Ngô Quốc Anh

September 17, 2008

L^2 differentiable?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 3 — Ngô Quốc Anh @ 13:04

Let and define .

a) Must be differentiable at 0?
b) Must have any differentiable points?
c) Let , show that exists and determine what it is.

Solutions.

a) No. For example, let for so that near zero. This but does not exist.

b) Yes. In fact, must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that which is true.

c) By the Schwarz inequality, $f^2(x) = \left(\int_0^x}g(t)\,dt\right)^2\le \left(\int_0^x 1\,dt\right) \left(\int_0^xg^2(t)\,dt\right).$

At least that’s it for Being careful about the other side, we determine that $0\le\phi(x) = f^2(x)\le|x|\left|\int_0^xg^2(t)\,dt\right|.$

But since is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that $\lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.$

Hence $\lim_{x\to0}\frac {\phi(x)}{x} = 0,$ so 