Martin Wanvik

Trondheim, Norway

Age: 31

Sep
30
awarded Explainer
Jul
26
awarded Enlightened
Jul
26
awarded Nice Answer
Feb
1
awarded Yearling
Feb
1
awarded Yearling
Nov
15
comment Consultation on point extreme
Your 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? Probability
Also, 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? Probability
Note 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 representations
Great! 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
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