Putnam 1980/B5 solution

41st Putnam 1980

Problem B5

R⁰⁺ is the non-negative reals. For α ≥ 0, C_α is the set of continuous functions f : [0, 1] → R⁰⁺ such that: (1) f is convex [ f(λx + μy) ≤ λf(x) + μf(y) for λ, μ ≥ 0 with λ + μ = 1]; (2) f is increasing; (3) f(1) - 2 f(2/3) + f(1/3) ≥ α ( f(2/3) - 2 f(1/3) + f(0) ). For which α is C_α is closed under pointwise multiplication?

Solution

Answer: α ≤ 1.

The function f(x) = x always belongs to C_α because it has f(1) - 2f(2/3) + f(1/3) = f(2/3) - 2f(1/3) + f(0) = 0. So if C_α is closed under pointwise multiplication, then f(x) = x² is also a member. But this requires 1 - 8/9 + 1/9 ≥ α (4/9 - 2/9 + 0) and hence α ≤ 1.

Now if f and g are convex and increasing, then the pointwise product f*g is also convex. For if x ≤ y, we have (f(y) - f(x))(g(y) - g(x)) ≥ 0 and hence λμ(f(x)g(y) + f(y)g(x)) ≤ λμ(f(x)g(x) + f(y)g(y)) (*).

But f*g(λx + μy) = f(λx + μy)g(λx + μy) ≤ (λf(x) + μf(y) )(λg(x) + μg(y) ) = λ²f(x)g(x) + λμ(f(x)g(y) + f(y)g(x) ) + μ²f(y)g(y). So, using (*), we get f*g(λx + μy) ≤ (λ² + λμ)f(x)g(x) + (λμ + μ²)f(y)g(y) = λ f*g(x) + μ f*g(y).

So it remains to show that if α ≤ 1 and f, g are in C_α, then f*g satisfies condition (3). This is not easy - one has to find the right collection of inequalities using the fact that f, g are increasing, convex and satisfy (3).

Since f(2/3) > f(1/3) and g(1) - 2g(2/3) + g(1/3) ≥ α (g(2/3) - 2g(1/3) + g(0) ), we have:

f(2/3) (g(1) - 2g(2/3) + g(1/3) ) ≥ α f(1/3) (g(2/3) - 2g(1/3) + g(0) ) (1)

Similarly, we have:

(f(1) - 2f(2/3) + f(1/3) ) g(2/3) ≥ α (f(2/3) - 2f(1/3) + f(0) ) g(1/3) (2)

Since f and g are convex, we have ( f(1) - f(2/3) ) ≥ ( f(2/3) - f(1/3) ) and ( g(1) - g(2/3) ) ≥ ( g(2/3) - g(1/3) ). So ( f(1) - f(2/3) ) ( g(1) - g(2/3) ≥ ( f(2/3) - f(1/3) ) ( g(2/3) - g(1/3) ). The rhs is non-negative and α ≤ 1, so ( f(2/3) - f(1/3) ) ( g(2/3) - g(1/3) ) ≥ α ( f(2/3) - f(1/3) ) ( g(2/3) - g(1/3) ). Hence:

( f(1) - f(2/3) ) ( g(1) - g(2/3) ≥ α ( f(2/3) - f(1/3) ) ( g(2/3) - g(1/3) ) (3)

Similarly:

( f(2/3) - f(1/3) ) ( g(2/3) - g(1/3) ) ≥ α ( f(1/3) - f(0) ) ( g(1/3) - g(0) ) (4)

Adding (1), (2), (3) and (4) gives the required result:

f(1)g(1) - 2f(2/3)g(2/3) + f(0)g(0) ≥ α ( f(2/3)g(2/3) - 2f(1/3)g(1/3) + f(0)g(0) ).

41st Putnam 1980