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 ƒ is continuously differentiable (ƒ in C¹([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.

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

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

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

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

Because ${\displaystyle \varphi '(a)=f'(a)-y>0}${\displaystyle \varphi '(a)=f'(a)-y>0}, we know ${\displaystyle \varphi }${\displaystyle \varphi } cannot attain its maximum value at ${\displaystyle a}${\displaystyle a}. (If it did, then ${\displaystyle (\varphi (t)-\varphi (a))/(t-a)\leq 0}${\displaystyle (\varphi (t)-\varphi (a))/(t-a)\leq 0} for all ${\displaystyle t\in (a,b]}${\displaystyle t\in (a,b]}, which implies ${\displaystyle \varphi '(a)\leq 0}${\displaystyle \varphi '(a)\leq 0}.)

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

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

Proof 2. The second proof is based on combining the mean value theorem and the intermediate value theorem.^{[1] [2]}

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

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

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

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

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.^[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 topologist's sine curve function:

${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0,\0円&{\text{for }}x=0.\end{cases}}}${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0,\0円&{\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 ${\displaystyle x\mapsto x^{2}\sin(1/x)}${\displaystyle 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.^[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. ^[4]

• This article incorporates material from Darboux's theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
• "Darboux theorem", Encyclopedia of Mathematics , EMS Press, 2001 [1994]

• Theorems in calculus
• Theory of continuous functions
• Theorems in real analysis

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