Darboux's theorem (analysis)

In mathematics, Darboux's theorem is a theorem in real analysis, named after Jean Gaston Darboux. It states that every function that results from the differentiation of another function has the intermediate value property: the image of an interval is also an interval.

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

Darboux's theorem
Let $$I$$ be a closed interval, $$f\colon I\to \R$$ be a real-valued differentiable function. Then $$f'$$ has the intermediate value property: If $$a$$ and $$b$$ are points in $$I$$ with $$ay>f'(b)$$. Let $$\varphi\colon I\to \R$$ such that $$\varphi(t)=f(t)-yt$$. If it is the case that $$f'(a) 0$$, we know $$\varphi$$ cannot attain its maximum value at $$a$$. (If it did, then $$ (\varphi(t)-\varphi(a))/(t-a) \leq 0 $$ for all $$ t \in (a,b] $$, which implies $$ \varphi'(a) \leq 0 $$.)

Likewise, because $$\varphi'(b)=f'(b)-y<0$$, we know $$\varphi$$ cannot attain its maximum value at $$b$$.

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

Proof 2. The second proof is based on combining the mean value theorem and the intermediate value theorem.

Define $$c = \frac{1}{2} (a + b)$$. For $$a \leq t \leq c,$$ define $$\alpha (t) = a$$ and $$\beta (t) = 2t - a$$. And for $$c \leq t \leq b,$$ define $$\alpha (t) = 2t - b$$ and $$\beta(t) = b$$.

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

Furthermore, $$g(t) \rightarrow {f}' (a)$$ when $$t \rightarrow a$$ and $$g(t) \rightarrow {f}' (b)$$ when $$t \rightarrow b$$; therefore, from the Intermediate Value Theorem, if $$y \in ({f}' (a), {f}' (b))$$ then, there exists $$t_0 \in (a, b)$$ such that $$g(t_0) = y$$. Let's fix $$t_0$$.

From the Mean Value Theorem, there exists a point $$x \in (\alpha (t_0), \beta (t_0))$$ such that $${f}'(x) = g(t_0)$$. Hence, $${f}' (x) = y$$.

Darboux function
A Darboux function is a real-valued function ƒ which has the "intermediate value property": for any two values a and b in the domain of ƒ, and any y between ƒ(a) and ƒ(b), there is some c between a and b with ƒ(c) = y. 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 topologist's sine curve function:
 * $$x \mapsto \begin{cases}\sin(1/x) & \text{for } x\ne 0, \\ 0 &\text{for } x=0. \end{cases}$$

By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function $$x \mapsto x^2\sin(1/x)$$ is a Darboux function even though it 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 ƒ on the real line can be written as the sum of two Darboux functions. 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. The Conway base 13 function is again an example.