If f is uniformly continuous in R. Then g (x, y) = f (x) -f (y) is uniformly continuous in R^2

Question

If $f$ is uniformly continuous in $\mathbb{R}$, then $g(x,y) = f(x) - f(y)$ is uniformly continuous in $\mathbb{R}^2$. Is this true or false?

I have no idea on how to approach this question, can someone help me?

Answer 1

Let's write out the definitions of what it means for a function $f: \mathbb{R} \rightarrow \mathbb{R}$ and a function $g:\mathbb{R^2} \rightarrow \mathbb{R}$ to be uniformly continuous.

The function $f: \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous if $\forall \varepsilon >0$, $\exists$ $\delta > 0$ such that $\forall x,y \in \mathbb{R}$ satisfying $|x-y| < \delta$ $\Rightarrow$ $|f(x)-f(y)|<\varepsilon$.

Then we also have that

The function $g: \mathbb{R^2} \rightarrow \mathbb{R}$ is uniformly continuous if $\forall \varepsilon > 0$, $\exists \delta > 0$ such that $\forall (x,y), (x',y') \in \mathbb{R^2}$ satisfying $|x-x'| + |y-y'| < \delta$ $\Rightarrow |g(x,y) - g(x',y')| < \varepsilon$

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So we know that $f$ is uniformly continuous, i.e. $\forall \varepsilon >0$, $\exists$ $\delta > 0$ such that

1. $\forall x,x' \in \mathbb{R}$ satisfying $|x-x'| < \delta_1$ $\Rightarrow$ $|f(x)-f(x')|<\frac{\varepsilon}{2}$
2. $\forall y,y' \in \mathbb{R}$ satisfying $|y-y'| < \delta_2$ $\Rightarrow$ $|f(y)-f(y')|<\frac{\varepsilon}{2}$

Now we consider $g(x,y) = f(x) - f(y)$, then $$|g(x,y) - g(x',y')| = |f(x) - f(y) - f(x')+f(y')| \leq |f(x) - f(x')| + |f(y) - f(y')|$$

So simply set $\frac{\delta}{2}:=\max(\delta_1,\delta_2)$, then for $|x-x'| + |y-y'|<\delta$, we have $$|g(x,y) - g(x',y')| = |f(x) - f(y) - f(x')+f(y')| $$$$\leq |f(x) - f(x')| + |f(y) - f(y')| $$$$\leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon$$ as required.

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Edit: with these problems, it is usually a matter of writing out the definitions and using the triangle inequality.