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Oct
20
comment Find asymptotics in a given form $n=(e+o(1))^{f(s)}$
@martycohen see edits to my answer
Oct
20
revised Find asymptotics in a given form $n=(e+o(1))^{f(s)}$
added 483 characters in body
Oct
20
answered Find asymptotics in a given form $n=(e+o(1))^{f(s)}$
Oct
19
comment spectral projection of an element in a C*-algebra
@MartinArgerami done!
Oct
19
answered spectral projection of an element in a C*-algebra
Oct
18
comment spectral projection of an element in a C*-algebra
Any normal operator $T$ gives rise to some spectral measure $E:\operatorname{Bor}(\sigma(T))\to\mathcal{P}(H)$ which maps Borel subsets of the spectrum of $T$ into orthogonal projections in $H$. If you take Borel subset $A\subset\sigma(T)$, then $E(A)$ is called a spectral projection. Search spectral theorem on this site.
Oct
14
reviewed Close how to solve $\int\frac{1}{1+x^4}dx$
Oct
14
reviewed Close Showing a set is a subset of another set
Oct
14
reviewed Close What is the value of X in this?
Oct
14
reviewed Close Perimeter & Area of a rectangle
Oct
14
reviewed Close How to evaluate $\int_0^\infty \frac{1}{x^n+1} dx$
Oct
14
reviewed Close Prove that the intersection of all maximal left ideals of a ring $R$ is a two sided ideal
Oct
14
revised Banach space, dual of its quotient, isometric isomorphism
deleted 4 characters in body
Oct
12
comment If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
@user136475, ok
Oct
12
revised If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
added 291 characters in body
Oct
12
comment If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
@user136475 see my comment above
Oct
12
comment If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
@DavidMitra, yes. Assume $x$ is not in the strong closure of $M$, then $d(x,M)>0$. Then we can find $\xi\in X^*$ s.t. $(\xi,x)\neq 0$ while $\xi|_M=0$. From weak convergence we then infer $(\xi,x)=\lim_n(\xi,x_n)=0$. Contradiction.
Oct
12
comment If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
M is a subspace, hence convex. By assumption it is closed
Oct
12
comment If a sequence converges weakly in a closed subspace $M$ of a Banach space, then the (strong) limit point is in $M$.
no, the Mazur's lemma I'm referring is the following: weak and strong closures of coincides for convex subsets of locally convex spaces.
Oct
12
comment Is $\mathcal{B}(H)$ complemented in $\ell_\infty(I, H)$
Wish I could know all this. Thank you!