# Weak Topology and the induced topology

Given a normed space $$E$$ with a subspace $$M$$, it is known that the weak topology on $$M$$ is the same as the induced topology of the weak topology on $$E$$. Why is this the case? From the Hahn-Banach theorem, we can extend the linear functionals on $$M$$ to $$E$$. So my intuition is that any element in the weak topology on $$M$$ is in the induced topology of the weak topology on $$E$$. But why does the other way also hold? I am not really clear how to work with a linear functional on $$E$$ which cannot be obtained by extending a linear functional on $$M$$.

• All linear functionals on $E$ can be obtained by extension. If $f$ is a linear functional on $E$, it's an extension of $f\lvert_M$. Commented Aug 1, 2020 at 13:33

The easiest way is to work with nets. A net $$\{x_{\lambda}\}_{\lambda\in\Lambda}$$ converges in the weak topology to $$x$$ if and only if $$f(x_{\lambda})\to f(x)$$ for every bounded linear functional $$f$$.
Now suppose we have a net $$(x_{\lambda})_{\lambda\in\Lambda}\subseteq M$$ which converges to $$x\in M$$ in the weak topology on $$M$$. This means $$f(x_{\lambda})\to f(x)$$ for every $$f\in M^*$$. We want to prove that $$x_{\lambda}\to x$$ also in the induced topology from $$E$$. So let $$f\in E^*$$. Then $$g:=f|_M\in M^*$$ and hence $$f(x_{\lambda})=g(x_{\lambda})\to g(x)=f(x)$$.
Conversely, suppose $$(x_{\lambda})\subseteq M$$ is a net which converges to $$x\in M$$ in the induced topology from $$E$$. This means we have $$f(x_{\lambda})\to f(x)$$ for every $$f\in E^*$$. We want to show $$x_{\lambda}\to x$$ in the weak topology on $$M$$. So let $$f\in M^*$$. By Hahn-Banach we can extend it to a bounded functional $$F\in E^*$$. Then by our assumption $$f(x_{\lambda})=F(x_{\lambda})\to F(x)=f(x)$$.