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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.

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    Your latest edit doesn't make sense. a) In "for any $(x,y) \in \mathbb{R}^n \times \mathbb{B}(x,\delta)$", $x$ can't be used on the right-hand side of an expression defining the possible values of $x$. The formulation in your comment under my question doesn't have this problem. b) Now that $\delta$ is used in the definition of $x$ and $y$, what does it mean to then say that $\delta \mapsto f(x,\theta_\delta)-f(y,\theta_\delta)$ is non-decreasing? It would have to be non-decreasing for fixed $x$ and $y$, but you now made those dependent on $\delta$.2012-05-02

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This is true, but perhaps more trivially so than you wanted it to be. Since $\delta \mapsto f(x,\theta_\delta)-f(y,\theta_\delta)$ is non-decreasing for every $(x,y) \in \mathbb{R}^n \times \mathbb{R}^n$, so is $\delta \mapsto f(y,\theta_\delta)-f(x,\theta_\delta)$, and thus it is in fact constant (with respect to $\delta$). Thus we can evaluate it at any fixed value of $\delta$, and as you already remarked, continuity in the first argument then implies the proposition.

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    Now I see.. I meant for any $x \in \mathbb{R}^n$ and $y \in \mathbb{B}(x,\delta)$2012-05-02