How do you prove $L^{\infty}$ is a C*-algebra?

Question

If we define on $L^{\infty}$ the essential supremum norm ($\| \|_{\infty}$), then how can I prove this norm is submultiplicative ($\| T_1T_2\|_{\infty}\leq \| T_1\|_{\infty}\|T_2 \|_{\infty}\, \forall T_1,T_2\in L^{\infty}$). And if we define the involution like $T\mapsto\bar{T}$\, (where $\bar{T}$ denote the complex conjugate), how can I prove the following equalities, that make $L^{\infty}$ a $C^*$-algebra: $$\| T\|_{\infty}=\| \bar{T}\|_{\infty}$$ $$\|\bar{T} T \|_{\infty}=\|T \|_{\infty}^2$$

Answer 1

Is a trivial consequence of: $$|T_1(x)T_2(x)|\le\|T_1\||T_2(x)|,$$ $$|T(x)|=|\overline{T(x)}|,$$ $$|T(x)\overline{T(x)}|= |T(x)|^2.$$

Answer 2

Since $\vert f(x)\vert=\vert\overline{f(x)}\vert$, it follows immediately that $\Vert f\Vert_\infty=\Vert f\Vert_\infty$.

For any $M\geq 0$, we have $$ \{x:\vert f(x)\vert>M\}=\{x:\vert f(x)\vert^2>M^2\} $$ and therefore $\vert f\overline f\vert_\infty=\vert f^2\vert_\infty=\vert f\vert^2_\infty$.