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2d
comment Compact operator
@MatthewDaws In fact, it's a verbatim of Conway's A Course in Functional Analysis, Chapter II, section 5, exercise 6.
2d
revised Compact operator
added 42 characters in body
Dec
13
comment pseudo inverse of a finite-to-one continuous map and measurability
Remark 3 seems wrong. Consider the Cantor set $\mathcal C\subseteq[0,1]$, and a nonmeasurable set $\mathcal N\subseteq[0,1]$ such that there's a bijection $f\colon\mathcal C\to\mathcal N$. We extend $f$ to $[0,1]$ such that $f(x)=x+1$ for all $x\not\in\mathcal C$. Note that the totality of $y$ such that $\#f^{-1}y=1$ isn't measurable. I doubt we should assume that $\pi$ is continuous.
Dec
13
comment Two generating meromorphic functions seperate points on a compact Riemann surface?
@wisefool So what?
Dec
7
asked Two generating meromorphic functions seperate points on a compact Riemann surface?
Nov
21
accepted Any real analytic Frobenius theorem used in the proof of integrable almost complex manifolds to arise from complex manifolds?
Nov
19
comment Background for reading Milnor's Morse Theory book
@tessellation Is it really self-contained? For example, it seems to me that in section 5, something on homology theory is used, and section 7 needs acquaintance with (relative) homotopy groups.
Nov
19
accepted $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
Nov
19
comment $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
I don't understand your Mazur-theorem argument, but the result follows from the fact that $L^2$ is reflexive and therefore by Banach-Alaoglu theorem, the unit ball is weakly compact, thus a lower semicontinuous function attains its minimum. The later perturbation argument could be simply replaced with applying Cauchy-Schwarz inequality on the set $\{\,x\,\vert\,g(x)<f(x)-\epsilon\,\}$, I think, and the fact that the functional takes minimum is also a consequence of this, without appealing to the general minimum-existence theorem.
Nov
15
answered Any real analytic Frobenius theorem used in the proof of integrable almost complex manifolds to arise from complex manifolds?
Nov
15
accepted Is the $\sigma$-finiteness condition necessary to ensure that $L^p(\mu)$ is reflexive?
Nov
15
comment Is the $\sigma$-finiteness condition necessary to ensure that $L^p(\mu)$ is reflexive?
It should be Folland's book. In addition, $L^1$ isn't reflexive in general even if $\mu$ is $\sigma$-finite, say, $(L^1(\mathbb R))^*=L^\infty(\mathbb R)$.
Nov
15
revised Any real analytic Frobenius theorem used in the proof of integrable almost complex manifolds to arise from complex manifolds?
edited tags; edited title
Nov
14
asked Is the $\sigma$-finiteness condition necessary to ensure that $L^p(\mu)$ is reflexive?
Nov
14
asked Any real analytic Frobenius theorem used in the proof of integrable almost complex manifolds to arise from complex manifolds?
Nov
14
revised $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
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Nov
14
revised $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
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Nov
14
revised $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
added 384 characters in body
Nov
14
asked $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
Oct
29
comment How to show that a measurable function on $R^d$ can be approximated by step functions?
@Groups Edited, thanks!