Prove that if $\frac{f'(x)}{x}$ is bounded, then $g := \frac{f(x)}{x}$ is uniformly continuous

Question

So, i've been trying to solve this problem.

If $f$ is a function from $[1,\infty)$ to $R$, such that $ \frac{f'(x)}{x}$ is bounded, then $ g(x) := \frac{f(x)}{x}$ is uniformly continuous in $[1,\infty)$. I have tried to adapt the proof of uniform continuity when the derivative of a function is bounded. This is what i have so far:
Let $x,y \in [1, \infty)$,then, by the mean value theorem we have that there is $z$ such that $f'(z) = \frac{f(x)-f(y)}{x-y}$. Then, by the boundness of $ \frac{f'(x)}{x}$, we have that there exists $M$ such that $\frac{|f'(z)|}{z} = \frac{|f(x)-f(y)|}{z |x-y|} < M$. So $|f(x)-f(y)|<M|x-y|z$ . Also we know that $|g(x)-g(y)| \leq |y \cdot f(x)-x \cdot f(y)|$

What i want is that when $|x-y| < \delta$, then $| \frac{f(x)}{x} - \frac{f(y)}{y}|< \varepsilon$ for some $\varepsilon >0$, that i haven't been able to define

Answer 1

Suppose $f'(x)/x\leq B$. Then $|g(x)-g(y)| = |g'(c)||x-y|,$ for $c:=c(x,y)\in(x,y)$.

Next, $|g'(c)|\leq \left|\frac{f'(c)}{c}\right|+\left|\frac{f(c)}{c^2}\right|\leq B+\left|\frac{f(c)-f(1)+f(1)}{c^2}\right|$.

Notice $|f(c)-1|/c^2\leq |f(c)-1|/|c-1|$ for $c\geq 1$. Now use the MVT again and the boundedness condition. Finally use the triangle inequality to conclude that $g$ is Lipschitz for $x,y\geq 1$, which concludes the proof.

Answer 2

Notice that $$g'(x) = \frac{f'(x)}{x} - \frac{f(x)}{x^2}$$

The first term is bounded by hypothesis. As for the second term, we have, for all $x>1$, $$|f(x)-f(1)|<Mz|x-1|<Mx^2$$ Therefore $$\left|\frac{f(x)}{x^2}\right| \le M + \frac{|f(1)|}{x^2} \le M + |f(1)| = N$$ Hence $g'$ is bounded and the result follows.

Answer 3

If $g(x)=\frac{f(x)}{x}$ then there is a $x<c<y$ such that

$g(x)-g(y)=g'(c)(y-x)=\frac{cf'(c)-f(c)}{c^2}(y-x)=\left(\frac{f'(c)}{c}-\frac{f(c)}{c^2}\right)(y-x)$

Since $\frac{f'(x)}{x}$ is bounded, $|f'(x)|<Mx$ for some $M<\infty$ so $\frac{f(c)-f(1)}{c-1}=f'(d)$ for some $1<d<c$, and since $|f'(d)|<Md<Mc,$ we have

$\frac{|f(c)|}{c^2}=\frac{f(1)+f'(d)(c-1)}{c^2}<f(1)+\frac{Mc(1-c)}{c^2}=f(1)+M\left(\frac{c-1}{c}\right)<f(1)+M$, so in fact $g$ is Lipschitz.