Show that a nondegenerate *-Banach algebra is a C*-algebra

Question

Takesaki in his operator theory says

A C*-algebra $M$ of operators on Hilbert space $H$ means a nondegenerate ( $\text {cl} (MH) = H$) $*-$ subalgebra of $B(H)$ which is closed under the uniform topology.

Suppose $M$ is a C*-algebra. Clearly $M$ is a $*-$ Banach algebra and $\text{cl}(MH)\subset H$. Just show that $H\subset \text{cl}(MH)$ . Let $\{u_i\}$ be a approximate identity of $M$. Thus $u_i\to1$(sot) and $\xi = \lim u_i\xi \in \text{cl} (MH)$ for $\xi\in H$.

For converse direction I know that $\|x\|^2=\|xx^*\|$ in general. but I would like to know the relation between nondegenerate and C*-property. Please give me a hint. Thanks.

Answer 1

Equality $\| x\|^2=\| x x^*\|$ holds for any bounded operator $x$ on $H$. Namely, let $\xi \in H$ be of norm $1$. Then $$ \| x\xi\|^2=\langle \xi,x^*x\xi\rangle\leq \| \xi\| \| x^* x\xi\|\leq \| x^* x\|\leq \| x^*\| \| x\|.$$ Since $x^{**}=x$ it follows easily that $\| x^*\|=\| x\|$. Now one has $$ \| x\|^2=\sup\{ \|x \xi\|^2; \xi\in H, \| \xi\|\leq 1\}\leq \sup\{ \| x^* x\xi\|; \xi\in H, \| \xi\|\leq 1\}=\| x^* x\|\leq \| x^*\| \| x\|=\| x\|^2.$$ Hence $\| x^* x\|=\| x\|^2$.