Let $Y$ and $W$ be two jointly distributed random variables; $Y$ takes values on $(y\_1,y\_2)$ and $W$ takes values on $(w\_1,w\_2)$. The conditional probability density of $W$ given $Y$ is given by $f\_{W|Y}$, assumed to be continuous and twice differentiable. 


Let $X$ be a continuous increasing function of random variable $W$. The expected value of $X$ conditional on $Y=y$ is given by:


$$\tilde{X}(y)\equiv\mathrm{E}\_{W|Y}[X(W)|Y=y]=\int\_{w\_1}^{w\_2} X(w)f\_{W|Y}(w|y)\mathrm{d}w $$


Which condition must be imposed on $f\_{W|Y}$ in order to have $\tilde{X}(y)$ finite and $\tilde{X}^{\prime}(y)>0$ for every $y$?


EDIT: I conjecture that if $f\_{W|Y}$ is such that $g(y)\equiv\mathrm{E}\_{W|Y}[W|Y=y]$ is increasing, then the result follows. But I could not show that formally in terms of fundamental conditions about $f\_{W|Y}$.



Sufficient condition for function of conditional probability density to be increasing
First here, some preliminary remarks: To avoid uninteresting technical complications, let us understand the term "increasing" in the non-strict sense of "non-decreasing". 


Related to this is the remark that the conditions of continuity and differentiability are inessential in this context, because nonsmooth functions can be approximated by smooth ones. 


Let us now proceed in accordance with these remarks. 


Note next that your conjecture is false. Informally, the reason is that you impose the monotonicity condition only on one function of $W$, namely $W$ itself, but then you expect this to imply the monotonicity of the conditional expectation for **all** increasing functions of $W$. Of course, this is not going to happen. Now formally, consider the following simple counterexample. Suppose that the conditional distribution of $W$ given $Y=y>0$ is $N(0,2/y^2)$. Then $E(W|y)=0$, which is constant and hence nondecreasing in $y$. However, $E(e^W|y)=e^{1/y^2}$ is decreasing in $y>0$. If you now change the conditional mean slightly from $0$, you can make $E(W|y)$ strictly increasing in $y$, while still keeping $E(e^W|y)$ decreasing in $y>0$. 


What is necessary and sufficient for $E(h(W)|y)$ to be increasing in $y$ for all increasing functions $h$ is that $E(h\_t(W)|y)=P(W>t|y)$ be increasing in $y$ for all increasing functions $h\_t$ of the special form given by the formula $h\_t(w):=1\_{w>t}$ for all $t\in[w\_1,w\_2]$. This is true because for any increasing (say left-continuous) function $h$ you can write 
$$h(w)=h(w\_1)+\int\_{w\_1}^{w\_2}dh(t)\,1\_{w>t}
=h(w\_1)h\_{w\_1}(w)+\int\_{w\_1}^{w\_2}dh(t)\,h\_t(w)
$$
for $w\in(w\_1,w\_2)$, so that $h$ is a mixture of the functions $h\_t$: 
$$h
=h(w\_1)h\_{w\_1}+\int\_{w\_1}^{w\_2}dh(t)\,h\_t 
$$
with nonnegative "coefficients". 


In turn, the condition that $P(W>t|y)$ be increasing in $y$ for each $t$ (which has just been shown necessary and sufficient for what you want) is known as the stochastic monotonicity (SM) condition, and the [SM is known to follow](https://en.wikipedia.org/wiki/Monotone_likelihood_ratio#First-order_stochastic_dominance) from the monotone likelihood ratio (MLR) condition, which latter means that 
$$\frac{f(w|z)}{f(w|y)}
$$
is increasing in $w$ if $z>y$; that is, 
$$f(w|z)f(v|y)\ge f(w|y)f(v|z)
$$
whenever $z>y$ and $w>v$. 


