When trying to prove that a recurrence $g(n)$ satisfies $g(n) = O(f(n))$, we sometimes are not able to find a valid $C$ to show the upper bound, so we try to prove it is $O(f(n-d))$ for some constant $d$ instead. 

If we are able to prove that $g(n) = O(f(n-d)$, does this mean that it is also $O(f(n))$? 
If so, does this go the other way too? That is, if $g(n)=O(f(n-d))$ for some $d$, then $g(n) = O(f(n))$ as well? 

edit 1: 
Here a valid $C$ means, $C$ such that $f(n) <= C\*g(n)$ for all $n > n0$ for some $n0$