I have the operator $T: l_2[0,\infty) \to l_2[0, \infty)$ defined by $T=S_l+S_r$. So $T(x_0,x_1,x_2,x_3,…)=(x_1, x_2+x_0,x_3+x_1,x_4+x_2,…)$.
I've shown that it's spectrum is a subset of $[-2,2]$.
What's left is to show is that $\left\lVert T \right\rVert =2$. It was easy to show that $\left\lVert T \right\rVert \leq 2$, but getting it equal to $2$ is tripping me up. Any help is appreciated.
Also, what is the best way to show the spectrum of T is indeed $[-2,2]$. I don't have much practice finding the continuous spectrum.
Best Answer
If you only want to see that the norm is $2$, consider $$ x=(\overbrace{1,\ldots,1}^{n},0,\ldots). $$ Then $$ Tx=(1,\overbrace{2,\ldots,2}^{n-1},1,0,\ldots). $$ So $$ \frac{\|Tx\|}{\|x\|}=\frac{\sqrt{4n-2}}{\sqrt{n}}=2\,\sqrt{1-\frac2n}. $$ As this can be done for any $n$, you get $\|T\|\geq2$.
Now, if you want to get the whole spectrum: fix $h,r\in\mathbb N$, and let $\lambda=2\cos\frac{h\pi}{r}$. For each $m\in\mathbb N$ let $$n=mr-1, \ \ s=mh=\frac{(n+1)h}r.$$ Consider $x^{(m)}\in \ell^2$ given by $$ x_k^{(m)}=\begin{cases} \sin\frac{ks\pi}{n+1},&\ 1\leq k\leq n\\ \ \\ 0,&\ k>n\end{cases} $$ Then (some trigonometry involved) $$ (Tx^{(m)})_k=\begin{cases} 2\cos\frac{s\pi}{n+1}\,\sin\frac{ks\pi}{n+1},&\ 1\leq k\leq n\\ \ \\ \sin\frac{ns\pi}{n+1},&\ k=n+1\\ \ \\ 0,&\ k>n+1 \end{cases} $$ As long as $h/r$ is not an integer, we have that $\|x^{(m)}\|\sim \sqrt {mr}$. Let $z^{(m)}=\frac{x^{(m)}}{\|x^{(m)}\|}$.
Since $2\cos\frac{s\pi}{n+1}=\lambda$, $$ \|Tz^{(m)}-\lambda z^{(m)}\|=\frac1{\|x^{(m)}\|}\,\left|\sin\frac{ns\pi}{n+1}\right|\leq\frac1{\|x^{(m)}\|}\,\xrightarrow[m\to\infty]{}0. $$ Thus $\lambda$ is an approximate eigenvalue for $T$. In other words, $$ \left\{2\cos q\pi:\ q\in\mathbb Q\right\}\subset\sigma(T). $$ As the spectrum is closed and the cosine continuous, $$ [-2,2]=\{2\cos t\pi:\ t\in\mathbb R\}\subset\sigma(T). $$