Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous with the following property: for every integer $n$, and every $x,y \in \mathbb{R}$ such that $|x|+|y|>n^2$ and $|x-y|<\frac{1}{n^2}$, we have $|f(x)-f(y)| <\frac{1}{n}$. Show that $f$ is uniformly continuous.

If I choose a set $A=\{(x,y)\in \mathbb{R} \times \mathbb{R}||x|+|y|>n^2$ and $|x-y|<\frac{1}{n^2}$}. Define a function $g(x,y)=\frac{|f(x)-f(y)|}{|x-y|}$. Clearly, $g$ is a continuous function. If $A$ is compact then $g$ is uniformly continuous. How do I show the compactness of $A$?

  • $\begingroup$ $A$ is an open set; it can't be compact. $\endgroup$ – B. Goddard Aug 1 at 13:33

Let $n>1$, be an integer, the restriction of $f$ $[-n^2,n^2]$ is uniformly continuous since it is compact. There exists $c$ such that for $x,y\in [-n^2,n^2]$ such that $|x-y|<c$ implies that $|f(x)-f(y))|<{1\over n}$, let $d=inf(c,{1\over n^2})$ and

$x,y\in\mathbb{R}$, if $x$ is not an element of $[-n^2,n^2]$, $|x|+|y|>n^2$ and $|f(x)-f(y)|<{1\over n}$.

If you take any integers $x,y$ either $x,y\in [-n^2,n^2]$ or either $|x|>n^2$ or $|y|>n^2$.

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