Timeline for If $f \in L^p(\Omega)$, then $f \in L^q(\Omega)$ for some $q < p + \epsilon$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 25 at 8:21 | comment | added | Pietro Majer | Let's also recall an elementary fact: if f is in $L^{p_1}$ and in $L^{p_2}$, it is every $L^p$ in between, by interpolation inequality | |
| Oct 25 at 8:12 | comment | added | Piero D'Ancona | Let me add that functions satisfying self-improvement properties like the one considered here are rather important in harmonic analysis (I am thinking of reverse Hölder classes) | |
| Oct 24 at 18:18 | comment | added | Jochen Wengenroth | @AlpinistKitten No need for apologies. You might consider asking on MathStackExchange before posting such a question here on MO. | |
| Oct 24 at 16:22 | comment | added | Nate Eldredge | Presumably you know that for each $q > p$, it is possible to find a function that is in $L^p$ but not in $L^q$ (as $|x|^{-\gamma}$ or the like). So for each $n$, we may find $f_n \in L^p \setminus L^{p+1/n}$; by replacing $f_n$ with $|f_n|$ we can assume $f_n \ge 0$, and by scaling $f_n$ we can assume $\|f_n\|_{L^p} \le 2^{-n}$. Now set $f = \sum_n f_n$. The sum converges in $L^p$, so we have $f \in L^p$, but $f \ge f_n \notin L^{p+1/n}$ for all $n$, so $f \notin L^q$ for any $q > p$. | |
| Oct 24 at 15:53 | comment | added | AlpinistKitten | @JochenWengenroth: my most sincere apologies for asking. | |
| Oct 24 at 15:50 | review | Close votes | |||
| Nov 2 at 3:03 | |||||
| Oct 24 at 15:44 | comment | added | AlpinistKitten | @OlivierBégassat: the choice of $\epsilon > 0$ will depend on $f$, as in the formulation of the question. If it belongs to the Lebesgue space with exponent $p+\epsilon$, then it will belong to all Lebesgue spaces with smaller exponent. | |
| Oct 24 at 15:40 | comment | added | Fedor Petrov | $f(t)=t^{-1}(\log t/2)^{-2}$ on $[0,1]$ belongs yo $L^1$ but not to $L^p$ with $p>1$ | |
| Oct 24 at 15:32 | comment | added | Jochen Glueck | If a proper vector space $V$ of a Banach space $X$ is itself a Banach space with respect to a stronger norm, then $V$ is meagre in $X$. Hence, a countable union of such spaces is meagre, too. Thus, $\bigcup_{q > p} L^q = \bigcup_{n \in \mathbb{N}} L^{p+\frac{1}{n}}$ is meagre in $L^p$. (Edit: Ok, the other @Jochen was again slightly faster than me. ;-) ) | |
| Oct 24 at 15:30 | comment | added | Jochen Wengenroth | That this is wrong can be seen by Baire's theorem. This is rather a little exercise than research level. | |
| Oct 24 at 15:20 | comment | added | Olivier Bégassat | The formulation is slightly unclear to me. When you write "for any $q$" do you mean for all $q \in [p, p + \epsilon)$ ? If so it should be sufficient to ask whether $f \in L^p$ implies that $f \in L^{p'}$ for some $p' > p$. | |
| Oct 24 at 14:57 | history | asked | AlpinistKitten | CC BY-SA 4.0 |