Does there exist a function with compact support and bounded discontinuous derivative?

Asked Feb 13 '25 at 01:57

Active Feb 13 '25 at 06:13

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In other word, suppose $f(x)$ has compact support, $f'(x)$ exists and is finite for all $x \in \mathbb{R}$. If $f'(x)$ is bounded, does it follow that $f'(x)$ is continuous?

There's a well-known example of a function with compact support and finite unbounded derivative: $g(x) = x^2 \sin \frac{1}{x^2}$ if $x \ne 0$ and $g(0)=0$. But it doesn't help.

edited Feb 13 '25 at 06:13

Dean Miller

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asked Feb 13 '25 at 01:57

Sergei Nikolaev

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Does thia answer your question? Examples of differentiable functions with bounded but discontinuous derivatives – Martin R Feb 13 '25 at 02:09
Yes, thank you! – Sergei Nikolaev Feb 13 '25 at 04:50
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@SergeiNikolaev Your example doesn't have compact support. You can take the example from the link of the first comment and multiply it by a continuously differentiable function with compact support that is equal to $1$ in a neighbourhood of $0$ to obtain your desired example. – Dean Miller Feb 13 '25 at 06:11

Does there exist a function with compact support and bounded discontinuous derivative?

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