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.

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.

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

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.