## Weak* convergent sequence uniformly bounded

Theorem 11 (Lax Functional Analysis): A weak* convergent sequence $\{u_n\}$ of points in a Banach space $U=X'$ is uniformly bounded.

We will need a previous Theorem 3: $X$ is a Banach space, $\{l_v\}$ a collection of bounded linear functionals such that at every point $x$ of $X$, $|l_v(x)|\leq M(x)$ for all $l_v$. Then there is a constant $c$ such that $|l_v|\leq c$ for all $l_v$.

Sketch of proof:

Weak* convergence means $\lim u_n(x)=u(x)$, thus there exists $N$ such that for all $n\geq N$, we have $|u_n(x)-u(x)|<1$, which in turns means $|u_n(x)|<1+|u(x)|$ via the triangle inequality. We have managed to bound the terms greater than equals to $N$.

For those terms less than $N$, we have $|u_n(x)|\leq\|u_n\|\|x\|$.

Thus, we may take $M(x)=\max\{\|u_1\|\|x\|,\dots,\|u_{N-1}\|\|x\|,1+|u(x)|\}$. The crucial thing is that $M(x)$ depends only on $x$, not $n$.

Then, use Theorem 3, we can conclude that $\|u_n\|\leq c$ for all $n$.

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