$ \{y_n \} $ does it converge weakly in $ X $ to $ y $?

Question

Let $(X,\|.\|)$ be a Banach space and $Y$ be the closed separable linear subspace of $X$.

Let $\{y_n\}$ be a sequence of $Y$ such that it converges weakly in $Y$ to $y$.

$ \{y_n \} $ does it converge weakly in $ X $ to $ y $?

An idea please

Answer 1

Yes, this is the case.

Key Lemma: Let $X$ be a normed space and $Y \subseteq X$ a linear subspace, then $X^* \subseteq Y^*$.

proof: Let $f \in X^*$, then $f: X \rightarrow \mathbb{C}$ linear and $\Vert f \Vert_{X^*} := \sup_{x \in X} \frac{\vert f(x) \vert}{\Vert x \Vert} < \infty$. Then since $Y \subseteq X$, the restriction of $f$ to $Y$ is also linear and we have $\Vert f \Vert_{Y^*} := \sup_{x \in Y} \frac{\vert f(x) \vert}{\Vert x \Vert} \leq \sup_{x \in X} \frac{\vert f(x) \vert}{\Vert x \Vert} < \infty$ i.e. $f \in Y^*$.

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Now assume $(y_n)$ converges weakly on $Y$, then we have $\forall f \in Y^*: f(y_n) \rightarrow f(y)$ and thus by the lemma: $\forall f \in X^*: f(y_n) \rightarrow f(y)$ i.e. weak convergence in $X$.

Answer 2

In fact the statement is true in both directions.

1. Let $y_n\rightharpoonup y$ in $X$ i.e. $\forall\varphi\in X^\ast$ $\varphi(y_n)\to\varphi(y)$. Supposing $\exists\varphi_0\in Y^\ast$ such that $\varphi_0(y_n)\not\to\varphi_0(y)$. By Hahn-Banach there exists $\psi\in X^\ast$ such that $\psi(Y)=\varphi_0$. Thus $\psi(y_n)=\varphi_0(y_n)\not\to\varphi_0(y)=\psi(y)$. A contradiction.

2. Let $y_n\rightharpoonup y$ in $Y$ i.e. $\forall\varphi\in Y^\ast$ $\varphi(y_n)\to\varphi(y)$. Supposing $\exists\varphi_0\in X^\ast$ such that $\varphi_0(y_n)\not\to\varphi_0(y)$. Then there exists $\psi=\varphi_0(Y)\in Y^\ast$ and $\psi(y_n)=\varphi_0(y_n)\not\to\varphi_0(y)=\psi(y)$. A contradiction.