Cfrgtkky

Classical Hardy's inequality (cfr. Hardy-Littlewood-Polya Inequalities, Theorem 327)

If $p>1$, $f(x) ge 0$ and $F(x)=int_0^xf(y), dy$ then

$$tag{H} int_0^infty left(frac{F(x)}{x}right)^p, dx < Cint_0^infty (f(x))^p, dx $$

unless $f equiv 0$. The best possibile constant is $C=left(frac{p}{p-1}right)^p$.

I would like to prove the statement in italic regarding the best constant. As already noted by Will Jagy here, the book suggests stress-testing the inequality with

$$f(x)=begin{cases} 0 & 0le x <1 \ x^{-alpha} & 1le x end{cases}$$

with $1/p< alpha < 1$, then have $alpha to 1/p$. If I do so I get for $C$ the lower bound

$$operatorname{lim sup}_{alpha to 1/p}frac{alpha p -1}{(1-alpha)^p}int_1^infty (x^{-alpha}-x^{-1})^p, dxle C$$

but now I find myself in trouble in computing that lim sup. Can someone lend me a hand, please?

UPDATE: A first attempt, based on an idea by Davide Giraudo, unfortunately failed. Davide pointed out that the claim would easily follow from

$$tag{!!} leftlvert int_1^infty (x^{-alpha}-x^{-1})^p, dx - int_1^infty x^{-alpha p }, dxrightrvert to 0quad text{as} alpha to 1/p. $$

But this is false in general: for example if $p=2$ we get

$$int_1^infty (x^{-2alpha} -x^{-2alpha} + 2x^{-alpha-1}-x^{-2}), dx to int_1^infty(2x^{-3/2}-x^{-2}), dx ne 0.$$

What you need isn't

$$lim_{alphasearrow1/p},leftlvert int_1^infty (x^{-alpha}-x^{-1})^pmathrm dx - int_1^infty x^{-alpha p }mathrm dxrightrvert=0$$

but

$$lim_{alphasearrow1/p}frac{int_1^infty (x^{-alpha}-x^{-1})^pmathrm dx}{int_1^infty x^{-alpha p }mathrm dx}=1;,$$

which is indeed the case, since as $alphasearrow1/p$, the integrals are more and more dominated by regions where $x^{-1}ll x^{-alpha}$. For arbitrary $bgt1$ and $1/pltalphalt1$, we have

$$
begin{eqnarray}
int_1^infty (x^{-alpha})^pmathrm dx
&gt&
int_1^infty (x^{-alpha}-x^{-1})^pmathrm dx
\
&gt&
int_b^infty (x^{-alpha}-x^{-1})^pmathrm dx
\
&gt&
int_b^infty (x^{-alpha}-b^{alpha-1}x^{-alpha})^pmathrm dx
\
&=&
(1-b^{alpha-1})^pint_b^infty (x^{-alpha})^pmathrm dx
\
&=&
(1-b^{alpha-1})^pb^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx
\
&=&
(b^{1/p-alpha}-b^{1/p-1})^pint_1^infty (x^{-alpha})^pmathrm dx;.
end{eqnarray}
$$

Then choosing $b=2^{1/beta}$ with $beta=sqrt{alpha-1/p}$ yields

$$int_1^infty (x^{-alpha})^pmathrm dx
gt
int_1^infty (x^{-alpha}-x^{-1})^pmathrm dx
gt
(2^{-beta}-2^{(1/p-1)/beta})^pint_1^infty (x^{-alpha})^pmathrm dx;.
$$

Since $betato0$ as $alphasearrow1/p$, the factor on the right goes to $1$, and thus

$$int_1^infty (x^{-alpha}-x^{-1})^pmathrm dxsimint_1^infty x^{-alpha p }mathrm dxquadtext{as}quadalphasearrow1/p$$

as required.

