lipschitz continuity from a constant c? Let $f : [1,\infty) \to \mathbb R$ be uniformly continuous. Show that there exists a number $C > 0$ such that
$$|f(x)|\le  Cx$$
for all $x \ge 1$.
 A: Hint: You mentioned $y=0$ and $f(y)=0$. You might be thinking of
$$\frac{|f(x)|}{x} = \frac{|f(x) - f(0)|}{x-0} \le C$$
in the case of Lipschitz continuous function. Indeed, it suffices to show 
$$(*)\ \ \ \ \frac{|f(x) - f(1)|}{x-1}\le C$$
for all $x> 1$ (That is, it does not matter what $f(1)$ is). Since 
$$\frac{|f(x)|}{x} \le \frac{|f(1)|}{x}+ \frac{|f(x) - f(1)|}{x}\le |f(1)| +\frac{|f(x) - f(1)|}{x-1} \le |f(1)| +C =: C' $$
(I have used $x\ge 1$ and $x>x-1$ in the second inequality) 
Now in order to show $(*)$, let $\epsilon =1$. Then there is $\delta>0$ so that 
$$|f(x) - f(y)|\le 1$$
whenever $|x-y|\le \delta$. Then try to break $[1, \infty)$ into 
$$[1, 1+\delta],\ [1+\delta, 1+2\delta], \ [1 + 2\delta, 1+3\delta], \cdots$$
note that $|f(x) - f(y)|\le 1$ whenever $x, y$ are in the same subintervals. 
A: Let us define $$\omega (f, \varepsilon ) =\sup\{|f(x) -f(y) |: x,y\geqslant 1 \wedge |x-y|\leqslant \varepsilon\} .$$ 
Now take any $x\in\mathbb{R} , x\geqslant 1$, let $k=\lfloor x\rfloor +1$ and let $l\in \mathbb{N}$ be such that $\frac{lx}{k} \leqslant 1\leqslant \frac{(l+1)x}{k}\ $, then we have 
$$\begin{split}
|f(x)| &= |f(x)-f(1)| +|f(1)| \\
&\leqslant \left|f(x) -f\left(\frac{(k-1)x}{k}\right)\right|+\left|f\left(\frac{(k-1)x}{k}\right) -f\left(\frac{(k-2)x}{k}\right)\right|+\cdots \\
&\ \ \ \ +\left|f\left(\frac{(l+1)x}{k}\right) -f(1)\right| +|f(1)| \\
&\leqslant (k-l)\omega (f, 1) +|f(1)|\\
&\leqslant (x-l+1) \omega (f, 1) +|f(1)| \\
&\leqslant x\omega (f, 1) +(|f(1)| +\omega (f,1)).
\end{split}$$
Thus $$\frac{|f(x)|}{x} \leqslant \omega (f,1) +|f(1)| +\omega (f, 1) =:C$$ for all $x\geqslant 1.$
