Helley's Theorem

# Helley's Theorem

Theorem 1 (Helley's Theorem): Let $X$ be a normed linear space. If $X$ is separable then every bounded sequence of continuous linear functionals in $X^*$ has a subsequence that weak-* converges in $X^*$ |

*Recall that if $X$ is a normed linear space then a sequence of continuous linear functionals $(\varphi_n)_{n=1}^{\infty}$ is said to weak-* converge to $\varphi$ if $(\varphi_n(x))_{n=1}^{\infty}$ converges to $\varphi(x)$ for every $x \in X$, that is, weak* converge is the same as pointwise convergence.*

**Proof:**Let $X$ be a normed linear space that is separable. Then $X$ has a countable and dense subset. Denote this subset by:

\begin{align} \quad \{ x_n \}_{n=1}^{\infty} \end{align}

- Now let $(\varphi_n)_{n=1}^{\infty}$ be a bounded sequence of linear functionals in $X^*$. Let:

\begin{align} \quad R = \sup_{n \geq 1} \| \varphi_n \| \end{align}

- Furthermore, the sequence of complex numbers $(\varphi_n(x_1))_{n=1}^{\infty}$ is bounded in $\mathbb{C}$. So there is an ascending subsequence of the natural numbers $(s(1, n))_{n=1}^{\infty}$ and an element $a_1 \in \mathbb{C}$ such that:

\begin{align} \quad \lim_{n \to \infty} \varphi_{s(1, n)} (x_1) = a_1 \end{align}

- Now consider the subsequence $(\varphi_{s(1, n)})_{n=1}^{\infty} \subseteq (\varphi_n)_{n=1}^{\infty}$ which is also bounded. Then the sequence of complex numbers $(\varphi_{s(1, n)}(x_2))_{n=1}^{\infty}$ is bounded in $\mathbb{C}$. So there is an ascending subsequence of the natural numbers $(s(2, n))_{n=1}^{\infty}$ and an element $a_2 \in \mathbb{C}$ such that:

\begin{align} \quad \lim_{n \to \infty} \varphi_{s(2, n)} (x_2) = a_2 \end{align}

- We continue on in this matter to obtain sequences of natural numbers and a sequence $(a_n)_{n=1}^{\infty}$ of complex numbers such that for each $k \in \mathbb{N}$ we have that:

\begin{align} \quad (s(k+1, n))_{n=1}^{\infty} \subseteq (s(k, n))_{n=1}^{\infty} \end{align}

(6)
\begin{align} \quad \lim_{n \to \infty} \varphi_{s(k, n)}(x_k) = a_k \end{align}

- For each $k \in \mathbb{N}$ let:

\begin{align} \quad n_k = s(k, k) \end{align}

- Then $(n_k)_{k=1}^{\infty}$ is a subsequence of $(s(j, n))_{n=1}^{\infty}$ for $k \geq j$ and so:

\begin{align} \quad \lim_{k \to \infty} \varphi_{n_k} (x_j) = a_j \end{align}

- We now aim to show that $(\varphi_{n_k}(x))_{n=1}$ converges for every $x \in X$. Let $\epsilon > 0$ be given and let $x \in X$. Since $\{ x_n : n \in \mathbb{N} \}$ is dense in $X$ there exists an $N \in \mathbb{N}$ such that:

\begin{align} \quad \| x - x_{N} \| < \epsilon \end{align}

- So for all $s, t \in \mathbb{N}$ we have by the triangle inequality that:

\begin{align} \quad \| \varphi_{n_s}(x) - \varphi_{n_t}(x) \| &= \| \varphi_{n_s}(x) - \varphi_{n_s}(x_N) + \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) + \varphi_{n_t}(x_N) - \varphi_{n_t}(x) \| \\ & \leq \| \varphi_{n_s}(x) - \varphi_{n_s}(x_N) \| + \| \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) \| + \| \varphi_{n_t}(x_N) - \varphi_{n_t}(x) \| \\ & \leq \| \varphi_{n_s} \| \| x - x_N \| + \| \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) \| + \| \varphi_{n_t} \| \| x_N - x \| \\ & < R\epsilon + \| \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) \| + R\epsilon \\ & < 2R\epsilon + \| \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) \| \end{align}

- Since $\| \varphi_{n_s}(x_N) - \varphi_{n_t}(x_N) \|$ can be made arbitrary small, we see from above that $(\varphi_{n_k}(x))_{k=1}^{\infty}$ is a Cauchy sequence of complex numbers and since $\mathbb{C}$ is complete, this sequence converges pointwise for every $x \in X$. So for each $x \in X$ we define $\varphi : X \to \mathbb{C}$ by:

\begin{align} \quad \varphi(x) = \lim_{k \to \infty} \varphi_{n_k}(x) \end{align}

- Clearly $\varphi$ is a linear functional. It is also continuous since:

\begin{align} \quad \| \varphi(x) \| \leq R \| x \| \end{align}

- And moreover, $(\varphi_{n_k})_{k=1}^{\infty}$ converges to $\varphi$ in the weak* topology.

- So if $X$ is separable, every bounded sequence of linear functionals in $X^*$ has a subsequence which weak* converges in $X^*$. $\blacksquare$