If we have the function
$f: I \to \mathbb{R}$ with derivative $f’: I \to \mathbb{R}$
such that:
$I \subseteq \mathbb{R}$
$f’ \in L^{1}(I)$
can we deduce that $f$ is uniformly continuous?
Of course it is true if $I
$ is compact since $f$ is continuous in $I$ and we can use Heine Cantor theorem; it is also true if $f’$ is bounded. But what happens when $I$ is not compact and $f’$ is unbounded and summable in $I$?
I tried to demonstrate that the implication holds but I failed, so I tried to find a counterexample taking $I$ as a countable union of non-overlapping intervals such that $ I $ is unbounded and $|I| < +\infty $ but I failed again. Can someone help me? Thank you
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Arianna
- 21
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2The answer depends on what notion of derivative you're using. Do you mean a pointwise a.e. derivative? Or a distributional/weak derivative? – Jose27 Dec 03 '22 at 12:56
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I mean a pointwise a.e. derivative – Arianna Dec 04 '22 at 15:00
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2Then the answer is no: Consider $f$ the indicator function of an interval. – Jose27 Dec 05 '22 at 02:04
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@Jose27, I think the OP meant to refer to a differentiable (rather than a.e. differentiable) function (otherwise there is no basis for claiming that either the function or the derivative will be bounded on a compact interval). It seems like a legitimate question. – Mikhail Katz Dec 06 '22 at 07:44
1 Answers
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HINT. It is enough to assume $I=[0, \infty)$ and you can also assume $f(0)=0$. You then have the formula $$\tag{1} f(x)=\int_0^x f'(t)\, dt.$$ Use this to study the increment $f(x)-f(y)$.
IMPORTANT REMARK. As Jose27 pointed out in comments to the main question, it is important to specify the notion of derivative we are talking about. The formula (1) holds if $f'$ is a distributional derivative, provided that $f'$ is an integrable function. It is actually an if and only if. That is, $$ \text{(1) holds} \iff f' \text{ is the distributional derivative of }f \text{ and }f'\text{ is integrable.}$$
In this case, it is said that $f$ is absolutely continuous.
This old question of mine deals exactly with this subject.

Giuseppe Negro
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