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Let $f:\mathbb R^d \to \mathbb R$ be convex and $$ F := \{x \in \mathbb R^d \mid f \text{ is not Fréchet differentiable at }x\}. $$

If $d=1$, then $F$ is countable. The proof in this case relies on the order of $\mathbb R$.

Is $F$ still countable if $d>1$?

Akira
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    If $g:\Bbb R\to\Bbb R$ is convex and you define $f:\Bbb R^2\to\Bbb R$ by $f(x,y)=g(x)$ then $f$ is convex... – David C. Ullrich Jun 29 '22 at 16:56
  • @DavidC.Ullrich I could not get your idea. Did you meant to reduce the multi-dimensional case to1D? Could you elaborate more? – Akira Jun 29 '22 at 23:15
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    $f : \mathbb R^2 \to \mathbb R$ defined by $f(x,y) = |x|$ is convex, but is non-differentiable on the uncountable line ${(x,y) : x=0}$. – GEdgar Jun 30 '22 at 00:10
  • Thank you @GEdgar so much. I have posted your example as an answer to remove this question from unanswered list. – Akira Jun 30 '22 at 00:43

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As @DavidC.Ullrich pointed in his comment, the result does not generalize to higher dimension. @GEdgar provided below example $$ f : \mathbb R^2 \to \mathbb R, (x,y) \mapsto |x|. $$

Then $f$ is not Fréchet differentiable on the set $\{(x,y) : x=0\}$ which is uncountable.

Akira
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