Problem 11812
(American Mathematical Monthly, Vol.122, January 2015) Proposed by C. Chiser (Romania).
Let f be a twice continuously differentiable function from [0, 1] into R. Let n be an integer greater than 1. Given thatPn−1
k=1f (k/n) = −(f (0) + f (1))/2, prove that
Z 1 0
f (x) dx
2
≤ 1 5!n4
Z 1 0
(f00(x))2dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By the second-order Euler-Maclaurin Formula, we have that for any g ∈ C2([0, n]),
n−1
X
k=0
g(k) = Z n
0
g(x) dx + B1(g(n) − g(0)) +B2
2 (g0(n) − g0(0)) −1 2
Z n 0
B2({x})g00(x) dx
where Bi is the ith Bernoulli number (B1 = 1/2, B2 = 1/6), B2(x) = x2− x + 1/6 is the second Bernoulli polynomial, and {x} = x − bxc is the fractional part of x.
Hence, by taking g(x) = f (x/n), we get
n−1
X
k=0
f (k/n) = n Z 1
0
f (x) dx +f (1) − f (0)
2 +f0(1) − f0(0)
12n − 1
2n Z 1
0
B2({nx})f00(x) dx
and by usingPn−1
k=1f (k/n) = −(f (0) + f (1))/2, we obtain Z 1
0
f (x) dx = 1 2n2
Z 1 0
B2({nx})f00(x) dx −f0(1) − f0(0) 12n2
= 1 2n2
Z 1 0
B2({nx}) − 1 6
f00(x) dx
= 1 2n2
Z 1 0
{nx}2− {nx} f00(x) dx.
Finally, by Cauchy-Schwarz inequality,
Z 1 0
f (x) dx
2
≤ 1 4n4
Z 1 0
{nx}2− {nx}2 dx ·
Z 1 0
(f00(x))2d = 1 120n4
Z 1 0
(f00(x))2dx, because
Z 1 0
{nx}2− {nx}2
dx = 1 n
Z n 0
{x}2− {x}2
dx = Z 1
0
x2− x2
dx = x5 5 −x4
2 +x3 3
1
0
= 1 30.
