Let $X$ be a NLS.Prove that the topology genrated by $\{x:\ |\phi(x-x_0)|<\epsilon\}$, $x_0\in X,\phi\in X^*,\epsilon>0$ is same as the norm topology.

Question

I call $V_{x_0,\phi,\epsilon}:=\{x:\ |\phi(x-x_0)|<\epsilon\}$, for $x_0\in X,\ \phi\in X^*,\ \epsilon>0$. Here $X^*$ denotes the space of all bounded linear functionals on $X$.

We donte $\tau_{\lVert\cdot\rVert}=$ norm topology and $\tau=$ topology generated by each $V_{x_0,\phi,\epsilon}$

Then $\displaystyle{V_{x_0,\phi,\epsilon}=\phi^{-1}(\phi(x_0)-\epsilon,\phi(x_0)+\epsilon)\in \tau_{\lVert\cdot\rVert}}$

Hence, $\tau\subseteq \tau_{\lVert\cdot\rVert}$

But I cannot prove the other way round i.e.$\tau_{\lVert\cdot\rVert}\subseteq \tau$ i.e. any open ball $B(x_0,\epsilon)$ in $\tau_{\lVert\cdot\rVert}$ is in $\tau$.

Can anyone help me to finish the part? Thanks for help in advance.

Answer 1

If $V$ is infinite-dimensional, this does not generate the norm topology. It generates the so-called weak topology on $V$.

Indeed, assume it does generate the norm topology, and consider the unit ball $B(0,1)$ which is norm-open. Then by our assumption, there exist $x_1, \ldots, x_n \in V$, $\phi_1, \ldots, \phi_n \in V^*$ and $\varepsilon_1, \ldots, \varepsilon_n > 0$ such that $$0 \in \bigcap_{k=1}^n\{x \in V : |\phi_k(x-x_k)|<\varepsilon_k\} \subseteq B(0,1).$$

Since $0$ is inside, in particular we have $|\phi_k(x_k)| < \varepsilon_k$ for all $1 \le k \le n$.

Since $V$ is infinite-dimensional, it cannot be $\bigcap_{k=1}^n \ker\phi_k = \{0\}$ (in fact, this intersection must be infinite-dimensional as well) so we can pick $y \ne 0$ such that $\phi_k(y)=0$ for all $1 \le k \le n$. Then for every $\lambda \in \Bbb{R}$ we have $$|\phi_k(\lambda y-x_k)| = |\phi_k(x_k)| <\varepsilon_k, \quad \text{ for all } 1\le k \le n$$ so $$\Bbb{R}y \subseteq \bigcap_{k=1}^n\{x \in V : |\phi_k(x-x_k)|<\varepsilon_k\} \subseteq B(0,1).$$ This is clearly a contradiction.

Answer 2

If $\mathcal{B}$ is a base of nieighbourhoods of zero for a topology $\tau$ then every set $A\in \mathcal{B}$ contains a linear subspace of finite codimension an hence cannot be a norm bounded unless the space $X$ is finite dimensional. Therefore the two topologies are not equivalent if $X$ is not finite dimensional.