Timeline for Bounding $L^p$ norms in terms of lower-order $L^q$ norms
Current License: CC BY-SA 4.0
12 events
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| May 5, 2020 at 17:22 | comment | added | JohnA | Yes, it seems I had forgotten to try the simplest of interpolation bounds. If @IosifPinelis is willing, I would suggest adding this to his answer, and I will accept this as the correct answer. | |
| May 5, 2020 at 13:43 | comment | added | Willie Wong | If $q < p < r$ there exists $\theta \in (0,1)$ such that $1/p = (1-\theta)/q + \theta/r$. Then $$ \int |f|^p = \int |f|^{(1-\theta)p} |f|^{\theta p} \leq \left( \int |f|^q \right)^{(1-\theta)p/q} \left( \int|f|^r\right)^{\theta p / r} $$ by Holder. So if $f_n \to f$ in $L^q$ and $f_n, f$ are uniformly bounded (say by $M$) in $L^r$, you have that $$ \int |f_n - f|^p \leq \left( \int |f_n - f|^q \right)^{(1-\theta)p/q} \left( 2M \right)^{\theta p} $$ using triangle inequality. | |
| May 4, 2020 at 21:07 | comment | added | JohnA | @WillieWong It seems I don't know enough about interpolation, then! What kind of interpolation inequalities give such explicit bounds? (If there is a standard reference, please feel free to share it.) | |
| May 4, 2020 at 20:53 | comment | added | Willie Wong | For your updated question: if you know that $f_n, f$ are bounded in $L^r$ for $r > p$, then you can directly interpolate to get convergence in $L^p$. (You don't need to assume $f_n \to f$ in $L^r$.) Iosif's example shows that this is sharp: with just $r = p$ this is not enough. | |
| May 4, 2020 at 18:19 | history | edited | JohnA | CC BY-SA 4.0 |
added 3 characters in body
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| May 4, 2020 at 17:44 | history | edited | JohnA | CC BY-SA 4.0 |
deleted 13 characters in body
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| May 4, 2020 at 13:03 | comment | added | ARG | PS: I think it's very unpolite that the downvoters and closers did not leave a comment... | |
| May 4, 2020 at 13:00 | comment | added | ARG | The interpolation inequalities for a $p$-norm only work if you know the $q$-norm and the $r$-norm with $q\leq p \leq r$. I would guess that taking $h = f-g$ (it does not matter what $f$ and $g$ are) given by $h(x) = \big( \frac{d}{dx} (\frac{1}{\ln^{[k]} x} ) \big)^{1/p}$ will get you a counterexample (here $\ln^{[k]}$ is an $k$-times iterated logartihm). The integrand of the $p$-norm is logarithmic [or an interation of such], while the integrand of q norms will be dominated by the power of $x^{-q/p}$. But given the amount of downvote, I suspect there is a textbook example. | |
| May 3, 2020 at 18:11 | review | Close votes | |||
| May 7, 2020 at 3:33 | |||||
| May 3, 2020 at 17:31 | history | edited | JohnA | CC BY-SA 4.0 |
made special case more specific
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| May 3, 2020 at 15:22 | answer | added | Iosif Pinelis | timeline score: 2 | |
| May 3, 2020 at 14:55 | history | asked | JohnA | CC BY-SA 4.0 |