Computing the best constant in classical Hardy's inequality












29














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



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










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  • 3




    Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
    – Giuseppe Negro
    Apr 1 '12 at 14:24












  • This is also an exercise in Rudin's book Real and Complex.
    – AD.
    Apr 3 '12 at 21:11










  • @joriki why does this prove that $frac{p}{p-1}$ is the best constant?
    – Guacho Perez
    Aug 3 '17 at 18:02
















29














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



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










share|cite|improve this question




















  • 3




    Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
    – Giuseppe Negro
    Apr 1 '12 at 14:24












  • This is also an exercise in Rudin's book Real and Complex.
    – AD.
    Apr 3 '12 at 21:11










  • @joriki why does this prove that $frac{p}{p-1}$ is the best constant?
    – Guacho Perez
    Aug 3 '17 at 18:02














29












29








29


16





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



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










share|cite|improve this question















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



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







analysis inequality






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edited Apr 13 '17 at 12:21









Community

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1










asked Apr 1 '12 at 14:15









Giuseppe NegroGiuseppe Negro

16.8k330122




16.8k330122








  • 3




    Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
    – Giuseppe Negro
    Apr 1 '12 at 14:24












  • This is also an exercise in Rudin's book Real and Complex.
    – AD.
    Apr 3 '12 at 21:11










  • @joriki why does this prove that $frac{p}{p-1}$ is the best constant?
    – Guacho Perez
    Aug 3 '17 at 18:02














  • 3




    Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
    – Giuseppe Negro
    Apr 1 '12 at 14:24












  • This is also an exercise in Rudin's book Real and Complex.
    – AD.
    Apr 3 '12 at 21:11










  • @joriki why does this prove that $frac{p}{p-1}$ is the best constant?
    – Guacho Perez
    Aug 3 '17 at 18:02








3




3




Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
– Giuseppe Negro
Apr 1 '12 at 14:24






Addendum: The proof is complete if it is shown that $operatorname{lim sup}_{alpha to 1/p} (ldots) ge left( frac{p}{p-1}right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that.
– Giuseppe Negro
Apr 1 '12 at 14:24














This is also an exercise in Rudin's book Real and Complex.
– AD.
Apr 3 '12 at 21:11




This is also an exercise in Rudin's book Real and Complex.
– AD.
Apr 3 '12 at 21:11












@joriki why does this prove that $frac{p}{p-1}$ is the best constant?
– Guacho Perez
Aug 3 '17 at 18:02




@joriki why does this prove that $frac{p}{p-1}$ is the best constant?
– Guacho Perez
Aug 3 '17 at 18:02










4 Answers
4






active

oldest

votes


















16





+200









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.






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  • In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
    – Antonio Vargas
    Apr 3 '12 at 18:15








  • 2




    @Antonio: Substitute $u=x/b$.
    – joriki
    Apr 3 '12 at 18:22










  • Ah, of course. Very nice, by the way.
    – Antonio Vargas
    Apr 3 '12 at 18:26






  • 1




    This is a master's solution and I learned a lot in working it out! Thank you very much.
    – Giuseppe Negro
    Apr 4 '12 at 10:07










  • @Giuseppe: Thanks; and you're welcome!
    – joriki
    Apr 4 '12 at 10:38





















1














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 $frac{p}{p-1}$.






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  • I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
    – Giuseppe Negro
    Dec 4 '18 at 11:15












  • Your answer inspired this question.
    – Giuseppe Negro
    Dec 4 '18 at 12:46










  • I corrected a typo in the formula for $F_A(x)$.
    – Giuseppe Negro
    Dec 29 '18 at 16:55










  • This answer of mine really is an appendix to yours. Maybe you will find it interesting.
    – Giuseppe Negro
    Dec 30 '18 at 9:03



















0















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 $(0, infty)$.






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  • Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
    – Giuseppe Negro
    Jun 14 '16 at 8:56






  • 1




    I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
    – Giuseppe Negro
    Jun 14 '16 at 9:00



















0














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, $alpha=1/p$.



Ell as a function of alpha



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.






