Does isometry preserves weak convergence

Question

Let $X$ and $Y$ be banach spaces and $f: X \to Y$ be a linear isometry, i.e. $\|f(a)-f(b)\|_Y= \| a-b\|_X$ for all $a,b\in X$. Then does $f$ preserves weak convergence, i.e. if $x_n \rightharpoonup x$ in X does this implies $f(x_n) \rightharpoonup f(x)$ in Y ?

Answer 1

Suppose that $f: X \to Y$ is a linear isometry and that $x_n \to x$ weakly in $X$. Take a bounded linear functional $\lambda: Y \to \mathbb C$ and define $\mu: X \to \mathbb C$ by $$\mu(x) = \lambda(f(x)), \,\,\,\, x \in X.$$ Note that for any $x,y\in X$ and $\alpha \in \mathbb C$, we have $$\mu(x + \alpha y) = \lambda(f(x+\alpha y)) = \lambda(f(x) + \alpha f(y)) = \lambda(f(x)) + \alpha \lambda(f(y)) = \mu(x) + \alpha \mu(y).$$ This shows that $\mu$ is a linear functional on $Y$. Next, for any $x \in X$, $$\lvert \mu(x)\rvert = \lvert \lambda(f(x)) \rvert \le \| \lambda \| \cdot\|f(x)\|_Y = \|\lambda \| \cdot \|x\|_X.$$ This shows that $\mu$ is bounded. Hence, by weak convergence, we have $$\mu(x_n) \to \mu(x).$$ But this shows that $$\lambda(f(x_n)) \to \lambda(f(x)).$$ Since $\lambda$ was an arbitrary bounded linear functional on $Y$, this shows that $f(x_n) \rightharpoonup f(x)$ in $Y$.