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 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.