Suppose $f(x)$ is defined for all $x$; then $f \circ \cos x$ is periodic and $2 \pi$ is a period. Conversely, if $g(x)$ is defined for all $x$ and is periodic, with $2 \pi$ as a period, can one find a function $f$ such that $g(x) = f \circ \cos x$?
Intuitively, I think that the answer is no, but would the following counterexample be valid: Because the domain of $f \circ \cos x$ is the range of $\cos x$, i.e. [-1,1], then the composition cannot be defined for all $x$ due to the domain?
If this one is not valid, is there a certain property of composition of functions that I am missing or not applying? Also, how could the hypothesis of the problem statement be rewritten so that it becomes true? Thanks in advance, anyone's help is greatly appreciated.
I am using the textbook Introduction to Analysis by Arthur Mattuck.
The last sentence in the second paragraph is mistaken: assuming $f$ has domain including $[-1,1]$, $f\circ \cos x$ has domain all of $\Bbb R$, so that is not a problem.
The statement that because $\cos x$ is even, $f \circ \cos x$ will be even and therefore we cannot represent all $2\pi $ periodic functions that way is correct, but symptomatic of something more basic: there exist $a,b\in [0, 2\pi )$ with $a \neq b, \cos a = \cos b$. This means that $f \circ \cos a = f \circ \cos b$ and only functions $g$ that respect that equality can be represented. Evenness just means we have lots of pairs $a,b$ like this. Given any bijection $h: [0,2\pi) \to $ some set $D$ we can extend $h$ to all of $\Bbb R$ by periodicity and find an $f: D \to \Bbb R$ so that $g(x)=f \circ h(x)$ by $f(y \in D)=g(h^{-1}(x))$