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 some number $B$ as desired.

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

Title	uniformly continuous on $ℝ$ is roughly linear
Canonical name	UniformlyContinuousOnmathbbRIsRoughlyLinear
Date of creation	2013-03-22 15:09:41
Last modified on	2013-03-22 15:09:41
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	22
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 26A15