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

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

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.
MC

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

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