# Continuity of a parameter-dependent function

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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I don't understand "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". There's one positive and one negative term; one is non-decreasing in $\delta$, the other non-increasing in $\delta$; then why should the maximum be non-decreasing? –  joriki May 2 '12 at 17:46
You are right: there is a typo in the question statement: for any (x,y) $\delta \mapsto f(x,\theta_\delta) - f(y,\theta_\delta)$ is non-decreasing. –  Adam May 2 '12 at 17:53
A minor point: How do you know the maximum exists? Shouldn't it be the supremum? –  joriki May 2 '12 at 18:07
I was thinking it exists because $y$ belongs to a compact set and the function is locally bounded. Am I wrong? –  Adam May 2 '12 at 18:12
Which compact set are you referring to? Doesn't $\mathbb B(x,\delta)$ refer to the open ball about $x$ with radius $\delta$? If it doesn't, you should introduce the notation; I think that's what it usually refers to. –  joriki May 2 '12 at 18:20

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.
Then I would say $\max_{y \in \mathbb{B}(x,\delta)}\{f(x,\theta_\delta)-f(y,\theta_\delta) \} \leq \max_{y \in \mathbb{B}(x,\delta)}\{f(x,\theta_{\bar{\delta}})-f(y,\theta_{\bar{\delta}}) \} < \epsilon$, for any $\bar{\delta} \geq \delta$. Do you think this works? –  Adam May 2 '12 at 19:29
@Adam: I think there was a misunderstanding. My point was that the expression inside the braces doesn't actually depend on $\delta$, so we have no just $\le$ but equality, and this holds for any $\bar\delta$, not just for $\bar{\delta} \geq \delta$. Perhaps you meant $y\in\mathbb B(x,\bar\delta)$ on the right-hand side? –  joriki May 2 '12 at 19:45
No, I didn't mean that. Notice that the proposition was like "[...] there exists $\delta$ such that...". Moreover, I think that the expression inside the brackets does depend on $\delta$. –  Adam May 2 '12 at 20:31
@Adam: If you disagree with my argument that the expression doesn't depend on $\delta$, you should point out where you think it goes wrong. There's little point in asking a question and then just ignoring the main argument in the answer. –  joriki May 2 '12 at 22:00
Sorry, I didn't want to be "not-kind". (1) Do you think my inequalities work? (2) I think that "the expression inside the brackets", i.e. $f(x,\theta_\delta) - f(y,\theta_\delta)$ does depend on $\delta$ because it's a $\theta_\delta$. In fact, if $f(x,\theta_\delta) - f(y,\theta_\delta)$ is not decreasing for decreasing $\delta$, then it can go to $\infty$ for small $\delta$s and the proposition would be false. –  Adam May 2 '12 at 22:44