Perhaps an application of Hardy's inequality

Question

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.

Answer 1

$\newcommand\la\lambda\newcommand\al\alpha$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.

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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 $\la\to\infty$.

Answer 2

Here's an elementary proof of a related inequality (with non-sharp constants), which may explain what Iosif said in his answer. For ease of typing, instead of $\lambda^{-k}$ I will just type $y \ll 1$.

Let $f$ be a $C^1$ function vanishing at $1$, then you have

$$ \int_y^1 (x f(x)^2)' = - y f(y)^2 $$

Expanding the left you get

$$ \int_y^1 f(x)^2 + 2 \int_y^1 x f(x) f'(x) = - y f(y)^2 $$

rearrange and use Cauchy-Schwarz, you get

$$ \int_y^1 f(x)^2 \leq 2\big(\int_y^1 x^{2-\alpha} f(x)^2\big)^{1/2} \big(\int_y^1 x^\alpha f'(x)^2 \big)^{1/2} + y f(y)^2 $$

Young's inequality makes it, for every $\epsilon > 0$:

$$ \int_y^1 f(x)^2 \leq \epsilon \int_y^1 x^{2-\alpha} f(x)^2 + \frac1{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

Since $\alpha < 2$, we have $x^{2-\alpha} \leq 1$, so we can absorb

$$ (1-\epsilon) \int_y^1 f(x)^2 \leq \frac{1}{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

To handle the boundary term, you can write

$$ f(y)^2 = \big( \int_y^1 f'(x) \big)^2 \leq (1-y) \int_y^1 f'(x)^2 \leq (1-y) y^{-\alpha} \int_y^1 x^\alpha f'(x)^2 $$

So you get, all things considered

$$ \int_y^1 f^2 \leq \frac{1}{1-\epsilon} \left( \frac{1}{\epsilon} + y^{1-\alpha}(1-y) \right) \int_y^1 x^\alpha (f')^2$$

Choosing $\epsilon = 1/2$ you get

$$ \int_{\lambda^{-k}}^1 f^2 \leq (4 + 2 \lambda^{-(1-\alpha)k}) \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 \leq 6 \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 $$

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So, to make this look like your original expression, if $C$ is allowed to be a function of $\lambda$, you can choose $C = 6 \lambda^p$ and the desired inequality will hold.

But you cannot recast this inequality in the form of $C\lambda^{-p}$ with a constant $C$ independent of $\lambda$, due to the number $4$ that shows up.