prove that $|f(x)|\le k|x|$ for some constant $k$

Question

If $f:\mathbb R \to \mathbb R$ is uniformly continuous and $f(0)=0$ then prove that $|f(x)|\le k|x|$ for some constant $k$

Answer 1

To complete the question and clear things I will add an answer.

The statement of the question is not true in general for it domain, the counterexample is any function like $f(x)=x^{1/n}, n>1$ that are uniformly continuous in it domain but where

$$\left|\frac{x^{1/n}}{x}\right|=\frac{1}{\left|x^{\frac{n-1}{n}}\right|}$$

is unbounded in it domain.

But it is true that for any uniformly continuous function we have that

$$\left|\frac{f(x)}{x}\right|\le K,\quad\forall x\in [C,+\infty)\text{ or }(-\infty,C]$$

(the answer to this last statement is in the duplicated question). The geometric intuition is that for some uniformly continuous function exist lines of the kind $g(x)=Kx$ where the image of the function is under these lines at some point $C$.