## You are here

HomeDarboux's theorem (analysis)

## Primary tabs

# Darboux’s theorem (analysis)

Let $f:[a,b]\to\mathbb{R}$ be a real-valued continuous function on $[a,b]$, which is differentiable on $(a,b)$, differentiable from the right at $a$, and differentiable from the left at $b$. Then $f^{{\prime}}$ satisfies the intermediate value theorem: for every $t$ between $f^{{\prime}}_{{+}}(a)$ and $f^{{\prime}}_{{-}}(b)$, there is some $x\in[a,b]$ such that $f^{{\prime}}(x)=t$.

Note that when $f$ is continuously differentiable ($f\in C^{1}([a,b])$), this is trivially true *by* the intermediate value theorem. But even when $f^{{\prime}}$ is *not* continuous, Darboux’s theorem places a severe restriction on what it can be.

Synonym:

intermediate value property of the derivative

Type of Math Object:

Theorem

Major Section:

Reference

## Mathematics Subject Classification

26A06*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Comments

## is the converse true?

is the convese of darboux's theorem true? i mean, if f(x) does not have any jump discontinuity, then is it certain that there exists a function F(x) such that F'(x) = f(x) ?

## Re: is the converse true?

But if f(x) does not have any jump on its domain, f(x) is integrable.

## Re: is the converse true?

Or I missing something?

## Re: is the converse true?

The original question asked is quite specialized.

We might think to try to construct the F

such that F' = f by

F(x) = \int_a^x f(t) dt + F(a)

This works as long as $f$ is integrable; apparently

there might be some $f$ that are not, for example:

f(x) = d/dx [ x^2 \sin(x^{-2}) ] =

This is a standard textbook example; f is not in Lebesgue

integrable on [0,1] because of the (non-jump) discontinuity

at 0. Of course one can construct F by taking limits

and integrating from (-epsilon, 1) or something like that.

But other than this simple counterexample,

I don't know the complete answer though.

## Re: is the converse true?

Thanks stevecheng, I misunderstood your question. Sorry.

perucho

## Re: is the converse true?

Hi Gorkem,

According to which stevecheng explained to me, it seems to be that Darboux's theorem deals with a kind of "weak" continuity (really I didn't know about that) that is not the conventional concept of continuity. So f(x) not necessarily is integrable and your question is not as trivial as I thought.

Sincerely,

perucho