Uniform continuity of functions of several arguments I came across the following phrase in my reading, and I would like to know what it exactly means.

Let $f:\Omega \times \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}$ such that $f(x,u,\xi)$ is continuous, uniformly in $u$ with respect to $\xi$. 

What does it mean by "uniformly in $u$ with respect to $\xi$?
 A: In the definition of continuity at a point $(x,u,\xi)$you begin with:

Given $\epsilon>0$ there is a $\delta>0$ such that...

In general $\delta$ will depend on $\epsilon$, $u$, $x$ and $\xi$. Uniformly with respect to $\xi$ means that you can choose $\delta$ independent of $\xi$.
A: A uniform continuity for a $g:\mathbb{R}\to\mathbb{R}$ would mean that:
 For each $\epsilon>0$ there is a $\delta>0$ such that  $|g(x)-g(y)|<\delta$ for each $|x-y|<\epsilon$. The point here is that $\delta$ is independent of specific points $x,y$ which is not the case for "regular" continuity.
If you change g with $G:\mathbb{R}^n\to\mathbb{R}$, then you have to change the absolute value $|.|$ with some distance function $D$ in $\mathbb{R}^n$. The rest remains pretty much the same:
For each $\epsilon>0$ there is a $\delta>0$ such that  $|G(x)-G(y)|<\delta$ for each $D(x,y)<\epsilon$. Note that the absolute value is still used for G(x)-G(y), because this one remains a number, not a vector like their arguments($x\in \mathbb{R}^n$ means $x=(x_0,x_1,...x_n)$).
Now for a function $f:\Omega \times \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}$ to be uniformly continuous with respect to third parameter (which is vector) this is the same as $G_{x,u}(
\xi) = f(x,u,\xi)$, i.e. for any fixed $x,u$ the function of one variable $G_{x,u}(
\xi)$ is uniformly continuous.
Finally, $f(x,u,\xi)$  uniformly in $u$ with respect to $ξ$, probably means that $u$ is also a function, i.e. $f(x,u,\xi) = f(x,u(\xi),\xi)$, i.e. 
For each $\epsilon>0$ there is a $\delta>0$ such that  $|f(x,u_1(\xi),\xi)-f(x,u_2(\xi),\xi)|<\delta$ for each $|u_1(\xi)-u_2(\xi)|<\epsilon$.
