How can you show this convergence for the supremum? Fix $\delta>0$ and consider the function $\Phi_\delta \colon \mathbb{R}^N \times \mathbb{R}^N \to \mathbb{R}$ defined by
$$\Phi_\delta(x,y)=w(x)-v(y)-\delta |x|^2-\delta |y|^2$$
where $w,v \in \mathcal{C}(\mathbb{R}^N)$ are both with sublinear growth.
then $\Phi_\delta$ is coercive so that we can find $R>0$ such that
$$\Phi_\delta(\bar x_\delta,\bar y_\delta)=\sup_{| x|, | y| < R} \Phi_\delta (x,y)=\sup_{x,y \in \mathbb{R}^N} \Phi_\delta (x,y)$$
How do you show rigorously that
$$\lim_{\delta \to 0} \sup_{x,y \in \mathbb{R}^N} \Phi_\delta (x,y) =\lim_{R \to \infty} \sup \left \{ [w(x)-v(y)] \colon x,y \in B_R  \right \} $$
This seems intuitive but how can you show it?
 A: Both limits in the equation exist because they are the limits of monotonic functions in $\delta,R$ respectively (note that $\Phi_{\delta} \geq \Phi_{\delta'}$ for $\delta<\delta'$).

We note that $$
w(x)-v(y) \geq \Phi_\delta(x,y) \quad \forall x,y \in \mathbb R^N , \delta>0
$$
Therefore, for any $R>0$, we have $$
\sup_{|x|,|y|<R} w(x)-v(y) \geq \sup_{|x|,|y|<R} \Phi_{\delta}(x,y) \quad \forall \delta>0,R>0
$$
Taking $R \to \infty$,$$
\lim_{R \to \infty} \sup_{|x|,|y|<R} w(x)-v(y) \geq  \lim_{R \to \infty} \sup_{|x|,|y|<R} \Phi_{\delta}(x,y) = \sup_{x,y \in \Bbb R^N} \Phi_{\delta}(x,y) \quad \forall \delta>0
$$
Since the LHS is independent of $\delta$, letting $\delta \to 0$ gives $$
\lim_{R \to \infty} \sup_{|x|,|y|<R} w(x)-v(y) \geq \lim_{\delta \to 0} \sup_{x,y \in \Bbb R^N} \Phi_{\delta}(x,y)
$$
which is one side of the inequality.

For the other, let $\epsilon>0,R>0$. For $\delta' = \frac{\epsilon}{2R^2}$ we know that $$
\delta(|x|^2+|y|^2) < 2 \delta R^2 < \epsilon \quad \forall |x|,|y|<R, \delta<\delta'
$$
In particular, it follows that $$
\Phi_{\delta}(x,y) \geq w(x)-v(y) - \epsilon \quad \forall |x|,|y|<R, \delta<\delta'
$$
and therefore $$
\sup_{x,y \in \mathbb R^N} \Phi_{\delta}(x,y) \geq \sup_{|x|,|y|<R} \Phi_{\delta}(x,y) \geq \left(\sup_{|x|,|y|<R} w(x)-v(y)\right) - \epsilon \quad \forall \delta < \delta'
$$
(The first inequality is obvious, the second follows from the previous line ,a supremum over the same set, and $\epsilon$ can come out of the supremum) The RHS is independent of $\delta$ and the equation holds for all $\delta$ close enough to $0$. Letting $\delta \to 0$ on the LHS gives $$
\lim_{\delta \to 0}\sup_{x,y \in \Bbb R^N} \Phi_{\delta}(x,y) \geq \left(\sup_{|x|,|y|<R} w(x)-v(y) \right)- \epsilon 
$$
Now, this inequality is true for all $R$ and the LHS is independent of $R$, so we can let $R \to \infty$ and get $$
\lim_{\delta \to 0}\sup_{x,y \in \Bbb R^N} \Phi_{\delta}(x,y) \geq \left(\lim_{R \to \infty} \sup_{|x|,|y|<R} w(x)-v(y)\right) - \epsilon 
$$
But this is true for all $\epsilon>0$! As a consequence, $$
\lim_{\delta \to 0}\sup_{x,y \in \Bbb R^N} \Phi_{\delta}(x,y) \geq \lim_{R \to \infty} \sup_{|x|,|y|<R} w(x)-v(y)
$$
which is the other direction, and was to be proved.

The main takeaway here from the proof is that the $\delta$ and $R$ can both can go to $0$ and $\infty$ at totally different rates, perhaps even independently, as long as it's compatible for this to occur. In this case, we observed that the difference between the two functions can be made small if $\delta R^2$ is small, and this proved the result.
