Does there exist a $f: \Bbb{ R} \to \Bbb{ R}$ which is differentiable,uniformly continuous and lim$_ {x \to \infty} f'(x)=\infty$?

Question

Does there exist a $f: \Bbb{R} \to \Bbb{R}$ which is differentiable, uniformly continuous and $$\lim_ {x \to \infty} f'(x)=\infty\ ?$$

I'm unable to find a counterexample.I know that if derivative is bounded then $f$ is uniformly continuous.

http://math.stackexchange.com/a/352341/4280 seems to come close — Henno Brandsma, Dec 12 '15 at 09:35
I don't think you can, but it should be possible to find a function where $f'(x)$ is unbounded as $x \to \infty$, but always coming back to $0$, and possibly going (very) negative, back and forth. An example might be $e^{-x^2}\cdot \sin(e^{x^6})$. — Arthur, Dec 12 '15 at 09:38
@Arthur The answer I linked to had that kind of function, I think. Not one whose derivative really tends to infinity. — Henno Brandsma, Dec 12 '15 at 09:39

score 3 · Answer 1 · answered Dec 12 '15 at 10:09

3

$\def\rr{\mathbb{R}}$Given any $f : \rr \to \rr$ that is differentiable and uniformly continuous with $f'(x) \to \infty$ as $x \to \infty$:

Let $δ > 0$ such that ( $|f(x)-f(y)| \le 1$ for any $x,y \in \rr$ such that $|x-y| \le δ$ ).

Let $m \in \rr$ such that $f'(x) > \frac{1}{δ}$ for any $x \ge m$.

Then $f(m+δ) - f(m) > δ\frac{1}{δ} = 1$ for any $x \ge m$ [by mean value theorem].

Contradiction.

Therefore no such $f$ as above exists.

answered Dec 12 '15 at 10:09

user21820

Note that this proof applied the uniform continuity condition for only one $ε$, so there cannot even be uniform $ε$-continuity for a single $ε > 0$. – user21820 Dec 12 '15 at 10:13

score 0 · Answer 2 · answered Dec 12 '15 at 09:34

If $|f'(x)|>c$ for all $x\in I$ ($I$ some interval), then $|f(x)-f(y)|>c|x-y|$, so $f$ can't be uniformly continuous.

2 Answers2