Showing existence of a Nash equilibrium in pure strategy
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.