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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






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    active

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    active

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    16





    +200









    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.






    share|cite|improve this answer























    • In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
      – Antonio Vargas
      Apr 3 '12 at 18:15








    • 2




      @Antonio: Substitute $u=x/b$.
      – joriki
      Apr 3 '12 at 18:22










    • Ah, of course. Very nice, by the way.
      – Antonio Vargas
      Apr 3 '12 at 18:26






    • 1




      This is a master's solution and I learned a lot in working it out! Thank you very much.
      – Giuseppe Negro
      Apr 4 '12 at 10:07










    • @Giuseppe: Thanks; and you're welcome!
      – joriki
      Apr 4 '12 at 10:38


















    16





    +200









    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.






    share|cite|improve this answer























    • In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
      – Antonio Vargas
      Apr 3 '12 at 18:15








    • 2




      @Antonio: Substitute $u=x/b$.
      – joriki
      Apr 3 '12 at 18:22










    • Ah, of course. Very nice, by the way.
      – Antonio Vargas
      Apr 3 '12 at 18:26






    • 1




      This is a master's solution and I learned a lot in working it out! Thank you very much.
      – Giuseppe Negro
      Apr 4 '12 at 10:07










    • @Giuseppe: Thanks; and you're welcome!
      – joriki
      Apr 4 '12 at 10:38
















    16





    +200







    16





    +200



    16




    +200




    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 3 '12 at 17:08

























    answered Apr 3 '12 at 17:03









    jorikijoriki

    170k10183343




    170k10183343












    • In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
      – Antonio Vargas
      Apr 3 '12 at 18:15








    • 2




      @Antonio: Substitute $u=x/b$.
      – joriki
      Apr 3 '12 at 18:22










    • Ah, of course. Very nice, by the way.
      – Antonio Vargas
      Apr 3 '12 at 18:26






    • 1




      This is a master's solution and I learned a lot in working it out! Thank you very much.
      – Giuseppe Negro
      Apr 4 '12 at 10:07










    • @Giuseppe: Thanks; and you're welcome!
      – joriki
      Apr 4 '12 at 10:38




















    • In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
      – Antonio Vargas
      Apr 3 '12 at 18:15








    • 2




      @Antonio: Substitute $u=x/b$.
      – joriki
      Apr 3 '12 at 18:22










    • Ah, of course. Very nice, by the way.
      – Antonio Vargas
      Apr 3 '12 at 18:26






    • 1




      This is a master's solution and I learned a lot in working it out! Thank you very much.
      – Giuseppe Negro
      Apr 4 '12 at 10:07










    • @Giuseppe: Thanks; and you're welcome!
      – joriki
      Apr 4 '12 at 10:38


















    In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
    – Antonio Vargas
    Apr 3 '12 at 18:15






    In moving from the fourth to the fifth line in the big eqnarray, how do you get $$int_b^infty (x^{-alpha})^pmathrm dx = b^{1-alpha p}int_1^infty (x^{-alpha})^pmathrm dx?$$
    – Antonio Vargas
    Apr 3 '12 at 18:15






    2




    2




    @Antonio: Substitute $u=x/b$.
    – joriki
    Apr 3 '12 at 18:22




    @Antonio: Substitute $u=x/b$.
    – joriki
    Apr 3 '12 at 18:22












    Ah, of course. Very nice, by the way.
    – Antonio Vargas
    Apr 3 '12 at 18:26




    Ah, of course. Very nice, by the way.
    – Antonio Vargas
    Apr 3 '12 at 18:26




    1




    1




    This is a master's solution and I learned a lot in working it out! Thank you very much.
    – Giuseppe Negro
    Apr 4 '12 at 10:07




    This is a master's solution and I learned a lot in working it out! Thank you very much.
    – Giuseppe Negro
    Apr 4 '12 at 10:07












    @Giuseppe: Thanks; and you're welcome!
    – joriki
    Apr 4 '12 at 10:38






    @Giuseppe: Thanks; and you're welcome!
    – joriki
    Apr 4 '12 at 10:38













    1














    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 $frac{p}{p-1}$.






    share|cite|improve this answer























    • I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
      – Giuseppe Negro
      Dec 4 '18 at 11:15












    • Your answer inspired this question.
      – Giuseppe Negro
      Dec 4 '18 at 12:46










    • I corrected a typo in the formula for $F_A(x)$.
      – Giuseppe Negro
      Dec 29 '18 at 16:55










    • This answer of mine really is an appendix to yours. Maybe you will find it interesting.
      – Giuseppe Negro
      Dec 30 '18 at 9:03
















    1














    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 $frac{p}{p-1}$.






    share|cite|improve this answer























    • I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
      – Giuseppe Negro
      Dec 4 '18 at 11:15












    • Your answer inspired this question.
      – Giuseppe Negro
      Dec 4 '18 at 12:46










    • I corrected a typo in the formula for $F_A(x)$.
      – Giuseppe Negro
      Dec 29 '18 at 16:55










    • This answer of mine really is an appendix to yours. Maybe you will find it interesting.
      – Giuseppe Negro
      Dec 30 '18 at 9:03














