Darboux function

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 ''ƒ'' is continuously differentiable (''ƒ'' in ''C''¹( 'a'',''b''), this is a consequence of the intermediate value theorem. But even when ''ƒ′'' 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 a, then for every $y$ between $f'(a)$ and $f'(b)$, there exists an $x$ in ,b/math> such that $f'(x)=y$.Apostol, Tom M.: Mathematical Analysis: A Modern Approach to Advanced Calculus, 2nd edition, Addison-Wesley Longman, Inc. (1974), page 112.Olsen, Lars: ''A New Proof of Darboux's Theorem'', Vol. 111, No. 8 (Oct., 2004) (pp. 713–715), The American Mathematical MonthlyRudin, Walter: Principles of Mathematical Analysis, 3rd edition, MacGraw-Hill, Inc. (1976), page 108

Proofs

Proof 1. The first proof is based on the extreme value theorem.

If $y$ equals $f'(a)$ or $f'(b)$, then setting $x$ equal to $a$ or $b$, respectively, gives the desired result. Now assume that $y$ is strictly between $f'(a)$ and $f'(b)$, and in particular that $f'(a)>y>f'(b)$. Let $\varphi\colon I\to \R$ such that $\varphi(t)=f(t)-yt$. If it is the case that f'(a) we adjust our below proof, instead asserting that $\varphi$ has its minimum on ,b/math>.

Since $\varphi$ is continuous on the closed interval ,b/math>, the maximum value of $\varphi$ on ,b/math> is attained at some point in ,b/math>, according to the extreme value theorem.

Because $\varphi'(a)=f'(a)-y> 0$, 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 (stationary points), 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 $c = \frac (a + b)$.
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 $g(t) = \frac$ with $a < t < b$.
$\, g$ is continuous in $(a, b)$.

Furthermore, $g(t) \rightarrow ' (a)$ when $t \rightarrow a$ and $g(t) \rightarrow ' (b)$ when $t \rightarrow b$; therefore, from the Intermediate Value Theorem, if $y \in (' (a), ' (b))$ 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 $'(x) = g(t_0)$.
Hence, $' (x) = y$.

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:
:$x \mapsto \begin\sin(1/x) & \text x\ne 0, \\ 0 &\text x=0. \end$

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.Bruckner, Andrew M: ''Differentiation of real functions'', 2 ed, page 6, American Mathematical Society, 1994 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.

Notes

External links

*
* {{SpringerEOM, title=Darboux theorem, id=p/d030190

Theorems in calculus
Theory of continuous functions
Theorems in real analysis
Articles containing proofs