The weak topology on $H$ is the weak* topology on $H^*$ pulled back via $\Phi$

Question

I'm reading the following in Analysis Now by Pedersen:

The map, $H$ a Hilbert space $$\Phi:H\to H^*: x\mapsto(\cdot\mid x)=[y\mapsto (y,x)]$$ is a conjugate linear isometry.

Then define the weak topology on $H$ as the topology induced by the family of functionals $$\{x\mapsto(x,y)\mid y\in H\}=\{\Phi(y)\mid y\in H\}$$ Now Pedersen says the following:

In follows that the weak topology on $H$ is the weak* topology on $H^*$ pulled back to $H$ via the map $\Phi$

but I don't quite understand what is meant by this. I would say this means that $\Phi$ is a homeomorphism between $H$ and $H^*$ both in their weak topologies.

But I don't quite see how this works. The weak* topology is induced by $$\{[x\mapsto (x,y)]\mapsto(z,y)\mid z\in H\}=\{\Phi(y)\mapsto(z,y)\mid z\in H\}$$ And now I just don't see how to proceed from here.

I would appreciate someone shedding some light on this quote.

Answer 1

The weak-* topology on $H^*$ is generated by the functionals $$\{h^* \mapsto h^*(x) \mid x \in H\}$$ If you pull-back this topology to $H$, you get the topology generated by $$\{y \mapsto \Phi(y)(x) \mid x \in H\} = \{y \mapsto (x,y) \mid x \in H\}$$ But this is the same as the weak topology which is generated by $$\{y \mapsto (y,x) \mid x \in H\}$$