Continuous dual of a weak topology

Continuity at zero supplies $f_1,\dots ,f_m\in X^{\ast }$ and $\epsilon >0$ such that $\mathrm{\mid }L(v)\mathrm{\mid }<1$ whenever ${\max }_j\mathrm{\mid }{f}_j(v)\mathrm{\mid }<\epsilon$. If every $f_j(v)=0$, every scalar multiple $tv$ is in this neighborhood. Then $\mathrm{\mid }tL(v)\mathrm{\mid }<1$ for all positive real $t$, forcing $L(v)=0$. For an empty list this already gives $L=0$.

Define $A:X\to {\mathbb{K}}^m$ by $Av=(f_1(v),\dots ,f_m(v))$. Step 1.1 makes $\mathrm{\ell }(Av)=L(v)$ a well-defined scalar-linear functional on $A(X)$. Choose a basis of this subspace and extend it to a basis of ${\mathbb{K}}^m$ by successively adding standard basis vectors when necessary; at most $m$ additions occur. Assign value zero on the added basis vectors. The resulting linear extension $\tilde{\mathrm{\ell }}$ has the form $\tilde{\mathrm{\ell }}(z)={\sum }_{j=1}^m{c}_j{z}_j$, where $c_j=\tilde{\mathrm{\ell }}(e_j)$. Only finite-dimensional basis choices occur.

By finite-dimensional boundedness $\mathrm{\ell }$ is bounded for the inherited norm, so factorization already gives norm boundedness of $L$. More explicitly $L={\sum }_j{c}_j{f}_j$ and $\mathrm{\mid }L(v)\mathrm{\mid }\le ({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\parallel }{f}_j\mathrm{\parallel })\mathrm{\parallel }v\mathrm{\parallel }$, hence $L\in X^{\ast }$. Conversely, for any $f\in X^{\ast }$ and any scalar disk around $f(x)$, its inverse image is a basic weak neighborhood, so $f$ is weakly continuous. Thus both inclusions hold. $\mathrm{\square }$