If $f$ is lipschitz, then $|f(x)| < C \left(1+|x|^λ\right)$

Question

I'm in a first course of analysis and we got this question and I wasn't able to figure it out. Any hint's are welcome.

If $f$ is lipschitz, then $\vert f(x)\vert<C(1+\vert x\vert ^λ)$ for some $C,\lambda>0$

We need to prove or disprove. First I tried some functions with bounded derivatives but all of them worked so no counterexample also my attempts for proof didn't get far.

Thank you.

Answer 1

Assuming $f$ is Lipschitz on the entire real line with a Lipschitz constant not necessarily equal to $\lambda > 0$, examine the ratio $$ {f(x) \over 1 + |x|^{\lambda}}. $$ as $|x| \rightarrow +\infty$. It may be easier to examine first $$ {f(x) - f(0) \over 1 + |x|^{\lambda}}. $$

Answer 2

I suppose you are considering a Lipschitz function $f:\mathbb R \to \mathbb R$. By definition there is a constant $L$ such that $|f(x)-f(y)| \leq L|x-y|$ for all $x,y$. Put $y=0$. We get $|f(x)| \leq |f(x)-f(0)|+|f(0)| \leq L|x|+|f(0)|\leq C(1+|x|)$ if $C=\max \{|f(0)|,L\}$.