Thursday, 12 December 2013

real analysis - Continuous functions on [0,1] with f(0) = f(1)







Consider the set C of real continuous functions defined on [0,1] such that f(0)=f(1). For each fC, we may consider the set A(f)={a[0,1]:f(x+a)=f(x) for some x[0,1a]}.




Then fCA(f)={1,,1n,,0}



It is easy to show that 1nA(f) for each fC by partitioning [0,1] in n parts of equal length and using intermediate value theorem for the auxiliary function given by g(x)=f(x+1n)f(x). Thus we have fCA(f){1,,1n,,0}.



How to prove the converse?

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