Let $f(z)$ be a function analytic in a domain containing the segment $[0,1]$ and satisfying $$ f(z+1)=azf(z)+p(z) $$ in that domain, where $a\in\mathbb{R}$ and $p$ is a polynomial. Show that $f$ can be analytically continued to a domain $\{z\in\mathbb{C}\,:\,|\Im z|<\varepsilon\}$ for some $\varepsilon>0$.
I don't quite know how to work with this. My attempt was to derive this equation until the polynomial disappears, but that gives me the equation (supposing $p$ has degree $(n-1)$) $$ f^{(n)}(z+1)=naf^{(n-1)}(z)+azf^{(n)}(z), $$ which I don't find helps much. I can't even think of a way to define this function which would just leave testing analyticity at each integer. Any help is greatly appreciated. Thank you
To begin with, suppose that $f(z)$ has been defined in the square $(-\epsilon,1+\epsilon)\times(\epsilon,\epsilon)$. (By choosing a finite open cover.)
By $f_1(z+1)=azf(z)+p(z)$, we define $f_1(z)$ in $(1-\epsilon,2+\epsilon)\times(\epsilon,\epsilon)$. Let's prove $f_1$ is a direct continuation of $f$. This follows from the fact that they are both analytic and that they agree on $(1-\epsilon,1+\epsilon)\times(-\epsilon,\epsilon)$.
Similarly, we can define $f_2,f_3,\ldots$ and $f_{-1},f_{-2},\ldots$ . Put $f_0=f$.
Finally, define $$F=\bigcup_{n\in\mathbb{Z}}f_n,$$ i.e. $$F(z)=f_n(z),\forall z\in(n-\epsilon,n+1+\epsilon)\times(-\epsilon,\epsilon).$$ It is clear that $F$ is an analytic continuation of $f$ which satisfies the condition.