If G is a secure PRG and G' is a function of G, is G' secure? For example: for secure PRG G if I have a G' s.t. G'(k) = G(k)||0, is G' necessarily secure? This question is not for the above example only but for any other possibilities as well. My thinking is that since G is secure (ADV is negligible and PRG is unpredictable), then no matter what I do with G to make G', G' is secure.
Hint: Consider the function $G'(k) = f(G(k))$, where $f(x) = 0$ for all $x$.  Clearly, $G'$ is a function of $G$.  Is $G'$ a secure PRG?