Problem 11884
(American Mathematical Monthly, Vol.123, January 2016) Proposed by C. Lupu and T. Lupu (Romania).
Let f be a real-valued function on [0, 1] such that f and its first two derivatives are continuous.
Prove that iff (1/2) = 0, then Z 1
0
(f′′(x))2dx ≥ 320
Z 1
0
f (x) dx
2 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Remark: this problem is a particular case of Problem 11756 with [−1, 1] replaced with [0, 1].
By the integral form of the remainder in the Taylor’s Theorem we have that if x ∈ [0, 1] then f (x) = f (1/2) + f′(1/2)(x − 1/2) +
Z x 1/2
f′′(t) (x − t) dt.
Since f (1/2) = 0, it follows that Z 1
0
f (x) dx = Z 1
0
Z x 1/2
f′′(t) (x − t) dt
! dx
= Z 1/2
x=0
Z 1/2 t=x
f′′(t) (t − x) dt dx + Z 1
x=1/2
Z x t=1/2
f′′(t) (x − t) dt dx
= Z 1/2
t=0
Z t x=0
f′′(t) (t − x) dx dt + Z 1
t=1/2
Z 1
x=t
f′′(t) (x − t) dx dt
= Z 1/2
t=0
f′′(t)
−(t − x)2 2
t
x=0
dt + Z 1
t=0
f′′(t) (x − t)2 2
1
x=t
dt
=1 2
Z 1/2 t=0
f′′(t)t2dt +1 2
Z 1
t=0
f′′(t)(1 − t)2dt
=1 2
Z 1
0
f′′(t) h(t) dt, where
h(t) =
t2 if t ∈ [0, 1/2]
(1 − t)2 if t ∈ [1/2, 1] . Hence, by the Cauchy-Schwarz inequality,
Z 1
0
f (x) dx
2
≤ 1 4
Z 1
0
(h(t))2dt · Z 1
0
(f′′(t))2dt = 1 320
Z 1
0
(f′′(t))2dt
which is equivalent to the desired inequality.
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