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I'm currently reading on operator norms of matrices on wikipedia and stumbled upon a simple question:

according to the article, for $n,m\in \mathbb{N}$ and for norms $\Vert \cdot \Vert_{\alpha}$, $\Vert \cdot \Vert_{\beta}$ on $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, the function $$\Vert \cdot \Vert: \mathbb{R}^{n\times m} \to \mathbb{R} \ , \ A\mapsto \sup{\{\Vert Ax \Vert_{\beta}: x \in \mathbb{R}^n, \Vert x \Vert_{\alpha}=1\}}$$ defines a matrix norm on the space $\mathbb{R}^{m \times n }$ of $m\times n$- matrices over $\mathbb{R}$. Now my question is: how do I know that the supremum of this set for a given matrix $A$ exists? Surely it would make sense if the set is finite, because then it would be bounded. I'm sure there's some vital information here that would explain this, but that I'm not aware of.

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  • $\begingroup$ In finite dimensions the sphere is compact. And the function $\Vert Ax \Vert_{\beta}$ is continuous. Hence the sup exists and in fact is a max in this case. $\endgroup$
    – lcv
    Commented Apr 3, 2022 at 23:23

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Let $x \in \mathbb R^n$ with $\|x\|_\alpha=1$, and write it as $x=x_1e_1+\cdots+x_ne_n$, where $e_1,\dots,e_n$ is the standard basis for $\Bbb R^n$. Since $\|\cdot\|_\alpha$ is equivalent to the norm $\|\cdot\|_\infty$, we can assume that there exists $c>0$ such that $\|\cdot\|_\infty \leq c\|\cdot\|_\alpha$. Then $$\begin{align*} \|Ax\|_\beta &= \|x_1Ae_1+\cdots+x_nAe_n\|_\beta \\ &\leq |x_1|\|Ae_1\|_\beta + \cdots |x_n|\|Ae_n\|_\beta \\ &\leq \|x\|_\infty (\|Ae_1\|_\beta + \cdots + \|Ae_n\|_\beta) \\ &\leq c(\|Ae_1\|_\beta + \cdots + \|Ae_n\|_\beta). \end{align*}$$ Thus $c(\|Ae_1\|_\beta + \cdots + \|Ae_n\|_\beta)$ is an upper bound for $\{\|Ax\|_\beta : x \in \mathbb R^n,\, \|x\|_\alpha=1\}$.

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