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I have a problem with this question.

If $f$ is a derivative function, exists in an open non empty subset A of $ℝ^n$ and it's continuous in a point of A does it implies the existence of a neighbourhood of that point in which $f$ is continuous?

I have worked on it for a bit but cannot prove it nord find a counterexample.

Thank you all in Advance!

Anne Bauval
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mathwannabe
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  • Welcome to MSE. Are you assuming that $f$ is differentiable at each point of $A$? – José Carlos Santos Oct 05 '23 at 09:08
  • Thank you. No, just existence in A and continuity in a ∈ A. – mathwannabe Oct 05 '23 at 09:29
  • Then you can take $n=1$, $A=\Bbb R$, and$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\&x&\mapsto&\begin{cases}x^2&\text{ if }x\in\Bbb Q\0&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is differentiable at $0$ and only at $0$. Since the domain of $f'$ is ${0}$, then $f'$ is continuous. But there is no neighborhood $V$ of $0$ such that $f|_V$ is continuous. – José Carlos Santos Oct 05 '23 at 09:41
  • @JoséCarlosSantos I think the OP means $f$ is a derivative (of another function like the one you called $f$), but on an open (non empty) subset. So, your example doesn't work. – Anne Bauval Oct 05 '23 at 09:51
  • @AnneBauval You are probably right. – José Carlos Santos Oct 05 '23 at 09:52
  • I found that I made some language mistakes, edit the question to match, sorry for It! The question I'm trying to answer requires that f is a derivative function that lives in an open subset. – mathwannabe Oct 05 '23 at 09:53

1 Answers1

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No, it is not necessarily continuous on an open neighborhood. We have counterexamples already for $n=1$.

First, let $g$ be differentiable on $\mathbb R$ with $g'$ bounded near $0$ but discontinuous at $0$ (see e.g. this answer for an example). By multiplying by a bump function and scaling if necessary we may assume $g\equiv 0$ on $\mathbb R\backslash (-1,1)$, and $g(x)\leq 1$ and $g'(x)\leq 1$ everywhere.

Then let $x_i\to 0$ be a sequence, with each $x_i$ distinct, so there are $r_i>0$ for which $(x_i-2r_i,x_i+2r_i)$ are pairwise disjoint and do not contain $0$. Note that we then have $r_i\to 0$ as well.

Let $h_i(x)=r_i^2 g(\frac{x-x_i}{r_i})$, so that each $h_i$ is supported on $(x_i-r_i,x_i+r_i)$, and let $h(x)=\sum_i h_i(x)$. Then clearly $h$ is differentiable everywhere away from $0$, and moreover, at $0$ we have, for $x\in (x_i-r_i,x_i+r_i)$, $r_i\leq |x-0|$, so

$$|h(x) -h(0)|=|h(x)-0|\leq r_i^2 \leq r_i|x-0|$$ and for $x$ anywhere else we have $h(x)=0$, so $h$ is differentiable at $0$ with $h'(0)= 0$. Moreover, $|h'(x)|\leq r_i$ for $x\in (x_i-r_i,x_i+r_i)$, and $h'(x)=0$ elsewhere, so $h'$ is continuous at $0$.

However, $h'$ is not continuous on any neighborhood of $0$, since each such neighborhood contains an interval $(x_i-r_i,x_i+r_i)$, on which $h'$, a scaled and translated version of $g'$, is not continuous at $x_i$.

The function $f=h'$ is therefore a counterexample to your question.

M W
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