Theorems on continuous embedding

Question

Let $X,Y$ be two normed spaces such that $X\hookrightarrow Y$.
Show the following:

1) If $A\subset X$ is closed in $Y$ then $A$ is closed in $X$;

2) If $X, Y$ are Banach spaces if $x_n$ tends weakly to $x \in X$ then $x_n$ tends weakly to $x \in Y$

3)If $X$ is reflexive and $x_n$ is bounded and weakly convergent to $x \in Y$, then $x \in X$ and $x_n$ converges weakly to $x \in X$.

4) If $X$ is reflexive the unitary closed ball in $X$ is closed in $Y$.

My definition is: $X \hookrightarrow Y$ if $X \subseteq Y$ and the map $J : X \to Y$ defined as $Jx=x$ is continuous: $||Jx||_Y=||x||_Y \le C||x||_X,\ \forall x \in X$, with $C$ a nonnegative constant. $J$ is assumed to be linear.

Answer 1

Proof:

1. If $A$ is closed in $Y$, then $J^{-1}(A)=A$ is closed in $X$ because $J$ is continuous.

2. Let $(x_n)$ be a sequence in $X$ such that $x_n$ converges weakly to $x$. Now, let $f\in Y^*$. Then $f\circ J\in X^*$. Hence $f(x_n)=(f\circ J)(x_n)$ converges to $(f\circ J)(x)=f(x)$. Therefore, $(x_n)$ converges weakly to $x$ in $Y$.

3. Since $X$ is reflexive space. Hence $X$ is a Banach space. Also $\parallel x\parallel _Y\leq C\parallel x\parallel _X$. Hence by the Bounded Inverse Theorem, both the norms on $X$ are equivalent. Hence $X$ is a closed convex subset of $Y$. Therefore, $X$ is weakly closed in $Y$. This implies that $x\in X$. Also with the help of Hahn-Banach extension theorem, we can verify that $(x_n)$ converges weakly to $x$ in $X$.

4. Since $X$ is reflexive, by Banach-Alaoglu theorem, the closed unit ball of $X$ is weakly closed in $X$. Since $X$ is weakly closed in $Y$, the closed unit ball of $X$ is also weakly closed in $Y$. Also the closed unit ball of $X$ is a convex set. This implies that the latter set is also closed in $Y$.