Problem 11814
(American Mathematical Monthly, Vol.122, January 2015) Proposed by C. Lupu (USA).
Let φ be a continuously differentiable function from [0, 1] into R, with φ(0) = 0 and φ(1) = 1, and suppose that φ0(x) 6= 0 for 0 ≤ x ≤ 1. Let f be a continuous function from [0, 1] into R, such that R1
0 f (x) dx =R1
0 φ(x)f (x) dx. Show that there exists t with 0 < t < 1 such thatRt
0φ(x)f (x) dx = 0.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let
g(s) =
φ0(s) (φ(s))2
Z s 0
φ(x)f (x) dx if s ∈ (0, 1], f (0)
2 if s = 0
(note that by Rolle’s theorem φ(s) 6= 0 for s ∈ (0, 1)).
The function g is continuous in [0, 1] because
lim
s→0+
φ0(s) (φ(s))2
Z s 0
φ(x)f (x) dx= φH 0(0) lim
s→0+
φ(s)f (s)
2φ(s)φ0(s)= f (0) 2 . Let
G(t) = Z t
0
g(s) ds then G(0) = 0 and
G(1) = Z 1
0
d ds
− 1 φ(s)
Z s 0
φ(x)f (x) dx
ds
=
− 1 φ(s)
·
Z s 0
φ(x)f (x) dx
1
0
+ Z 1
0
1 φ(s)
d ds
Z s 0
φ(x)f (x) dx
ds
= − 1 φ(1)
Z 1 0
φ(x)f (x) dx + lim
s→0+
1 φ(s)
Z s 0
φ(x)f (x) dx + Z 1
0
f (s) ds
= − 1 φ(1)
Z 1 0
φ(x)f (x) dx + lim
s→0+
φ(s)f (s) φ0(s) +
Z 1 0
f (s) ds
= − 1 φ(1)
Z 1 0
φ(x)f (x) dx +φ(0)f (0) φ0(0) +
Z 1 0
f (s) ds = 0.
Therefore, by Rolle’s theorem, there exists t ∈ (0, 1) such that G0(t) = g(t) = 0. Since φ(t) 6= 0 and φ0(t) 6= 0 for t ∈ (0, 1), it follows that
Z t 0
φ(x)f (x) dx = 0.
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