Putnam 1993

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Problem B4

K(x, y), f(x) and g(x) are positive and continuous for x, y ∈ [0, 1]. ∫01 f(y) K(x, y) dy = g(x) and ∫01 g(y) K(x, y) dy = f(x) for all x ∈ [0, 1]. Show that f = g on [0, 1].

 

Solution

Moderately easy.

Let h = min { f(x)/g(x): x ∈ [0, 1] }, k = min{ g(x)/f(x): x ∈ [0, 1] }. Assume h < k. Then for all x, g(x)/f(x) ≥ k ≥ h, or g(x) - h f(x) ≥ 0. Since [0, 1] is compact, the value h must be attained. So take x0 such that f(x0) = h g(x0). Then we have 0 = f(x0) - h g(x0) = ∫ K(x0, y) (g(y) - h f(y) ) dy. But K is positive, and g(y) - h f(y) is continuous and non-negative, so if we had g(y) - h f(y) > 0 for any y, then there would be an interval on which g(y) - h f(y) > 0 and the integral would be non-zero. Hence g(y) - h f(y) = 0 for all y. But now we have f(x) = ∫ K(x, y) g(y) dy = ∫ K(x, y) h f(y) dy = h g(x) = h2 f(x). Hence h = 1, and f(x) = g(x).

 


 

Putnam 1993

© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998