Let 1/2 < α ∈ R, the reals. Show that there is no function f : [0, 1] → R such that f(x) = 1 + α ∫x1 f(t) f(t - x) dt for all x ∈ [0, 1].
Solution
Suppose there is such a function. Let K = ∫01 f(x) dx. Then K = 1 + α ∫01∫x1 f(t) f(t - x) dt dx.
Interchanging the order of integration gives ∫01∫x1 f(t) f(t - x) dt dx = ∫01∫0t f(t) f(t - x) dx dt = ∫01 f(t) ∫0t f(x) dx dt (*).
Put g(x) = ∫0x f(t) dt. Then g'(x) = f(x), so (*) gives ∫01 g'(t) g(t) dt = 1/2 g(1)2 - 1/2 g(0)2. But g(1) = K and g(0) = 0. Thus we have K = 1 + α/2 K2, or rearranging (K - 1/α)2 = -2/α2 (α - 1/2). But that is impossible for α > 1/2.
© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002