Ngô Quốc Anh

May 1, 2008

An integral inequality involving differentiable functions with continous derivatives

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 16:31

In this entry, we address the following question

Let f : [0,1] \to \mathbb R be a derivable function, with a continuous derivative f' on [0,1]. Prove that if f\left(\frac{1}{2}\right)=0, then

\displaystyle\int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2.

Solution. The best possible constant comes from the 2-norm of \frac{1}{2}-x. Since

\displaystyle \int_0^1 \left(\frac{1}{2}-x\right)^2 = \frac{1}{12},

this constant is indeed 12. We get equality if and only if f'(x) is a real multiple of \frac{1}{2}-x.

First, write

\displaystyle f(t)=\int_\frac{1}{2}^t f'(x)dx.

Define g(x)=f'(x). The LHS is

\displaystyle \int_0^1 (g(x))^2dx,

and the RHS becomes

\displaystyle\left(\int_0^1 \int_{\frac12}^t g(x)dxdt\right)^2= \left(\int_0^1 \int_x^{\frac12}g(x)dtdx\right)^2 =\left(\int_0^1 \left(\frac12-x\right)g(x)dx\right)^2.

In the language of inner products, we want to show that

\displaystyle \|g\|^2 \geqslant 12 (g \cdot h)^2,


\displaystyle h(x)=\frac{1}{2}-x.

The standard C-S inequality gives

\displaystyle (g\cdot h)^2\leqslant \|g\|^2\|h\|^2=\frac{1}{12}\|g\|^2,

and we are done.

Source: RMO 2008, Grade 12, Problem 2.


1 Comment »

  1. Very nice proof, NQA.

    Comment by Minh Can — January 23, 2011 @ 8:59

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