Every Weakly Convergent Sequence in X is Norm Bounded

Theorem 1: Let $X$ be a normed linear space. If $(x_n)$ weakly converges to $x \in X$ then there exists an $M > 0$ such that $\| x_n \| \leq M$ for every $n \in \mathbb{N}$.

• Proof: Let $(x_n)$ weakly converge to $x$. Then for every $f \in X^*$:

(1)

\begin{align} \quad \lim_{n \to \infty} f(x_n) = f(x) \end{align}

• In particular, for each $n \in \mathbb{N}$, we have that for each $f \in X^*$:

(2)

\begin{align} \quad \lim_{n \to \infty} \hat{x}_n (f) = \hat{x}(f) \end{align}

• That is, $(\hat{x}_n)$ converges pointwise to $\hat{x}$ on $X^*$. Since $X^* = B(X, \mathbb{R})$ is a Banach space and since for every $f \in X^*$ we have that:

(3)

\begin{align} \quad \sup_{n \geq 1} | \hat{x}_n(f) | = \sup_{n \geq 1} |f(x_n)| < \infty \end{align}

• (Since $f(x_n)$ converges to $f(x)$). We have by The Uniform Boundedness Principle that:

(4)

\begin{align} \quad \sup_{n \geq 1} \| \hat{x}_n \| < \infty \end{align}

• On the The Natural Embedding J we saw that $\| \hat{x}_n \| = \| x_n \|$ for every $n \in \mathbb{N}$ and so $\sup_{n \geq 1} \| x_n \| < \infty$. So there exists an $M > 0$ such that $\| x_n \| \leq M$ for all $n \in \mathbb{N}$. $\blacksquare$