I find myself confused about the following thought. Consider the right-shift operator $R$ on $\ell^2(\mathbb Z).$ It has spectrum $\{|z| = 1\}.$ Now pick $\lambda>1.$ Then if $y=(R-\lambda)x,$ we have $$ y_n = x_{n-1} - \lambda x_n, x = (..., x_{-1}, x_0, x_1 , ...) $$ Then if $x_n$ decays fast as $n\to +\infty,$ in particular $x_n\lambda^n \to 0,$ we have $$ x_0 = y_1 + \lambda y_2 + \lambda^2 y_3 + ... \space(*) $$
Now pick a value $y\in \ell^2 (\mathbb Z)$ with strictly positive elements, but decays slowly at a slow rate, say $y_n = 1/|n|.$ Let $y^{(n)}$ be the finite truncation of $y,$ making all elements except for the $(-n)$-th all the way to the $n$-th zero. Then $y^{(n)} \to y$ strongly on $\ell^2(\mathbb Z).$
The inverse $(R-\lambda)^{-1}$ exists and is bounded, since $\lambda$ is not in the spectrum. Let $\xi_0$ be the map $\ell^2(\mathbb Z) \to \mathbb C : x\mapsto x_0.$ Then $T = \xi_0 \circ (R-\lambda)^{-1}$ is a bounded linear functional. Note that whenever $y$ has only finitely many nonzero elements, then there exists $N$ such that $x_n = 0$ for $n>N.$ (Otherwise, $x$ grows exponentially in one direction, and cannot be in $\ell^2(\mathbb Z).$) Therefore, the value $Ty$ is given by $(*)$ above. Thus, $$ Ty^{(n)} \to \infty \text{ as }n \to \infty $$ since the series $(*)$ diverges for such a slowly decaying sequence $y.$ However, $T$ is bounded, and $Ty_n \to Ty,$ contradiction.
So, what is wrong here?
$$ \begin{split} (R-\lambda I)=-\lambda(I-\frac{1}{\lambda}R)\\ (R-\lambda I)^{-1}=-\frac{1}{\lambda}\sum_{k=0}^\infty \lambda^{-k}R^k \end{split} $$ with the latter series justified since $\|\frac{1}{\lambda}R\|<1$. From this $$ x_n=-\sum_{k=0}^\infty\lambda^{-k-1}y_{n-k}=-\frac{1}{\lambda}y_n-\frac{1}{\lambda^2}y_{n-1}-\frac{1}{\lambda^3}y_{n-2}-\frac{1}{\lambda^4}y_{n-3}-\ldots $$ which is entirely the other direction from $(*)$ in the original post. It is also straightforward to verify that this is consistent with $x_n-\lambda x_{n+1}=y_{n+1}$.