Which is the relationship between weak convergence and pointwise convergence?

Question

In one of my indipendent works at functional analysis course have to come up with an explicit way of telling which is the relationshpip between weak and pointwise convergence for $C(K)$ where $K$ is a compact Hausdorff space. I searched some articles but didn't find anything which is enough helpfull. The theory behind is kind of complicated and I don't think I have the necessary background to understand it. Can somebody guide me to get to the right point? Till now I have been working with Eberlein-Smulian theorem and weak compactness.

Answer 1

Are you talking about convergence of sequences? Without thinking about it further I don't know what happens for convergence of nets, because the proof below (from Proposition 19.3.1 of Semadeni's classic book Banach spaces of continuous functions) uses the dominated convergence theorem (which is proved in Semadeni's book as Theorem 19.2.6, but also on Wikipedia), which for all I know might not have an analogue for nets.

Besides the dominated convergence theorem, the following proof uses also the principle of uniform boundedness and the Riesz representation theorem (Theorem 18.4.1 in Semadeni's book).

Let $K$ be a compact Hausdorff space, $f\in C(K)$ and $(f_n)$ a sequence in $C(K)$. The following are equivalent:

1. $f_n$ converges weakly to $f$.
2. $f_n$ converges pointwise to $f$ and $\sup_{n\in\mathbb{N}}\Vert f_n\Vert<\infty$.

Proof: If $(1)$ holds, then for each $k\in K$ we have $f_n(k) = \delta_k(f_n)\longrightarrow \delta_k(f) = f(k)$, so that $f_n$ converges pointwise to $f$. Moreover, the fact that $\sup_{n\in\mathbb{N}}\Vert f_n\Vert<\infty$ follows from the principle of uniform boundedness. Thus $(1) \Rightarrow (2)$.

Conversely, suppose $(2)$ holds and let $x^\ast\in C(K)^\ast$. Then, by the Riesz representation theorem, there exists a unique Radon measure $\mu_{x^\ast}$ on $K$ such that $$ x^\ast(g) = \int_K g \,d\mu_{x^\ast} $$ for every $g\in C(K)$. Moreover, the functions $f_n$ are dominated by the constant function $h: K\longrightarrow \mathbb{R}$ satisfying $h(k) = \sup_{n\in\mathbb{N}}\Vert f_n\Vert$ for every $k\in K$; that is, $\vert f_n(k)\vert\leq h(k)$ for every $k\in K$. It follows then by the dominated convergence theorem that for every $x^\ast\in C(K)$ we have $$ x^\ast(f_n) = \int_K f_n \,d\mu_{x^\ast} \longrightarrow \int_K f \,d\mu_{x^\ast} = x^\ast(f)\,. $$ Since $x^\ast$ was arbitrary, it follows that $f_n$ converges weakly to $f$, hence $(2)\Rightarrow (1)$.

Answer 2

A sequence $f_n \in C(K)$ converges weakly to $f\in C(K)$ if $$ w(f_n) \to w(f) \quad\forall w\in C(K)^{\ast} $$ In particular, if $x \in K$, then $w_x \in C(K)^{\ast}$, where $$ w_x(f) := f(x) $$ Hence, if $f_n \xrightarrow{w} f$, then $$ f_n(x) \to f(x) \quad\forall x\in K $$ Hence, $f_n$ converges to $f$ pointwise.

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The converse is not true, however. Take $K = [0,1]$ and $$ f_n(x) = n^2xe^{-nx} \text{ and } f \equiv 0 $$ Then $f_n(x) \to f(x)$ for all $x\in [0,1]$. However, if $$ w(f) := \int_0^1 f(x)dx $$ then $w\in C(K)^{\ast}$ and $$ w(f_n) = 1 - (1/e^n) - (n/e^n) $$ So $$ \lim_{n\to \infty} w(f_n) = 1 \neq 0 = w(f) $$ Hence, $f_n$ does not converge weakly to $f$.