    1












    1








    1






    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 $frac{p}{p-1}$.






    share|cite|improve this answer














    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 $frac{p}{p-1}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 6:53

























    answered Nov 30 '18 at 18:31









    Oliver DiazOliver Diaz

    1926




    1926












    • I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
      – Giuseppe Negro
      Dec 4 '18 at 11:15












    • Your answer inspired this question.
      – Giuseppe Negro
      Dec 4 '18 at 12:46










    • I corrected a typo in the formula for $F_A(x)$.
      – Giuseppe Negro
      Dec 29 '18 at 16:55










    • This answer of mine really is an appendix to yours. Maybe you will find it interesting.
      – Giuseppe Negro
      Dec 30 '18 at 9:03


















    • I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
      – Giuseppe Negro
      Dec 4 '18 at 11:15












    • Your answer inspired this question.
      – Giuseppe Negro
      Dec 4 '18 at 12:46










    • I corrected a typo in the formula for $F_A(x)$.
      – Giuseppe Negro
      Dec 29 '18 at 16:55










    • This answer of mine really is an appendix to yours. Maybe you will find it interesting.
      – Giuseppe Negro
      Dec 30 '18 at 9:03
















    I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
    – Giuseppe Negro
    Dec 4 '18 at 11:15






    I thank you for this post, no matter it comes late, I don't think that they changed the inequality in these years! At a first glance, it seems to me that the idea of this computation is akin to the one of the main post, with the difference that you truncate in space rather than playing with the exponent. This is interesting.
    – Giuseppe Negro
    Dec 4 '18 at 11:15














    Your answer inspired this question.
    – Giuseppe Negro
    Dec 4 '18 at 12:46




    Your answer inspired this question.
    – Giuseppe Negro
    Dec 4 '18 at 12:46












    I corrected a typo in the formula for $F_A(x)$.
    – Giuseppe Negro
    Dec 29 '18 at 16:55




    I corrected a typo in the formula for $F_A(x)$.
    – Giuseppe Negro
    Dec 29 '18 at 16:55












    This answer of mine really is an appendix to yours. Maybe you will find it interesting.
    – Giuseppe Negro
    Dec 30 '18 at 9:03




    This answer of mine really is an appendix to yours. Maybe you will find it interesting.
    – Giuseppe Negro
    Dec 30 '18 at 9:03











    0















    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 $(0, infty)$.






    share|cite|improve this answer























    • Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
      – Giuseppe Negro
      Jun 14 '16 at 8:56






    • 1




      I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
      – Giuseppe Negro
      Jun 14 '16 at 9:00
















    0















    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 $(0, infty)$.






    share|cite|improve this answer























    • Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
      – Giuseppe Negro
      Jun 14 '16 at 8:56






    • 1




      I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
      – Giuseppe Negro
      Jun 14 '16 at 9:00














    0












    0








    0







    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 $(0, infty)$.






    share|cite|improve this answer















    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 $(0, infty)$.







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    share|cite|improve this answer








    edited Apr 13 '17 at 12:20









    Community

    1




    1










    answered Jun 13 '16 at 16:33









    user345872user345872

    117419




    117419












    • Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
      – Giuseppe Negro
      Jun 14 '16 at 8:56






    • 1




      I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
      – Giuseppe Negro
      Jun 14 '16 at 9:00


















    • Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
      – Giuseppe Negro
      Jun 14 '16 at 8:56






    • 1




      I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
      – Giuseppe Negro
      Jun 14 '16 at 9:00
















    Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
    – Giuseppe Negro
    Jun 14 '16 at 8:56




    Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong.
    – Giuseppe Negro
    Jun 14 '16 at 8:56




    1




    1




    I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
    – Giuseppe Negro
    Jun 14 '16 at 9:00




    I think I have found an error. I suspect that it is not true that $| Tf|_p=frac{p}{p-1}$. Instead, it is correct that $| Tf|_p = left(frac{p}{p-1}right)^{frac1p}$.
    – Giuseppe Negro
    Jun 14 '16 at 9:00











    0














    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, $alpha=1/p$.



    Ell as a function of alpha



    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.






    share|cite|improve this answer




























      0














      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, $alpha=1/p$.



      Ell as a function of alpha



      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.






      share|cite|improve this answer


























        0












        0








        0






        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, $alpha=1/p$.



        Ell as a function of alpha



        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.






        share|cite|improve this answer














        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, $alpha=1/p$.



        Ell as a function of alpha



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 9:01

























        answered Dec 30 '18 at 1:44









        Giuseppe NegroGiuseppe Negro

        16.8k330122




        16.8k330122






























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