Unfortunately this was just word of mouth, and I might be misremembering.
O(n^2 * ln(n)).
Waxing theoretically, there is some integer-valued function f such that optimal matrix multiplication is
Ω(n^2 * f(n)).
We can ask whether f is O(1). I think your statement is equivalent to saying that f is o(n^ε) for any ε > 0.
Yes you're right it would be true. We can see this by just constructing a sequence of algorithms A_i that are implementations of the putative optimal O(n^2 * ln(n)) algorithm except artificially slowed down by ever smaller polynomial overheads. But the hypothesis I'm describing above holds that no such O(n^2 * ln(n)) algorithm exists.
> Waxing theoretically, there is some integer-valued function f such that optimal matrix multiplication is...
The hypothesis I'm describing is that no optimal algorithm exists. Instead, the claim is that there is an infinite sequence of ever better algorithms (specified by ever longer sets of instructions, and with ever larger constant overheads).
> Don Knuth: As you say, I've come to believe that P = N P, namely that there does exist an integer M and an algorithm that will solve every n-bit problem belonging to the class N P in nM elementary steps.
> Some of my reasoning is admittedly naïve: It's hard to believe that P ≠ N P and that so many brilliant people have failed to discover why. On the other hand if you imagine a number M that's finite but incredibly large—like say the number 10 3 discussed in my paper on "coping with finiteness"—then there's a humongous number of possible algorithms that do nM bitwise or addition or shift operations on n given bits, and it's really hard to believe that all of those algorithms fail.
> My main point, however, is that I don't believe that the equality P = N P will turn out to be helpful even if it is proved, because such a proof will almost surely be nonconstructive. Although I think M probably exists, I also think human beings will never know such a value. I even suspect that nobody will even know an upper bound on M.