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( might not be a research-level question... if it violates any term of this website I will delete it ASAP )

I'm wondering if there is any way to show the boundedness/unboundedness of operators like $(\partial_x + f )^{-1}$ or $(H\partial_x +f)^{-1}$, such as in $L^2 (\mathbb{R})$. (Here $H$ is the Hilbert transform, $f$ is a bounded smooth function, but possibly we need more assumptions on that...? ) Could anyone give me a clue or tell me if there is any literature to follow?

Thanks in advance!

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    $\begingroup$ Operators involving derivatives are, as a rule, never bounded. $\endgroup$
    – Christian Remling
    Commented Aug 27, 2016 at 20:52
  • $\begingroup$ Oh sorry I just found out that I made a typo in the question...It should be the inverses of those operators. Thanks anyway! $\endgroup$
    – alby
    Commented Aug 27, 2016 at 21:48
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    $\begingroup$ 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). $\endgroup$
    – Igor Khavkine
    Commented Aug 27, 2016 at 22:22
  • $\begingroup$ 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) $\endgroup$
    – alby
    Commented Aug 27, 2016 at 23:12
  • $\begingroup$ 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. $\endgroup$
    – Christian Remling
    Commented Aug 28, 2016 at 0:23

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