# Weak Convergence and Hahn-Banach Theorm

Kreyzig proves the following lemma:

Let $(x_n)$ be a weakly convergent sequence in a normed space $X$, then the weak limit $x$ of $(x_n)$ is unique.

He begins by supposing $(x_n)$ converges weakly to two limits, say $x$ and $y$. By supposition, this means $f(x_n) \rightarrow f(x)$ and $f(x_n) \rightarrow f(y)$. $f$ is linear. So $$f(x) - f(y) = f(x-y) = 0$$

Then Kreyzig uses Hahn-Banach theorem to assert this implies $x-y = 0$. I don't understand why this doesn't just follow from $f$ being linear, don't linear maps between normed spaces need to preserve 0 as they do for inner product spaces?

• Yes, linear maps preserve $0$, but the crux is the other direction. Not every $z$ that is mapped to $0$ must itself be $0$. But by Hahn-Banach, every $z$ that is mapped to $0$ by all continuous linear functionals must itself be $0$. – Daniel Fischer Feb 13 '17 at 11:36
• @DanielFischer thanks! – yoshi Feb 13 '17 at 16:23

## 1 Answer

If $x-y$ is not zero, you can define on $vect(x-y)$ a linear function $g$ such that $g(x-y)=1$. Hahn Banach implies that you can extend $g$ to a bounded function $f$ defined on $X$.You have $f(x-y)=g(x-y)=1$. Contradiction.

• What do you mean by $vect()$? – mavavilj Apr 9 '18 at 8:01