The question given in my Real Analysis textbook is as follows:
Let there be f: $\mathbb{R}\to\mathbb{R}$ a periodic function with period T and a $\in$ $\mathbb{R^*}$.
Prove the function $x → f (ax)$ is periodic and determine its period. Lastly, prove that the period is the answer previously obtained.
My take on this would be as follows:
By the definition given by my textbook, a function is periodic if $f(x)=f(x+T)$ and we already know that f is periodic.
Lets note $ f(ax) = f_a(x)$
Let's firstly consider $a>0$, for all $x \in \mathbb{R}$ we have: $f_a(x + (T/a)) = f(ax+T)= f(ax)=f_a(x)$
Then we consider the same thing for $a<0$
$f_a(x+(T/|a|)) = f(ax-T)= f(ax-T+T)=f(ax)=f_a(x)$
So for all $a\neq 0$, $f_a(x+(T/|a|)) = f_a(x)$
// which shows that the function is indeed periodic so we have proved the first part of the question. I hope this proof is correct :')
So the period is $T/|a|$
Now we have to prove that the period is indeed $T/|a|$
I have some ideas for this but in not sure how to properly express them mathematically. So by definition we know that the period of a periodic function has to be the smallest $T>0$.
So maybe here we can prove by contrapositive? so assume that there is another period $T^\prime$ which is greater than zero but smaller than $T/|a|$ such that $f_a(x+T^\prime)=f_a(x)$ Now we need to show why this would be impossible but I'm not quite sure how to continue. So how would I finish this proof?
Thank you in advance for all your help!