# Is there proof that the exact exchange correlation functional is an analytic function?

Mathematics show that some problems cannot be solved analytically. Could this be true for the exact exchange correlation functional?

Most textbooks state that the form of the exact exchange correlation functional is unknown. It is said to be an unknown mathematical object. Does this mean that said functional is an analytical function? And if so, what is the proof that shows that the functional is analytical?

• What do you mean by analytic? Nov 27 '16 at 3:46
• Then I think the answer is no, since there is derivative discontinuity for the exact functional. Nov 28 '16 at 10:41
• @user26143 Sorry. I meant this one: mathworld.wolfram.com/Analytic.html Not: en.wikipedia.org/wiki/Analytic_function Nov 28 '16 at 13:10
• I guess the answer depends on the scope of "known functions". If we restrict to the elementary functions, most likely it is not (I do NOT know any PROOF for this point). It has been shown finding exact functional is NP-hard problem, e.g. nature.com/nphys/journal/v5/n10/full/nphys1415.html Nov 28 '16 at 14:41