Relationship between weak and nuclear topologies

Question

Let $(X,\Vert\cdot\Vert)$ be a Banach space. Is it always true that $X$ equipped with the weak topology $\sigma(X,X')$ is a nuclear space?

Answer 1

Yes, this is true. A locally convex space is nuclear if, for every continuous seminorm $p$ on $X$, there is another one $q\ge p$ such that the canonical map between the local Banach spaces $X_q\to X_p$ is a nuclear operator. Here, $X_q$ is the completion of the normed space $X/\{q=0\}$ with the quotient norm and the canonical map is the extension to the completions of the map $I_{q,p}:X/\{q=0\} \to X/\{p=0\}$ note that $\{q=0\}$ is contained in $\{p=0\}$). For the weak topology $\sigma(X,X')$ the typical continuous seminorms are of the form $p(x)=\max\{|f(x)|: f\in E\}$ where $E\subseteq X'$ is finite. This implies that $X/\{p=0\}$ is finite dimensional (and hence already complete). Since every linear map between finite dimensional normed spaces is nuclear, this implies the nuclearity of $(X,\sigma(X',X))$ for every locally convex space.

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There are much more important nuclear locally convex spaces in analysis, like spaces $C^\infty(\Omega)$ of smooth functions on an open subset of $\mathbb R^d$ or $H(\Omega)$ of holomorphic functions on an open subset of $\mathbb C$, the Schartz space $\mathscr S(\mathbb R^d)$, its dual space of tempered distributions, and many more spaces of distribution theory.