Prove that if $f$ is differential in $(0,\infty)$ and $f'(x)>x$ for every $x>0$, $f$ is not uniformly continuous in $(0,\infty)$.

Question

$f$ is differential in $(0,\infty)$ and $f'(x)>x$ for every $x>0$. I need to prove that $f$ is not uniformly continuous in $(0,\infty)$.

Steps:

1. Let $y>x>0$. Since $f$ is differential in $(0,\infty)$, it is also differential in $[x,y] \subset (0,\infty)$. Since $f$ is differential in the interval, it is also continuous at $[x,y]$.

2. According to the Fundamental theorem of calculus, there exists $c\in(x,y)$ for which $f'(c) = \frac{f(y)-f(x)}{y-x}>c>x \implies f(y)-f(x) > (y-x)\cdot x$

3. Still no success here. I was trying to use the information I got in point 2. I tried proving the negation of the definition of uniform continuity. There exists $\epsilon>0$, in a way that for every $\delta>0$, there exist $y,x$ in a way that $|y-x|<\delta$ also brings $|f(y)-f(x)|>=\epsilon$, but that only got me this far: $|f(y)-f(x)|\geq f(y)-f(x) \geq (y-x)\cdot x$

How can I prove point 3? Or is there a different way to use the information from point 2?

Answer 1

Suppose $f$ is uniformly continuous. There exists $c$ such that $|x-y|<c$ implies that $|f(x)-f(y)|<1$. Take $x$ such that ${xc\over 4}>1$. Let $y>x+c$ and $z$ such that $|y-z|=c/2=$, you have $y,z>x$. There exists $w\in [y,z]$ such that $|f(z)-f(y)|=f'(w)|y-z|\geq w|y-z|=wc/2>xc/2>1$. Contradiction.

Answer 2

Given any $\epsilon>0$ and $\delta>0$, consider for example $x = \dfrac{4\epsilon} \delta$ and $y = x +\dfrac\delta 2$

so $|y-x|= y-x =\dfrac\delta 2\lt \delta$.

You have $f'(z)\gt z \ge x$ for $z \ge x$

and thus $|f(y)-f(x)| = f(y)-f(x) \gt (y-x)x =\dfrac\delta 2 \times \dfrac{4\epsilon} \delta = 2 \epsilon \gt \epsilon$.

So you do not have uniform continuity