I am given a linear operator $T$ on $l^2$ that maps an element $x = (\xi_k)_{k\ge1}$ to $Tx = (\xi_2, \xi_1, \xi_3, \cdots)$. (It just swaps the first two elements). I now need to calculate and classify the spectrum of this operator.
I started by calculating $(T - \lambda I)^{-1} =: R_{\lambda}$. I believe that in this case this is $R_{\lambda}x = (\xi_2 + \lambda, \xi_1 + \lambda, \xi_3 + \lambda, \cdots...) = (T + \lambda I) x$. Now I didn't really have an ansatz so I just tried to calculate wether $R_\lambda$ is bounded
$$ ||R_\lambda x||^2 = ||(T + \lambda I)x||^2 = \sum_{k=1}^{\infty} |(\xi_2 + \lambda, \xi_1 + \lambda, \xi_3 + \lambda, ...)|^2 $$
The way I see it, this sum only converges for $\lambda = 0$. All other lambdas would turn the sequence into a non-null sequence, so the sum can't converge anymore. What this tells me is that every $\lambda \in \mathbb{C} \setminus \{0\}$ is part of the continuous spectrum of $T$. However we learned a theorem that the spectrum of a bounded operator on a Banach space is contained in the disk $|\lambda| \le ||T||$. Afaik $l^2$ is a banach space and $||T|| \le 1$, but the calculated spectrum is definitely outside this disk.
So what am I missing? It is probably something obvious, as I said I'm just starting to try to understand spectral theory.
You are making a basic mistake. You are writing $\lambda I x$ as $(\lambda, \lambda,...)$. But it is $\lambda (\xi_1,\xi_2,...)=(\lambda\xi_1,\lambda \xi_2,...)$
Also you formula for $R_{\lambda}$ is wrong. The solution of $(T-\lambda I) x=y$ is not $x=(T+\lambda I) y$.