User Chaos | chat.stackexchange.com

Chaos

Oct 27, 2021 06:39

Of course that for any $\eta,\xi\in S'(\mathbb R)$ we have that $\langle \xi,P_m\eta\rangle=\langle P_m\xi,\eta\rangle$ where the angle brackets denote the bilinear pairing between test functions and distributions

Chaos

Oct 27, 2021 06:38

for instance the Hermite functions. We know that the linear span of that family is dense in the adjoint space $S'(\mathbb R)$ of tempered distributions. If I define $P_m$ to be the orthogonal projection on the span of $(e_1,...,e_n)$ can I say that $P_m$ is self adjoint in $S'(\mathbb R)$?

Chaos

Oct 27, 2021 06:38

Hello everyone, I apologice for the silly question but's really early in the morning ahaha: Suppose we have an orthonormal family $(e_n)$ of functions in the Schwartz space of rapidly decreasing functions $S(\mathbb R)$,

Chaos

Mar 31, 2021 06:18

@AndrewMicallef that looks like the second moment of a Gaussian random variable with mean $a$ and variance $1/2\lambda$

Chaos

Mar 31, 2021 06:15

Excuse me @TedShifrin, you said that to me? I don't know if you were discussing something above!

Chaos

Mar 31, 2021 06:08

In particular I would like to know if there's a Feynman.Kac representation for the solution

Chaos

Mar 31, 2021 06:07

hey guys good morning, I was wondering do you know about some reference treating the (approximately) heat equation with nonlinear source, something like $\partial_t u=\sigma(t)\partial_{xx} u +b(u)$ where $b$ is bounded and Lipschitz

Chaos

Feb 18, 2021 17:02

@Thorgott that makes sense, thanks! I am getting a bit lost with all this considerations, I work in Stochatic Analysis and I decided to follow a course in Quantum field theory. Bad idea ahaha

Chaos

Feb 18, 2021 16:57

@AkivaWeinberger oh shit you are right! ahah thanks!

Chaos

Feb 18, 2021 16:55

is this some physicist convention?

Chaos

Feb 18, 2021 16:54

there are some "*" missing... but I was wondering isn't a homeomorphism a bijection by definition? what's the point of saying "if it's a bijection then it's a *-isomorphism"?

Chaos

Feb 18, 2021 16:53

Hi guys, good afternoon. Hope you can give me an answer for this, in general (see Wikipedia for instance) an "homeomorphism" is defined as a "topological isomorphism", and this of course implies that it's a bijection. In a book about Quantum Mechanics it's stated that a -homemophism (between C-algebras ) is a called *-isomorphism if it's also a bijection.

Chaos

Jan 7, 2021 17:17

Hi guys, on simple question, is there any version of the Wierstrass theorem (extreme value theorem) where the compactness can be understood in some weaker sense, namely compact w.r.t to the weak topology?

RScrlli

Jul 6, 2020 09:12

Hello everyone, I was wondering, do you know of some extension of Minkowski inequality to the case of negative exponents?

RScrlli

May 7, 2020 15:03

Hey guys I have a question; If we have an orthonormal basis $(e_i)$ on a Hilbert space , can we say $\langle f,g\rangle =\sum_{i=1}^{\infty} \langle f, e_i\rangle \langle g, e_i\rangle$? This looks like a sort of Parseval identity,

RScrlli

Feb 26, 2020 16:53

Hey guys a simple question, is there any difference between "polynomial chaos" and "homogeneous chaos" in the context of Wiener chaos?

RScrlli

Jan 31, 2020 17:45

6

Q: How to show this is not a martingale.

Assume we have the following stochastic process: $$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$ where $(B)_{t\geq 0}$ is a Brownian Motion. I have to show that $X_t$ is not a martingale. I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process ...

RScrlli

Jan 31, 2020 17:45

Hi guys, would you mind to take a look at my question and give me some thought? Thanks! math.stackexchange.com/questions/3528142/…

RScrlli

Nov 15, 2019 09:41

does the author intend to say that $(S,\mathcal S)$ is a measurable space and $S$ must be metric?

RScrlli

Nov 15, 2019 09:40

hey guys good morning, I am studying the Skorkhod's theorem and I have a doubt, the statement starts by saying "Let be $P_n$,$n\in\mathbb N$ a sequence of probability measures defined on a metric space $(S,\mathcal S)$. I might be wrong, but by definition $(S,\mathcal S)$ is a measurable space, it can happen that $S$ is metric, but this statements seems off to me.

RScrlli

Nov 7, 2019 14:54

can I still say that they differ UP to a set with zero-measure?

