converse of Darboux’s theorem
(analysis) is not true∗
Gorkem†
2013-03-21 23:51:55
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 0 = 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
(
1
cos(ln x) if x > 0
f (x) =
.
x
0
if x ≤ 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(ln b). 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 0 (x) = f (x) on R. Then consider the function G(x) = sin(ln x) which is
defined on (0, ∞). We have G0 (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 F (x) = lim sup G(x) + c = 1 + c
x→0+
x→0+
∗ hConverseOfDarbouxsTheoremanalysisIsNotTruei created: h2013-03-21i by: hGorkemi
version: h39973i Privacy setting: h1i hExamplei h26A06i
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1
and
lim inf F (x) = lim inf G(x) + c = −1 + c
x→0+
x→0+
which contradicts the differentiability of F at 0.
2