$X$ compact, $C(X)$ equipped with inner product, evaluation maps continuous, prove $X$ is finite

Question

I'm working on the following problem from Conway V.4.

Let $X$ be compact and supppose there is a norm on $C(X)$ that is given by an inner product making $C(X)$ into a Hilbert space such that for every $x \in X$ the functional $\Lambda_x : f\mapsto f(x)$ is continuous with respect to the Hilbert space norm. Show $X$ is finite.

Idea on approach: I think we need to assume $X$ is infinite and come to some sort of contradiction.

Facts noted:

• $\text{ball}{(C(X))}$ is weakly compact.

• $X^* \subset C(X)$, $X^*$ is definitely a subspace, and if it is closed under the norm induced by the inner product we have $X^*$ is Hilbert and hence reflexive. (Since I am not sure of the closedness of $X^*$ under the norm, I am not sure if this could be useful.)

• The hypothesis is allowing us to extend the notion of a weak-$*$ topology on $X^*$ to $C(X)$.

• The continuity of the $\Lambda_x$'s and Riesz Rep theorem imply that there is a $g \in C(X)$ such that $\Lambda_x(f) = \langle f,g \rangle = |f(x)| \leq M \langle f,f\rangle^{1/2}$.

I'm not entirely sure how to put these pieces together. I was hoping to take a $X \supset \{x_n\}$ with $x_n \to x$, and by weak compactness we have for any sequence $\{f_j\} \subset \text{ball}(C(X))$ there is a subsequence such that $f_{j_k}$ converges weakly to some $f \in \text{ball}(C(X))$, which by our hypotheses implies that $f_{j_k}(x_n) \to f(x_n)$ as $k \to\infty$. We also know that $f_{j_k}(x_n) \to f_{j_k}(x)$ as $n\to \infty$. But there is nothing contradictory I can see coming from this observation. Any hints would be appreciated. Thanks.

Edit: Also not entirely sure how $X$ compact fits in exactly, as I don't think I've taken much advantage of that fact in my observations except for claiming there is a convergent sequence.

Answer 1

Hint: Use the uniform boundedness principle to show that the Hilbert space norm on $C(X)$ must be equivalent to the sup norm. There are probably many approaches you can take after that, but one approach is that it would suffice to show that $C(X)$ (with the sup norm) is not reflexive if $X$ is infinite.

A full solution is hidden below.

The functionals $\Lambda_x$ are pointwise bounded (since every $f\in C(X)$ is a bounded function), and therefore must be uniformly bounded. This means exactly that the identity map $C(X)\to C(X)$ is bounded where the domain has our Hilbert space norm and the codomain has the sup norm. By the open mapping theorem, this implies the Hilbert space norm is equivalent to the sup norm on $C(X)$.

Since any Hilbert space is reflexive, this means $C(X)$ is reflexive in the sup norm. By the Riesz-Markov theorem, the dual of $C(X)$ is the space $M(X)$ of regular complex Borel measures on $X$, acting on $C(X)$ by integration. If $X$ is infinite, there exists a non-isolated point $x\in X$, and then the functional $\mu\mapsto\mu(\{x\})$ is a bounded functional on $M(X)$ which is not given by integration of any continuous function on $X$. So in that case $C(X)$ is not reflexive, and so $X$ must be finite.