# Darboux's theorem (analysis)

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Darboux's theorem is a theorem in real analysis, named after Jean Gaston Darboux. It states that all functions that result from the differentiation of other functions have the intermediate value property: the image of an interval is also an interval.

When f is continuously differentiable (f in C1([a,b])), this is a consequence of the intermediate value theorem. But even when f′ is not continuous, Darboux's theorem places a severe restriction on what it can be.

## Darboux's theorem

Let ${\displaystyle I}$ be a closed interval, ${\displaystyle f\colon I\to \mathbb {R} }$ a real-valued differentiable function. Then ${\displaystyle f'}$ has the intermediate value property: If ${\displaystyle a}$ and ${\displaystyle b}$ are points in ${\displaystyle I}$ with ${\displaystyle a, then for every ${\displaystyle y}$ between ${\displaystyle f'(a)}$ and ${\displaystyle f'(b)}$, there exists an ${\displaystyle x}$ in ${\displaystyle (a,b)}$ such that ${\displaystyle f'(x)=y}$.[1][2][3]

## Proof

If ${\displaystyle y}$ equals ${\displaystyle f'(a)}$ or ${\displaystyle f'(b)}$, then setting ${\displaystyle x}$ equal to ${\displaystyle a}$ or ${\displaystyle b}$, respectively, works. Now assume that ${\displaystyle y}$ is strictly between ${\displaystyle f'(a)}$ and ${\displaystyle f'(b)}$, and in particular that ${\displaystyle f'(a)>y>f'(b)}$. Let ${\displaystyle \phi \colon I\to \mathbb {R} }$ such that ${\displaystyle \phi (t)=f(t)-yt}$.If it is the case that ${\displaystyle f'(a) we adjust our below proof, instead asserting that ${\displaystyle \phi }$ has its minimum on ${\displaystyle [a,b]}$.

Since ${\displaystyle \phi }$ is continuous on the closed interval ${\displaystyle [a,b]}$, the maximum value of ${\displaystyle \phi }$ on ${\displaystyle [a,b]}$,is attained at some point in ${\displaystyle [a,b]}$, according to the extreme value theorem.

Because ${\displaystyle \phi '(a)=f'(a)-y>0}$, we know ${\displaystyle \phi }$ cannot attain its maximum value at ${\displaystyle a}$. Likewise, because ${\displaystyle \phi '(b)=f'(b)-y<0}$, we know ${\displaystyle \phi }$ cannot attain its maximum value at ${\displaystyle b}$.

Therefore ${\displaystyle \phi }$ must attain its maximum value at some point ${\displaystyle x\in (a,b)}$. Hence, by Fermat's theorem, ${\displaystyle \phi '(x)=0}$, i.e. ${\displaystyle f'(x)=y}$.

Another proof can be given by combining the mean value theorem and the intermediate value theorem.[1][2]

In fact, let's take ${\displaystyle c={\frac {1}{2}}(a+b)}$. For ${\displaystyle a\leq t\leq c,}$ define ${\displaystyle \alpha (t)=a}$ and ${\displaystyle \beta (t)=2t-a}$. And for ${\displaystyle c\leq t\leq b,}$ define ${\displaystyle \alpha (t)=2t-b}$ and ${\displaystyle \beta (t)=b}$.

Thus, for ${\displaystyle t\in (a,b)}$ we have ${\displaystyle a\leq \alpha (t)<\beta (t)\leq b}$. Now, define ${\displaystyle g(t)={\frac {(f\circ \beta )(t)-(f\circ \alpha )(t)}{\beta (t)-\alpha (t)}}}$ with ${\displaystyle a. ${\displaystyle \,g}$ is continuous in ${\displaystyle (a,b)}$.

Furthermore, ${\displaystyle g(t)\rightarrow {f}'(a)}$ when ${\displaystyle t\rightarrow a}$ and ${\displaystyle g(t)\rightarrow {f}'(b)}$ when ${\displaystyle t\rightarrow b}$, and, therefore, from the Intermediate Value Theorem, if ${\displaystyle y\in ({f}'(a),{f}'(b))}$ then, there exists ${\displaystyle t_{0}\in (a,b)}$ such that ${\displaystyle g(t_{0})=y}$. Let's fix ${\displaystyle t_{0}}$.

From the Mean Value Theorem, there exists a point ${\displaystyle x\in (\alpha (t_{0}),\beta (t_{0}))}$ such that ${\displaystyle {f}'(x)=g(t_{0})}$. Hence, ${\displaystyle {f}'(x)=y}$.

## Darboux function

A Darboux function is a real-valued function f which has the "intermediate value property": for any two values a and b in the domain of f, and any y between f(a) and f(b), there is some c between a and b with f(c) = y.[4] By the intermediate value theorem, every continuous function on a real interval is a Darboux function. Darboux's contribution was to show that there are discontinuous Darboux functions.

Every discontinuity of a Darboux function is essential, that is, at any point of discontinuity, at least one of the left hand and right hand limits does not exist.

An example of a Darboux function that is discontinuous at one point, is the function

${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0\\0&{\text{for }}x=0\end{cases}}}$.

By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function ${\displaystyle x\mapsto x^{2}\sin(1/x)}$ is a Darboux function that is not continuous at one point.

An example of a Darboux function that is nowhere continuous is the Conway base 13 function.

Darboux functions are a quite general class of functions. It turns out that any real-valued function f on the real line can be written as the sum of two Darboux functions.[5] This implies in particular that the class of Darboux functions is not closed under addition.

A strongly Darboux function is one for which the image of every (non-empty) open interval is the whole real line. Such functions exist and are Darboux but nowhere continuous.[4]

## Notes

1. ^ a b Apostol, Tom M.: Mathematical Analysis: A Modern Approach to Advanced Calculus, 2nd edition, Addison-Wesley Longman, Inc. (1974), page 112.
2. ^ a b Olsen, Lars: A New Proof of Darboux's Theorem, Vol. 111, No. 8 (Oct., 2004) (pp. 713–715), The American Mathematical Monthly
3. ^ Rudin, Walter: Principles of Mathematical Analysis, 3rd edition, MacGraw-Hill, Inc. (1976), page 108
4. ^ a b Ciesielski, Krzysztof (1997). Set theory for the working mathematician. London Mathematical Society Student Texts. 39. Cambridge: Cambridge University Press. pp. 106–111. ISBN 0-521-59441-3. Zbl 0938.03067.
5. ^ Bruckner, Andrew M: Differentiation of real functions, 2 ed, page 6, American Mathematical Society, 1994