3

Let $f$ be a continuously differentiable real-valued function on $[0,b]$, where $b>0$, with $f(0)=0$. Prove that $$\int\limits_0^b\frac{f(x)^2}{x^2}dx\leq4\int\limits_0^b f'(x)^2dx.$$ Thank you!

Anton
  • 75

1 Answers1

2

First, It follows from the differentiability that $\mathrm{LHS}$ converges (why?).

Now integrate by part (how?) and apply Cauchy-Schwarz inequality, we have

$$\int_0^b\frac{f(x)^2}{x^2}dx=-\frac{f(b)^2}b+2\int_0^b\frac{f(x)f'(x)}xdx\le 2\,\left(\int_0^b\frac{f(x)^2}{x^2}dx\int_0^bf'(x)^2dx\right)^{1/2}$$

The required inequality follows.

Yai0Phah
  • 9,733
  • $\lim\limits_{x\to0+}\frac{f(x)}{x}=\lim\limits_{x\to0+}\frac{f(0)-f(x)}{0-x}=f'(0)<\infty$ – Anton Aug 07 '13 at 16:28
  • $$\int\limits_0^b\frac{f(x)^2}{x^2}dx=\left[\begin{array}{lll}u=f(x)^2 & \Rightarrow & du=2f(x)f'(x)\dv=\frac{1}{x^2} & \Rightarrow & v = -\frac{1}{x} \end{array}\right] = \left. -\frac{f(x)^2}{x} \right|_0^b+2\int\limits_0^b\frac{f(x)f'(x)}{x}dx$$

    And $\lim\limits_{x\to0+}\frac{f(x)^2}{x}=f'(0)f(0)=0$, so

    $$\int\limits_0^b\frac{f(x)^2}{x^2}dx=-\frac{f(b)^2}{b}+2\int\limits_0^b\frac{f(x)f'(x)}{x}dx$$

     – Anton Aug 07 '13 at 16:41