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

Question

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 1

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.

Answer 2

The function $f(x) -\frac{x^2}{2} $ is increasing. So $$f\left( n+\frac{1}{n} \right)-\frac{1}{2}\left( n+\frac{1}{n} \right)^2 -f(n) +\frac{n^2}{2} >0$$ hence $$f\left( n+\frac{1}{n}\right) -f(n) >1+\frac{1}{2n^2 }$$ hence $f$ is not uniformly continuous.

Answer 3

I like @FisiaiLusia's answer, but I'll offer a proof by contraditcion that works with the basic $\epsilon-\delta$ definition: $\forall \epsilon\ \forall x\ \exists \delta\ \forall y\ \ |y-x|<\delta\implies |f(y)-f(x)|<\epsilon$

Take $\epsilon=1$ and the appropriate $\delta$, then set $x=\frac 2\delta$. Taking $y$ such that,

$$\frac \delta 2<y-x<\delta \implies y>x=\frac 2\delta \implies f'(y) > \frac 2\delta$$

Taking $f(y)=f(x) + \int _x^y f'(t)\ dt \implies f(y)-f(x) > \frac 2\delta(y-x) > 1 = \epsilon$ which is a contradiction.

EDIT

Without using integrals you can use the mean value theorem in the last step: $\frac {f(y)-f(x)}{y-x} = f'(\xi) > \xi > x = \frac 2 \delta \implies f(y)-f(x) > 1$