A non-negative solution $f$ that satisfies $Qf = \alpha f$ is indeed also smaller than $1$ where $Q$ is conservative $Q$ matrix Let $\alpha>0$ and $Q=(q(x,y))_{x,y \in E}$ such that $E$ is a countable state space and $Q$ is conservative $Q$-matrix with transition matrix $P=(p(x,y))_{x,y \in E}$ where $p(x,x)=0$ for any $x \in E$. Further define $c(x) = \lvert q(x,x)\rvert$ such that $p(x,y)c(x)=q(x,y)$ for any $y \neq x$
Show that a non-negative solution $f$ on $E$ that satisfies $Qf = \alpha f$ is indeed also $\leq  1$
Approach thus far:
Let $f\geq 0$ be non-negative solution on $E$, then we have that for any $x\in E$:
$ \sum\limits_{y\in E}q(x,y)f(y)=\alpha f(x)\implies f(x)= \sum\limits_{y\in E\setminus \{x\}}\frac{c(x)}{\alpha+c(x)}f(y)p(x,y)\leq \sum\limits_{y\in E\setminus \{x\}}f(y)p(x,y)$
I believe there must be some kind of contradiction here, if I assume that $f\geq 0$ and that there exists $x^*$ such that $f(x^*)> 1$ but I cannot seem to find it. Any ideas?
Edit: A conservative Q-matrix is one such that all entries are finite, the off-diagonals elements are positive and the row sums are equal to zero, i.e. $\sum\limits_{s\in E}q(r,s)=0$ for any $s \in E$
 A: Since I already did quite a bit to squeeze into a comment, here is the proof for a special case, and some examples to argue why some more assumptions may be needed.
Something much more general holds for rate matrices for CTMCs, namely, their eigenvalues, if they exist, should be all non-positive in a reasonable domain (for instance, rate bounded reversible rate matrices). Does $f$ in the question belong to a reasonable class of functions? (say, bounded functions or at least summable functions?)
Otherwise, consider $q(i, j) = 4^{-(j-i)^+}$ and $q(i, i) = -1/3$, and consider the functions $f(i) = 2^i$. Then, we have that $\alpha = 2/3$ is an eigenvalue, as $\sum q(i, j) f(j) = (-1/3) 2^i + 2^i (1/2 + 1/4 + ...) = (2/3) 2^i$.
Thus, let us assume that the conjectured $f$ is bounded.
You already have the main idea in your attempt. Let $\epsilon > 0$, and let $x^*$ be such that $f(x^*) > \sup f - \epsilon > 0$. For notational convenience, let $-q(x^*, x^*) := \lambda > 0$.
We now have
$ \sum\limits_{y\in E}q(x^*,y)f(y)=\alpha f(x) \implies \sum_{y \in E \setminus \{x^*\}} q(x^*,y) f(y) = (\alpha + \lambda) f(x^*)$
However, the left hand side is a non-negative quantity upper bounded by $\lambda \sup_{x \in E} f(x)$ (which one can observe by bounding each $f(y)$ by the sup, and then using the definition of a  conservative matrix), so we have $\lambda \sup f \geq (\alpha + \lambda) (\sup f - \epsilon)$. Now, simplifying, we have:
$\lambda \epsilon \geq \alpha (\sup f - \epsilon)$
which, if there is a uniform bound on $\lambda$s, gives you a nice contradiction, by picking $\epsilon$ small enough.
If there is no uniform bound on $\lambda$s, similar to the counterexample before, let $-q(i, i) = q(i, i+1) = \alpha_i$ where $\alpha_i$ solves the equation below, and let $f(i) = 2 - 1/i$, which gives
$\alpha_i ((2 - 1/(i+1)) - (2- 1/i)) = \alpha (2 - 1/i)$.
Note that this answer does not use the $p(x,x)= 0$, or anything about $p$, since as written, the question does not seem to really use it, so there could be some slackness there.
