User blog comment:Alemagno12/An extremely fast-growing OCF/@comment-5029411-20170726203426/@comment-5529393-20170727055319

Oh, I see. Well, I do see where your second statement can follow from your first, since if $$f(a) = \Omega_a$$ then $$f^{\omega}(0) = \psi(\psi_I(0))$$. So the difficult part is that first part. It's clear that the "natural" extension of $$f(n) = \Omega_n$$ for finite n is obviously $$f(a) = \Omega_a$$ for arbitrary ordinals a, but that's not going to work for arbitrary ordinal functions. This is something I was trying to explain to SimplicityAboveAll, without success; if you already have an ordinal notation in place that stretches far enough, than generally one can define "natural extensions" one at a time in an ad hoc fashion, but only because the existing ordinal notation makes such extensions natural. It's not something that you can define in general.