Why if $f$ is uniformly continuous on $[1,\infty )$ then $\frac{|f(x)|}{x}$ is bounded?

Question

Suppose that $f : [1, ∞) → \mathbb{R}$ is uniformly continuous on $[1, ∞)$ .

Prove that there exists $M > 0$ such that $$\frac{|f(x)|}{x}<M \; \; \; \; \forall x\geq 1$$

First, I noticed that $f,x$ both continuous on $[1, ∞)$ and $x\neq 0$ on $[1, ∞)$, so $\frac{f(x)}{x}$ is continuous on $[1, ∞)$.

Then, $\forall \varepsilon > 0, \exists \delta > 0$ s.t "$\forall x$ that satisfies $|x-x_{0}|<\delta $ we have $\left | \frac{f(x)}{x}-\frac{f(x_{0})}{x_{0}} \right |< \varepsilon $."

This implies, $\left | \left |\frac{f(x)}{x} \right |-\left |\frac{f(x_{0})}{x_{0}} \right | \right |= \left |\frac{|f(x)|}{x}-\frac{|f(x_{0})|}{x_{0}} \right |< \varepsilon $

So, $\frac{|f(x)|}{x}<\frac{|f(x_{0})|}{x_{0}}+\varepsilon \leq |f(x_{0})|+\varepsilon $

But, I'm stuck here because it doesn't seem the right way to get the $M$.

Can I get some enlightenment for this problem? (What to note, and where to start)

Answer 1

Fix $\epsilon>0$ and Partition $[1,\infty)$ into a countable union of open intervals $U_i$ of width $\delta$ such that for any $x,y\in U_i$, $|f(x)-f(y)|<\epsilon$. Thus by triangle inequality

$$|f(x)-f(1)|= |f(x)-f(x-\epsilon)+f(x-\epsilon)-f(x-2\epsilon)+...-f(1)|\leq \frac{x}{\epsilon}\epsilon=x.$$

The result follows after rearranging the above and using the fact that $x\geq 1$.

Answer 2

Actually for uniformly continuous funcitons, one has $|f(x)|\leq Ax+B$ for $A,B>0$ and sufficiently large $x>0$, and so $|f(x)|/x$ is bounded.

Look at here.