Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
$\lim\limits_{x\to0+}\frac{f(x)}{x}=\lim\limits_{x\to0+}\frac{f(0)-f(x)}{0-x}=f'‌​(0)<\infty$ –  Anton Aug 7 '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 7 '13 at 16:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.