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I've been puzzling over this for some time now, and can't quite make my intuitions precise. I need to find the resolvent set and spectrum of the operator

$$ Lu=i\frac{du}{dx} $$ taken to be (densely) defined over $L^2([0,\infty))$, and eventually its adjoint, which is the same operator but on the set (assuming I've computed it correctly):

$$ \text{Dom}(L^*)=\{v\in L^2[0,\infty):v(0)=0\} $$

The pointwise spectrum of $L$ is easy to compute: $(\lambda I-L)u=0$ is a simple ODE, whose solution along with the $L^2$ condition implies

$$ \sigma_p(L)=\{\lambda=a+bi\in\Bbb{C}:b<0\}, $$the lower half-plane. Now intuitively, since $\sigma(L)$ is a closed set, I expect $\rho(L)$ to be the upper half-plane, the residual spectrum to to be the real axis, and the continuous spectrum to be empty. However I'm struggling to show this rigorously since $L$ is an unbounded operator, and all the theory I find pertains to bounded linear operators.

So here are my questions:

  • Is there a (somewhat) direct way to characterize the residual spectrum of an operator without explicitly computing $(\lambda I-L)^{-1}$? Or am I just being a whimp and should find $(\lambda I-L)^{-1}$?
  • The adjoint clearly has no pointwise spectrum because of the domain restriction. Does this mean its resolvent set will be all of $\Bbb{C}$ or am I missing something?

Feel free to direct me to your favorite book on spectral/operator theory if necessary - I can't seem to find answers to my questions after looking in 4 or 5. Thanks!

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You need to specify the domain of $L$ more precisely, as the spectrum depends on it very much, e.g. the resolvent set is empty unless $L$ is a closed operator. I assume henceforth that $\text{dom}L = \{f \in L^{2}[0,\infty): f' \in L^2[0,\infty)\}$. In that case, if I haven't made any mistake, $\text{dom}L^{\ast}=\{f \in L^{2}[0,\infty): f(0)=0, f' \in L^2[0,\infty)\}$; you missed the integrability of derivative. Note also that the condition $f(0)=0$ makes sense, since if $f' \in L^2[0,\infty)$ (so $f$ is a member of Sobolev space $H^{1}[0,\infty)$) then it is a continuous function (absolutely continuous one, actually).

Let's now turn to the computation of spectrum. As you have noted, lower half-plane is a subset of the spectrum of $L$. This means that the upper half-plane is contained in the spectrum of $L^{\ast}$, since $\sigma(L^{\ast}) = \overline{\sigma(L)}$. We will now show that are no other elements of the spectrum of $L^{\ast}$. Pick then any element of the lower half-plane ($-i$ would do so we restrict to this case; in fact, it is a general fact about symmetric operators that checking for $-i$ suffices) and try to prove that $-i - L^{\ast}$ is invertible. First, easy integration by parts shows that $\|(-i-L^{\ast})v\|\geqslant \|v\|$, so that it is bounded below. It will therefore be enough to prove that the range is dense. But this is quite easy, since for compactly supported smooth functions we just need to solve some easy ODE and check that the solution is in the domain of $L^{\ast}$. This ODE can be even solved successfully for any $L^2$-function, so you can write an explicit formula for the resolvent*. This means that the spectrum of $L^{\ast}$ is exactly equal to the upper half-plane, so the spectrum of $L$ is equal to the lower half-plane.

*Generally, it may sometimes be managed in the following way: $-iL^{\ast}$ is probably a generator of some semigroup (it looks like left shift) and then the resolvent can be calculated as a Laplace transform of the semigroup. I hope that helps.

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  • $\begingroup$ To avoid possible misinterpretation of the statement "$\textrm{dom}\,L = \{f \in L^{2}[0,\infty): f' \in L^2[0,\infty)\}$" I would say something like "here $f'$ is understood as distributional (weak) derivative;" or, alternatively, "$f$ is required to be absolutely continuous." Otherwise one might think that a function like $f=\chi_{[0,1]}$ qualifies. $\endgroup$ – 75064 May 4 '13 at 17:02
  • $\begingroup$ I'm a still a little confused - am I correct in thinking that $L^*$ does not have a point spectrum because $L^*v=\lambda v$ has no solutions with $v(0)=0$? So the spectrum of $L^*$ is continuous...? $\endgroup$ – icurays1 May 4 '13 at 20:10
  • $\begingroup$ @user75064, you are, of course, right, I should have made that point clear. icurays1, you are right about the emptiness of the point spectrum of $L^{\ast}$. However, for elements $\lambda$ in the spectrum that are not real, the operator $\lambda - L^{\ast}$ is bounded below so is injective and has closed range; the spectrum has to be then residual. As for the real line, I'm not sure, but I would put my money on the continuous part. $\endgroup$ – Mateusz Wasilewski May 5 '13 at 9:38

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