50
1 Preliminaries
taking n large enough, the two intervals
x 0 , x 0 +
π
n
and
x 1 , x 1 +
π
n
are located
in the interval [a, b], but they are mutually disjoint, and in the interval
x 0 , x 0 +
π
n
,
the inequality f (ξ ) > a 0 holds, moreover in the interval
x 1 , x 1 +
π
n
, the inequality
f (ζ ) < b 0 holds. At this moment, choosing the function η(x) as follows
η
(x) =
⎧
⎨
⎩
sin
2
[n(x − x 0 )] x ∈
x 0 , x 0 +
π
n
cos
2
[n(x − x 1 )] x ∈
x 1 , x 1 +
π
n
0
x is at the rest points in the interval [a, b]
(1.5.17)
Obviously, the function η(x) =
x
a η
(x)dx is continuous, has continuous
derivative η
(x), and
η(a) = 0
η(b) =
b
a
η
(x)dx =
x 0 +
π
n
x 0
sin
2
[n(x − x 0 )]dx −
x 1 +
π
n
x 1
sin
2
[n(x − x 0 )]dx = 0
According to the conditions of lemma, there is
b
a
f (x)η
(x)dx = 0
But then, there is
b
a
f (x)η
(x)dx =
x0+
π
n
x0
f (x) sin
2 [n(x − x 0 )]dx −
x1+
π
n
x1
f (x) sin
2 [n(x − x 0 )]dx
> (a 0 − b 0 )
π
n
0
sin
2 nxdx > 0
(1.5.18)
This contradicts Eq. (1.5.15), therefore f (x) must be a constant. Quod erat
demonstrandum.
Proof 2 Using integration by parts for Eq. (1.5.15), we obtain
b
a
f (x)η
(x)dx = [ f (x)η(x)]|
b
a −
b
a
f
(x)η(x)dx = −
b
a
f
(x)η(x)dx = 0
It follows from Lemma 1.5.2 that f
(x) ≡ 0, the integration gives f (x) ≡ C.
Quod erat demonstrandum.
Lemma 1.5.8 Let a function f (x) be continuously derivable in the interval [a, b],
arbitrary function η(x) has the n − 1th continuous derivative in the interval [a, b],
and η
(k)
(a) = η
(k)
(b) = 0 (k = 1, 2, . . . , n − 1), the integral