Mixing
P-norm of the Poisson Kernel
$begingroup$
If we have $P_r(theta)=sum_{n=-infty}^{infty}r^{|n|}e^{intheta}=frac{1-r^2}{1-2rcos(theta)+r^2},$
Is there a quick way to compute the norm $|P_r|_p$, or an upper estimate?
complex-analysis analysis harmonic-functions
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
$endgroup$
add a comment |
$begingroup$
If we have $P_r(theta)=sum_{n=-infty}^{infty}r^{|n|}e^{intheta}=frac{1-r^2}{1-2rcos(theta)+r^2},$
Is there a quick way to compute the norm $|P_r|_p$, or an upper estimate?
complex-analysis analysis harmonic-functions
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
$endgroup$
add a comment |
$begingroup$
If we have $P_r(theta)=sum_{n=-infty}^{infty}r^{|n|}e^{intheta}=frac{1-r^2}{1-2rcos(theta)+r^2},$
Is there a quick way to compute the norm $|P_r|_p$, or an upper estimate?
complex-analysis analysis harmonic-functions
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
$endgroup$
If we have $P_r(theta)=sum_{n=-infty}^{infty}r^{|n|}e^{intheta}=frac{1-r^2}{1-2rcos(theta)+r^2},$
Is there a quick way to compute the norm $|P_r|_p$, or an upper estimate?
complex-analysis analysis harmonic-functions
complex-analysis analysis harmonic-functions
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
asked Jan 26 at 13:19
Mark_HoffmanMark_Hoffman
665515
665515
add a comment |
add a comment |
1 Answer
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"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.
An upper estimate? Certainly. It's clear that $$||P_r||_infty=frac{1-r^2}{1-2r+r^2}=frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2pi$ as usual it's clear that $$||P_r||_ple||P_r||_infty.$$
Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $int P_r=1$) and $$||P_r||_p^ple||P_r||_1||P_r||_infty^{p-1}.$$
Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_ple c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_pge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
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1 Answer
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$begingroup$
"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.
An upper estimate? Certainly. It's clear that $$||P_r||_infty=frac{1-r^2}{1-2r+r^2}=frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2pi$ as usual it's clear that $$||P_r||_ple||P_r||_infty.$$
Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $int P_r=1$) and $$||P_r||_p^ple||P_r||_1||P_r||_infty^{p-1}.$$
Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_ple c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_pge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
$endgroup$
add a comment |
$begingroup$
"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.
An upper estimate? Certainly. It's clear that $$||P_r||_infty=frac{1-r^2}{1-2r+r^2}=frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2pi$ as usual it's clear that $$||P_r||_ple||P_r||_infty.$$
Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $int P_r=1$) and $$||P_r||_p^ple||P_r||_1||P_r||_infty^{p-1}.$$
Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_ple c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_pge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
$endgroup$
add a comment |
$begingroup$
"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.
An upper estimate? Certainly. It's clear that $$||P_r||_infty=frac{1-r^2}{1-2r+r^2}=frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2pi$ as usual it's clear that $$||P_r||_ple||P_r||_infty.$$
Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $int P_r=1$) and $$||P_r||_p^ple||P_r||_1||P_r||_infty^{p-1}.$$
Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_ple c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_pge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
$endgroup$
"Compute" $||P_r||_p$? II doubt it; in fact I doubt that a closed form exists.
An upper estimate? Certainly. It's clear that $$||P_r||_infty=frac{1-r^2}{1-2r+r^2}=frac{1+r}{1-r},$$and assuming we're talking about the normalized Lebesgue measure $dt/2pi$ as usual it's clear that $$||P_r||_ple||P_r||_infty.$$
Of course that's a hugely bad upper estimate. You could get something marginally better by noting that $||P_r||_1=1$ (since $P_r>0$ and the Fourier series shows that $int P_r=1$) and $$||P_r||_p^ple||P_r||_1||P_r||_infty^{p-1}.$$
Edit: I find it surprising that in fact it's easy to show that such a crude estimate is best possible, within a constant multiple. We have $||P_r||_ple c/(1-r)^{1-1/p}$ and we want to show that $||P_r||_pge c/(1-r)^{1-1/p}$ of course with a different constant. Parseval shows that the second inequality holds for $p=2$, and now $$||P_r||_2^2le||P_r||_p||P_r||_{p'}$$gives the lower estimate on $||P_r||_p$ from the upper estimate on $||P_r||_{p'}$.
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
edited Jan 26 at 17:13
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
answered Jan 26 at 16:04
David C. UllrichDavid C. Ullrich
61.6k43994
61.6k43994
add a comment |
add a comment |
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