Let $X$ be a locally compact hausdorff space. Then $X$ is normal. Let $K$ be a compact subset of $X$. Denote $C_c(X)$ the set of continuous functions valued in complex number with compact support. Now under sup norm $C_c(X)$ is directed limit of $C_K(X)$ which is the set of continuous functions of $X$ vanishing outside $K$.
In Lang's real and functional analysis, Chpt IX, Lemma 1.1, let $\lambda:C_c(X)\to C$ be a positive linear map.(i.e. For all $C_c(X)\ni f\geq 0$, then $\lambda(f)\geq 0$.) Then $\lambda$ is bounded on $C_K(X)$ for any compact $K$.
The proof is done by picking a function $g\in C_K(X)$ with $supp(g)\supset K$ and $g|_K=1$ by normality and $0\leq g\leq 1$. Then the proof is reduced to real case. Say $f\in C_K(X)$ s.t. $b=sup_{x\in K}|f(x)|$. Clearly $bg\pm f\geq 0$. So $|\lambda(f)|\leq ||f||_\infty\lambda(g)$. Then the book claim $\lambda(g)$ is the desired bound.
$\textbf{Q1:}$ My guess is that the book is assuming $\lambda$ is complex linear for positive linear functional. Is this correct?
$\textbf{Q:}$ Why $\lambda(g)$ is the bound of the $\lambda$? I will assume $f=f_x+if_y$ with $f_x,f_y\in C_K(X)$. $|\lambda(f)|\leq \sqrt{(\lambda(f_x))^2+(\lambda(f_y))^2}$. Pick $b_x,b_y$ be corresponding $sup|f_x|,sup|f_y|$. Then it follows that $|\lambda(f)|\leq\lambda(g)\sqrt{b_x^2+b_y^2}$. However, $b_x^2+b_y^2\geq sup_x|f(x)|$.(There is no reason that $f_x,f_y$ achieve the max at the same point $x$.) So my bound is $2\lambda(g)$ instead. I could not achieve the bound $\lambda(g)$ by this estimation. Did I miss anything?
Best Answer
The proof definitely works when $f$ is real-valued. For complex-valued $f \in C_K(X)$, choose $\alpha \in \mathbb{C}$ with $|\alpha| = 1$ so that $\lambda(f) = \alpha |\lambda(f)|$ holds. (I.e., decompose $\lambda(f)$ in polar form.) Then
$$|\lambda(f)| = |\lambda(\alpha^{-1}f)| = |\lambda(\operatorname{Re}(\alpha^{-1}f))| \leq \|\operatorname{Re}(\alpha^{-1}f)\|_{\infty}\lambda(g) \leq \|f\|_{\infty}\lambda(g). $$
The second equality follows because $\lambda(\alpha^{-1}f)$ has zero imaginary part.