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Let $X$ be a Banach space, and let $f : X \to \mathbb{R}\cup\{+\infty\}$ be proper convex. Assume that $f$ is locally bounded above at $x \in X$, so that $f$ is locally Lipschitz at $x$. Assume also that $\partial f(x) = \{y\}$ is a singleton, so that $f$ is Gâteaux differentiable at $x$.

My question: Is the condition of local Lipschitz continuity strong enough to give Fréchet differentiability?

Jas Ter
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  • I think so. See Clarke (1983), propositions 2.2.6 and 2.2.7. I’m not able to read them carefully to give an answer with certainty at the moment, but I think those should tell you what you need. If I’m correct, then there is also probably a simpler way to establish the result you want. – Theoretical Economist Sep 03 '19 at 10:48
  • @TheoreticalEconomist, which article/book is that? – Jas Ter Sep 03 '19 at 11:13
  • Optimisation and Nonsmooth Analysis. – Theoretical Economist Sep 03 '19 at 11:21
  • Reading that source, I cannot see right away how that solves the problem: Prop. 2.2.6 is just the well-known fact that local bounded above implies Lipschitz for convex functions, while 2.2.7 affirms that for convex function, two notions of generalized gradients coincide with the subdifferential and Gâteaux differentiability, respectively. There could be more here. – Jas Ter Sep 03 '19 at 11:51
  • Whoops. I meant proposition 2.2.4, not 2.2.6. Is that better? – Theoretical Economist Sep 03 '19 at 11:53
  • So, these results seem to strengthen Gâteaux differentiability to "strict differentiability". But is this the same as Fréchet differentiability? – Jas Ter Sep 03 '19 at 12:26
  • Strict differentiability should be more demanding than Fréchet differentiability, though do double-check the definitions. – Theoretical Economist Sep 03 '19 at 12:34
  • To me it seems it's the opposite! I will check carefully. Thanks! – Jas Ter Sep 03 '19 at 12:54
  • Aha. It seems you're right. My mistake. Apologies for leading you down a deadend! – Theoretical Economist Sep 03 '19 at 13:02
  • I learned something. – Jas Ter Sep 03 '19 at 13:43
  • @JasTer The following implications hold: Continuously Frechet differentiability $\Rightarrow $ Strict differentiability $\Rightarrow$ Frechet differentiability. So, the proposition in Clarke's book solves your problem – John D Sep 04 '19 at 05:58
  • @JohnD If this is true, that's really wonderful! This also holds in infinite dimensions? – Jas Ter Sep 04 '19 at 06:51
  • @JasTer Ups, sorry. My concept of strict differentiability is different from the one in Clarke's book. For me, $f$ is strictly differentiable at $\bar{x}$ (Mordukhovich 2005 volume 1) if there exists $\nabla f(\bar{x})\in L \in X^*$ such that $$\lim_{x\to \bar{x}, u \to \bar{x},x\ne u}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u) }{|x-u|}=0.$$ You have to check the relationship between this concept and that one of Clarke. – John D Sep 04 '19 at 07:46
  • @JohnD Thanks! As far as I can see, the Clarke's strict diffability does not imply Fr'echet. (I believe it would be stated somewhere, if that was the case!) I will study your definition, too. – Jas Ter Sep 04 '19 at 07:49
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    @JasTer I think you need the Asplund condition in the space, but is just a hunch. See Proposition 2.18 in Mordukhovich's book volume 1 – John D Sep 04 '19 at 10:22

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