Let $X$ be a normed space and $Y$ be a subspace of $X$. Denote $X^*$, $Y^*$ the dual space of $X$, $Y$ respectively, and for $\varphi\in X^*$, denote $\varphi|_Y$ the restriction of $\varphi$ on $Y$.
Is it true that if $\varphi_n$ converges weakly to $\varphi$ in $X^*$ then $\varphi_n|_Y$ converges weakly to $\varphi|_Y$ in $Y^*$?
I managed to prove this assertion only in case that $X$ is reflexive: in this case, we know that $\varphi_n$ converges weakly to $\varphi$ in $X^*$ iff $\varphi_n(x)\to\varphi(x)$ in $X$ for every $x\in X$.
But I am not able to prove the assertion for normed spaces, not necessarily reflexive.
Thanks for any help.
Best Answer
Let us first prove the following lemma, which is quite interesting on its own.
Lemma: Let $X,Y$ be Banach spaces and let $T:X \rightarrow Y$ be a (norm) continuous linear map. Then $T$ is weak-to-weak continuous.
Proof: Let $(x_a)$ be a net in $X$ which weakly converges to some $x \in X$. We want to show that $T(x_a)$ weakly converges to $T(x)$ in $Y$. Take any $y^* \in Y^*$. Then $(y^* \circ T) \in X^*$ (composition of two continuous linear mappings). Then, as $x_a$ is weakly convergent to $x$, we get $$y^*(T(x_a)) = (y^* \circ T) (x_a) \rightarrow (y^* \circ T) (x) = y^*(T(x))$$ and $T$ is indeed weak-to-weak continuous. $\square$
Now lets return to your question. In this case we have that $T: X^* \rightarrow Y^*$ is the restriction mapping $T^*(\varphi) = \varphi \restriction Y$. It is easy to check that $T$ is continuous and linear. Hence you got the statement you want from the Lemma above.