§4. Differentiable functions
263
Corollary 2. Let f be a real function defined and differentiable on an interval I; for f to be increasing in I it is necessary and sufficient that l' (x) ~ 0
for all x E I; for it to be strictly increasing, it suffices that f'(x) > 0 for all
x E I. If f'(x) > 0 in I and if f is of class CP on I, then the inverse map
g: f(1) ~ I is of class CPo
The fact that 1'(x) ~ 0 if f is increasing is obvious. It is equally clear
(Theorem 18) that f is strictly increasing if f'(x) > 0, and then f maps
I bijectively onto an interval J = f(1) with an inverse map 9 : J ~ I,
differentiable according to the rule (D 5) of nO 15. Further, at corresponding
points x E I and y = f(x) E J, one has
g'(y) = 1/f'(x) = 1/f'[g(y)],
so that g is of class C1 if f is; if f is of class C2, then 1/1' is of class C1,
and the preceding formula shows that g' is too, so that 9 is of class C 2 , and
so on.
Note that the derivative of a strictly increasing function can well vanish
at some points, as is the case for the function x 3 on 1R, for example.
Formula (1) has an important consequence when the derivative l' is
bounded on I. Since 11'(c)1 ~ Ilf'lll, the least upper bound over I of the
numbers If' (x) I, we have
(16.5)
If(b) - f(a)1 ~ 1If'111·lb - al
for all a, bEl. This is the inequality of the mean" an expression whose
significance stems from the theory of integration (Chap. V, nO 11). Note that
it is not the uniform norm of l' on I which really appears; it is that of f' over
the compact interval [a, b], and this is finite if, for example, l' is continuous
(nO 9, Theorem 11).
This argument supposes that f has real values, but the result extends to
functions with complex values, thanks to the following trick.
Suppose that one wants to prove that a given complex number u is ~ A,
where A > 0 is given. If such is the case, then Izul ~ Aizi for all z E C and
so
(16.6)
I Re(zu) I ~ Aizi
for all z E Co
If, conversely, this condition is fulfilled, it applies to z = ii, whence lui ~ A
since Re( uii) = Re(luI 2 ) = lul 2 .
This point established, let us return to a complex function f that is everywhere differentiable on an interval I, and put u = f(b) - f(a). For z E C we
then have
IRe(zu) I = IRe{z[/(b) - l(a)]}1 = IRe[zl(b))- Re[zl(a)) I.
The function Iz(x) = Re[zl(x)) is differentiable, like I, and
263
Corollary 2. Let f be a real function defined and differentiable on an interval I; for f to be increasing in I it is necessary and sufficient that l' (x) ~ 0
for all x E I; for it to be strictly increasing, it suffices that f'(x) > 0 for all
x E I. If f'(x) > 0 in I and if f is of class CP on I, then the inverse map
g: f(1) ~ I is of class CPo
The fact that 1'(x) ~ 0 if f is increasing is obvious. It is equally clear
(Theorem 18) that f is strictly increasing if f'(x) > 0, and then f maps
I bijectively onto an interval J = f(1) with an inverse map 9 : J ~ I,
differentiable according to the rule (D 5) of nO 15. Further, at corresponding
points x E I and y = f(x) E J, one has
g'(y) = 1/f'(x) = 1/f'[g(y)],
so that g is of class C1 if f is; if f is of class C2, then 1/1' is of class C1,
and the preceding formula shows that g' is too, so that 9 is of class C 2 , and
so on.
Note that the derivative of a strictly increasing function can well vanish
at some points, as is the case for the function x 3 on 1R, for example.
Formula (1) has an important consequence when the derivative l' is
bounded on I. Since 11'(c)1 ~ Ilf'lll, the least upper bound over I of the
numbers If' (x) I, we have
(16.5)
If(b) - f(a)1 ~ 1If'111·lb - al
for all a, bEl. This is the inequality of the mean" an expression whose
significance stems from the theory of integration (Chap. V, nO 11). Note that
it is not the uniform norm of l' on I which really appears; it is that of f' over
the compact interval [a, b], and this is finite if, for example, l' is continuous
(nO 9, Theorem 11).
This argument supposes that f has real values, but the result extends to
functions with complex values, thanks to the following trick.
Suppose that one wants to prove that a given complex number u is ~ A,
where A > 0 is given. If such is the case, then Izul ~ Aizi for all z E C and
so
(16.6)
I Re(zu) I ~ Aizi
for all z E Co
If, conversely, this condition is fulfilled, it applies to z = ii, whence lui ~ A
since Re( uii) = Re(luI 2 ) = lul 2 .
This point established, let us return to a complex function f that is everywhere differentiable on an interval I, and put u = f(b) - f(a). For z E C we
then have
IRe(zu) I = IRe{z[/(b) - l(a)]}1 = IRe[zl(b))- Re[zl(a)) I.
The function Iz(x) = Re[zl(x)) is differentiable, like I, and