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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 $M$.

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

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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 $f(x_{\lambda})=F(x_{\lambda})\to F(x)=f(x)$.

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