The difference between $(\partial f)^{-1}$ and $\partial f^*$ on non-reflexive space.

Question

Let $X$ be a Banach space, $f:X\to \Bbb R\cup\{\infty\}$ be a proper function. The multifunction $(\partial f)^{-1}$ is defined by $$ (\partial f)^{-1}(x^*) = \{ x\in X : x^* \in \partial f(x) \}. $$

It can be shown that $(\partial f)^{-1} = \partial f^*$ if $X$ is reflexive. Of course, $f^*$ is the Fenchel conjugate of $f$.

For a non-reflexive $X$, has anyone seen a concrete example of when $$ (\partial f)^{-1}(x^*) \subsetneq \partial f^*(x^*)? $$

More specifically, I am interested in the case that $(\partial f)^{-1}(x^*)=\emptyset$ but $\partial f^*(x^*)$ is non-empty.

Answer 1

The following should work:

Let $X=c_0$ and $f=\tfrac{1}{2}\|\cdot\|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=\ell_1$.
Now take a sequence in $\ell_1$ such as $x^* =(1/2^n)_{n\geq 1}$.
Then $\partial f^*(x^*)=\{(1,1,\ldots)_{n\geq 1}\}$
but $(\partial f)^{-1}=\varnothing$ (because $(1,1,\ldots)\notin c_0$).