# 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.

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.

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.

• Okay I have misunderstood the question. Can I show that the vector addition and scalar multiplication is continuous in this weak topology? Commented Jun 25, 2021 at 18:28
• @DeltaEpsilon Yes, you can. For addition try to show that $$+\left(V_{x_0,\phi,\frac\varepsilon2} \times V_{y_0,\phi,\frac\varepsilon2}\right) \subseteq V_{x_0+y_0, \phi,\varepsilon}$$ and then think about what if $x_0+y_0 \subseteq V_{z_0,\phi,\varepsilon}$. Similar for scalar multiplication. Commented Jun 25, 2021 at 18:52
• Okay I understand. Thanks. Basically, I'm absolutely new to this topic, so maybe I'm asking some basic questions. Commented Jun 25, 2021 at 19:04

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.