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

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.

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

• 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. – BigbearZzz Feb 4 at 16:38
• 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. – max_zorn Feb 4 at 17:36