uniformly continuous on is roughly linear

Theorem 1

Uniformly continuous functions defined on [a,) for a>0 are roughly linear. More precisely, if f:[a,)R then there exists B such that |f(x)|Bx for xa.

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

Let xa, choose n to be the smallest positive integer such that x(n+1)δ. Then


so that we have

|f(x)| |f(x)-f(a+nδ)|+i=1n|f(a+iδ)-f(a+(i-1)δ)|+|f(a)| (1)
n+1+|f(a)|. (2)


|f(x)|x |f(a)|+n+1nδ (3)
|f(a)|nδ+n+1nδ. (4)

As n, the rhs converges to 1δ. Hence, the sequence defined by bn=|f(a)|nδ+n+1nδ is boundedPlanetmathPlanetmathPlanetmath by some number B as desired.

Note we can extend this result to f:[0,) if f is differentiableMathworldPlanetmathPlanetmath at 0.

Title uniformly continuous on 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