Expert: Sherry Wallin - 1/13/2011

Question
Let f: [0,1] --> R be a differentiable function with a continuous derivative such that f(0)= f(1)= -1/6. Prove that integral from 0 to 1 (f'(x))square dx greater than or equal to 2 integral from 0 to 1 f(x) dx + 1/4

Answer
Tony Chu~

Hi Tony~

My bad, I misread the original problem, I thought it said that f'(0) = f'(1) = -1/6. If that were the case then my argument is absolutely correct because the slope (f'(x)) would be constant, and in fact f(x) would be exactly (-1/6)x + c.

For some reason your teachers example is not here, what I see is:  (f(x) could be  . Then, c=1/2, but f'(x) is not 0 for all x). I can't really respond to what I can't see. If you would like to send your question with that info intact I would be happy to examine it.

As for the function f(x)= (-1/6)+h(x)x^r (x-1)^s, when you evaluate f(0) you get (-1/6)+h(0)*0*(-1)^s -> f(0) = -1/6. And when you evaluate f(1) you get (-1/6)+h(1)*1^r(1-1)^s = -1/6+ 0 = -1/6, so indeed this form of f(x) generates the required relationship. Try some other number like 1/2 and see what happens in f(x), f(1/2) = -1/6 + h(1/2)*(1/2)^r(1/2-1)^s = -1/6 + h(1/2)*(1/2)^r*(-1/2)^s ~= -1/6 unless h(1/2) = 0. So what you need to establish is that whatever the integral from 0 to 1 (f'(x))square dx is, it is always greater than or equal to 2*integral from 0 to 1 f(x) dx + 1/4.
The integral(0,1)[f'(x)]^2 dx can be integrated in the following way:
Let u = f'(x), then du = f"(x) dx. Now let the integral from (0,1) be u^2 du = (u^3)/3 evaluated over (0,1) = 1/3 and on the right hand side by the fundamental theorem of calculus is 2*[F(1)-F(0)]+(1/4) =
-1/6-(-1/6) + 1/4 = 1/4.

Math Prof