Is the conjugate function of a strictly convex function continuously differentiable?

Question

I saw a claim that for function $f$ which is strictly convex, we know that $f(z) - y^Tz$ has a unique minimizer over $z$ and it must be $\nabla f^{*}(y)$. It would be $\nabla f^{*}(y)$ instead of $\partial f^{*}(y)$ only if $f^{*}(y)$ is continuously differentiable, right? How can I show that $f^{*}(y)$ is always continuously differentiable?

Answer 1

Fix an arbitrary point $y$. Note that by strict convexity of $f$,

$$ \underset{x}{\mathrm{argmax}} \left\{ \langle x, y \rangle - f(x) \right\} = \underset{x}{\mathrm{argmin}} \left\{ f(x) - \langle x, y \rangle \right\} $$

is a unique element (call it $x_{\star})$. At the same time, we have $$ \bar{x} \in \underset{x}{\mathrm{argmin}} \{f(x) - \langle x, y \rangle \} \Leftrightarrow \bar{x} \in \partial f^*(y) $$

This means that $\partial f^*(y) = \{ x_{\star} \}$, which is a singleton. Therefore, $f^*$ is differentiable.