3
$\begingroup$

Let $u(x),x\in R_+$ be a non-negative decreasing smooth function with compact support $[0,L]$, I want to know the following inequality is true? $a\in (0,1)$ $$\int_0^\infty \frac{1}{1+x}u^{1+a}dx \le \epsilon \int_0^\infty |u_x|^2dx+C_\epsilon \int_0^\infty u^{2a}dx,$$ where we need $C_\epsilon$ is independent of $L$!!!

P.S. when the left hand is $\int_0^\infty \frac{1}{(1+x)^\beta}u^{1+a}dx$ with $\beta>1$, the inequality is true, I want to consider the case $\beta=1$.

$\endgroup$
  • 1
    $\begingroup$ How can a decreasing function have compact support? $\endgroup$ – Alexandre Eremenko Dec 23 '13 at 6:51
  • $\begingroup$ I think he means that the support of $u$ is assumed to be $[0,L]$, which is compact. $\endgroup$ – user5678 Dec 23 '13 at 10:27
  • $\begingroup$ Apparently he means that $u(x)\ge 0$, for all $x\in\mathbb R_+$ and $u(x)=0$, for $x\ge 0$. $\endgroup$ – smyrlis Dec 23 '13 at 10:51
  • $\begingroup$ @user44565: Can you explain how it is proved for $\beta>1$? $\endgroup$ – smyrlis Dec 23 '13 at 10:52
  • $\begingroup$ For $\beta > 1$, I believe it is proved via an integration by parts and then Young's inequality. $\endgroup$ – user5678 Dec 23 '13 at 11:37
2
$\begingroup$

It is false for every $\epsilon > 0$. The family of functions to use is this: $$ u(x):= \left\{ \begin{aligned} & (L-x)^{\frac1{1-a}}, && 0\leq x \leq L, \\ & 0, && x>L. \end{aligned} \right. $$ Note that $u\in C^1(\mathbb R_+)$. (To get something smooth, mollify this $u$.)

Note also that, for $0\leq x \leq L$, $$(u'(x))^2 = \frac{1}{1-a} u^{2a}(x) = \frac{1}{1-a}(L-x)^{\frac{2a}{1-a}},$$ so the integrals on the right side are the same, up to a constant depending on $a$. Meanwhile, after an integration by parts, the integral on the left side is $$ \frac{1+a}{1-a} \int_0^L (L-x)^{\frac{2a}{1-a}} \log(1+x) \, dx, $$ which is again almost the same, except for the logarithmic factor. But if you make $L$ very large, then most of the contribution of this integral is for $x$ large, and then the log factor is a large weight. So as $L\to \infty$, the left side swamps the right side, regardless of the constant.

$\endgroup$
  • $\begingroup$ This counterexample is very nice. Thank you for you help. $\endgroup$ – user44565 Dec 24 '13 at 1:18
  • $\begingroup$ Oh, sorry, I am a new user. $\endgroup$ – user44565 Dec 24 '13 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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