1
$\begingroup$

I have this exercise (Hardy's inequality)

Show that every integrable function $f:(0,T) \to \mathbb R$ satisfies the inequality $$\int_0^T \left\{\frac{1}{x}\int_0^xf(u)\,du\right\}^2dx \le 4\int_0^T f^2(x)\,dx,$$ and show, moreover, that the constant 4 cannot be replaced with any smaller value.

The book says to deepen our understanding of the above inequality , we might also see if we can confirm that the constant $4$ is the best one can do. The author uses a so-called stress testing method to do that. The test function is the power map $x \mapsto x^\alpha$. When we substitute this function into an inequality of the form

$$ \int_0^T \left\{\frac{1}{x}\int_0^xf(u)\,du\right\}^2dx \le C\int_0^T f^2(x)\,dx, \tag{1}$$

we see that it implies

$$\frac{1}{(\alpha+1)^2(2\alpha+1)}\le \frac{C}{2\alpha+1} \text{for all $\alpha$ such that $2\alpha+1>0$.} \tag{2}$$

Now, by letting $\alpha \to -1/2$, we see that for the bound $(1)$ to hold in general one must have $C\ge 4$.

My questions are:

  1. How substituting the map $x \mapsto x^\alpha$ into (1) gives (2)?

  2. Why letting $\alpha \to 1/2$ implies that for the bound $(1)$ to hold in general one must have $C\ge 4$?

$\endgroup$
1
$\begingroup$

(2) follows from (1) by simply setting $f(x)=x^{\alpha}$. I'll write one part out for you: $$ \int_0^T\left\{\frac{1}{x}\int_0^x u^{\alpha}\,\mathrm{d} u\right\}^2\,\mathrm{d} x = \int_0^T \frac{1}{x^2}\cdot \left(\frac{x^{\alpha+1}}{\alpha+1}\right)^2\,\mathrm{d} x = \frac{1}{(\alpha+1)^2}\int_0^T x^{2\alpha}\,\mathrm{d} x = \frac{1}{(\alpha+1)^2(2\alpha+1)}\cdot T^{2\alpha+1} $$ If you write out the other side your self, you will see that you can divide by $T^{2\alpha+1}$ on both sides. I don't know why the assumption $2\alpha+1>0$ is made, as I believe it is not necessary as long as $2\alpha+1\neq 0$, $\alpha+1\neq0$, $\alpha\neq 0$ (where we used the latter in the integrating).

For the second part of your question, multiply the equation by $(2\alpha+1)$ to get $$ \frac{1}{(\alpha+1)^2} \leq C $$ if $2\alpha+1>0$. This is minimised under $2\alpha+1>0$ if $\alpha=-1/2$ which leads to $4\leq C$.
If, however, $2\alpha+1<0$, the sign flips and you don't get any information about a lower bound for $C$.

$\endgroup$
  • $\begingroup$ I think your last two sentences already answered why the assumption $2\alpha+1 > 0$ is made. $\endgroup$ – user547265 Jun 25 '18 at 19:54
  • $\begingroup$ yeah ok, you're right $\endgroup$ – Stan Tendijck Jun 25 '18 at 20:34
1
$\begingroup$

Assume $2\alpha+1>0$, so none of the integrals of powers I'm about to discuss diverge at $x=0$.For $f(x)=x^\alpha$ the inequality becomes $\int_0^T\frac{x^{2\alpha}dx}{(\alpha+1)^2}\le C\int_0^T x^{2\alpha}dx$. Cancelling the $\int_0^T x^{2\alpha }dx$ factor gives (2). It simplifies to $C\ge (\alpha+1)^{-2}$ for any $\alpha >-\frac{1}{2}$. This requires $C\ge 4$.

$\endgroup$

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.