Show that Z 1 −1 (f00(x))2dx = 10 Z 1 −1 f (x) dx 2

Problem 11548

(American Mathematical Monthly, Vol.118, January 2011) Proposed by Cezar Lupu (Romania) and Tudorel Lupu (Romania).

Let f be a twice-differentiable real-valued function with continuous second derivative, and suppose that f (0) = 0. Show that

Z 1

−1

(f⁰⁰(x))²dx = 10

Z 1

−1

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.

Let

g(x) =

 (x + 1)² for x ∈ [−1, 0], (x − 1)² for x ∈ [0, 1]

then

g(−1) = g(1) = g⁰(−1) = g⁰(1) = 0, g(0) = 1, and g⁰⁰(x) = 2 for x ∈ [−1, −1] \ {0}, and

Z 1

−1

(g(x))² dx = Z 0

−1

(x + 1)⁴dx + Z 1

0

(x − 1)⁴dx = 2 5. Therefore, since f (0) = 0,

Z 1 0

g(x)f⁰⁰(x) dx = [g(x)f⁰(x)]¹₀− Z 1

0

g⁰(x)f⁰(x) dx

= −f⁰(0) − [g⁰(x)f (x)]¹₀+ Z 1

0

g⁰⁰(x)f (x) dx = −f⁰(0) + 2 Z 1

0

f (x) dx.

In a similar way, we have that Z 0

−1

g(x)f⁰⁰(x) dx = f⁰(0) + 2 Z 1

0

f (x) dx.

Hence, by Cauchy-Schwarz inequality, 2

5· Z 1

−1

(f⁰⁰(x))² dx = Z 1

−1

(g(x))² dx · Z 1

−1

(f⁰⁰(x))² dx ≥

Z 1

−1

g(x)f⁰⁰(x) dx

²

=

 2

Z 1

−1

f (x) dx

² ,

and the result follows immediately.