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?
Best Answer
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.