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.

Author: mathtuition88


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.