weak convergence in subspace of dual space

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.

Solution 1:

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.