# uniformly continuous on $\mathbb{R}$ is roughly linear

###### Theorem 1

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

Proof: By continuity we can choose $\delta>0$ such that $|x-y|\leq\delta$ implies $|f(x)-f(y)|<1$.

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

 $\displaystyle|f(x)|$ $\displaystyle\leq$ $\displaystyle|f(x)-f(a+n\delta)|+\sum_{i=1}^{n}|f(a+i\delta)-f(a+(i-1)\delta)|% +|f(a)|$ (1) $\displaystyle\leq$ $\displaystyle n+1+|f(a)|.$ (2)

Therefore,

 $\displaystyle\frac{|f(x)|}{x}$ $\displaystyle\leq$ $\displaystyle\frac{|f(a)|+n+1}{n\delta}$ (3) $\displaystyle\leq$ $\displaystyle\frac{|f(a)|}{n\delta}+\frac{n+1}{n\delta}.$ (4)

As $n\to\infty$, 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 $f:[0,\infty)\to\mathbb{R}$ if $f$ is differentiable   at 0.

Title uniformly continuous on $\mathbb{R}$ is roughly linear UniformlyContinuousOnmathbbRIsRoughlyLinear 2013-03-22 15:09:41 2013-03-22 15:09:41 Mathprof (13753) Mathprof (13753) 22 Mathprof (13753) Theorem msc 26A15