Set-up: Suppose there exist two real valued functions $V_{A}(K_{A})$ and $V_{B}(K_{B})$ with $K_{A},K_{B} \in \mathbb{R}_{\geq 0}$. The optimal $K_{A}, K_{B}$, solve the necessary first-order conditions : $$FOC \ 1A:\hspace{1cm} \frac{\partial V_{A}(K_{A})}{\partial K_{A}} =\int_{K_{A}}^{\infty} (\epsilon-2K_{A})f_{A}(\epsilon)d\epsilon-m=0$$ $$FOC \ 1B:\hspace{1cm} \frac{\partial V_{B}(K_{B})}{\partial K_{B}}=\int_{K_{B}}^{\infty} (\epsilon-2K_{B})f_{B}(\epsilon)d\epsilon-m=0$$ where $f_{A}(\epsilon)$ and $f_{B}(\epsilon)$ are continuous probability density functions over $\mathbb{R}_{\geq 0}$ and $F_{A}(K) \leq F_{B}(K)$ for all $K$ (i.e. $F_{A}$ first order stochastically dominates $F_{B}$). Finally, constant $m \in \mathbb{R}_{\geq 0}$.
Question: Can one tell - without explicitely solving for $K_{A}^{*},K_{B}^{*}$ - whether $K_{A}^{*}$ is always larger/smaller than $K_{B}^{*}$? Stated otherwise, is there a way to sign $K^{*}_{A}-K^{*}_{B}$?
Integrating by parts gives $$ -\int_K^{\infty} (\varepsilon-2K)d(1-F(\varepsilon)) - m = -[(\varepsilon-2K)(1-F(\varepsilon)]_{K}^\infty + \int_K^\infty 1-F(\varepsilon) d \varepsilon-m $$ which equals $$ K(1-F(K)) + \int_K^\infty 1-F(\varepsilon) d \varepsilon - m = 0. $$ Evaluate the FONC for the smaller of the two: $$ K_A(1-F_A(K_A)) + \int_{K_A}^\infty 1-F_A(\varepsilon) d \varepsilon - m = 0. $$ But $F_B\ge F_A$ for all $\varepsilon$, so $1-F_B \le 1-F_A$ for all $\varepsilon$, so that the FONC for B evaluated at the solution for A is: $$ K_A(1-F_B(K_A)) + \int_{K_A}^\infty 1-F_B(\varepsilon)d\varepsilon-m \le 0 $$ If the problem is well posed, the FONC is decreasing in the control and you have only one solution. So if the FONC for $F_B$ is negative at the solution for $F_A$, that means you have gone "too far", and need to reduce your control to get the FONC back to zero (sketch a picture to see what I mean if that's not obvious). That means $K_B^* \le K_A^*$.