Let 1/2 < α ∈ R, the reals. Show that there is no function f **:** [0, 1] → R such that f(x) = 1 + α ∫_{x}^{1} f(t) f(t - x) dt for all x ∈ [0, 1].

**Solution**

Suppose there is such a function. Let K = ∫_{0}^{1} f(x) dx. Then K = 1 + α ∫_{0}^{1}∫_{x}^{1} f(t) f(t - x) dt dx.

Interchanging the order of integration gives ∫_{0}^{1}∫_{x}^{1} f(t) f(t - x) dt dx = ∫_{0}^{1}∫_{0}^{t} f(t) f(t - x) dx dt = ∫_{0}^{1} f(t) ∫_{0}^{t} f(x) dx dt (*).

Put g(x) = ∫_{0}^{x} f(t) dt. Then g'(x) = f(x), so (*) gives ∫_{0}^{1} 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 K^{2}, or rearranging (K - 1/α)^{2} = -2/α^{2} (α - 1/2). But that is impossible for α > 1/2.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002