Does space(n) have a complete language? Actually the following was in a complexity cource final exam :
if A is SPACE(n) hard then A is also PSPACE-hard
(this is supposed to be shown by padding...i don't know how exactly)
i think that it is false (if space(n) has a complete language then it is trivially false) because of the proof of TBQF being PSPACE-complete (with some minor modifications, i think that it can be shown that there is a language which is space(n) hard (not nessesarily complete) that it is decided by a O(n^k) for some small value of k). I am not sure though!!