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I have a question regarding operators and I found an answer but I do not want to type the full answer. If you want to do it I can give you the pieces of the answer (see below). Once the first well-written answer is posted I will accept it. (The answer below is incomplete.)

Let $T:\ell^2\to\ell^2$ be the right-shift operator $T(x_1,x_2,x_3,\dots)=(0,x_1,x_2,\dots)$. Then its adjoint is the left-shift operator $T^*(x_1,x_2,x_3,\dots)=(x_2,x_3,x_4,\dots)$.

Question: How to prove that the self-adjoint operator $T+T^*$ does not have eigenvalues? $$(T+T^*)(x_1,x_2,x_3,\dots)=(x_2,x_1+x_3,x_2+x_4,\dots,x_n+x_{n+2},\dots)$$

Answer in Pieces: First, suppose that $x=(x_n)_n$ is an eigenvector of $T+T^*$ corresponding to some eigenvalue $a\in\mathbb{C}$. In order to find the closed form of $x_n$ and the justification of its uniqueness see this question (click here).

Second, to prove that the sequence $(x_n)_n$ is not an eigenvector it is enough to prove that $\lim x_n\neq0$ see the hint in this question.

Finally, the final piece is in the comments and answer of this question.

ViktorStein
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Chilote
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Suppose that $x:=(x_1,x_2,...)$ is an eigenvector of the operator $T+T^*$ and $\lambda$ its respective eigenvalue (we are assuming that $\sigma(T+T^*)\neq \emptyset$ otherwise we have nothing to say). From the equation $$(T+T^*)x=\lambda x$$ follow the following equations $$x_2=\lambda x_1,x_1+x_3=\lambda x_2,...,x_n+x_{n+2}=\lambda x_{n+1},....$$ These imply $$x_2=\lambda x_1, x_3=(\lambda^2-1)x_1,x_4=(\lambda^3-2\lambda)x_1,..., x_n=P_n(\lambda)x_1,...$$ where $P_n(\lambda)$ is a polynomial in $\lambda$ of degree $n-1$ satisfying the recursive relation $$P_n(\lambda)=\lambda P_{n-1}(\lambda)-P_{n-2}(\lambda)$$ for $n\geqslant 2$. With convention let $P_{0}(\lambda)=0$ and $P_1(\lambda)=1$. Since $x\in l^2(\mathbb{N})$ then $$||x||^2_{l^2}=|x_1|^2+|x_2|^2+...+|x_n|^2+...=|x_1|^2(|P_1(\lambda)|^2+|P_2(\lambda)|^2+...+|P_n(\lambda)|^2+...)<\infty$$ It is desirable to get a formula for $P_n(\lambda)$ and then getting necessary conditions on $\lambda$ so that the infinite sum above is finite. But we have a recursive relation of second degree together with two initial conditions. So in principle you should be able to get a formula for $P_n(\lambda)$.

Arian
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    Hi @Arian, thanks for your answer but of course this is something that I also can do. As it is now, I cannot consider it as a full answer. Actually, it is what I did: see https://math.stackexchange.com/q/2844980 – Chilote Jul 08 '18 at 21:00
  • I see. Thnx for letting me know. Well it seems we have to deal with a really cumbersome expression for $P_n$. – Arian Jul 08 '18 at 21:05
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    @Chilote You are just asking to check that there are no solutions to a second order linear recurrence with constant coefficients except for linear combinations of exponentials? It is simple, just as in ODE: construct a solution to a given "initial value problem" using the exponentials, and then show that any two solutions to that initial value problem must be equal (by subtracting them and checking that they solve the initial value problem with zero initial data). An equivalent proof is to use linear algebra and the Jordan normal form. – Ian Jul 08 '18 at 21:08
  • Thanks, @Ian, I got it. – Chilote Jul 08 '18 at 22:03
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    $P_n(\lambda) = U_{n-1}(\lambda/2)$ where $U_n$ are the Chebyshev polynomials of the second kind. But I'm pretty sure there won't be any $\lambda$ for which $\sum_n P_n(\lambda)^2$ converges. – Robert Israel Jul 09 '18 at 01:47
  • @Arian I finally got the proof (with a lot of help) that $T+T^*$ does not have eigenvalues. Since you started the proof I want to give you the pieces so for sake of completeness you can complete your answer (only if you are willing to do it) and I can accept it. First: in order to find the closed form of $x_n$ and the justification of its uniqueness see https://math.stackexchange.com/q/2844980 Then, to prove that the sequence $(x_n)$ is not an eigenvector it is enough to prove that $\lim x_n\neq 0$ see the hint in https://math.stackexchange.com/q/2845045 – Chilote Jul 09 '18 at 02:17
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    The final piece is in the comments of the question and the answer of https://math.stackexchange.com/q/2845126 – Chilote Jul 09 '18 at 02:18