A simple proof of a sharp Hardy-type inequality for positive sequences

Fix $x>0$. Write

By Cauchy--Schwarz,

Now, since $t \mapsto (t^2+x^2)^{-1}$ is decreasing, we have

Therefore,

Set

Then

We now choose $x>0$ to minimize the right-hand side. The expression $x A + B/x$ attains its minimum at $x = \sqrt{B/A}$, and the minimum value is

Hence

Squaring gives

which is exactly the desired inequality.

To see that the constant $\pi^2$ is optimal, take

A straightforward asymptotic computation gives, as $\varepsilon \to 0^+$,

Thus the ratio

So any constant smaller than $\pi^2$ would fail for sufficiently small $\varepsilon$. The constant is therefore best possible.

This is essentially Carleson's inequality for sequences: the parameter $x$ acts as a cut-off, and the integral comparison gives exactly the sharp constant after optimization.

Proof.

Posons

Fixons $x>0$. Pour tout $n\ge1$, écrivons

Appliquons l'inégalité de Hölder généralisée avec les exposants

Il vient

Par conséquent,

Comme la fonction

est positive et décroissante sur $[0,+\infty)$,

Ainsi

Le membre de droite est minimal lorsque

Pour ce choix,

D'où

En élevant au carré,

∎

A convenient way to recover the sharp constant is to use the one-sided Fourier spectrum. Let

For any $x>0$,

and Cauchy--Schwarz gives

Since

Parseval yields

Choosing

gives the one-sided Agmon inequality

Here the factor $1/2$ comes from the fact that only positive Fourier frequencies occur.

Now

Therefore

and hence