Английская Википедия:Darboux's theorem (analysis)
Шаблон:Short description 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.
Darboux's theorem
Let <math>I</math> be a closed interval, <math>f\colon I\to \R</math> be a real-valued differentiable function. Then <math>f'</math> has the intermediate value property: If <math>a</math> and <math>b</math> are points in <math>I</math> with <math>a<b</math>, then for every <math>y</math> between <math>f'(a)</math> and <math>f'(b)</math>, there exists an <math>x</math> in <math>[a,b]</math> such that <math>f'(x)=y</math>.[1][2][3]
Proofs
Proof 1. The first proof is based on the extreme value theorem.
If <math>y</math> equals <math>f'(a)</math> or <math>f'(b)</math>, then setting <math>x</math> equal to <math>a</math> or <math>b</math>, respectively, gives the desired result. Now assume that <math>y</math> is strictly between <math>f'(a)</math> and <math>f'(b)</math>, and in particular that <math>f'(a)>y>f'(b)</math>. Let <math>\varphi\colon I\to \R</math> such that <math>\varphi(t)=f(t)-yt</math>. If it is the case that <math>f'(a)<y<f'(b)</math> we adjust our below proof, instead asserting that <math>\varphi</math> has its minimum on <math>[a,b]</math>.
Since <math>\varphi</math> is continuous on the closed interval <math>[a,b]</math>, the maximum value of <math>\varphi</math> on <math>[a,b]</math> is attained at some point in <math>[a,b]</math>, according to the extreme value theorem.
Because <math>\varphi'(a)=f'(a)-y> 0</math>, we know <math>\varphi</math> cannot attain its maximum value at <math>a</math>. (If it did, then <math> (\varphi(t)-\varphi(a))/(t-a) \leq 0 </math> for all <math> t \in (a,b] </math>, which implies <math> \varphi'(a) \leq 0 </math>.)
Likewise, because <math>\varphi'(b)=f'(b)-y<0</math>, we know <math>\varphi</math> cannot attain its maximum value at <math>b</math>.
Therefore, <math>\varphi</math> must attain its maximum value at some point <math>x\in(a,b)</math>. Hence, by Fermat's theorem, <math>\varphi'(x)=0</math>, i.e. <math>f'(x)=y</math>.
Proof 2. The second proof is based on combining the mean value theorem and the intermediate value theorem.[1][2]
Define <math>c = \frac{1}{2} (a + b)</math>. For <math>a \leq t \leq c,</math> define <math>\alpha (t) = a</math> and <math>\beta (t) = 2t - a</math>. And for <math>c \leq t \leq b,</math> define <math>\alpha (t) = 2t - b</math> and <math>\beta(t) = b</math>.
Thus, for <math>t \in (a,b)</math> we have <math>a \leq \alpha (t) < \beta (t) \leq b</math>. Now, define <math>g(t) = \frac{(f \circ \beta)(t) - (f \circ \alpha)(t)}{\beta(t) - \alpha(t)}</math> with <math>a < t < b</math>. <math>\, g</math> is continuous in <math>(a, b)</math>.
Furthermore, <math>g(t) \rightarrow {f}' (a)</math> when <math>t \rightarrow a</math> and <math>g(t) \rightarrow {f}' (b)</math> when <math>t \rightarrow b</math>; therefore, from the Intermediate Value Theorem, if <math>y \in ({f}' (a), {f}' (b))</math> then, there exists <math>t_0 \in (a, b)</math> such that <math>g(t_0) = y</math>. Let's fix <math>t_0</math>.
From the Mean Value Theorem, there exists a point <math>x \in (\alpha (t_0), \beta (t_0))</math> such that <math>{f}'(x) = g(t_0)</math>. Hence, <math>{f}' (x) = y</math>.
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.[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:
- <math>x \mapsto \begin{cases}\sin(1/x) & \text{for } x\ne 0, \\ 0 &\text{for } x=0. \end{cases}</math>
By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function <math>x \mapsto x^2\sin(1/x)</math> 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. The Conway base 13 function is again an example.[4]
Notes
External links
- ↑ 1,0 1,1 Apostol, Tom M.: Mathematical Analysis: A Modern Approach to Advanced Calculus, 2nd edition, Addison-Wesley Longman, Inc. (1974), page 112.
- ↑ 2,0 2,1 Olsen, Lars: A New Proof of Darboux's Theorem, Vol. 111, No. 8 (Oct., 2004) (pp. 713–715), The American Mathematical Monthly
- ↑ Rudin, Walter: Principles of Mathematical Analysis, 3rd edition, MacGraw-Hill, Inc. (1976), page 108
- ↑ 4,0 4,1 Шаблон:Cite book
- ↑ Bruckner, Andrew M: Differentiation of real functions, 2 ed, page 6, American Mathematical Society, 1994
