# Martin Wanvik

Trondheim, Norway

Age: 32

 Feb 1 awarded Yearling Feb 1 awarded Yearling Sep 30 awarded Explainer Jul 26 awarded Enlightened Jul 26 awarded Nice Answer Nov 15 comment Consultation on point extremeYour definition of an extreme point is wrong as it stands. An extreme point of a convex set $C$ is a point $x \in C$ that cannot be written $x = ty + (1-t)z$ with $t \in (0,1)$, $y,z \in C$ and $z \neq y$. Nov 15 comment What is the limit of a sequence of events? ProbabilityAlso, the sets $B_n$ you've described are not pairwise disjoint. You may want to define $B_3 \setminus A_1 \cup A_2 = B_3 \setminus A_2$ instead (note that $A_2 \cap A_1 = A_1$, and so on...) Nov 15 comment What is the limit of a sequence of events? ProbabilityNote that you're not considering a limit of a sequence of events here at all. You're considering the limit of an ordinary sequence of numbers - $P(A_n)$ is a real number for each $n$. Aug 1 revised What is the connection between strong norms and norms coming from scalar products (in pre-hilbert spaces)?Changed < and > to \langle and \rangle, and || (double vertical pipes) to \|. Corrected spelling of the name Schwarz. Jun 19 revised Can we print an array like this?deleted 6 characters in body Jun 19 answered Can we print an array like this? Jun 11 revised Rank and determinant of $D$ , an $n\times n$ real matrix, $n\ge 2$Improved formatting of the counterexample to proposition (1). Jun 11 comment Rank and determinant of $D$ , an $n\times n$ real matrix, $n\ge 2$@Mex: How about #2? (And #4 may be correct?) Jun 11 comment Rank and determinant of $D$ , an $n\times n$ real matrix, $n\ge 2$@Mex: Yes, if it has non-zero determinant, then it has full rank (a fact that has been mentioned twice in the other answers). And yes, #3 is correct, but it is not the only one... Jun 11 answered Rank and determinant of $D$ , an $n\times n$ real matrix, $n\ge 2$ Jun 8 awarded Constituent Jun 8 awarded Caucus Jun 7 comment Bounded linear functionals and representationsGreat! Thanks, Nik. For posterity, I'll add the exact reference to Takesaki's book: theorem 4.2 and proposition 4.6 in chapter III (I actually did ask someone else about this prior to posting here on MO, and was pointed in this direction - what I failed to notice, however, was the equality $\| \omega \| = \| \phi \|$, and ended up with the estimate $\| \varphi \| \geq |\varphi(1)| = |\langle \pi(v)\xi,\xi \rangle|$, rather than $\| \varphi \| \geq \omega(1) = \| \xi\|^2$). Jun 6 awarded Supporter Jun 6 awarded Student