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Oct 22 at 5:59 answer added D.W. timeline score: 2
Oct 22 at 5:23 comment added D.W. I understand your statement about the $L_p$ norm now, thank you for explaining.
Oct 22 at 3:32 comment added cgmil @D.W. In general it's not true, but in this case it is. The integrand would be raised to the power $p$, but the integrand is only 0 or 1 at every $t$, so you're always computing $0^p$ or $1^p$ which does not depend on $p$ for $p > 1$. Second, if I assume finite discontinuities, then the function can be represented in a computer via just the points of discontinuity. That's what we will have in practice.
Oct 21 at 21:17 comment added D.W. How are the functions specified in the input (in finite length)? It takes infinitely many bits to specify a function on $[0,1]$.
Oct 21 at 21:15 comment added D.W. I don't think it's true that the $L_p$ norm is the $L_1$ norm raised to the power $1/p$. Are you sure about that statement?
Oct 21 at 21:09 history asked cgmil CC BY-SA 4.0