Problem 11756
(American Mathematical Monthly, Vol.121, February 2014) Proposed by P. Perfetti (Italy).
Let f be a function from[1, 1] to R with continuous derivatives of all orders up to 2n + 2. Given f(0) = f′′(0) = · · · = f(2n+2)(0) = 0, prove
(4n + 5)((2n + 2)!)2 2
Z 1
−1
f(x) dx
2
≤ Z 1
−1
f(2n+2)(x)2 dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By the integral form of the remainder in the Taylor’s Theorem we have that if x ∈ [−1, 1] then f(x) = P2n(x) + R2n(x) where
P2n(x) =
2n
X
k=0
f(k)(0)
k! xk and R2n(x) = Z x
0
f(2n+2)(t)
(2n + 1)! (x − t)2n+1dt.
Since f (0) = f′′(0) = · · · = f(2n+2)(0) = 0, it follows that P2n is an odd function and Z 1
−1
P2n(x) dx = 0.
Therefore Z 1
−1
f(x) dx = Z 1
−1
R2n(x) dx
= Z 0
x=−1
Z 0 t=x
f(2n+2)(t)
(2n + 1)! (t − x)2n+1dt dx+ Z 1
x=0
Z x
t=0
f(2n+2)(t)
(2n + 1)! (x − t)2n+1dt dx
= Z 0
t=−1
Z t
x=−1
f(2n+2)(t)
(2n + 1)! (t − x)2n+1dx dt+ Z 1
t=0
Z 1 x=t
f(2n+2)(t)
(2n + 1)! (x − t)2n+1dx dt
= Z 0
t=−1
f(2n+2)(t) (2n + 1)!
−(t − x)2n+2 2n + 2
t
x=−1
dt+ Z 1
t=0
f(2n+2)(t) (2n + 1)!
(x − t)2n+2 2n + 2
1
x=t
dt
= Z 0
−1
f(2n+2)(t)
(2n + 2)! (t + 1)2n+2dt+ Z 1
0
f(2n+2)(t)
(2n + 2)! (1 − t)2n+2dt
= Z 1
−1
f(2n+2)(t) (2n + 2)! h(t) dt, where
h(t) =
(t + 1)2n+2 if t ∈ [−1, 0]
(1 − t)2n+2 if t ∈ [0, 1] . Hence, by the Cauchy-Schwarz inequality,
Z 1
−1
f(x) dx
2
≤ Z 1
−1
(h(t))2dt · Z 1
−1
(f(2n+2)(t))2
((2n + 2)!)2 dt= 2
4n + 5· 1 ((2n + 2)!)2
Z 1
−1
f(2n+2)(t)2
dt
which is equivalent to the desired inequality.
