Problem 12205
(American Mathematical Monthly, Vol.127, October 2020) Proposed by C. Chiser (Romania).
Find the minimum value of R1
0 x2(f′(x))2dx R1
0 x2(f (x))2dx
over all nonzero continuously differentiable functions f: [0, 1] → R with f (1) = 0.
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) = xf (x) then g ∈ C1([0, 1]), it is non-identically zero and g(0) = g(1) = 0.
Moreover, by integrating by parts we get Z 1
0
x2(f′(x))2dx= Z 1
0
(g′(x) − f (x))2dx
= Z 1
0
(g′(x))2dx −2 Z 1
0
f(x) d(g(x)) + Z 1
0
(f (x))2dx
= Z 1
0
(g′(x))2dx −2[f (x)g(x)]10+ Z 1
0
x d(f (x))2) + Z 1
0
(f (x))2dx
= Z 1
0
(g′(x))2dx+ [x(f (x))2]10− Z 1
0
(f (x))2dx+ Z 1
0
(f (x))2dx
= Z 1
0
(g′(x))2dx.
Hence, by Wirtinger’s inequality, R1
0 x2(f′(x))2dx R1
0 x2(f (x))2dx = R1
0(g′(x))2dx R1
0(g(x))2dx ≥π2.
Let f (x) =sin(πx)x (extended continuously at 0) then f (1) = 0. Therefore g(x) = sin(πx) and R1
0 x2(f′(x))2dx R1
0 x2(f (x))2dx = R1
0(g′(x))2dx R1
0(g(x))2dx = R1
0(π cos(πx))2dx R1
0(sin(πx))2dx = π2.
Hence we may conclude that the desired minimum value is π2.
