Timeline for Boundedness for resolvents of perturbed operators
Current License: CC BY-SA 3.0
18 events
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| Aug 28, 2016 at 18:18 | comment | added | Christian Remling | On the other hand, $id/dx$ and $Hd/dx$ (are self-adjoint on suitable domains and) have spectrum $\mathbb R$ and $[0,\infty)$, respectively (maybe I'm off by a sign here). This means that if $f=c + g$, $c$ outside the spectrum and $g$ small in a suitable sense, then your resolvent will be bounded. | |
| Aug 28, 2016 at 18:15 | comment | added | Christian Remling | The whole set-up really seems unnatural to me as both operators are unbounded for $f=0$, so they won't be bounded for small $f$ (for example, a relatively compact perturbation). What kind of class of functions $f$ did you have in mind? | |
| Aug 28, 2016 at 13:57 | comment | added | alby | @IgorKhavkine My apologize for that - just edited the question. Thanks! | |
| Aug 28, 2016 at 13:55 | history | edited | alby | CC BY-SA 3.0 |
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| Aug 28, 2016 at 6:35 | comment | added | Igor Khavkine | @Ever_99, if your $H$ denotes the Hilbert transform, rather than a constant or a function of $x$, you should state that explicitly. That information significantly changes part of your question. | |
| Aug 28, 2016 at 5:36 | history | edited | alby | CC BY-SA 3.0 |
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| Aug 28, 2016 at 2:49 | comment | added | fedja | OK, that may be trickier. Let me think... | |
| Aug 28, 2016 at 1:57 | comment | added | alby | @fedja Thanks! It's just that I still need to deal with the operator with the Hilbert transform together with the function $f$ | |
| Aug 28, 2016 at 1:55 | comment | added | alby | @ChristianRemling Thanks! But how can one deal with the operator with the Hilbert transform and a general function $f$ ? | |
| Aug 28, 2016 at 1:20 | comment | added | fedja | Why don't you just solve the first order ODE as usual (integrating factors, etc.). That will give you a fairly explicit kernel for the inverse and it may happen that one of the standard (un)boundedness tests will apply to your particular case. | |
| Aug 28, 2016 at 0:23 | comment | added | Christian Remling | Now (= your comment) you're asking if $-1$ is outside the spectrum of $d/dx$, which is true because this operator (on a suitable domain) has purely imaginary spectrum. | |
| Aug 27, 2016 at 23:12 | comment | added | alby | Thanks! Do you mean that same thing happens for operators like $(\partial_x +1 )^{-1}$ ? ( I can assume $f$ is real-valued in this question actually) | |
| Aug 27, 2016 at 22:22 | comment | added | Igor Khavkine | The operator $(-i\partial_x + 3)^{-1}$ is not bounded, where $3$ could also be any real number, though it becomes bounded when $3$ is replaced by any non-real complex number. You can check this explicitly if you write its fundamental solution as an integral operator with kernel $G(x,x')$, where $(-i\partial_x +3) G(x,x') = \delta(x-x')$. The same goes for your differential operators, if you can get some information about their fundamental solutions (look up the "variation of constants" formula, if you don't know it already). | |
| Aug 27, 2016 at 21:48 | comment | added | alby | Oh sorry I just found out that I made a typo in the question...It should be the inverses of those operators. Thanks anyway! | |
| Aug 27, 2016 at 21:44 | history | edited | alby | CC BY-SA 3.0 |
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| Aug 27, 2016 at 20:58 | review | First posts | |||
| Aug 27, 2016 at 22:42 | |||||
| Aug 27, 2016 at 20:52 | comment | added | Christian Remling | Operators involving derivatives are, as a rule, never bounded. | |
| Aug 27, 2016 at 20:47 | history | asked | alby | CC BY-SA 3.0 |