Let $X$ be a normed linear space. Suppose that fro each $f \in X$ there is bounded linear functional $T \in X^*$ for which $T(f) =||f||$ and $||T||_* =1$
i Prove that if $\{f_n\}$ converges weakly in $X$ to both $f_1$ and $f_2$ then $f_1=f_2$
ii Prove that if $\{f_n\} \rightharpoonup f$ in $X$ then $||f||
\leq \liminf ||f_n||$
Proof i
Assume $f_n$ converge weakly in $X$ to both $f_1$ and $f_2$
$f_n \rightharpoonup f_1 \text{ (using definition of weak convergence) }\Leftrightarrow T(f_n) \to T(f)$ for all $T \in X^*$ where $X^*$ is dual space of $X$
Based problem given $T(f)=||f||$ we have $||f_n|| \to ||f_1||$
Using similar approach for $f_2$ we have $||f_n|| \to ||f_2|$
Now consider $||f_1 -f_2||$
\begin{align*}
||f_1 -f_2|| = &||f_n – f_n + f_1-f_2||\\
&=||(f_n -f_2) + (f_1-f_n)||\\
&\leq ||f_n -f_2||+||f_1-f_n|| (by triangle inequality)\\
& \to 0
\end{align*}
Therefore $||f_1-f_2|| =0 \Rightarrow f_1=f_2$
Am I correct on this part ( i have not use the fact $||T||_*=1$)
Proof ii
Any help on the second part?
Thanks
Best Answer
By dual representation of the norm and the continuity of $| \, |$you have $$\|f\|=\sup_{\|\varphi\|=1} |\varphi(f)|=\sup_{\|\varphi\|=1} lim_{n \to \infty} |\varphi(f_n)|\leq \liminf \sup_{\|\varphi\|=1} |\varphi(f_n)|\leq\liminf \|f_n\|$$