I would be greatful to someone famililar with operator theory to verify if my though process and calculations are correct.
Consider $x \in \mathbb{Z}_{+}$ and consider the following finite difference operator
\begin{equation} \Delta_1[f(x)] = f(x+1)-f(x) \end{equation}
where $f: \mathbb{Z}_{+}\mapsto \mathbb{R}_+$.
I'm interest in computing the eigenvalues and eigenfunctions of $\Delta_1$. In particular I'll try the following eigenfunction
\begin{equation} e(x)=x^n, \quad n \in \mathbb{Z}_{>} \end{equation}
from the defintion of $\Delta_1$ it follows that
\begin{equation} \Delta_1[e(x)] = \Delta_1[x^n] = x^{n+1}-x^{n} = x x^{n}-x^{n} = x^{n}(x-1) \end{equation}
in other words the eigenvalue $\lambda$ would be $\lambda = (x-1)$ and the eigenfunction $e(x) = x^{n}$
Are the steps correct or is the computation of the eigenvalue not correct as it depends upon $x$?
Thank you all in advance!
Your variable is $n$, so your computation is correct. And you can show that those are exactly the possible eigenfunctions: that is, if $\Delta_1 f=\lambda f$, then you get (by evaluating at $n$) $$ f(n+1)-f(n)=\lambda f(n). $$ That is, $f(n+1)=(\lambda+1)f(n)$. Iterating this you get $$ f(n)=(\lambda+1)^{n-1}f(1). $$ So, for any given $\lambda\in\mathbb R_+$ there is an eigenfunction $f$, given by $$f(n)=(\lambda+1)^{n-1}.$$ The computation also shows that the eigenspace for each $\lambda$ is one-dimensional.