RScrlli

Nov 7, 2019 14:53

mainly because I cannot say that they differ in a zero-measure set, because $M$ is not measurable

RScrlli

Nov 7, 2019 14:52

hi guys, I was thinking, let's say you have a measure space that is not complete, then let $N$ be a measurable set with zero measure, and let $M\subset N$ be a set that is not measurable. Then can I say that if $f:=g+1_{M}$, then $f=g$ a.e.?

RScrlli

Oct 17, 2019 08:46

could it be the span?

RScrlli

Oct 17, 2019 08:46

what does $\{\cdot\}$ mean in this context?

RScrlli

Oct 17, 2019 08:46

I am not familiar with this notation

RScrlli

Oct 17, 2019 08:45

Morning guys, when defining "maximal" Hilbert space my textbook says, "the idea behind maximality is that we can obtain $\mathfrak H$ as a limit of a finite-dimensional projections, i.e. $$\mathfrak H= \bigoplus_{k\in\mathbb N} \{e_k\}$$"

RScrlli

Oct 15, 2019 10:20

actually the condition is $F\subset \bar F$ implies implies $\bar F^{\perp}\subset F^{\perp}$

RScrlli

Oct 15, 2019 10:19

I've made a type

RScrlli

Oct 15, 2019 10:18

Lets say we have a linear subspace $F$ of a Hilbert space $\mathfrak H$.
How could I prove that $\bar F\subset (F^{\perp})^{\perp}$ (the closure of $F$ is a subset of the "double" orthogonal complement) implies $\bar F^{\perp} \subset F^{\perp}$?

RScrlli

Oct 15, 2019 10:18

hey guys, good morning!

RScrlli

Oct 3, 2019 09:10

@MatheinBoulomenos I see! Thank you very much!

RScrlli

Oct 3, 2019 09:06

yes ! @MatheinBoulomenos I should have clarify that

RScrlli

Oct 3, 2019 09:00

I understand that the sigma finiteness of $\mu$ implies that there's an exhausting sequences with finite measure, but I don't quite follow the reasoning to make that assertion

RScrlli

Oct 3, 2019 08:58

then my textbook says that because of $\sigma$-finiteness of $\mu$ is easy to see that $\sigma(\delta)=\mathcal A$

RScrlli

Oct 3, 2019 08:57

$\delta:=\{A\in\mathcal A:\mu(A)<\infty\}$

RScrlli

Oct 3, 2019 08:56

Good morning guys, suppose we have a $\sigma$-finite measure space $(X,\mathcal A,\mu)$ and let $\delta:=\{A\in\mathcal A:\mu(A)<\infty}$

RScrlli

Sep 11, 2019 14:07

@AlessandroCodenotti @SayanChattopadhyay You are right, I absolutely forgot that detail, thanks guys!

RScrlli

Sep 11, 2019 11:04

One could easily show that all $x\in \mathbb{R}$ is a Borel set, by simply noticing that $$x=\bigcap_{j\in\mathbb{N}} (x-\frac 1 j,x+\frac 1j)$$
is the countable intersection of Borel sets, and since the Borel $\sigma$-algebra is $\cap$-stable the statement follows.

And since we know that the Borel $\sigma$-algebra is stable under countable unions (intersections), I don't seem to understand how the Cantor set can have subsets that are not Borel sets.
I've seen numerous answers in the site proving that statement by using the fact that the cardinality of the Cantor set is $2^{\aleph_0}$ (a…

RScrlli

Sep 9, 2019 10:24

how would you interpret the following notation: $u\in\mathcal{L}^1(\mathcal{G})$ where $\mathcal{G}$ is a sigma algebra? Does it means that the function $u$ is integrable (with respect to some measure) on the sigma algebra $\mathcal{G}$?

RScrlli

Sep 9, 2019 08:05

morning guys

RScrlli

Sep 6, 2019 15:49

Hello guys, if I say that a function $u:X\rightarrow \mathbb{R} \in \mathcal{L}(\mu)$ (the function is $\mu$-integrable), one should understand that the function is ($\mu$)-integrable over all the domain $X$, could it happen that a function is integrable on a subset of the domain?

RScrlli

Sep 5, 2019 16:31

have a nice day guys

RScrlli

Sep 5, 2019 16:31

Ok guys I am leaving, thanks again @TedShifrin and I'll check your book and the videos!

RScrlli

Sep 5, 2019 16:26

still morning ? where are u? hahaha @TedShifrin

RScrlli

Sep 5, 2019 16:25

I will check them out for sure @TedShifrin.

RScrlli

Sep 5, 2019 16:24

@TedShifrin Indeed, could you give us a link?

RScrlli

Sep 5, 2019 16:22

I know, just joking!

RScrlli

Sep 5, 2019 16:22

ahahah

RScrlli

Sep 5, 2019 16:22

I had a hard time dealing with baby rudin, I don't want to meet Mr. Grandpa

Chaos

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