Consider a function $f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}_{\geq 0}$ continuous in the $1^{st}$ argument, locally bounded in the $2^{nd}$ one.
We consider a $\delta$-neighborhood of $x \in \mathbb{R}^n$, i.e. $\mathbb{B}(x,\delta)$ (closed ball with center $x$ and radius $\delta$), and we see the $2^{nd}$ argument as a $\delta$-dependent parameter $\theta_\delta \in \mathbb{R}^m$. We study
$ \max_{y \in \mathbb{B}(x,\delta)} \{ f(x,\theta_\delta)-f(y,\theta_\delta) \} $
We assume that for any $x \in \mathbb{R}^n$ we have $ \delta \mapsto f(x,\theta_\delta)-f(y,\theta_\delta) $ non-decreasing if $y \in \mathbb{B}(x,\delta)$. Therefore, for any $x \in \mathbb{R}^n$, also $\delta \mapsto \max_{y \in \mathbb{B}(x,\delta)} \{ f(x,\theta_\delta)-f(y,\theta_\delta) \}$ is non-decreasing.
Say if the following proposition is true (if not, find a counterexample).
$ \forall \epsilon > 0 \ \exists \delta >0 \text{ such that } \max_{y \in \mathbb{B}(x,\delta)} \{ f(x,\theta_\delta)-f(y,\theta_\delta) \} < \epsilon $
Note: the proposition would hold for fixed $\delta$-independent $\theta$ as $x \mapsto f(x,\theta)$ is continuous. I suspect the proposition is true also for $\theta_\delta$ because the smaller $\delta$ is the smaller $\max_{y \in \mathbb{B}(x,\delta)} \{ f(x,\theta_\delta)-f(y,\theta_\delta) \}$ is.