P vs NP and OWFS It is known (simple HW exercise) that:
If P = NP, that OWFs (one way functions) can not exist.
It is also known that there is a Universal OWF:
  namely, there is a function f:
    s.t. if any OWF exists, then f is a OWF.
  [This is a standard result of concatenating many functions.]
Question:
  Is the following question open:
    (P != NP) => (OWFs exist) ?
[And what is known about this question?]
Thanks 
 A: P != NP does not imply anything about the existence of one-way functions. From Goldwasser and Bellare's "Lecture Notes on Cryptography":

However, the above mentioned necessary condition (e.g.: P != NP) is not a sufficient one. P != NP only implies that the encryption scheme is hard to break in the worst case. It does not rule-out the possibility that the encryption scheme is easy to break in almost all cases. In fact, one can easily construct "encryption schemes" for which the breaking problem is NP-complete and yet there exist an efficient breaking algorithm that succeeds on 99% of the cases. Hence, worse-case hardness is a poor measure of security.

Also, two of Impagliazzo's worlds where P != NP, Heuristica and Pessiland, have no one-way functions while two others, Minicrypt and Cryptomania, do.
A: Complexity theory has a another notion of one-way functions: worst-case one-way functions. 
It is known that $P \ne NP$ if and only if worst-case one-way functions exist.
Worst-case one-way functions exist if and only if there is injective length-increasing polynomial-time computable function that can not be  inverted in  polynomial-time.
Reference:
Alan L. Selman. A survey of one-way functions in complexity theory. Mathematical systems theory, 25(3):203–221, 1992.
