Friday, 22 August 2014

real analysis - 1-periodic and continuous is uniformly continuous

Let f be a 1-periodic function that is continuous on R. Then f is uniformly continuous on R.



Indeed, since f is continuous on R, it is continuous in [0,1] which is compact; therefore f is uniformly continuous there, that is, for ε>0 there exists δ(ε)>0 such that for all x,y[0,1] |xy|<δ|f(x)f(y)|<ε. Now, let x,y be two points of R such that |xy|<δ(ε/2). We distinguish two cases:



Case 1: there exists an integer n between x,y. Then without loss of generality, we may assume that $x<1

Case 2: there is no integer between x,y. Then without loss of generality (periodicity) we may assume that x,y(0,1) and we have nothing to prove.



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