Proving the periodicity of function

Question

 

I just wanted  a verification the proof i have written

Problem statement: If f(x) is periodic with period T, then f(ax+b) is periodic with period T/a, a>0

My procedure:

1) f(x)=f(x+T); (since f(x) is periodic with period T)

2) Replacing x by ax+b, f(ax+b)= f(ax+b+T)(since f(x) is periodic with T and this means T being the smallest no. for the repetition of value f(x) ) = f(a[x+T/a]+b)

3) 2) shows that x+T/a is the point of repetition of value f(ax+b) for any x belonging to domain and could be the period of f(ax+b),for T/a really to be the period , it has to be the smallest no. for repetition of f(ax+b)

4) Let us assume that a period for f(ax+b) exists T1(say) and ∗∗T1<(T/a), then f(a(x+T1)+b)=f(ax+b)=f(ax+aT1+b) (we assumed that T1 is the period)**

5) from 2) f(ax+b)= f(ax+b+T) = f(ax+aT1+b)(from 4) )= f(ax+(aT1/T)T+b), but f(ax+b)=f(ax+b+T) shows that T is the period of f(x)(see prob. statement and 1) ) thus in f(ax+(aT1/T)T+b) , since T is the period of f(x), (aT1/T)≥1 or T1≥(T/a), thus contradicting our assumption T1 as a period and thus T/a is the period of f(ax+b).

Did I do correctly?

Answer 1

The proof is incorrect. Having T as a period of f does not imply T is the smallest period. For example, the sine function is 4pi - periodic, but 4pi is not the smallest period of sine; rather, 2pi is the smallest period.

 

To show that the function g(x) := f(ax + b) has period T/a, you must show g(x + T/a) = g(x) for all x, i.e., f(a(x + T/a) + b) = f(ax + b) for all x. We have

 

f(a(x + T/a) + b) = f(ax + T + b) = f((ax + b) + T) = f(ax + b) for all x;

 

the last equality follows from the fact that f has period T.