If $f$ is a uniformly continous function, then $|f(x)|\leq a|x|+b$

Question

Suppose $f$ is a uniformly continous function. Prove that there exists $a,b\in\mathbb{R}$ such that for any $x$: $$|f(x)|\leq a|x|+b$$

I proved it for a Lipschitz function with constant $k$ and taking $y=0$:

$$|\frac{f(x)-f(0)}{x}|\leq k$$ $$|f(x)|-|f(0)|\leq|f(x)-f(0)\leq k|x|$$ $$|f(x)|\leq k|x|+|f(0)|$$

Therefore, for a Lipschitz function, this follows taking $a = k,b=|f(0)|$, but I just don't find how to get this for a general value using only the definition of uniformly continous function.

Answer 1

Hint: Pick any $\epsilon>0$. Then pick $\delta>0$ according to the definition of unifom continuity. Then what can you say by induction about $f(\pm k\frac\delta2)-f(0)$ with $k\in\mathbb N$? What about the values inbetween?

Answer 2

Suppose that $f(x)$ is defined on $\mathbb{R}$.

$f(x)$ uniformly continuous

$\Rightarrow\forall\epsilon>0,\exists\delta>0,\forall x_1,x_2\in\mathbb{R}:|x_1-x_2|\leq\delta,|f(x_1)-f(x_2)|<\epsilon$, then fix $\epsilon=1$.

$\forall x\in\mathbb{R},\exists k\in\mathbb{N}^+$, such that $\frac{|x|}{\delta}\leq k<\frac{|x|}{\delta}+1$, then $\frac{|x|}{k}\leq\delta$.

Note that $f(x)=\sum_{i=1}^k[f(\frac{i}{k}x)-f(\frac{i-1}{k}x)]+f(0)$, so $$|f(x)|\leq\sum_{i=1}^k|f(\frac{i}{k}x)-f(\frac{i-1}{k}x)|+|f(0)|<k+|f(0)|<\frac{1}{\delta}|x|+1+|f(0)|$$ Set $a=\frac{1}{\delta},b=1+|f(0)|$.