weak topology of $X$ and the weak-* topology of bidual of $X$

Question

Given a normed vector space $X$, is the weak topology of $X$ the subspace topology of the weak-star topology of $X^{**}$?

Answer 1

Yes. We have the canonical isometric embedding $\text{ev}:X\to X^{**}$ where $x\mapsto\text{ev}_x$ and $\text{ev}_x$ is defined by $\text{ev}_x(\phi)=\phi(x)$ for all $\phi\in X^*$.

Take a net $\{\chi_i\}\subset X^{**}$ and a point $\chi\in X^{**}$. Then $\chi_i\to\chi$ in the weak-star topology of $X^{**}$ iff $\chi_i(\phi)\to\chi(\phi)$ for all $\phi\in X^*$.

Now take a net $\{x_i\}\subset X$ and a point $x\in X$. Then $x_i\to x$ in the weak topology iff $\phi(x_i)\to\phi(x)$ for all $\phi\in X^*$, i.e. iff $\text{ev}_{x_i}(\phi)\to\text{ev}_x(\phi)$ for all $\phi$, i.e. iff $\text{ev}_{x_i}\to\text{ev}_x$ in the weak star topology of $X^{**}$.

Comment: Note that it is not true that $X$ is closed in the weak topology as a subspace of $X^{**}$. On the contrary, Goldstine's theorem tells us that the closed unit ball of $X$ is weak-star dense in the closed unit ball of $X^{**}$.