Let R be the reals. Show that there is at most one continuous function f : [0, 1]2 → R satisfying f(x, y) = 1 + ∫0x∫0y f(s, t) dt ds.
Solution
If there are two, then their difference d(x, y) satisfies d(x, y) = ∫0x∫0y d(s, t) dt ds (*).
The domain is compact and d is continuous, so |d| has some upper bound k. Now applying (*) gives |d(x, y)| ≤ k xy. Applying (*) again gives |d(x, y)| ≤ k x2/2 y2/2. So by a simple induction |d(x, y)| ≤ k xn/n! yn/n! . Hence |d(x, y)| ≤ k/(n! n!) for all n, and so d(x, y) = 0.
© John Scholes
jscholes@kalva.demon.co.uk
25 Feb 2002