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If $E\subseteq \mathbb R^d$ is convex and $f:E\to\mathbb R$ is a convex function, is it true that :

  1. For all $x\in E$, if $f$ has partial derivatives at $x$ along all directions $u$ such that $x+tu\in E$ for small enough $t>0$ (i.e., $f$ is Gateaux-differentiable at $x$), then $f$ is differentiable (in the Fréchet sense) at $x$ ?

  2. If $f$ is differentiable on $E$, then its gradient is continuous on $E$, even at boundary points ?

When $E$ is open, the answer is YES to both, but I am mostly interested in what happens at the boudary.

NB : when I say that $f$ is differentiable (in the Fréchet sense) at $x\in E$, I mean that there exists a vector $g\in\mathbb R^d$ such that $$\frac{f(y)-f(x)-g^\top (y-x)}{\|y-x\|} \xrightarrow[y\to x]{} 0$$ In that case, there is only one such vector that is parallel to $E$, i.e., in the span of $E-E$, and that's what I call the gradient of $f$ at $x$.

TrivialPursuit
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1 Answers1

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This is a counterexample for the first point:

Take $E = [0,\infty)^2$ and set $$ f(x,y) = \frac{-x\,y}{\sqrt{x^2 + y^2}}. $$ Then, $f$ is convex and differentiable at all points in $E \setminus \{0\}$. At $0$, all the partial derivatives exist. However, $f$ is not differentiable in $0$.

Edit: And this disproves the second point: Again, $E = [0,\infty)^2$ and $$f(x,y) = x^2 \, \exp(-y/x^2).$$ Then, $f$ is convex and everywhere differentiable. However, $$f'(0,0) = (0,0),\quad f'(x,0) = (2\,x, -1)$$ for $x > 0$.

gerw
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