Are bounded domain functions periodic

Answer 1

No. You can always "cheat" to prove this by simply giving the function's domain a bound.

Ex: f: [0,1] --> R

I simply defined the function to have a bounded domain from 0 to 1 mapping to the codomain of the set of real numbers. The function itself can be almost anything, periodic or not.

Another way to "cheat" is to simply recognize that all functions having a domain of R are bounded functions, by definition, in the complex plane, C.

(Technically, you would say a non-compact Hermitian symmetric space has a bounded domain in a complex vector space.) Obviously, those functions include non-periodic functions as well.