I am having a problem with this exercise. Could someone help?
Suppose a∈(0,1) is a real number which is not of the form 1n for any natural number n
n. Find a function f which is continuous on [0,1] and such that f(0)=f(1) but which does not satisfy f(x)=f(x+a) for any x with x, x+a∈[0,1].
I noticed that this condition is satisfied if and only if f(x)≥f(0)
Thank you in advance
Answer
Look at f(x)=sin(2πx). For which values of a can you find an x∈[0,1] with x+a∈[0,1] and f(x)=f(x+a)? In particular, if you additionally require a>12, can such an a exist at all?
Once you've answered that you've solved your problem for some values of a. Which are those?
A general solution can be found in the answers to Universal Chord Theorem (Link found by the user who asked the question). To quote f(x)=sin2(πxa)−x sin2(πa)
is a solution. This works because f(x)=f(x+a) implies asin2(πa)=0 and thus a=1n for some n∈N. The answers to the linked questions also prove that a≠1n for every n∈N is a necessary condition for a solution to exist.
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