Let $f:[1,\infty)\rightarrow \mathbb{R}$ such that $\frac{f'(x)}{x}$ is bounded on $[1,\infty)$, then $\frac{f(x)}{x}$ is uniformly continuous on $[1,\infty)$
Proof: Since $\frac{f'(x)}{x}$ is bounded on $[1,\infty)$, there exist $M>0$, such that: $\left | \frac{f'(\alpha)}{\alpha} \right |\leq M$, for all $\alpha \in [1,\infty).$ Let $x,y\in [1,\infty)$ sucht that if $(x,y)\subset [1,\infty]$, then by mean value theorem, there exist $\alpha \in (x,y)$ such that:
$\frac{f'(\alpha)}{\alpha}(x-y)=\frac{f(x)}{x}-\frac{f(y)}{y}$, implies: $\left | \frac{f(x)}{x}-\frac{f(y)}{y} \right |=\left | \frac{f'(\alpha)}{\alpha}(x-y) \right |\leq M|x-y|$, thus $\frac{f(x)}{x}$ is a Lipschitz function, implies that is uniformly continuous on $[1,\infty)$.
I would like to know if the proof is ok, likewise, I would like to receive a suggestion on how to weaken the hypothesis.