antonio
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 Sep24 awarded Autobiographer May5 awarded Critic May3 revised First derivative test and uniqueness of local extrema Added "given the assumptions for f", kinda implicit before May3 comment First derivative test and uniqueness of local extrema "rework these lines to prove [...]", without changing the assumptions, i.e. "Let $f(x)$ be differentiable everywhere and have a minimum at $x^∗$.", but I will make it more explicit. May3 comment First derivative test and uniqueness of local extrema I don't get you. I wrote: "[...] to get a minimum, it is necessary that $f'(x^*)=0$ [...]", not sufficient. As for the uniqueness of the local extremum, the assumption is that we do have a minimum (or a maximum), but we don't know if it is unique. May2 revised First derivative test and uniqueness of local extrema labels May2 asked First derivative test and uniqueness of local extrema May2 accepted Taking a limit for a modified difference quotient without L'Hopital May1 asked Taking a limit for a modified difference quotient without L'Hopital Apr23 comment Class term with Kuratowski pair Anyway I can only set $y$ properties outside the set definition. So in full it should be $\{(x,y)\mid x\in A\} \land y\in A$. I wonder if the properties of $y$ can be given inside the set definition and if this makes sense in general. Apr11 accepted Class term with Kuratowski pair Apr11 asked Class term with Kuratowski pair May26 comment Prove the supremum of the set of affine functions is convex I am no expert, anyway I have seen that in convex analysis often they use the notion of 'proper function', which is $+\infty$ outside its 'effective domain'. I see often notations like $\sup\limits_{i\in I} f_i$ or $\sup\limits_{y\in A} f(x,y)$ without a definition. Can you address me to a formal definition? May26 awarded Scholar May26 accepted Pointwise supremum of a convex function collection May26 comment Pointwise supremum of a convex function collection +1. As the function is defined 'proper' it can take $+\infty$ values outside its 'effective domain'. As you observe it works in this case. Also the related sup value is different. If some $f_i$ is $+\infty$ so is the sup; otherwise it would be the sup of the $f_i$ defined at $x$. May26 comment Prove the supremum of the set of affine functions is convex Well, I suppose there are two possibilities: one is defining $$g(x) = \sup \{g_i(x) \mid i \in I, x\in \mathrm{dom}\, g_i \}$$ So, for every $x^0$, I include only the $g_i(x^0)$ value when the function is defined at $x^0$, like to say that, if a function is not defined at point, then the function does not exist there and so no value is (can be) included in the set. Another possibility is to assume the function is $+\infty$ when not defined. I don't know if there is some established convention ruling out these options. May26 revised Pointwise supremum of a convex function collection added 3 characters in body May26 comment Pointwise supremum of a convex function collection @Did: sorry when you say "takes into account every index i", do you mean that every $f_i(x)$ has the same domain? So I cannot have a situation where, for some $x^0,i,j$, $\;x^0 \in \mathrm{dom}\, f_i$ and $x^0 \not\in \mathrm{dom}\, f_j$. May26 awarded Editor