Reputation
634
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
4 16
Newest
 Yearling
Impact
~16k people reached

1d
asked What is known about the space of measure-preserving transformations?
Apr
14
revised For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
edited body; edited title
Apr
14
comment For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
@MartinArgerami, damn, you're right, I'll edit it right away. I think everything is correct now, including the definition of strong resolvent.
Apr
14
comment For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
@MartinArgerami, problem 27 actually.
Apr
14
comment For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
@MartinArgerami, I'm not sure, but that's the statement of the problem. And by ${}^{-1}$, I think it's meant $\frac{1}{T_n-I}$ and so on rather than the inverse (although I guess that is the inverse). But given the statement it seems it's assumed that it is invertible.
Apr
14
revised For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
added 158 characters in body
Apr
14
comment For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
@JonWarneke, it's not a pesky question at all, and I guess I shouldn't have assumed everyone knows the notation. I'm just going through Simon and Reed and that's what they use. $s$ indeed stands for strong convergence or convergence in the strong operator topology. So in this case $\|(T_n - I)^{-1}(x) - (T - I)^{-1}(x) \| \to 0$ $\forall x$.
Apr
14
revised For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
added 50 characters in body
Apr
14
revised For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
deleted 65 characters in body
Apr
13
asked For $\{T_n\}$ and $T$ positive and self-adjoint, show $T_n \stackrel{SR}{\to} T$ iff $(T_n + I)^{-1} \stackrel {s}{\to} (T + I)^{-1}$
Apr
11
comment If a map $C:X\rightarrow U$ maps every weakly convergent sequence into strongly convergent
What is $\overline{T(A)}$ here? Did you mean $\overline{C(A)}$?
Mar
22
accepted If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete
Mar
22
comment If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete
But the metric you defined above is the same as he uses in the book, isn't it?
Mar
22
comment If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete
Ah, so the statement that $M$ is complete as a metric space?
Mar
22
comment If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete
Where did you get this Hopf-Rinow theorem? Do Carmo has a different version, and I wanted to restrict myself to his machinery. Is there a way to do it differently? Hmm, upon further inspection, this would seem to follow from his Proposition 2.6, right?
Mar
21
awarded  Yearling
Mar
21
asked If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete
Feb
19
comment Using direct sums, construct an inseparable Hilbert space with an uncountable orthonormal basis
Hmm, it's not a direct sum, though.
Feb
19
comment Using direct sums, construct an inseparable Hilbert space with an uncountable orthonormal basis
@user251257, thanks, I did, although they don't really make a distinction between countable and uncountable sums, they just say infinite. I wonder whether that's because it holds for both or because the article isn't entirely correct and precise.
Feb
19
comment Using direct sums, construct an inseparable Hilbert space with an uncountable orthonormal basis
@user251257, why would that be a Hilbert space, though? I know that you can take a countable direct sum of Hilbert spaces, but I'm not sure there's any result that says you can do the same uncountably to get another Hilbert space.