This statement is true or not?

Question

Someone posted a problem as follows

Prove that the following condition is necessary and sufficient for $f$ to be uniformly continuous onan interval $I$: Given $\epsilon>0$, there is $G>0$ such that for every $x_1,x_2\in I$, $x_1\ne x_2$, $$\left|\frac{f(x_1)-f(x_2)}{x_1-x_2}\right|>G \text{ implies }|f(x_1)-f(x_2)|<\epsilon $$

But I think this may be not true. Here is a counterexample: Consider $f(x)=x$. Obviously, $f(x)$ is uniformly continuos over any interval $I$ (even if an infinite interval), but we can not find such $G>0$ that $|f(x_1)-f(x_2)|<\varepsilon$, since $$\left|\frac{f(x_1)-f(x_2)}{x_1-x_2}\right|\equiv1.$$

Answer 1

One direction: Assume $f$ is uniformly continuos, i.e., for every $\epsilon>0$, exists $\delta>0$, such that for all $x_1,x_2\in I$ with $|x_1-x_2|<\delta$, we have $|f(x_1)-f(x_2)|<\epsilon$.

With $\delta$ as above, let $G=\frac{2\epsilon}\delta$. Assume we have $x_1<x_2$ with $\frac{|f(x_2)-f(x_1)|}{x_2-x_1}>G$. Two possibilies arise:

• $x_2-x_1<\delta$. Then we immediately conclude $|f(x-1)-f(x_1)|<\epsilon$.
• Or $x_2-x_1\ge \delta$. Let $n=1+\lfloor\frac{x_2-x_1}\delta\rfloor\ge 2$. Then $(n-1)\delta\le x_2-x_1<n\delta$ and we can partition $[x_1,x_2]$ into $n$ parts of size $<\delta$, i.e., for $0\le i\le n$, let $t_i=x_1+frac in(x_2-x_1)$. Then $|f(t_{i+1})-f(t_i)|<\epsilon$ for all $i$, hence $$G<\frac{|f(x_2)-f(x_1)|}{x_2-x_1}\le \frac{\sum|f(t_{i+1}-f(t_i)|}{x_2-x_1}<\frac{n\epsilon}{(n-1)\delta}<\frac{2\epsilon}\delta,$$ contradiction

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Other direction: Assume $f$ has the property of the problem statement, i.e., given $\epsilon>0$, there is $G$ such that etc. Let $\delta=\frac\epsilon G$. Consider $x_1,x_2$ with $|x_1-x_2|<\delta$. If $x_1=x_2$ or $|f(x_1)-f(x_2)|<\epsilon$, we are done. Hence we may assume $x_1\ne x_2$ and $|f(x_1)-f(x_2)|\ge\epsilon$, and by the given property, $$ \frac{|f(x_2)-f(x_1)|}{|x_2-x_1|}> \frac\epsilon\delta=G,$$ and so $|f(x_1)-f(x_2)|<\epsilon$ ("contradiction", or "as desired")