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Suppose that $f$ is differentiable on $[a, b]$. Prove that if $f'(a) < c < f'(b)$, then $f'(x) = c$ for some $x$ in $(a,b)$. (This result is known as Darboux's theorem.)

Source: Spivak's Calculus Ch.11 - Significance of the Derivative.

I have a hard time picturing this or interpreting it... What are the implications of this?

Similarly, for another theorem introduced in the same chapter, I am having the problem of seeing the significance of this...

Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x = a$. Suppose, moreover, that $\lim \limits_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and $f'(a) = \lim \limits_{x \to a} f'(x)$.

Doesn't this mean that $f'$ is just continuous at point $a$?

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2 Answers 2

up vote 6 down vote accepted

One says that a function $f: I \rightarrow \mathbb{R}$ has the Intermediate Value Property (IVP) if for all $a < b \in I$ and all $L$ in between $f(a)$ and $f(b)$, there exists $c \in (a,b)$ such that $f(c) = L$.

Certainly the most famous and important class of functions which satisfy IVP are the continuous functions: this is the content of the Intermediate Value Theorem.

Darboux's Theorem gives a second class of functions which satisfy IVP: derivatives. Note well: not the differentiable functions but the functions which are of the form $f'$ for some other function $f$.

This is certainly an interesting result, if somewhat subtle: most of the differentiable functions one meets in freshman calculus actually have continuous derivatives, and in this case the fact that $f'$ satisfies IVP follows from the Intermediate Value Theorem. But indeed there are derivatives which are discontinuous, even rather badly: namely there are differentiable functions $f: [a,b] \rightarrow \mathbb{R}$ such that $f': [a,b] \rightarrow \mathbb{R}$ exists but is unbounded. There are also differentiable functions with a derivative $f'$ which is bounded but nevertheless not Riemann integrable.

One can view the significance of Darboux's Theorem as follows: it says that a derivative can be discontinuous but cannot have a jump discontinuity, i.e., a discontinuity in which the one-sided limits exist but are different (and also not a removable discontinuity, when the limit exists but is not equal to the value at the point). This is an interesting contrast to monotone functions, which also need not be continuous but can have only jump discontinuities.

In terms of actual applications of Darboux's Theorem...it seems they are rather few. As I mentioned in my (Spivak calculus) class, you can use Darboux's Theorem together with the (deep!) theorem that every continuous function admits an antiderivative to prove the Intermediate Value Theorem...but this is a strange way to prove IVT.

About the second result you mention: I admit to being somewhat perplexed as to what that is doing there: it is really not a standard textbook result. I seem to recall that this theorem gets used for something later on in the book, though: let's wait and see. (I am almost halfway through my year long course and we have currently covered about the first 13 chapters.)

I have some lecture notes for this course I'm teaching -- you're more than welcome to take a look at them. Here is the main course page. This handout on differentiation and this handout including the "Monotone Jump Theorem" (on discontinuities of monotone functions) are directly relevant to your question.

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a.k.a. function having Darboux property or Darboux function –  Martin Sleziak Dec 3 '11 at 6:59
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@Martin: yes, definitely. At the research level "Darboux function" is probably more commonly used, but in a context of teaching undergraduates I prefer the more transparent IVP... –  Pete L. Clark Dec 3 '11 at 17:41
    
Though not a general result, Darboux comes in handy for this problem: Let $f\in C^1 ( [0,1], \mathbb{R} ) $ be such that $\int^1_0 f(x) dx = \int^1_0 x f(x) dx = 1.$ Prove that there exists $ c \in [0,1] $ such that $f'(c) = 6.$ –  Ragib Zaman Dec 4 '11 at 4:21

Recall the Intermediate Value Theorem: it says that if $f$ is continuous on $[a,b]$, and $c$ is some number between $f(a)$ and $f(b)$, then there is some $x\in(a,b)$ such that $f(x)=c$. In other words, a continuous function can’t get from $f(a)$ to $f(b)$ without passing through every intermediate value.

Now the derivative of a function need not be continuous, so the Intermediate Value Theorem doesn’t guarantee that if $c$ is between $f\;'(a)$ and $f\;'(b)$, then there is some $x\in(a,b)$ such that $f\;'(x)=c$. None the less, this statement is true: if $f$ is differentiable on an interval, its derivative $f\;'$ has this same intermediate value property possessed by all continuous functions. In other words, a derivative can’t get from $f\;'(a)$ to $f\;'(b)$ without passing through every intermediate value even if it isn’t continuous. This is the content of Darboux’s theorem.

You’re right that the conclusion of the second theorem that you quote is that $f\;'$ is continuous at $a$. The theorem could be rephrased as follows:

If $f$ is continuous at $a$, there are $b<a$ and $c>a$ such that $f\;'(x)$ exists for all $x\in(b,a)\cup(a,c)$, and $\lim\limits_{x\to a}f\;'(x)$ exists, then $f\;'$ is continuous at $a$.

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Dear Brian, One could also say that "$f'$ exists and is continuous at $a$", and maybe the reason for the phrasing in the OP is precisely to emphasize that the assumptions force $f'(a)$ to exist (which it might not, a priori). Regards, –  Matt E Dec 3 '11 at 6:45
    
@Matt: One could indeed say explicitly that $f\;'$ exists as well as noting that it’s continuous at $a$, and there might be sound pædagogical reasons to do so, but it’s certainly not necessary: continuity of $f\;'$ at $a$ entails existence of $f\;'(a)$. –  Brian M. Scott Dec 3 '11 at 7:12
    
I agree with Matt E: of course what Brian writes is correct, but I think the issue here is mostly pedagogical. The content of the conclusion is in fact that $f'(a)$ exists; if so, then -- by Darboux's Theorem -- given that $\lim_{x \rightarrow a} f'(x)$ exists, $f'$ must be continuous at $a$. –  Pete L. Clark Dec 3 '11 at 17:49
    
@Pete: In the abstract the issue is indeed pædagogical, but for once that wasn’t my main concern: here I was responding to the OP’s specific question. –  Brian M. Scott Dec 3 '11 at 17:54
    
@Brian: Fair enough. –  Pete L. Clark Dec 3 '11 at 18:49

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