(Proof Verification) Continuous image of a compact set must be bounded.

Question

Let $D\subseteq \mathbb{E}^n$ and suppose that $\mathbf{f}:D\to \mathbb{E}^m$. Show that if $D$ is compact and if $\mathbf{f}$ is continuous on $D$ then $\mathbf{f}$ is bounded on $D$ (that is, $M=\sup\{\mathbf{f}(\mathbf{x}):\mathbf{x}\in D\}<\infty$ and there is a point $\mathbf{x_o}\in D$ such that $\mathbf{f}(\mathbf{x})=M$.

To prove that $\mathbf{f}$ is bounded, I used the Extreme Value Theorem which states if $f:D\to\mathbb{R}$ is continuous then $f$ achieves both a maximum value and a minimum value on $D$ (not sure if this also works for vector-valued functions $\mathbf{f}$ but this is what is provided to me in my book).

Since each component of $\mathbf{f} = (f_1,f_2,\cdots,f_m)$ is also continuous, I can use EVT to find maximum values, namely, $M_1,M_2, \cdots, M_m$ for each $f_1,f_2,\cdots,f_m$ respectively such that $f_i(\mathbf{x}) \leq M_i$ for all $\mathbf{x}\in D$.

Therefore, \begin{align} \|\mathbf{f}(\mathbf{x})\| &= \sqrt{f_1(\mathbf{x})^2 + \cdots + f_m(\mathbf{x})^2} \\ &\leq \sqrt{M_1^2 + \cdots + M_m^2} \\ &= M \end{align} and the function $\mathbf{f}$ must be bounded. But I am not sure if the $M$ that I found is $M=\sup\{\mathbf{f}(\mathbf{x}):\mathbf{x}\in D\}<\infty$.

How should I prove that there exists $\mathbf{x_0}$ such that $\mathbf{f}(\mathbf{x_0})=M$?

Answer 1

You can't expect for your $M$ to be the sup-norm of $\mathbf{f}$ (i.e. $\sup_{\mathbf{x} \in D} \lVert \mathbf{f}(\mathbf{x}) \rVert$). In fact, you can't expect to find any nice expression for the sup-norm of a function. The best you can do is an existence proof.

If $D$ is compact, then $\lVert \mathbf{f}(\mathbf{x}) \rVert$ achieves its maximum on $D$ since it is a continuous function on a compact domain. You can use the Extreme Value Theorem here to see this or just go through the proof of the Extreme Value Theorem again:

Let $\lVert \mathbf{f}(\mathbf{x}_1) \rVert, \lVert \mathbf{f}(\mathbf{x}_2) \rVert, \lVert \mathbf{f}(\mathbf{x}_3) \rVert, \ldots$ be a sequence that tends to $\sup_{\mathbf{x} \in D} \lVert \mathbf{f}(\mathbf{x}) \rVert$ (we don't assume this latter value is finite at the moment). Since $D$ is compact, there is a convergent subsequence $\mathbf{x}_{n_1}, \mathbf{x}_{n_2}, \mathbf{x}_{n_3} \to \mathbf{x}_0 \in D$. Then, by continuity and since a subsequence of a convergent sequence has the same limit,

$$ \lVert \mathbf{f}(\mathbf{x}_0) \rVert = \left\lVert \mathbf{f}\left( \lim_k \mathbf{x}_{n_k} \right) \right\rVert = \lim_k \lVert \mathbf{f}(\mathbf{x}_{n_k}) \rVert = \sup_{\mathbf{x} \in D} \lVert \mathbf{f}(\mathbf{x}) \rVert. $$