41st Putnam 1980

Problem A6

Let R be the reals and C the set of all functions f : [0, 1] → R with a continuous derivative and satisfying f(0) = 0, f(1) = 1. Find infC01 | f '(x) - f(x) | dx.

Solution

If it were not for the condition f(0) = 0, we could arrange for the infimum to be zero (which is obviously the lowest possible value), by taking f(x) = ex-1. It is tempting to think that we can adjust this solution (and keep zero inf) by taking f(x) = ex-1 for [ε, 1] and kx for [0, ε]. However, this does not work because ∫0ε f '(x) dx = f(ε) = 1/e.

We need the following (familiar) trick: (f(x) e-x)' = (f '(x) - f(x) ) e-x. So ∫01 | f '(x) - f(x) | dx = ∫01 | ex (f(x) e-x)' | dx ≥ ∫01 | (f(x) e-x)' | dx ≥ ∫01 (f(x) e-x)' dx = f(1) e-1 - f(0) e0 = 1/e.

Finally, we just need to improve slightly our opening example. The integral of f(x) over [0, ε] will be arbitrarily small for small ε. But derivative is discontinuous at ε. However, we can fix that by taking f(x) to be an arbitrary monotonic increasing function with the required values at 0 and ε and with the required derivative at ε.