real analysis - Let $f$ be a differentiable function and for all $x$ $f'(x)>x$, prove $f$ isn't uniformly continuous

Suppose $f:(0,\infty)\to \mathbb R$ is differentiable and $f'(x)>x$.

Prove that $f$ isn't uniformly continuous in $(0,\infty)$.

Hint, prove first that for all $y>x>0$ we have $f(y)-f(x)\ge (y-x)x$.

From the first line I understand the function is always increasing and never has an extremum.

The hint looks like it is directly from Lagrange's MVT, but the definition doesn't really work with $\mathbb R$ and $\infty$, and using the definition of a derivative with $\displaystyle\lim_{y\to x}\frac {f(y)-f(x)}{y-x}=f'(x)$ I don't see how to loose the limit.

The next step would probably be to show the function isn't Lipschitz, since a uniformly continuous function pass Lipschitz definition: $f(y)-f(x)\le M(y-x)$ for some $M\ge 0$ but here it will never hold since for all $x$ we have $f(y)-f(x)\ge x(y-x)$

Answer

Proof of the hint

Let $y>x>0$.

By the mean value theorem, there is some $\beta\in (x,y)$ such that $\displaystyle \frac{f(y)-f(x)}{y-x}=f'(\beta)$

By assumption, $f'(\beta)>\beta>x$

Hence $\displaystyle \frac{f(y)-f(x)}{y-x}>x$

Proof of the claim

Suppose that for $\epsilon=1$, there is some $\delta$ such that $|x-y|\leq \delta\implies |f(x)-f(y)|\leq 1$

Choose $N$ an integer such that $n\geq N \implies \frac{1}n\leq \delta$

For $n\geq N$, consider $x_n=n^2$ and $y_n=n^2+\frac1n$

By the previous lemma, and since $|x_n-y_n|\leq \delta$, we have $|y_n-x_n|x_n\leq 1$

That is $n\leq 1$

Contradiction.