Let $A$ be a $C^*$-algebra. $M(A)$ denotes the set of all multipliers of $A$, i.e. $m\in M(A)$ means that there is a map $m^*:A\to A$ such that $m(a)^*b=a^*m^*(b)$ for all $a,b\in A$.
I want to know why $*:M(A)\to M(A),\; m\mapsto m^*$ defines an involution on $M(A)$. I already know that $(m^*)^*=m$ and $(m\circ n)^*=n^*\circ m^*$ for every $m,n\in M(A)$.
But why is $(\lambda m+\mu n)^*=\overline{\lambda}m^*+\overline{\mu} n^*$ for every $\lambda,\mu\in\mathbb{C}$,$m,n\in M(A)$? I think it is an easy calculation, but maybe I have seen too many stars today and now I'm confused. But i'm stuck.
My try: $((\lambda m+\mu n)(a))^*b=\overline{\lambda}m(a)^*b+\overline{\mu}n(a)^*b=\overline{\lambda}a^*m^*(b)+ \overline{\mu}a^*n^*(b)$ for every $b\in A$. But I don't know how to continue.
You are almost done: \begin{align} ((\lambda m+\mu n)(a))^*b&=\overline{\lambda}m(a)^*b+\overline{\mu}n(a)^*b=\overline{\lambda}a^*m^*(b)+ \overline{\mu}a^*n^*(b)=a^*(\bar\lambda m^*(b)+\bar\mu n^*(b))\\ &=a^*(\bar\lambda m^* +\bar\mu n^*)(b). \end{align} But now you know that $((\lambda m+\mu n)(a))^*b=a^*(\lambda m+\mu n)^*(b)$. If you look at the right-hand sides, you have shown that $$ a^*(\bar\lambda m^* +\bar\mu n^*)(b)=a^*(\lambda m+\mu n)^*(b) $$ for all $a,b$, which implies $\bar\lambda m^* +\bar\mu n^*=(\lambda m+\mu n)^*$.