# An interpolation type inequality

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 $\beta=1$.

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

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.