uniformly continuous on $ℝ$ is roughly linear

Theorem 1

Uniformly continuous functions defined on $\mathrm{[}a\mathrm{,}\mathrm{\infty }\mathrm{)}$ for $a\mathrm{>}\mathrm{0}$ are roughly linear. More precisely, if $f\mathrm{:}\mathrm{[}a\mathrm{,}\mathrm{\infty }\mathrm{)}\mathrm{\to }\mathrm{R}$ then there exists$B$ such that $\mathrm{|}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{|}\mathrm{\le }B\mathit{}x$ for $x\mathrm{\ge }a$.

Proof: By continuity we can choose $\delta >0$ such that $|x-y|\le \delta$ implies .

Let $x\ge a$, choose $n$ to be the smallest positive integer such that$x\le (n+1)\delta$. Then

$$f(x)-f(a)=f(x)-f(a+n\delta )+\sum _{i=1}^{n}f(a+i\delta )-f(a+(i-1)\delta )$$

so that we have

$|f(x)|$$\le$$|f(x)-f(a+n\delta )|+{\displaystyle \sum _{i=1}^n}|f(a+i\delta )-f(a+(i-1)\delta )|+|f(a)|$(1)

$\le$$n+1+|f(a)|.$(2)

Therefore,

${\displaystyle \frac{|f(x)|}{x}}$$\le$${\displaystyle \frac{|f(a)|+n+1}{n\delta }}$(3)

$\le$${\displaystyle \frac{|f(a)|}{n\delta }}+{\displaystyle \frac{n+1}{n\delta }}.$(4)

As $n\to \mathrm{\infty }$, the rhs converges to $\frac{1}{\delta }$.Hence, the sequence defined by$b_n=\frac{|f(a)|}{n\delta }+\frac{n+1}{n\delta }$ is bounded by somenumber $B$ as desired.

Note we can extend this result to $f:[0,\mathrm{\infty })\to ℝ$ if $f$ is differentiable at 0.