Problem 11946
(American Mathematical Monthly, Vol.123, December 2016) Proposed by M. Omarjee (France).
Letf be a twice differentiable function from [0, 1] to R with f′′continuous on[0, 1] andR2/3
1/3 f (x)dx = 0. Prove
4860
Z 1
0
f (x)dx
2
≤11 Z 1
0
(f′′(x))2dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let g(x) be the piecewise differentiable function defined as
g(x) :=
x if x ∈ [0, 1/3), 1 − 2x if x ∈ [1/3, 2/3), x − 1 if x ∈ [2/3, 1), and let
G(x) :=
Z x 0
g(t)dt =
x2
2 if x ∈ [0, 1/3),
−x2+ x −16 if x ∈ [1/3, 2/3),
x2
2 −x + 12 if x ∈ [2/3, 1).
SinceR2/3
1/3 f (x) dx = 0, it follows that Z 1
0
f (x) dx = Z 1/3
0
f (x) dx − 2 Z 2/3
1/3
f (x) dx + Z 1
2/3
f (x) dx = Z 1
0
f (x)g′(x) dx.
Hence, by integrating by parts, we obtain Z 1
0
f (x) dx = Z 1
0
f (x) d(g(x)) = [f (x)g(x)]10− Z 1
0
g(x) d(f (x)) = − Z 1
0
g(x)f′(x) dx
= − Z 1
0
f′(x) d(G(x)) = − [f′(x)G(x)]10+ Z 1
0
f′′(x)G(x)dx = Z 1
0
f′′(x)G(x)dx.
Finally, by Cauchy-Schwarz inequality, we get
Z 1
0
f (x) dx
2
≤ Z 1
0
(G(x))2dx · Z 1
0
(f′′(x))2dx = 11 4860
Z 1
0
(f′′(x))2dx because
Z 1
0
(G(x))2dx = 2 Z 1/3
0
x4 4 dx + 2
Z 1/2 1/3
−x2+ x − 1 6
2
dx = 11 4860.
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