# How to prove function $f(x,y)=\frac{1}{xy}$ is not uniformly continuous?

Here I consider uniform continuity of functions in $\mathbb{R}^n$. Take a function of two variables for example.

We said that $f(x,y)$ is uniformly continuous if for any $\epsilon>0$, we can find a $\delta>0$ [depends on $\epsilon$ only] such that $|f(x_1,y_1)-f(x_2,y_2)|<\epsilon$ whenever $|x_1-x_2|+|y_1-y_2|<\delta$. Hence to find a counter example. We need is to show \begin{align*} \exists\epsilon_0>0\,\,\text{such that}\,\,\forall\delta>0,\,\,\exists(x_1,x_2)\in \mathbb{R}^2\,\,\text{with}\,\,|x_1-x_2|<\delta\,\,\text{and}\,\,|f(x_1)-f(x_2)|\ge\epsilon_0 \end{align*}

Now consider $f(x,y)=\frac{1}{xy}$, I was told that it is not uniformly continuous and I have trouble show this. Could anyone help for a solution using the definition I wrote?

• This function is not continuous at the origin $(0,0)$ or on the $x$ or $y$ axis. Thus it clearly isn't continuous over the whole of $\mathbb R^2$ . If it's not continuous, then it is not uniformly continuous. – user860374 Oct 2 '15 at 5:00
• Perhaps he means restricting the domain to $\{(x,y) : xy\ne 0\}$. – Math1000 Oct 2 '15 at 5:01
• @DJS The function is continuous, it's just not defined everywhere in $\Bbb R^2$. – Arthur Oct 2 '15 at 5:02

You have $$f(x,1)-f(2x,1)=\frac{1}{x}-\frac{1}{2x}=\frac{1}{2x}$$ and $$\lim\limits_{x \to 0^+} \frac{1}{2x}=+\infty$$ Hence the function cannot be uniformly continuous.

• Nicely done. ${}$ – Rudy the Reindeer Oct 2 '15 at 5:19
• Yes it is good. But it would have been better if you could answered directly from the definition, I mean the $\epsilon-\delta$ criterion. – math101 Oct 2 '15 at 5:24

I am actually algebraist, but I will see at this problem in following way:

For the sequence $x_1,x_2,x_3,\cdots$ given by $$\Big{(}\frac{1}{10},\frac{1}{10}\Big{)}, \Big{(}\frac{1}{10^2},\frac{1}{10^2}\Big{)}, \Big{(}\frac{1}{10^3},\frac{1}{10^3}\Big{)},\cdots$$ observe two things (better to see through picture)

1) Is the difference $|x_1-x_2|$, $|x_2-x_3|$, $|x_3-x_4|$, $\cdots$ becoming close to $0$?

2) what about the corresponding difference $|f(x_1)-f(x_2)|$, $|f(x_2)-f(x_3)|$, $\cdots$?

Can you see something "bad" is happening?