Bounded linear operators on Hilbert space is a $C^*$ algebra Let $H$ be a Hilbert space, and $B(H\to H)$ is the algebra of boulded linear operators on this space, with adjoint map $T\mapsto T^*$ and the operator norm, I want to show that $B(H\to H)$ is a unital C$^*$ algebra. I can prove that $\|T\|=\|T^*\|$, but how to show the $C^*$ identity, that is $\|T^*T\|=\|T\|\|T^*\|$? 
 A: We have 
$$ \Vert T^*T \Vert = \sup \{\langle T^*T v,w\rangle |\: \Vert v\Vert = 1, \Vert w\Vert = 1 \} =\\= \sup \{\langle T v,Tw\rangle | \: \Vert v\Vert = 1, \Vert w\Vert = 1  \} = \Vert T \Vert^2$$
As you already know $\Vert T\Vert = \Vert T^*\Vert$, so we have $\Vert T^*T \Vert = \Vert T^*\Vert \Vert T\Vert$.
Remark: I used that, for any $T\in B(H\to H)$, 
$$\Vert T \Vert = \sup \{\langle T v,w\rangle |\: \Vert v\Vert = 1, \Vert w\Vert = 1 \}$$
This is easy to prove. Here are the details: 
If $\Vert T \Vert=0$, then $T=0$ and so we have $\Vert T \Vert=0=\sup \{\langle T v,w\rangle |\: \Vert v\Vert = 1, \Vert w\Vert = 1 \}$. 
Suppose $\Vert T \Vert\neq 0$. Note that 
$$\Vert T\Vert = \sup\{\Vert Tv\Vert \:|\: \Vert v \Vert=1\} = \sup\{\Vert Tv\Vert \:|\: \Vert v \Vert=1 \textrm { and } \Vert Tv\Vert \neq 0 \}= \\ =\sup\left \{\left \langle  Tv, \frac{Tv}{\Vert Tv\Vert} \right \rangle \:|\: \Vert v \Vert=1 \textrm { and } \Vert Tv\Vert \neq 0 \right\} \leq \\ \leq \sup \{\langle T v,w\rangle |\: \Vert v\Vert = 1, \Vert w\Vert = 1 \} \leq \Vert T\Vert$$
So we have
$$\Vert T \Vert = \sup \{\langle T v,w\rangle |\: \Vert v\Vert = 1, \Vert w\Vert = 1 \}$$
