Problem 11555
(American Mathematical Monthly, Vol.118, February 2011) Proposed by Duong Viet Thong (Vietnam).
Let f be a continuous real-valued function on [0, 1] such thatR1
0 f (x) dx = 0. Prove that there exists c in the interval (0, 1) such that
c2f (c) = Z c
0
(x + x2)f (x) dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let
F (x) = Z x
0
(t2+ (1 − x)t)f (t) dt then F is differentiable in [0, 1] and
F0(x) = (x2+ (1 − x)x)f (x) − Z x
0
tf (t) dt = xf (x) − Z x
0
tf (t) dt.
It suffices to prove that there are a, b ∈ (0, 1] such that F0(a) > 0 and F0(b) < 0.
In fact, since F0 is continuous, it follows that there is d between a and b such that F0(d) = F0(0) = 0 and by the Flett’s Mean Value Theorem there exists c ∈ (0, d) such that F (c) − F (0) = cF0(c) that
is Z c
0
(t2+ (1 − c)t)f (t) dt = c2f (c) − c Z c
0
tf (t) dt and the desired equality is verified.
Since f is a continuous in [0, 1], it follows that there exist xM, xm∈ [0, 1] such that M := max
x∈[0,1]
f (x) = f (xM) and m := max
x∈[0,1]
f (x) = f (xm).
By hypothesis,R1
0 f (x) dx = 0, and if f is not identically zero (otherwise the thesis is trivial) then M > 0 and m < 0. Moreover, if xM > 0 then let a = xM and
F0(a) ≥ M xM − M Z xM
0
t dt = M xM(1 − xM/2) > 0.
On the other hand, if xM = 0 then, by continuity, there is 0 < a ≤ 1 such that f (x) ≥ 3M/4 for x ∈ [0, a] and
F0(a) ≥ 3
4M a − M Z a
0
t dt =1
4M a(3 − 2a) > 0.
In a similar way, if xm> 0 then let b = xmand F0(b) ≤ mxm− m
Z xm
0
t dt = mxm(1 − xm/2) < 0.
On the other hand, if xm = 0 then, by continuity, there is 0 < b ≤ 1 such that f (x) ≤ 3m/4 for x ∈ [0, b] and
F0(b) ≤3
4mb − m Z b
0
t dt = 1
4mb(3 − 2b) < 0.
