Uniform continuity implies existence of increasing continuous function

Question

In the book that I’m reading, the author makes the following assertion, which I was not able to prove:

If $c:\mathbb R \times \mathbb R\to \mathbb R$ is a continuous function on a compact set (i.e $c$ restricted to a compact set) and hence uniformly continuous, then there exists an increasing continuous function $w:\mathbb R_+ \to \mathbb R_+$, with $w(0)=0$ and such that $$ \mid c(x,y) - c(x’,y’) \mid \leq w(d(x,x’) + d(y,y’)) $$

Can anyone prove that this is in fact true?

P.s: Note that $d$ is a metric.

Answer 1

This is not true in general. Consider $c(x,y) = x^2$ and the points $(x, y), (x', y')$ such that $d(x, x') = d(y, y') =a$ in which case the RHS is 0. But $x^2 - x'^2$ is not always 0.

Answer 2

A friend of mine showed me the solution, so here it is:

I'll sove this for $\mathbb R$ instead of $\mathbb R^2$ for sake of simplicity, because the proof is the same. So, I'll change $c:\mathbb R \times \mathbb R \to \mathbb R$ for $f:\mathbb R\to \mathbb R$, also continuous.

Let $K$ be the compact and $$ w(z) := \sup_{x,y:d(x,y)\leq z} |f(x) - f(y)|$$ Hence, $|f(x) - f(y)|\leq w(d(x,y))$. Also, it's clear that $w$ is an increasing function. We have to prove that $w$ is continuous, and $w(0)=0$.

It's clear that $w(0) = 0$, since $d(x,y) =0 \iff x=y$.

The only thing left to prove is the continuity, which we'll prove for $w$ at 0, and a similar proof can be done to other points.

Note that since $f$ is continuous on a compact set, it is then uniformly continuous. Therefore, $$\forall \epsilon>0, \exists \delta>0, \quad \text{s.t} \quad d(x,y) \leq \delta \implies |f(x) - f(y)| \leq \epsilon$$

Taking the $\sup$ , we obtain that: $$ \sup_{x,y:d(x,y)\leq \delta}|f(x) - f(y)| = w(\delta) \leq \epsilon $$

This means that $d(x,y)\leq \delta \implies w(\delta) \leq \epsilon$, hence, $\lim_{\delta \to 0}w(\delta) = 0$.

Which proves that $w(0) =0$m and that $w$ is continuous on $0$. The only thing left to do is to prove the continuity for the rest of the points, which I could not do actually.