If $P \neq NP$ why doesn't this prove the existence of OWF? i know that $P = NP \Rightarrow$ non existence of OWF.
but i don't understand why the claim:  $P \neq NP \Rightarrow$ existence of OWF is wrong?

An intuitive answer would be enough.
an answer from  Goldwasser and Bellare's lecture notes. 
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, worst-case hardness is a poor measure of security. Security requires hardness on most cases or at least average-case hardness. Hence, a necessary condition for the existence of secure encryption schemes is the existence of languages in NP which are hard on the average. Furthermore, P≠ NP is not known to imply the existence of languages in NP which are hard on the average.