Is $K$ is convex?

Question

Consider the sup-normed Banach space $C(S)$, which designates the space of continuous functions over a compact Hausdorff Space $S$.

Let $K=\{\mu\in A^{\perp}:||\mu||\leqslant 1\}$.

I tried to prove $K$ is convex.

If $0<\alpha<1$ and $\mu_1,\mu_2\in K$

Defining:

$\beta=\alpha\mu_1+(1-\alpha)\mu_2$

Then it needs to be proven $\beta\in K$ for $K$ to be convex.

$\int f d\beta=\int f d\alpha\mu_1+\int fd(1-\alpha)\mu_2=0$

$||\beta||=||\alpha\mu_1+(1-\alpha)\mu_2||\leqslant\alpha||\mu_1||+(1-\alpha)||\mu_2||\leqslant \alpha\times 1+ (1-\alpha)\times 1=1 $

Then $\beta\in K$ which implies that $K$ is convex.

Question:

Is my proof right? If not, how should I prove $K$ to be convex?

Thanks in advance!