Divergence of squared sum of Chebyshev Polynomials $\equiv L+R$ has empty point spectrum

Question

The Chebyshev Polynomials of the second kind $U_n$ are the solutions of the differential equation $$(1-x^2)U_n''(x)-3xU_n'(x)+n(n+2)U_n(x)=0$$

Alternatively they are defined inductively:

$$U_0(x)=1 \qquad U_1(x)=2x \qquad U_{n+1}(x)=2xU_n(x) - U_{n-1}(x)$$

My question is whether or not

$$\sum_n |U_n(x/2)|^2\tag{1}$$

diverges for all $x \in \mathbb C$. The convergence of (1) is equivalent to the operator $L+R$ on $\mathscr l^2 (\mathbb N)$ having an eigenvalue $x$, where $L$ is the left shift and $R$ the right shift operator. The correspondence is explained in this answer. (This is the reason behind the functional analysis tag.)

Since $L+R$ is hermitian, bounded on a Hilbert space and has norm $≤2$, the spectrum is a subset of $[-2,2]$, and as such the sum automatically diverges if $x \notin [-2,2]$.

Experimentally it looks like as if the sum diverges on $[-2,2]$, but I would be interested in an angle allowing a proof or evidence for points at which it does converge.

Answer 1

As you pointed out, you are equivalently asking if the self-adjoint difference operator $$ (Ju)_n = \begin{cases} u_{n+1} + u_{n-1} & n\ge 1 \\ u_1 & n=0 \end{cases} $$ on $\ell^2(\{0,1,2, \ldots \})$ can have $x$ as an eigenvalue, and the answer to this is no, for any $x\in\mathbb C$.

For $-2< x< 2$, we cannot have an eigenvalue for the simple reason that the difference equation $u_{n+1}+u_{n-1}=xu_n$ has a solution basis of the form $e^{\pm in\alpha}$ (and here $2\cos\alpha =x$, but we don't need this); in particular, no non-trivial solution is in $\ell^2$. For $x=\pm 2$, the situation is similar: we have one constant in size solution and one polynomially growing solution, and again no linear combination of these is in $\ell^2$.

As you already observed yourself, for $x\notin [-2,2]$, we are outside the spectrum since $\|J\|=2$, and such an $x$ is certainly not an eigenvalue (in fact, the spectrum equals $[-2,2]$, and it is purely absolutely continuous).

Answer 2

if $a_n = 0$ for $n < 0$, $a_0 = 1$ and $a_{n+1} = 2 x a_n - a_{n-1} $ for $n > 1$ then by multiplying by $z^{-n}$ and summing we get the Z-transform :

$$\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^{-n} = \sum_{n=0}^\infty \delta(n) z^{-n} = 1$$

and we have also $\sum_{n=0}^\infty (a_{n+1} - 2 x a_n + a_{n-1}) z^{-n} = (2x + z^{-1} + z) \sum_{n=0}^\infty a_n z^{-n} $

so with $A(z) =\sum_{n=0}^\infty a_n z^{-n}$ : $$A(z) = \frac{1}{2x + z^{-1} + z} = \frac{z}{1 -2x z + z^2} = \frac{4z}{(z-2x+2\sqrt{x^2-1})(z-2x+2\sqrt{x^2-1})} = \frac{bz}{z-2x+2\sqrt{x^2-1}} + \frac{cz}{z-2x-2\sqrt{x^2-1}}$$

(for some constants $b,c$ obtained in general by partial fraction decomposition)

so for $n \ge 0$ :

$$a_n = B (2x-2\sqrt{x^2-1})^{n} + C (2x+2\sqrt{x^2-1})^{n} $$

the solutions for other initial conditions are easily obtained by linearity and shifting,

so the convergence of $\sum_n |a_n|^2$ depends on if those two roots $2x\pm 2\sqrt{x^2-1}$ are of modulus strictly $< |1|$

(don't worry if I made some mistakes the idea is there)