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I have a question about the fine structure of convex functions. Convex functions behave very regular in the interior of their domain of definition (e.g. they are locally Lipschitz continuous there) but otherwise some weird things can happen. My question concerns convex functions that possess at most one subgradient at each point. Let's fix notation:

Let $X$ be a Banach space and $J:X\to\mathbb{R}\cup\{\infty\}$ be a convex, extended valued function. Denote by $\DeclareMathOperator{\dom}{dom}\dom J = \{x\ :\ J(x)<\infty\}$ and assume that the subdifferential $\partial J$ of $J$ is at most single valued and denote its unique element by $\nabla J(x)$ (if it exists). Moreover, denote the Gâteaux directional derivative at $x$ in direction $h$ by $DJ(x;h)$. My question is:

Does $x,y\in \dom J$ imply that $$\langle \nabla J(x),y-x\rangle = DJ(x;y-x)\ ?$$

Some background: I would like to state that in the above framework for some non-strictly convex $J$ there exist $x,y\in\dom J$ such that $\langle \nabla J(x),y-x\rangle = J(y)-J(x)$. It is clear that one gets $x$ and $y$ such that for $\lambda\in]0,1]$ it holds that $$ \frac{J(\lambda y + (1-\lambda)x) - J(x)}{\lambda}=J(y) - J(x) $$ which implies $DJ(x,y-x) = J(y) - J(x)$.

However, there exists a pathological convex function such that its subdifferential at some point is single valued although it is not Gâteaux differentiable there (Example 4.2.6 in Borwein and Vanderwerffs "Convex Functions: Constructions, Characterizations and Counterexamples", see here). However, I assume a wee bit more, namely that the subdifferential is at most single valued everywhere (but probably this already rules out some pathological things…).

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  • $\begingroup$ What about $f(x) = -1(1-|x|^2)^{1/2}$ on $|x| \le 1$ and $f(x)=+\infty$ elsewhere? This function is in fact differentiable at $x$ for all $|x| < 1$, but $\partial f(x) = \emptyset$ for all $|x| \ge 1$---the badness comes from $1 \in \dom f$--- does this break your setup? $\endgroup$
    – Suvrit
    Commented Dec 20, 2013 at 17:26
  • $\begingroup$ Well, I am interested in the case where $\partial f(x)$ is a singleton for all $x$ in $\mathrm{dom} f$. In your example, the subdifferential is empty precisely where the directional derivative does not exist. $\endgroup$
    – Dirk
    Commented Dec 20, 2013 at 18:02
  • $\begingroup$ Did you ever find an answer or insight to this question? It is interesting. $\endgroup$
    – ABIM
    Commented Mar 21, 2019 at 20:52
  • $\begingroup$ Nope, no update from my side... $\endgroup$
    – Dirk
    Commented Mar 21, 2019 at 21:33
  • $\begingroup$ Is there an example of a function $f:\mathbb{R}^n\to\bar{\mathbb{R}}$ such that $\mathrm{dom} f \neq \mathbb{R}^n$ but $\partial f$ is single valued on $\mathrm{dom} f$? I don't see how it's possible - if the boundary of $\mathrm{dom}f$ is non empty then the subdifferential must be multivalued there. $\endgroup$
    – giorgi nguyen
    Commented May 28, 2022 at 0:16

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You can take $X = \ell^\infty$ (actually $\ell^p$ with $p > 2$ should work) and define $$ f(x) = \frac12 \| x \|_{\ell^2}^2. $$ Note that $f$ equals $\infty$ on $\ell^\infty \setminus \ell^2$. One can check that $f$ is convex (clear), proper (clear) and lower semicontinuous (via Fatou). However, $f$ is everywhere discontinuous (e.g., continuity at $0$ would give the embedding $\ell^\infty \hookrightarrow \ell^2$ which is absurd).

One can check that $$ \partial f(x) = \{x\} $$ (with the canonical embedding $\ell^1 \hookrightarrow (\ell^\infty)^*$) if $x \in \ell^1$ and on $\ell^\infty \setminus \ell^1$ the subdifferential is empty.

The discontinuity of $f$ yields that $f$ is nowhere Gâteaux differentiable, consequently your desired equality cannot hold.

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    $\begingroup$ Oh, that was quite simple! Thanks! And this took you ten and a half years? 😉 $\endgroup$
    – Dirk
    Commented Jul 16 at 19:21

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