
> 
> Let $f \in H\_{0}^{1}(0,1)$ and $\lambda >0$ big enough. Consider $0 <\alpha < 1$ and some $k > 0$. I would like to show the following inequality
> $$
> \int\_{\lambda^{-k}}^{1}|f(x)|^{2}dx \leq C\lambda^{-p}\int\_{\lambda^{-k}}^{1}x^{\alpha}|f^{\prime}(x)|^{2}dx
> $$
> for some $p> 0$. Here, $C> 0$ is constant.
> 
> 
> 


* I want to use the following Hardy's Inequalities:


Let $-\infty \leq a \leq b \leq \infty$, $g(x) \geq 0$,$h(x)\geq0$. Then the following statements are equivalent:
$$
\bigg(\int\_{a}^{b}|Qu(x)|^{2}g(x)^{2}dx\bigg)^{\frac{1}{2}} \leq C\bigg(\int\_{a}^{b}|u(x)|^{2}h(x)^{2}dx\bigg)^{\frac{1}{2}},
$$
$$
K = \sup\_{x \in [a,b]}\bigg(\int\_{a}^{x}[g(t)]^{2}dt\bigg)^{\frac{1}{2}}\bigg(\int\_{x}^{b}[h(t)]^{-2}dt\bigg)^{\frac{1}{2}} < \infty ,
$$
where $Qu(x) = \int\_{x}^{b}u(s)ds$. Moreover, the best constant $C$ satisfies $K \leq C \leq 2K$.


* How $f(1) = 0$, because $f \in H\_{0}^{1}(0,1)$, then
$$
|f(x)| = |f(1) - f(x)|= \bigg|\int\_{x}^{1}f^{\prime}(s)ds\bigg|= |Qf^{\prime}(x)|.
$$


Consider $h(x) = x^{\alpha/2}$, $g(x) =1$, $a=\lambda^{-k}$ and $b=1$.Thus, according to the inequality above, we would have to show that
$$
\sup\_{x \in [\lambda^{-k},1]}\bigg(\int\_{\lambda^{-k}}^{x}dt\bigg)^{\frac{1}{2}}\bigg(\int\_{x}^{1}t^{-\alpha}dt\bigg)^{\frac{1}{2}} < \infty
$$
But it is clear that the supreme of the above product is finite. My big question is:
After applying Hardy's inequality, how do I get $\lambda^{-p}$ to appear in front of the integral.



Perhaps an application of Hardy's inequality
$\newcommand\la\lambda\newcommand\al\alpha$If $C$ is allowed to depend on $\lambda$, just take $C=2\lambda^p$.


If $C$ is not allowed to depend on $\lambda$, take any nonzero $f\in H\_{0}^{1}(0,1)$ and let $\lambda\to\infty$. Then the left-hand side of your desired inequality will go to $\int\_0^1|f(x)|^2\,dx>0$ whereas its right-hand side will go to $0$, so that your desired inequality will fail to hold.




---


Let $f(x)=x(1-x)$ and let $\la\to\infty$. Then
$$\int\_{\la^{-k}}^1|f(x)|^2\,dx\to\frac1{30}$$
and
$$\int\_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx\to h(\al):=\frac{\alpha ^2+\alpha +2}{\alpha ^3+6 \alpha ^2+11 \alpha +6},$$
so that the constant factor $L$ in the inequality
$$\int\_{\la^{-k}}^1|f(x)|^2\,dx
\le L\,\int\_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx$$
must be $\ge\dfrac1{30h(\al)}$. So, $L$ cannot go to $0$ as $\la\to\infty$.


