Let $e_i$ be the standard bais vectors of $\ell^2(\mathbb{N)}$ and let $S$ denote the right shift operator on $\ell^2(\mathbb{N)}$, i.e. $Se_i= e_{i+1}$. Now the operator $T = S + S^*$ is self-adjoint and I need to prove that the spectrum of $T$ is equal to $[-2,2]$. I have already shown that there are uniquely determined polynomials $p_n$ of degree $n$ that satisfy the recurrence relation $xp_n(x) = p_{n+1}(x) + p_{n-1}(x)$ and moreover, that $p_n(x) = U_n(\frac{x}{2})$, where $U_n(x)$ is the Chebyshev polynomial of the second kind.
Now, in the last part of my exercise, I have also shown that the Chebyshev polynomials are orthogonal with respect to the measure $\sqrt{1-x^2}dx$ on $[-1,1]$ and they now ask me to conclude that $\sigma(T) = [-2,2]$ but I don't know how this follows from this fact.
Hint: Let $H$ denote the Hilbert space of functions over $[-1,1]$ with finite norm, under the inner-product defined by the measure $\sqrt{1 - x^2}dx$. Define the bounded linear map $\Phi:\ell^2(\Bbb N) \to H$ by $\Phi(e_i) = U_i$; note that $\Phi$ is an isometric isomorphism of Hilbert spaces.
The map $\Phi T \Phi^{-1}$ has the same spectrum as $T$, and by the recurrence relation that you have proven we have $$ [\Phi T\Phi^{-1}](f)(x) = \frac x2 \cdot f(x). $$