converse of Darboux’s theorem (analysis) is not true
Darboux’ theorem says that, if f:ℝ→ℝ has an antiderivative, than f has to satisfy the intermediate value property, namely, for any a<b, for any number C with f(a)<C<f(b) or f(b)<C<f(a), there exists a c∈(a,b) such that f(c)=C. With this theorem, we understand that if f does not satisfy the intermediate value property, then no function F satisfies F′=f on ℝ.
Now, we will give an example to show that the converse is not true, i.e., a function that satisfies the intermediate value property might still have no antiderivative.
Let
f(x)={1xcos(lnx)ifx>00ifx≤0. |
First let us see that f satisfies the intermediate value property. Let a<b. If 0<a or b≤0, the property is satisfied, since f is continuous on (-∞,0] and (0,∞). If a≤0<b, we have f(a)=0 and f(b)=(1/b)cos(lnb). Let C be between f(a) and (b). Let a0=exp(-2πk0+π) for some k0 large enough such that a0<b. Then f(a0)=0=f(a), and since f is continuous on (a0,b), we must have a c∈(a0,b) with f(c)=C.
Assume, for a contradiction that there exists a differentiable function F such that F′(x)=f(x) on ℝ. Then consider the function G(x)=sin(lnx) which is defined on (0,∞).
We have G′(x)=f(x) on (0,∞), and since it is a an open connected set, we must have F(x)=G(x)+c on (0,∞) for some c∈ℝ. But then, we have
lim sup |
and
which contradicts the differentiability of at .
Title | converse of Darboux’s theorem (analysis![]() |
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Canonical name | ConverseOfDarbouxsTheoremanalysisIsNotTrue |
Date of creation | 2013-03-22 17:33:51 |
Last modified on | 2013-03-22 17:33:51 |
Owner | Gorkem (3644) |
Last modified by | Gorkem (3644) |
Numerical id | 6 |
Author | Gorkem (3644) |
Entry type | Example |
Classification | msc 26A06 |