An exercise in C*-algebra

Question

Let $A$ be a C*-algebra, $\phi$ be a pure state and $L=\{a\in A:\phi(a^{\ast}a)=0\}$, how to prove that $L+L^*\subseteq ker\phi$. ($L^*=\{a^{\ast}: a\in L\}$)

I think it is an easy exercise, but......could someone show me the answer or just give me some hints? Thanks.

Answer 1

To see that $L\subset\ker\phi$, if $a\in L$ then by Kadison's inequality $$ 0\leq|\phi(a)|^2=\phi(a)^*\phi(a)\leq\phi(a^*a)=0, $$ so $\phi(a)=0$. And now for $b\in L^*$, then $b^*\in L$, so $\phi(b)=\overline{\phi(b^*)}=0$ (this last part is just the fact that the kernel of a selfadjoint functional contains adjoints; it is an ideal, in fact).