Let $f$ a real function defined in $[1,\infty]$ such that $\frac{f'(x)}{x}$ is bounded in the domain of $x$. Show that function $\frac{f(x)}{x}$ is uniformly continuous in $[1,\infty]$. Can you weaken hypotheses? I was thinking to use the next theorem:
Theorem: If $f'$ exist and is bounded in $(-\infty,\infty)$ then $f$ is uniformly continuous in $(-\infty,\infty)$. But I don't know how to apply it to resolve this.
Lets us first establish that the bound on $\frac{f'(x)}{x}$ for $x \geq 1$implies $|f'(x)|\leq Cx$. Then we calculate: $$ |f(x)|=|\int_1^xf'(t)dt+f(1)|\leq C\int_1^x tdt+|f(1)|=\frac{1}{2}Cx^2+K $$ with $K=-\frac{C}{2}+|f(1)|$. Define $p(x)=\frac{f(x)}{x}$. Then: $$ p'(x)=\frac{f'(x)}{x}-\frac{f(x)}{x^2} $$ The first term is bounded by assumption. The second term is bounded for $x \geq 1$ since $$ |\frac{f(x)}{x^2}|\leq \frac{\frac{1}{2}Cx^2+K}{x^2} \leq C_2 <\infty $$ Overall, we have $$ |p'(x)|\leq C+C_2<\infty $$ and you can use your theorem to conclude.