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 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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