How to show a function is continuous but not uniformly continuous.

Question

I'm given a function $f:\Bbb R^2 \mapsto \Bbb R; f(x,y) = sinx + x^3y^2.$ Show that $f$ is continuous, but not uniformly continuous.

I'm using the text Undergraduate Topology by Kasriel, which defines continuous mappings as satisfying the following conditions at a point $x_0 \in X$ where $(X,d)$ and $(Y, \rho)$ are metric spaces:

For each $\epsilon > 0,\exists \ \delta > 0 $ such that if $x \in X$ and $d(x, x_0) < \delta $, then $ \rho (f(x),f(x_0)) < \epsilon$

Uniform continuity is defined as:

Let $f:(X,d) \mapsto (Y,\rho)$ be a mapping. $f$ is said to be uniformly continuous on $X$ provided that for each $\epsilon > 0$, there is a $\delta > 0$ such that if $x_1, x_2 \in X$, and $d(x_1, x_2) < \delta$, then $\rho(f(x_1), f(x_2)) < \epsilon$.

As usual, I'm having difficulty understanding how to apply the definitions to my intuition. Can anyone give me advice on how to begin the proof?

Answer 1

You hopefully have a pretty good intuition for continuous functions: their graphs don't have any tears or holes in them. Uniform continuity is somewhat more subtle, but still manageable: while arbitrary continuous functions can have graphs which are stretched out arbitrarily, uniformly continuous functions have a bounded quantity of stretching between any two points sufficiently close together. This is perhaps too vague to be useful, but it's what's in my head; more practically, you might try investigating which of the single-variable functions $\sin x,\sin(1/x),1/x,$ and $\sqrt{x}$ are uniformly continuous on $(0,\infty)$.

With respect to your particular problem, I'll assume you can show $f$ is continuous-this follows from standard theorems on sums and product of continuous functions. So the challenge is to show it's not uniformly continuous: formally, that there's some $\epsilon$ such that no $\delta$ guarantees that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$, and informally that there's some amount of stretching that occurs between points arbitrarily close together.

Usually, any $\epsilon$ will work for these things. So try $\epsilon=1$: we have to find for every $\delta$ an $x$ and a $y$ within $\delta$ of each other but with $|f(x)-f(y)|>1$. The latter is $|\sin x_1+x_1^3x_2^2-\sin y_1-y_1^3y_2^2|$. Now we just throw away as much of the complexity in this expression as possible. It will be bigger than $1$ if, for instance, $x_1^3x_2^2-y_1^3y_2^2>3$, using the bound $|\sin|\leq 1$. This will, in turn, be true if $x_2=y_2=1$ (just random simplification again) and $x_1^3-y_1^3>3$. This finally factors as $(x_1-y_1)(x_1^2+x_1y_1+y_1^2)$, and so we've reduced showing $f$ is not uniformly continuous to finding reals $x_1>y_1$ with $x_1-y_1<\delta$ but $(x_1^2+x_1y_1+y_1^2)$ sufficiently big. But if $x_1$ and $y_1$ are both positive and close together, this is around $3x_1^2$, which is certainly unbounded! Hopefully you can now fill in the last details and have an idea of the game played around this type of problem.