Point Evaluations are Continuous Linear Functionals on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

Theorem

Let $K$ be a compact Hausdorff space.

Let $\map \CC K$ be the space of complex-valued continuous functions of compact Hausdorff space.

For $k \in K$, define $\delta_k : \map \CC K \to \C$ by:

$\map {\delta_k} f = \map f k$

for each $f \in \map \CC K$.

Then $\delta_k$ is a continuous linear functional on $\map \CC K$ and further:

$\norm {\delta_k}_{\map \CC {K}^\ast} = 1$

Proof

Let $k \in K$.

$\delta_k$ is clearly linear, so we need to show that it is continuous.

From Continuity of Linear Functionals, it is enough to show that $\delta_k$ is bounded.

For each $f \in \map \CC K$ we have:

$\cmod {\map f k} \le \sup_{x \in K} \cmod {\map f x}$

That is:

$\cmod {\map {\delta_k} f} \le \norm f_\infty$

If $K$ is a singleton, then we have $\cmod {\map {\delta_k} f} = \norm f_\infty$ for each $f \in \map \CC K$ and hence $\norm {\delta_k}_{\map \CC {K}^\ast} = 1$.

If $K$ is not a singleton, pick distinct points $k, \ell \in K$.

From Urysohn's Lemma, there exists a continuous function $f : K \to \closedint 0 1 \subseteq \C$ with $\map f \ell = 0$ and $\map f k = 1$.

For this $f \in \map \CC K$, we have:

$\cmod {\map {\delta_k} f} = \cmod {\map f k} = 1 = \sup_{x \in K} \cmod {\map f x} = \norm f_\infty$

Hence $\norm {\delta_k}_{\map \CC K^\ast} = 1$ in this case as well.

$\blacksquare$