64th Putnam 2003

Show that ∫₀¹ ∫₀¹ |f(x) + f(y)| dx dy ≥ ∫₀¹ |f(x)| dx for any continuous real-valued function on [0,1].

Solution

Let A be the set of values with f(x) ≥ 0, B the values with f(x) < 0. Put I_A = ∫_A f(x) dx, I_B = -∫_B f(x) dx, a = |A|, b = |B|. Then rhs = I_A + I_B. We have ∫_A ∫_A |f(x) + f(y)| dx dy = ∫ (af(x) + ∫ f(y) dy) dx = ∫ (af(x) + I_A) dx = 2aI_A. Similarly ∫_B ∫_B = 2bI_B.

wlog a ≥ b. If I_A ≥ I_B, then (a-b)(I_A-I_B) ≥ 0, so aI_A + bI_B ≥ bI_A + aI_B and hence 2aI_A + 2bI_B ≥ (a+b)(I_A+I_B) = I_A + I_B = rhs. So we are already done.

So assume I_A < I_B. Then ∫_x∈A ∫_y∈B ≥ ∫_x∈A ∫_y∈B -f(x) - f(y) dy dx = ∫_x∈A(-bf(x) + I_B) dx = aI_B - bI_A. But 2bIB + (aI_B - bI_A) = (a+b)I_B + b(I_B - I_A) ≥ I_B and 2aI_A ≥ (a+b)I_A = I_A. So we have lhs ≥ rhs.

64th Putnam 2003

© John Scholes
jscholes@kalva.demon.co.uk
8 Dec 2003
Last corrected/updated 8 Dec 03