Problem 12088
(American Mathematical Monthly, Vol.126, January 2019) Proposed by F. Stanescu (Romania).
Let k be a positive integer with k≥ 2, and let f : [0, 1] → R be a function with continuous k-th derivative. Suppose f(k)(x) ≥ 0 for all x ∈ [0, 1], and suppose f(i)(0) = 0 for all i ∈ {0, 1, . . . , k − 2}.
Prove
Z 1 0
xk−1f(1 − x) dx ≤ (k − 1)!k!
(2k − 1)!
Z 1 0
f(x) dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since f(i)(0) = 0 for i = 0, 1, . . . , k − 2, by integration by parts, Z 1
0
xk−1f(1 − x) dx = xk
k f(1 − x)
1
0
+ Z 1
0
xk
k f(1)(1 − x) dx = Z 1
0
xk
k f(1)(1 − x) dx
=
xk+1
k(k + 1)f(1)(1 − x)
1
0
+ Z 1
0
xk+1
k(k + 1)f(2)(1 − x) dx = Z 1
0
xk+1
k(k + 1)f(2)(1 − x) dx
=
x2k−1
k(k + 1) · · · (2k − 1)f(k−1)(1 − x)
1
0
+ Z 1
0
x2k−1
k(k + 1) · · · (2k − 1)f(k)(1 − x) dx
= (k − 1)!
(2k − 1)!
f(k−1)(0) + Z 1
0
x2k−1f(k)(1 − x) dx
.
On the other hand, by the integral form of the remainder in Taylor’s Theorem, we have that for x∈ [0, 1],
f(x) = f (0) + f′(0)x + · · · +f(k−1)(0) (k − 1)! xk−1+
Z x
0
(x − t)k−1
(k − 1)! f(k)(t) dt
=f(k−1)(0) (k − 1)! xk−1+
Z x
0
(x − t)k−1
(k − 1)! f(k)(t) dt, which implies
Z 1 0
f(x) dx =f(k−1)(0) (k − 1)!
Z 1 0
xk−1dx+ Z 1
0
Z x
0
(x − t)k−1
(k − 1)! f(k)(t) dt
dx
=f(k−1)(0)
k! +
Z 1 0
f(k)(t)
Z 1 t
(x − t)k−1 (k − 1)! dx
dt
= 1 k!
f(k−1)(0) + Z 1
0
(1 − t)kf(k)(t) dt
= 1 k!
f(k−1)(0) + Z 1
0
xkf(k)(1 − x) dx
.
Hence, the given inequality is equivalent to (k − 1)!
(2k − 1)!
f(k−1)(0) + Z 1
0
x2k−1f(k)(1 − x) dx
≤ (k − 1)!
(2k − 1)!
f(k−1)(0) + Z 1
0
xkf(k)(1 − x) dx
that is
Z 1 0
x2k−1f(k)(1 − x) dx ≤ Z 1
0
xkf(k)(1 − x) dx
which holds because f(k)(x) ≥ 0 and x2k−1≤ xk for x ∈ [0, 1].
