# show operator norm submultiplicative

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

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

• So you think nice proofs have to be complicated? – celtschk Jul 3 '13 at 21:20

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