A graphical interpretation of the failure of $x^2$ to be uniformly continuous

Question

This question has been bugging me forever.. so let me go ahead and ask it

How to understand graphically why $x \to x^2$ is not uniformly continuous?

I am interested in insights and special ways of looking at the behavior of such functions. Thanks!

Answer 1

Let's first work through the formal definitions and such.

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

Now fix any $\varepsilon>0$ and suppose there was such a $\delta$. Let $x$ be a variable for the moment but define $y=x+\frac{\delta}{2}$. Then clearly $|x-y|<\delta$, thus by assumtion $$|f(x)-f(y)|=|x^2-x^2+\delta x+\frac{\delta^2}{4}|=\delta |x+\frac{\delta}{4}|<\varepsilon.$$ Obviously the latter statement is false since $x$ could be chosen arbitrarily large.

Clearly we have shown that the function is not uniformly continuous. The problem is that we still had a choice in $x$ and this resulted in $f(x)$ lying arbitrarily far away from $f(y)$ even though $y$ was close to $x$. Uniform continuity expresses that if $x$ and $y$ are at a distance less than $\delta$, then the distance between $f(x)$ and $f(y)$ is less than $\varepsilon$. You can visualize this as follows:

Consider an arbitrary interval of the form $(x-\delta,x+\delta)$ in the domain (you can move this around by varying $x$). Uniform continuity says that for any value $y$ that you choose in this interval, we have that $f(y)\in (f(x)-\varepsilon, f(x)+\varepsilon).$ Hence the largest distance in the set $f((x-\delta,x+\delta))$ is less than $2\varepsilon$. In particular, this distance is independent of $x$, however this is clearly false for $x\mapsto x^2$.