Weak convergence to the same point implies equality.

Theorem: (Royden, Real Analysis 4th ed., Section 8.2 Problem 25)

Let X be a normed linear space such that for each \(f \in X\) there corresponds a bounded linear functional \(T(f)=\|f\|\) and \(\|T\|_*=1\).

(i) Prove that if \(\{f_n\}\) converges weakly to both \(f_1\) and \(f_2\) then \(f_1=f_2\).

(ii) Prove that if \(\{f_n\}\to f\) in X then \(\|f\|\le \liminf \|f_n\|\)

Proof:

(i) Let \(T\in X^*\) be such that \(T(f_1-f_2)=\|f_1-f_2\|\) (we could further require \(\|T\|_*=1\) but it won’t be useful here).

$$ \|f_1-f_2\| = \lim T(f_1-f_n+f_n-f_2) = \lim T(f_1-f_n)+\lim T(f_n-f_2) = 0 $$

(ii) Let T be the bounded linear functional such that \(T(f)=\|f\|\) and \(\|T\|_* = 1\). Note that because \(\|T\|_*=1\) entails that for all \(x\in X\) we have \(|T(x)|\le \|x\|\). Then by the continuity of bounded linear functionals

$$ \|f\|=T(f) = \lim T(f_n) = \liminf T(f_n) \le \liminf \|f_n\|$$