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1d
revised Question regarding the dual space of $H_0^1(\Omega)$
spelling in title
1d
answered Two continuous functions with connected images
1d
comment “Isomorphy” in mathematical texts
Closely related: What word means "the property of being holomorphic"?
2d
comment Prove $f(x)=x$ is Lebesgue integrable on $[0,1]$
@MichaelHardy: So, I guess you haven't read Royden :-) I suppose his definition has the advantage of not requiring you to first define "Lebesgue measurable function".
2d
comment Prove $f(x)=x$ is Lebesgue integrable on $[0,1]$
Hint: consider simple functions that look like staircases.
2d
comment Proof of Nesbitt's Inequality?
Something seems to be wrong with the Wolfram Alpha link.
2d
comment A few questions on the properties of $\mathbb{R} ^ {[0,1]}$
Every compact space $Y$ is locally compact. To show $Y$ is locally compact, we need to show that every point of $Y$ has a neighborhood which is compact. But if $Y$ is compact, for each point we can take that neighborhood to be $Y$ itself. Not every locally compact space is compact; $\mathbb{Z}$ is a counterexample.
2d
comment A few questions on the properties of $\mathbb{R} ^ {[0,1]}$
I formatted your questions as a numbered list, and removed the algebraic-topology tag which is not applicable.
2d
revised A few questions on the properties of $\mathbb{R} ^ {[0,1]}$
format
2d
comment Selfadjointness of the Dirac operator on the infinite-dimensional Hilbert space
I'm confused by your reference to boundary conditions. I thought you wanted to your operator to act on functions from $\mathbb{R}^3$ to $\mathbb{C}^4$. $\mathbb{R}^3$ has no boundary so there are no boundary conditions. (It might in general be necessary to impose growth or integrability conditions, however.) I don't see how it would make sense for $D$ to act on functions defined on a bounded (one-dimensional) interval.
2d
comment Selfadjointness of the Dirac operator on the infinite-dimensional Hilbert space
This is true on its face but I think it misses the point. $D$ will certainly be an unbounded operator so we should not expect it to be defined everywhere on the Hilbert space $H$, but instead on some dense subspace (its domain). Part of the question generally is making an appropriate choice for the domain, but in this case I think the natural choice is going to be the Sobolev space $H^1(\mathbb{R}^3; \mathbb{C}^4)$. To say it in the terms I used in the answer you link, this operator will be self-adjoint but not Hermitian. We should not expect to use Hellinger-Toeplitz.
Oct
18
answered absolute continuity - Dirac measure with respect to gaussian measure
Oct
18
comment Convergence in distribution example
No, your graph isn't correct for $n > 1$. For instance, take $n=3$. Then $P(Y_3/3 \le 1/2) = P(Y_3 \le 3/2) = 1/3$, but according to your graph it would be 0. Try again to draw the correct graph, and I think it will become more clear why this is true.
Oct
18
revised Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
edited tags
Oct
17
comment Is there a dense set of positive measure which does not contain any open set?
As you have edited the question, can you also edit the title to reflect your new question?
Oct
17
revised Is there a dense set of positive measure which does not contain any open set?
edited tags
Oct
17
comment Spectral Measures: Integration of Product
@Freeze_S: Your $F$ is not a cumulative distribution function as it is not continuous from the right. And if you try to use it to define $E$ as stated, you will again find it is not countably additive. The intervals $(1/n, \infty)$ all have measure 0, so countable additivity forces $E((0,\infty)) = 0$. Since $E((-\infty, 0]) = 0$ also this cannot be reconciled with $E(\mathbb{R})=1$.
Oct
15
revised Dose “optional stopping theorem” imply “optional sampling theorem”?
tags
Oct
15
comment A few questions on $\omega_1$
$0$ is an isolated point, which means that the set $\{0\}$ is open. It is also closed (as is any finite set in $\omega_1$ or any other Hausdorff space) and so its complement is open. Thus we have written $\omega_1$ as a union of two nonempty disjoint open sets, so it is not connected.
Oct
14
revised Why isn't $C[0,1]$ a Banach space in this unusual norm?
clarify title