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 $C\ge 4$?


(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 $C$.

  • $\begingroup$ 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 $\endgroup$ Jun 25 '18 at 20:34

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 $C\ge 4$.


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