Continuity everywhere+partial derivatives in one point$\implies$Total differentiability in this point?

Question

Let $f\colon\mathbb{R}^{n}\to\mathbb{R}$ be continuous everywhere. If we know that in one single point $x_0$ all the partial derivatives exist and are linear, $\partial_vf(x_0)=L(v)$, does this imply that $f$ is totally differentiable in $x_0$?

In a previous question, user Etienne gave an example that showed that the continuity of $f$ everywhere rather than just in $x_0$ is necessary.

Do you mean directional derivatives? – dfeuer Jul 20 '13 at 00:20 — dfeuer, Jul 20 '13 at 00:20
What do you mean by the previous Question here? – Kumar Sep 02 '20 at 09:00 — Kumar, Sep 02 '20 at 09:00
@Kumar https://math.stackexchange.com/q/447104/87023 – Chris Culter Sep 02 '20 at 10:08 — Chris Culter, Sep 02 '20 at 10:08

score 2 · Accepted Answer · answered Jul 20 '13 at 04:25

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The answer is still "no". In fact, we can smooth out Etienne's example from the previous question.

Let $B\colon\mathbb{R}\to[0,1]$ be a smooth bump function supported on $[0,2]$ such that $B(1) = 1$. Now let $f(x,y)=xB(y/x^2)$.

answered Jul 20 '13 at 04:25

Chris Culter

26,806

1

What do you mean by the previous Question here? – Kumar Sep 02 '20 at 09:00
1

@Kumar https://math.stackexchange.com/a/447197/87023 – Chris Culter Sep 02 '20 at 10:08

Continuity everywhere+partial derivatives in one point$\implies$Total differentiability in this point?

1 Answers1