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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.

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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 $(1,1,\ldots)\notin c_0$).

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  • $\begingroup$ Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $\partial f^*(x^*)=\{(1,1,\ldots)_{n\geq 1}\}$? I've never done calculation like this before. $\endgroup$ – BigbearZzz Feb 4 at 16:38
  • $\begingroup$ Write $J=\partial \tfrac{1}{2}\|\cdot\|^2$. Then $(Jz)_n = \|z\|\mathrm{Sign}(z_n)$ by the chain rule, where $\mathrm{Sign}$ is the sign function, but with $\mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $\ell_1$ norm on $\mathbb{R}^2$ - this example is nothing but that generalization. $\endgroup$ – max_zorn Feb 4 at 17:36

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