Timeline for A min-max approximation
Current License: CC BY-SA 4.0
7 events
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| Oct 2, 2018 at 13:52 | comment | added | Paata Ivanishvili | I have updated the answer. It should not play a role that the infimum is taken with respect to domain $x_{0}<x_{1}$ or $x_{0}\leq x_{1}$ because everything is continuous and if it happens that in the second case the infimum is attained (notice that by compactness it is always attained) for some $x^{*}_{0}$ and $x^{*}_{1}$ such that $x^{*}_{0}=x^{*}_{1}$ then by taking the infimum over the domain $x_{0}<x_{1}$ you can take sequence $x_{0}^{k}, x_{1}^{k}$ such that $x_{0}^{k}<x_{1}^{k}$ and $\lim_{k \to \infty} x_{j}^{k} \to x_{j}^{*}$ for each $j=0,1$ then by continuouty you get the result. | |
| Oct 2, 2018 at 13:47 | history | edited | Paata Ivanishvili | CC BY-SA 4.0 |
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| Oct 2, 2018 at 2:28 | comment | added | Paata Ivanishvili | It does not matter we have $<$ or $\leq$ because of the $\inf$. | |
| Oct 2, 2018 at 2:17 | comment | added | user521337 | you say "beacuse of the symmetry of RHS, we can assume $x_0<x_1<...<x_n$ ..." is that really true or do you mean $x_0\le x_1\le...\le x_n$ ? | |
| Oct 2, 2018 at 2:14 | history | edited | Paata Ivanishvili | CC BY-SA 4.0 |
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| Oct 2, 2018 at 2:12 | comment | added | user521337 | there are typo s ... in your very first equality of the statement you want to prove, you've two $\inf$ s and then so on at each and every step ... | |
| Oct 2, 2018 at 2:06 | history | answered | Paata Ivanishvili | CC BY-SA 4.0 |