finite limit implying uniform continuity

Theorem.  If the real function f is continuousMathworldPlanetmath on the interval[0,)  and the limit  limxf(x)  exists as a finite number a, then f is uniformly continuousPlanetmathPlanetmath on that interval.

Proof.  Let  ε>0.  According to the limit condition, there is a positive number M such that

|f(x)-a|<ε2x>M. (1)

The functionMathworldPlanetmath is continuous on the finite interval  [0,M+1];  hence f is also uniformly continuous on this compactPlanetmathPlanetmath interval.  Consequently, there is a positive number  δ<1  such that

|f(x1)-f(x2)|<εx1,x2[0,M+1]with|x1-x2|<δ. (2)

Let x1,x2 be nonnegative numbers and  |x1-x2|<δ.  Then  |x1-x2|<1  and thus both numbers either belong to  [0,M+1]  or are greater than M.  In the latter case, by (1) we have

|f(x1)-f(x2)|=|f(x1)-a+a-f(x2)||f(x1)-a|+|f(x2)-a|<ε2+ε2=ε. (3)

So, one of the conditions (2) and (3) is always in , whence the assertion is true.

Title finite limit implying uniform continuity
Canonical name FiniteLimitImplyingUniformContinuity
Date of creation 2013-03-22 19:00:20
Last modified on 2013-03-22 19:00:20
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 5
Author pahio (2872)
Entry type Theorem
Classification msc 26A15