent. Let the limit of $\{f_{n}(x)\}$ be $f(x)$, i.e. f(x) = $\lim \limits_{n \to \infty} f_{n}(x)$. Hence, we obtain a function $f(x)$ on $X$. On the other hand, a Cauchy sequence in a normed space must be bounded, i.e. there exists $M>0$ so that $\| f_{n} \|_{\infty,vector} \leq M$. In other words, for each $x \in X$, $\| f_{n}(x) \|_{2} \leq \| f_{n} \|_{\infty,vector} \leq M$. Taking $n \to \infty$, we find $\| f(x) \|_{2} = \lim \limits_{n \to \infty} \| f_{n}(x) \|_{2} \leq M$, for each $x \in X$ by the continuity of the norm. This shows that $f(x)$ is a bounded function on $X$ and thus $f \in B(X,\mathbb{R}^{d})$. Furthermore, for each $x \in X$, and $n \geq N$, $\| f_{n}(x) - f(x) \|_{2} = \lim \limits_{m \to \infty}\| f_{n}(x) - f_{m}(x) \|_{2} \leq \epsilon$.
This shows that for $n \geq N$, $\| f_{n} - f \|_{\infty,vector} \leq \epsilon$. We find $f$ is the limit of $\{f_{n}\}$ in $B(X,\mathbb{R}^{d})$: This proves that $\{f_{n}\}$ is convergent in $B(X,\mathbb{R}^{d})$.
We can also again show that the space of continuous functions on $X$, call it $C(X,\mathbb{R}^{d})$, is a closed subset of $B(X,\mathbb{R}^{d})$.
Proof: Let f be a limit point of $C(X,\mathbb{R}^{d})$, then f is a bounded real vector-valued function. There exists $\{f_{n}\}$ in $C(X,\mathbb{R}^{d})$ so that $\{f_{n}\}$ converges to f in $B(X,\mathbb{R}^{d})$. To show $f \in C(X,\mathbb{R}^{d})$, we need to show that f is a continuous function.
Given $\epsilon>0$, we can choose $N>0$ so that $\| f_{N} - f \|_{\infty,vector}<\frac{\epsilon}{3}$. Since $f_{N}$ is uniformly continuous, there exists $\delta>0$, so that if $d(x,y)<\delta$, $\| f_{N}(x) - f_{N}(y) \|_{2}<\frac{\epsilon}{3}$ where $d$ is a metric on $X$. If $d(x,y)<\delta$, we see
$\| f(x) - f(y) \|_{2} \leq \| f(x) - f_{N}(x) \|_{2} + \| f_{N}(x) - f_{N}(y) \|_{2} + \| f_{N}(y) - f(y) \|_{2}$
$\leq 2 \| f - f_{N} \|_{\infty,vector} + \frac{\epsilon}{3} < \epsilon$.
This shows that f is uniformly continuous on X and hence continuous.
Now I've got two questions.
Why does it not matter which norm we use on $\mathbb{R^{d}}$ to define the supremum norm in the vector-valued case? I think the definition makes sense since we know that all norms are equivalent on $\mathbb{R^{d}}$, so that the boundedness of the range w.r.t the Euclidean norm can be used to bound the range with any other norm. The crucial step in the proof was that $\mathbb{R^{d}}$ is complete with the Euclidean norm such that the pointwise limit exists. Is $\mathbb{R^{d}}$ with other norms also complete simply by the equivalence of norms?
How can I show the same results for functions with range $G \subset \mathbb{R^{d}}$ where $G$ is closed? I think the proof in general works fine since a closed subset of a complete space is also complete, but I don't see how I can show that the subset of functions with range $G$ is a vector subspace of $C(X,\mathbb{R}^{d})$. How do we know that the space is closed by addition and scalar multiplication?
Edit:
To be more precise regarding 2): How can we show that the set of continuous functions $f: [a,b] \to G$ where $G \subset \mathbb{R^{d}}$ is closed is a vector space. I think it should be a subspace of $C(X,\mathbb{R}^{d})$. But how can we show that it is closed under addition and scalar multiplication?
Thanks very much for any help and suggestions!