Showing existence of a Nash equilibrium in pure strategy Consider a game with $N$ players, each indexed by $i=1,...,N$. Every player $i$ has to choose a $J\times 1$ vector of actions $a_i\equiv (a_{i,1},...,a_{i,J})$ where each $a_{i,j}$ can be zero or one. For each player $i$, let $a_{-i}$ denote the actions of the other players.
The payoff of each player $i$ is
$$
u_i(a_i, a_{-i}; \theta)+v_i(a_i; \delta)
$$
where $u_i$ and $v_i$ are some parametric functions of the parameters $\theta$ and $\delta$. Moreover $v_i$ is monotone decreasing in $\sum_{j=1}^J a_{i,j}$.
A pure strategy Nash equilibrium (PSNE) of the game is $a^*\equiv (a_1^*,...,a_N^*)$ solving
$$
 a_i^*\in \arg\max_{a_i\in \{0,1\}^J} u_i(a_i, a^*_{-i}; \theta)+v_i(a_i; \delta) \quad \forall i=1,...,N
$$
Question: Suppose that I'm able to show that a PSNE exists for $\theta=\theta_0$ and $\delta=\delta_0$ where $\theta_0$ and $\delta_0$ are some specific real values of the parameters $\theta$ and $\delta$. Can I conclude that a PSNE exists for $\theta=\theta_0$ and any value of $\delta$?
In particular, here, I'm wondering whether the claim may follow from the fact that $v_i(a_i; \delta)$ enters additively, does not depend on $a_{-i}$, and it is monotone.
No. Consider the matching pennies game
$$ \begin{array}{c |c}
&amp; H &amp; T  &amp; \\ \hline
H &amp; (1,-1) &amp; (-1,1)\\ 
T &amp; (-1,1) &amp; (1,-1)\\
\end{array} $$
which we know no PSNE exists.
Define the function
$v_i(\cdot,\cdot)$ by
$$
v_i(H,0) = 5 \\
v_i(x,y) = 0 \quad \forall (x,y) \not = (H,0)
$$
At $\delta = 0$, we have the game
$$ \begin{array}{c |c}
&amp; H &amp; T  &amp; \\ \hline
H &amp; (5,4) &amp; (-4,1)\\ 
T &amp; (-1,5) &amp; (1,-1)\\
\end{array} $$
which has a PSNE of $(H,H)$.
For any other value of $\delta$, we're back in the original game so no PSNE exists.
Redefine $H = 0, T = 1$ and you have your monotone decreasing requirement.