It's been a long time since the original post, but here is yet another way to show that $frac{p}{p-1}$, $1<p<infty$, is the best constant in Hardy's inequality. This follows A suggestion in Exercise 3.14 in Rudin's book of Real and Complex Analysis. Take
$$ f_A(x)=x^{-1/p}mathbf{1}_{(1,A]}(x)$$
A straight forward computation shows that $|f_A|^p_p=log(A)$. On the other hand, the Hardy transform of $f_A$ is
$$ F_A(x)=frac{p}{p-1}frac{1}{x}Big((min(A,x))^{1-tfrac{1}{p}}-1Big) $$
Then
$$
begin{align*}
|F_A|^p_p &= Big(frac{p}{p-1}Big)^pBig( int^A_1(x^{-1/p}-x^{-1})^p,dx + frac{(1-A^{frac{1-p}{p}})^p}{p-1}Big)
end{align*}
$$
We now normalize $f_A$, by the factor $frac{1}{|f_A|_p}f_A$ to obtain
$$frac{|F_A|_p}{|f_A|_p} geq Big(frac{p}{p-1}Big)left(frac{int^A_1(x^{-1/p}-x^{-1})^p,dx}{log(A)}right)^{1/p}xrightarrow{Arightarrowinfty}frac{p}{p-1}$$
From this, it follows immediately that $C:=frac{p}{p-1}$ is the best constant in Hardy's inequality. In terms of operators, as user user345872 suggested above, The operator norm of the Hardy transform $fmapsto frac{1}{x}int^x_0f(t),dt$ in $L((0,infty),dx)$ is indeed $frac{p}{p-1}$.

We have the operator $T: L^p(mathbb{R}^+) to L^p(mathbb{R}^+)$ with $p in (1, infty)$, defined by$$(Tf)(x) := {1over x} int_0^x f(t),dt.$$Calculate $|T|$.

For the operator $T$ defined above, the operator norm is $p/(p - 1)$. We will also note that this is also a bounded operator for $p = infty$, but not for $p = 1$.

Assume $1 < p < infty$, and let $q$ be the dual exponent, $1/p + 1/q = 1$. By the theorem often referred to as "converse Hölder,"$$|Tf|_p = sup_{|g|_q = 1}left|int_0^infty (Tf)(x)g(x),dxright|.$$So, assume that$|g|_q = 1$,begin{align*} left| int_0^infty (Tf)(x)g(x),dxright| & le int_0^infty |Tf(x)||g(x)|,dx le int_0^infty int_0^x {1over x}|f(t)||g(x)|,dt,dx \ & = int_0^infty int_0^1 |f(ux)||g(x)|,du,dx = int_0^1 int_0^infty |f(ux)||g(x)|,dx,du \ & le int_0^1 left(int_0^infty |f(ux)|^pdxright)^{1over p} left(int_0^infty |g(x)|^q dxright)^{1over q}du \ & = int_0^1 u^{-{1over p}}|f|_p |g|_q du = {pover{p - 1}}|f|_p.end{align*}So that gives us that the operator norm is at most $p/(p - 1)$. To show that this is tight, let $f(x) = 1$ on $(0, 1]$ and zero otherwise. We have $|f|_p = 1$ for all $p$. We can then compute that$$(Tf)(x) = begin{cases} 1 & 0 < x le 1 \ {1over x} & x > 1end{cases}$$and by direct computation, $|Tf|_p = p/(p - 1)$ for $p > 1$.

The same example also shows that we can have $f in L^1$ but $Tf notin L^1$, so we must restrict to $p > 1$. However, it is straightforward to show that $T$ is bounded from $L^infty to L^infty$ with norm $1$. Note that the range of the operator in that case is contained within the bounded continuous functions on $(0, infty)$.

This is an extended comment to the answer of Olivier Diaz. Let
$$
Hf(x):=frac1 xint_0^x f(y), dy.$$
I computed the limit
$$
ell=lim_{Ato infty} frac{ | Hf_A^alpha|_p^p}{|f_A^alpha|_p^p}, $$
where $f_A^alpha:=x^{-alpha}mathbf 1_{{1le xle A}}$, for $alpha in (0, 1)$.

When $alpha<1/p$,
$$
ell=(1-alpha)^{-p} +frac{1-alpha p}{(1-alpha)(p-1)}.$$
When $alpha >1/p$,
$$
ell=frac{alpha p -1}{(1-alpha)^p}int_1^infty (x^{-alpha}-x^{-1})^p, dx.$$
When $alpha=1/p$, as in Olivier's answer,
$$
ell=left(frac{p}{p-1}right)^p.$$
The following picture shows the graph of $ell$ as a function of $alpha$ (for $p=4$). The peak occurs at $alpha=0.25$, that is, $alpha=1/p$.

The fact that $ell$ is maximal at $alpha=1/p$ is to be expected; as Olivier shown in his answer, $f^{1/p}_A$ attains the supremum of the ratio $|Hg|_p/|g|_p$, in the limit $Ato infty$. However, this picture also contains another, new, piece of information; it shows that $alpha=1/p$ is the only value of $alpha$ that attains such supremum. In other words, among all powers $x^{-alpha}$, and in a vague sense,

$$
x^{-1/p}text{ is the only maximizer to the Hardy inequality.}$$

See this great answer of David C. Ullrich.