### 34th Putnam 1973

Problem B4

f is defined on the closed interval [0, 1], f(0) = 0, and f has a continuous derivative with values in (0, 1]. By considering the inverse f -1 or otherwise, show that ( ∫01 f(x) dx )2 ≥ ∫01 f(x)3 dx. Give an example where we have equality.

Solution

We might suspect that there is nothing special about the upper limit 1. So put g(y) = ( ∫0y f(x) dx )2 - ∫0y f(x)3 dx. Then certainly g(0) = 0. So the obvious approach is to show that g'(y) ≥ 0. Differentiating, we get 2 f(y) ∫0y f(x) dx - f(y)3. We know that f(y) ≥ 0 (because it is 0 at y = 0 and has non-negative derivative). So we need to show that 2 ∫0y f(x) dx - f(y)2 ≥ 0. Certainly it is 0 at y = 0, so we try differentiating again, hoping that the derivative will be non-negative. Differentiating gives 2 f(y) - 2 f(y) f '(y). But that is non-negative, because f(y) ≥ 0 and f '(y) ≤ 1.

It is easy to see that f(x) = x gives equality. [Careful, f(x) = 0 is not an acceptable solution because the derivative is not strictly positive.]

[The suggestion about f-1 was probably intended as follows. Let if y = f(x), let x = g(y). Then ∫ f(x) dx = ∫ y dx/dy dy = ∫ y g'(y) dy. Now think of the square as a double integral and use g'(y) ≥ 1. But the straightforward solution above seems easier.]