This question considers a special class of problems in (NP,P-samplable). The question is:Do there exists a problem $(L,\mu) \in \mbox{(NP,P-samplable)}$ such that:$L$ is $\rm{NP}$-complete, and $L$ is hard on instances sampled by $\mu$; That is, for any $\rm{BPP}$ algorithm $A$, any polynomial $p(\cdot)$, and any all sufficiently large $n$:   $\Pr_{x \leftarrow \mu(1^n)}[A(1^n,x) = \chi_L(x)] < 1/p(n)$. (Here, $\chi_L(\cdot)$ denotes the characteristic function of $L$.)Your answer can be based on any reasonable assumption, like: $\rm{P} \ne \rm{NP}$;one-way functions (or permutations) exist;factoring is hard;etc.Another question is based on the following quote from On Basing One-Way Functions on NP-Hardness:More than two decades ago, Brassard (Relativized Cryptography) observed that the inverting task associated with a one-way permutation cannot be NP-hard, unless NP = coNP.Is my understanding correct: It is saying that, there does not exist polynomial-time-computable permutations whose inverting is worst-case NP-hard and hard-on-average (unless NP = coNP)?
NP-Complete Hard-on-Average Problems
cc.complexity theory;np hardness;cr.crypto security;one way function;average case complexity
If there exist some NP language L that's hard on the average on some distribution $\mu$, then by applying the standard reduction $f$ from L to 3SAT you get that 3SAT is hard on $f(\mu)$.Your understanding is wrong. By saying that the inverting task  for a function f is NP hard, one means that there is a polynomial-time algorithm A such that A can solve 3SAT on every input given black-box access to an inverter I that inverts f on a non-negligible fraction of the inputs.