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Let $(X,\|.\|)$ be a separable Banach space and $\mathcal{C}(X)$ be the collection of all nonempty, closed and convex subsets of $X$. For any $C$ in $\mathcal{C}(X)$ we set $$ s(x^*, C) := \sup_{x\in C}{\langle x^*, x \rangle}\qquad;\qquad \|C\| := \sup_{x\in C} \|x\| $$ to define for the set $C$ respectively its support function $s(., C)$ and radius $\|C\|$.

On $\mathcal{C}(X)$ we use the following convergence : a net $\{C_\alpha\}$ convergent scalarly to $C_\infty$ in $\mathcal{C}(X)$ if $$ \lim_\alpha s(x^*, C_\alpha)=s(x^*, C_\infty)\qquad \forall x^*\in X^* $$

Can we say that the scalar convergence in $\mathcal{C}(X)$ is topologizable ?

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    $\begingroup$ Isn't it just the initial topology with respect to the functions $s(x^*,-)$? So a prebase would be given by sets of the form $\{C\text{ such that }s(x^*,C)\in U\}$ for $x\in X^*$ and $U$ a open set of $\overline{\mathbb R}$. $\endgroup$
    – Pierre PC
    Commented Jun 19, 2020 at 18:13
  • $\begingroup$ Do you know the Wijsman topology? $\endgroup$
    – Made
    Commented Jun 19, 2020 at 18:45
  • $\begingroup$ I just read the definition on Wikipedia. Is my comment not enough to answer your question? $\endgroup$
    – Pierre PC
    Commented Jun 19, 2020 at 20:03

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