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We had in our lecture on numerical analysis the following: Let $\mathrm{Lin}(X,Y)$ be the set of all linear maps $X\rightarrow Y$. Let $A\in\mathrm{Lin}(\mathbb R^l,\mathbb R^n)$ and $B\in\mathrm{Lin}(\mathbb R^n,\mathbb R^m)$ and $\|C\|_{op}:=\max_{\|x\|\leq1}\|C(x)\|$.

Then our lecturer followed $\|A\circ B\|_{op}\leq\|A\|_{op}\cdot\|B\|_{op}$. So he didn't prove it and so I've tried it by my own.

My attempt:

$$\begin{aligned} \|A\circ B\|_{\mathrm{op}} &= \max_{\|x\| < 1}\|(A \circ B)(x)\| \\ & \leq \max_{\|x\| < 1} \|A\|_{\mathrm{op}} \|B(x)\| \\ &= \|A\|_{\mathrm{op}} \max_{\|x\|\leq1}\|B(x)\|\\ &= \|A\|_{\mathrm{op}}\|B\|_{\mathrm{op}} \end{aligned}$$

But this seems too easy. I am really interested in a nice proof so anybody could help? Thanks a lot!

Rodrigo de Azevedo
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user58679
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2 Answers2

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$$ ||AB||=\max_{x \ne 0} \frac{||ABx||}{||x||}=\max_{Bx \ne 0}\frac{||ABx|| }{||x||} =\max_{ Bx\ne 0}\frac{||ABx||}{||Bx||} \frac{||Bx||}{||x||}\le \max_{y \ne 0} \frac{||Ay||}{||y||} \max_{x \ne 0} \frac{||Bx||}{||x||} $$ It shows that: $$ ||AB||\le ||A|| ||B|| $$

dresden
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    @Ramita actually I think the inclusion goes in the other direction: $Bx\neq0\Longrightarrow x\neq0$, thus in general $\max_{x\neq 0}(...)\ge\max_{Bx\neq0}(...)$. Nonetheless, the $x\neq0$ such that $Bx=0$ correspond to vanishing norm, which only achieve the maximum if the maximum is $0$; so we can restrict the maximisation to $x\neq0$ that are also such that $Bx\neq0$. – glS Feb 23 '21 at 23:07
  • Absolutely, thanks, I deleted the comment. – N. Pullbacki Feb 26 '21 at 01:45
  • I feel dumb, I don't see at all why $\max_{x\neq0} \frac{|ABx|}{|x|} = \max_{Bx \neq 0}\frac{|ABx|}{|x|}$ – César VB Jan 12 '24 at 04:00
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This is strange to look for a nicer proof for such a simple problem. Your solution is correct and as short as it could be.

celtschk
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Norbert
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