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I use Emacs and R for financial research


14h
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.
Apr
11
accepted Class term with Kuratowski pair
Apr
11
asked Class term with Kuratowski pair
May
26
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?
May
26
awarded  Scholar
May
26
accepted Pointwise supremum of a convex function collection
May
26
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 $x$.
May
26
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.
May
26
revised Pointwise supremum of a convex function collection
added 3 characters in body
May
26
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 $x^0 \not\in \mathrm{dom}\, f_j$.
May
26
awarded  Editor
May
26
revised Pointwise supremum of a convex function collection
deleted 3 characters in body
May
26
comment Prove the supremum of the set of affine functions is convex
I am wondering, when $g_i$ have non-identical domains, if the claim holds and can still be proved via epigraphs. Pointwise supremum of a convex function collection.
May
26
asked Pointwise supremum of a convex function collection
May
3
awarded  Supporter
May
3
comment $g(\theta):= f(\theta {\bf y} + (1 − \theta){\bf x})$, find $g'$.
@PeterTamaroff: Transposition as the gradient is a vector. I am in doubt probably $\frac{\mathrm{d}f}{\mathrm{d}z}$ should be transposed too.
May
2
awarded  Student
May
2
asked $g(\theta):= f(\theta {\bf y} + (1 − \theta){\bf x})$, find $g'$.