Let $B(H)$ be a space of bounded linear operator on Hilbert space $H$.
Where operator norm is defined as follows
$$\lVert A\rVert_{op} = \sup_{\lVert v\rVert = 1}\lVert Av\rVert$$
Show that for $A,B \in B(H)$ the inequality below is true:
$$\lVert AB\rVert_{op} \leq \lVert A\rVert_{op} \lVert B\rVert_{op}$$
My attempt:
$$\forall_{v\in H, \lVert v\rVert=1} \lVert (AB)v\rVert = \lVert A(Bv)\rVert = \lVert \lVert Bv \rVert \cdot A(\frac{Bv}{\lVert Bv\rVert})\rVert =
\lVert Bv\rVert \cdot \lVert A(\frac{Bv}{\lVert Bv\rVert})\rVert \leq \lVert B \rVert_{op} \lVert A\rVert_{op}$$
$\Rightarrow \lVert AB \rVert_{op} \leq \lVert B \rVert_{op} \lVert A\rVert_{op}$
Is it correct?
My concern is division by $\lVert Bv \rVert$ under operator $A$ right after second equality sign. Is it allowed, if not how can I fix it?
Best Answer
Note that you don't need to use the division by $||Bv||$ at all because you know that both $A$ and $B$ are bounded. Thus, we have: $$||ABv|| = ||A(Bv)|| \le ||A||||Bv||\le ||A||||B||||v||$$ so that: $$\frac{||ABv||}{||v||} \le ||A||||B||$$ so we can take the $\sup_{||v||=1}$ in the left hand side of the above equation. Note that I'm obviously assuming $v\neq 0$, once this is a trivial case of the inequality.