# If $f_n \overset{\star}{\rightharpoonup} f$ in $\sigma(E^\star, E)$, then $\|f\| \le \liminf \|f_n\|$

I'm trying to prove this result. Could you have a check on my proof?

Let $$(E, | \cdot|)$$ be a normed linear space and $$E^\star$$ its topological dual. Let $$\sigma(E^\star, E)$$ be the weak$$^\star$$ topology on $$E^\star$$. Let $$f\in E^\star$$ and $$(f_n)$$ be a sequence in $$E^\star$$ such that $$f_n \overset{\star}{\rightharpoonup} f$$ in $$\sigma(E^\star, E)$$. Then $$\|f\| \le \liminf \|f_n\|$$.

My attempt: Let $$B := \{x \in X \mid |x|=1\}$$ be the unit sphere. We have $$\lim_n \langle f_n, x \rangle = \langle f, x \rangle$$ for all $$x\in X$$. So $$\sup_{x\in B} \lim_n \langle f_n, x \rangle = \sup_{x\in B} \langle f, x \rangle = \|f\|.$$

Hence it suffices to show that $$\sup_{x\in B} \lim_n \langle f_n, x \rangle \le \liminf_n \|f_n\|.$$

In fact, we have $$\langle f_n, x \rangle \le \sup_{x\in B} \langle f_n, x \rangle = \|f_n\|$$ and thus $$\lim_n \langle f_n, x \rangle = \liminf_n \langle f_n, x \rangle \le \liminf_n \|f_n\|.$$ The claim then follows by taking the supremum on both sides of above inequality.

• What is your question? Jan 26, 2022 at 23:25
• @uniquesolution I forgot to include the question. I ask for a proof verification. Jan 26, 2022 at 23:27
• Your proof is OK. Jan 26, 2022 at 23:29