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.

How can I prove that the constant in classical Hardy's inequality is optimal?

$$\int_0^{\infty}\left(\frac{1}{x}\int_0^xf(s)ds\right)^p dx\leq \left(\frac{p}{p-1}\right)^p\int_0^{\infty}(f(x))^pdx,$$ where $f\geq0$ and $f\in L^p(0,\infty)$.

This inequality fails for $p=1$ and $p=\infty$ ?

share|improve this question
    
You may want to look at this, especially Davide's answer: math.stackexchange.com/questions/83946/… –  Paul Jan 15 '12 at 22:30

1 Answer 1

In this form you need $p > 1.$ For the constant, take small $\epsilon > 0$ and define $$ f(x) = x^{(-1/p) - \epsilon}, \; \; x \geq 1 $$ but $$ f(x) = 0, \; 0 \leq x < 1. $$

share|improve this answer
6  
To find this function, you can equate the integrands, take the $p$-th root, multiply through by $x$ and differentiate with respect to $x$. This yields a differential equation with solution $x^{-1/p}$. Then you just need to take functions in $L^p(0,\infty)$ that are arbitrarily close to that. –  joriki Jan 15 '12 at 22:41
3  
@joriki, it was much easier than that, to find this function I just looked at page 243 in Hardy, Littlewood, and Polya, Inequalities. –  Will Jagy Jan 15 '12 at 22:45
    
:-) ${}{}{}{}{}$ –  joriki Jan 15 '12 at 23:05

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.