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I can't remember now and never gave measure theory a proper study, but isn't there a happy result that you can quotient out by a space of functions with zero in
by ndriscoll 3y ago
I can't remember now and never gave measure theory a proper study, but isn't there a happy result that you can quotient out by a space of functions with zero integral to get an almost everywhere differentiable function or something? Maybe that's for L2:continuous, not continuous:differentiable?
Even a result like f(x) = nice(x) + evil(x) with |evil| < epsilon should be "happy enough" right?
Edit: I may have been misremembering the Lebesgue decomposition theorem which is not quite so nice, as the singular part doesn't just go away.
- yorwba 3y agoFor a "happy enough" approximation to the Weierstrass function, you can cut off the series once a^n/(1 - a) < epsilon and get a function that's differentiable everywhere. But it'll have an extremely large number of local optima, which is still bad for gradient descent.