Timeline for Characterizing pseudo-differential operators as a subalgebra of continuous endomorphisms of tempered distributions
Current License: CC BY-SA 4.0
7 events
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| Jul 10, 2018 at 9:09 | comment | added | Bazin | @mcd You are right, the inclusion holds true only for $m\le 0.$ We can get that $\text{Op}\Sigma^0$ is bounded on $L^p$. Then we may define $\mathcal W^{s,p}=\{u\in \mathscr S'(\mathbb R^n), \forall a\in \Sigma^s, (\text{Op} a) u\in L^p\}$ and hopefully prove that Op$(\Sigma^m)$ sends continuously $\mathcal W^{s,p}$ into $\mathcal W^{s-m,p}$. | |
| Jul 10, 2018 at 8:56 | comment | added | mcd | @Bazin for $m > 0$ the inclusion is not true, since you have "global" $S_{1,0}^m$ estimates; but the Shubin symbol are allowed to grow in $x$, for example the harmonic oscillator is not in $S^2_{1,0}$, but obviously in $\Sigma^1$. | |
| Jul 9, 2018 at 19:16 | comment | added | Bazin | @Saal Hardali You mean certainly the $L^p$ boundedness. It is a delicate matter since some classes of pseudo-differential operators are bounded on $L^2$ and not on $L^p$. Here it is simpler since $\Sigma^m\subset S^{2m}_{1,0}$ and an operator with symbol in the larger set is indeed bounded from $W^{s,p}$ into $W^{s-2m,p}$. Maybe I was too quick for the completeness, but I do not see why it would be different, only the Sobolev filtration is different with the definition of the spaces $\mathscr H^m$ in my answer. | |
| Jul 9, 2018 at 17:49 | comment | added | Saal Hardali | hmmm, i'm not sure I understand your point about the boundness, could you say a few words about the boundness? This is the part i'm least sure about. Also the completeness isn't at all obvious to me for the second class of symbols you propose. | |
| Jul 9, 2018 at 17:26 | comment | added | Bazin | @Saal Hardali Considering the harmonic Oscillator $\mathcal H=-\Delta_x+\vert x\vert^2$, you can define the scale of Hilbert spaces $\mathscr H^m$ as the temperate distributions $u$ such that $\mathcal H^m u\in L^2$ and you can prove that $\mathscr H^m$ is also the set of $u$ such that $(\text{Op}a) u$ belongs to $L^2$ for any $a\in \Sigma^m.$ Another point is to prove that $\text{Op}\Sigma^0$ is included in the bounded operators on $L^p$ for $1<p<+\infty$, but it is a consequence of the same result for $S^0$. | |
| Jul 9, 2018 at 12:16 | comment | added | Saal Hardali | Could you elaborate on why the second algebra you give satisfies 3? it doesn't seem obvious but maybe im missing something. | |
| Jul 9, 2018 at 12:12 | history | answered | Bazin | CC BY-SA 4.0 |