Suppose the following maximization:
$$v_t = \arg \max_{v} E_{v_{t-1}} 1\{S(x) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}\ln f(x;v) = max_{v} E_{v_{t-1}} 1\{S(x) \geq \gamma\} W(u;v_{t-1}) \ln f(x;v),$$
where $f(\cdot; v)$ is some parametric family and $W(u;v_{t-1})$ is the likelihood ratio, $S(x)$ is an arbitrary function.
I want to prove that,
$$E_{v_{t-1}} 1\{S(x) \geq \gamma\} \ln \frac{f(x;v_{t})}{f(x;v_{t-1})} \geq 0.$$
I tried the following:
Due to maximization:
$$E_{v_{t-1}} 1\{S(x) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}\ln f(x;v_{t}) \geq E_{v_{t-1}} 1\{S(x) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}\ln f(x;v_{t-1}).$$
Thus,
$$E_{v_{t-1}} 1\{S(x) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}\ln f(x;v_t) - E_{v_{t-1}} 1\{S(X) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}\ln f(x;v_{t-1}) \geq 0 \implies $$
$$\sum_{x}1\{S(x) \geq \gamma\} \frac{f(x;u)}{f(x;v_{t-1})}(f(x;v_{t-1})\ln\frac{ f(x;v_t)}{ f(x;v_{t-1})}) \geq 0.$$
I need to get rid of the $\frac{f(x;u)}{f(x;v_{t-1})}$ and the proof is complete. But keep in mind that $\frac{f(x;u)}{f(x;v_{t-1})} \; \neq \; \textrm{constant}$.