Problem 11918
(American Mathematical Monthly, Vol.123, June-July 2016) Proposed by Le Van Phu Cuong (Vietnam).
Letf be n times continuously differentiable on [0, 1], with f (1/2) = 0 and f(i)(1/2) = 0 when i is even and less thann. Prove
Z 1 0
f (x) dx
2
≤ 1
(2n + 1)4n(n!)2 Z 1
0
f(n)(x)2
dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. If g ∈ Cn([0, 1]), then by integrating by parts we obtain Z a
0
g(x) dx =
n−1X
i=0
(−1)iai+1g(i)(a)
(i + 1)! +(−1)n n!
Z a 0
xng(n)(x) dx
for any a ∈ (0, 1). Hence Z 1/2
0
f (x) dx =
n−1X
i=0
(−1)if(i)(1/2)
2i+1(i + 1)! +(−1)n n!
Z 1/2 0
xnf(n)(x) dx,
and
Z 1 1/2
f (x) dx = Z 1/2
0
f (1 − x) dx =
n−1X
i=0
f(i)(1/2) 2i+1(i + 1)!+ 1
n!
Z 1/2 0
xnf(n)(1 − x) dx.
Since f(i)(1/2) = 0 when i is even and less than n, it follows that Z 1
0
f (x) dx = Z 1/2
0
f (x) dx+
Z 1 1/2
f (x) dx = 1
n! (−1)n Z 1/2
0
xnf(n)(x) dx + Z 1/2
0
xnf(n)(1 − x) dx
! .
Finally, by Cauchy-Schwarz inequality,
Z 1 0
f (x) dx
2
≤ 2
(n!)2
Z 1/2
0
xnf(n)(x) dx
!2
+ Z 1/2
0
xnf(n)(1 − x) dx
!2
≤ 2
(n!)2 Z 1/2
0
x2ndx Z 1/2
0
f(n)(x)2
dx + Z 1/2
0
x2ndx Z 1/2
0
f(n)(1 − x)2
dx
!
≤ 1
(2n + 1)4n(n!)2 Z 1
0
f(n)(x)2
dx,
where we used Z 1/2
0
x2ndx = 1
(2n + 1)22n+1 and
Z 1/2 0
f(n)(1 − x)2
dx = Z 1
1/2
f(n)(x)2
dx.
