I found this question. I am going to just copy it here again.
Ley $K$ be a compact Hausdorff space and let $X = C(K)$ be the space of continuous functions defined on $K$ with values in $\mathbb{K}$ alongside the supremum norm $\|\cdot\|_{\infty}$. Show that $L \in X^*$ and $\|L\| = L(\textbf{1})$, if and only if $L$ is a positive operator, that is, $L(f) \geq 0$ when $f \geq 0$. By $\textbf{1} \in X$ we are refering to the constant function $1$.
The proof provided by a user is the following:
Let $a$ be a positive element such that $I(a)<0$. For a small enough $\beta$, $\|1-\beta a\|< 1$, but we have $I(1-\beta a)=I(1)-\beta I(a)>I(1)=\|I\|\geq \|I\| \|1-\beta a\|$, hence we get a contradiction, then $I$ is positive.
I understand that proof, but there is one piece missing. The fact that $K$ was a compact Hausdorff space was never used. My question is then, how does that fact play in?
I think it plays in, in that it gives boundedness of $I$, namely, since $L\in X^*$, then $L<\infty$. But I am not sure of that. Any hint would be appreciated.
You're using it explicitly to get that $\|1-\beta a\|$ is strictly smaller than $1$. Say $K=\mathbb{R}$ (and we're considering the set of bounded continuous functions on $K$, so that the norm makes sense) and $a(x)=\exp(-|x|)$, then $\|1-\beta a\|\geq 1-\lim_{x\to \infty}a(x)=1$, so that part of the argument collapses, even though $a$ is strictly positive.