Proving that prBPP = prP is as hard as proving that “almost NP” is not contained in P/poly

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What circuit lower bounds are necessary in order to prove that promise-BPP=promise-P? We show that the recent breakthrough result of Murray and Williams (STOC 2018) can be used to show a dramatic strengthening of the previously-known answer to this question. Specifically, we show that if promise-BPP=promise-P, then NTIME[nf(n)]⊈P/poly, for essentially any f(n)=ω(1).We also prove a technical strengthening of this result. Specifically, we show that if promise-BPP=promise-P, then for essentially any s:N→N it holds that NTIME[sO(1)∘sO(1)]⊈SIZE[s]. Moreover, we show that size-s circuits fail to compute the “hard” function in any interval of length poly(s(poly(n))). The proof of this result uses tools of Murray and Williams, but relies on a different proof strategy. (Their proof strategy yields three compositions of s instead of two, and does not yield the guarantee of failure in any small interval.)