Does there exist a $g$ such that $g$ is continuous on $(0,1)\times (0,1)\subset \mathbb R^2$ and that $g$ is integrable with respect to $\lambda_2$ but $\int g(x,y) d\lambda (y)=\infty$ for some $x\in (0,1)$. The hint to find such $g$ was to consider $y^{\phi(x)-1}$ for some appropriate $\phi$ so I was trying to construct that. I'm also thinking about the example that you gave, which didn't seem to work as $\int_0^1 \frac 1{(x^2+y^2)}d\lambda (y)=\log (1+\sqrt{x^2+1})-\log x$, which is $\lt \infty$ for every $x\in (0,1)$.
there is sqrt in denominator of the integrand in the last line.
I thought I had created an example like that: $y^{x-2}=g(x,y)$ but I realized that this is wrong as $\int g(x,y) d\lambda (y)=\infty$ for every $x$ and that violates Fubini's theorem.
Suppose that $E$ is Borel sigma algebra $B(\mathbb R)$, then any hints to show that the set {$(x,y)\in \mathbb R^2: x+y\in E\}$ is in $B(\mathbb R^2)$?
Basically, if f is any continuous map from R^2 to R. Then, f^{-1}(U): U is open in R is open in R^2, i.e., f^{-1}(U)= $V_1\times V_2$, where V_1 and V_2 are open in R^2. The set {E in B(R): $f^{-1}(E)\in B(R^2)$} is a sigma algebra containing all open sets in R so contains B(R). Hence, f^{-1}(E) is in B(R^2) for every Borel set E in R.
uh, if f is continuous from R^2 to R, you generally won't have that the inverse image of an open subset of R is a cartesian product of open subsets of R. for example if f(x,y) = y-x then f^{-1} (0, infty) aka "the stuff above the line y = x" is not a cartesian product. e.g. (1,2) is in there and (-2,-1) is in there but (1,-1) is not.
i'm not sure it helps to think about basic open sets in R^2. isn't it enough just to know that f^{-1}(U) is open?
well if ted isn't going to melt down about that i guess i have to. really?
there's a general phenomenon of a basis or subbasis of a topology often being enough to consider for various theorems, particularly about continuous maps, open maps, etc. but the topologies themselves tend to be much bigger.
think about the open unit disc. if it were a cartesian product it would have to be (-1,1) x (-1, 1) because of how it projects to the factors.
or to turn that into something maybe more useful, if an open subset G of R^2 both contains the open unit disc and has the form G = UxV, then G contains (-1,1)x(-1,1).
Suppose that W is open in R^2. Then, for any w in W, there is a basis element $U_x\times V_x$ (where U_x and V_x are open in R) such that $x\in U\times V\subset W$. So $W=\cup_x (U_x\times V_x)$
i agree with that. the issue is that an arbitrary union of cartesian products is not guaranteed to be a cartesian product, in the same way that a union of two rectangles might not be a rectangle
Suppose that $X, Y$ are given non empty sets. Let $C, D$ be collection of subsets of $X$ and $Y$ respectively. Suppose that there exists sequences $\{C_n\}$ (of elements of $C$) and $\{D_n\}$ (of elements of $D$) such that $\cup_n C_n= X, \cup_n D_n=Y$. Then, $M(C\times D)=M(C)\otimes M(D)$, whe...
Let me consider $(\Omega, F,(F_n)_n, \Bbb{P})$. Let $(X_n)_n$ be an adapted process at $0$. I need to show that $(X_n)_n$ is a martingale iff $\Bbb{E}(X_T)=0$ for all bounded stopping times $T$.
My idea was the following:
$\Rightarrow$ Let me assume $X_n$ is a martingale and pick a bounded stop...
In Step 2 of Theorem 22.1 from Munkres' Topology:
Let $p:X\to Y$ be a quotient map; let $A$ be a subspace of $X$ that is saturated with respect to $p$; let $q:A\to p(A)$ be the map obtained by restricting $p$.
(1) If $A$ is either open or closed in $X$, then $q$ is a quotient map.
(2) If $p$ is ...
$A= \cup_{y\in S}p^{-1}(y)$ where $S$ is some subset of Y.
But when we say- A is subspace of X. This means that A has subspace topology inherited from X. Why is A union of fibers of p is at all useful?
It seems redundant as far as the proof is concerned.
Part i): Given: A open in X. It follows that p(A) is open in Y (because p is quotient map). Take any $U\subset p(A)$ open (in subspace topology of Y). Then there exists $V\subset Y$ open such that $U=p(A)\cap V$. $q^{-1}(U)=A\cap p^{-1}(V)$. which is open in $A$. So $q$ is continuous. Now, let $q^{-1}(W)\subset A$ be any saturated open set, then by surjectivity of $q$, we have $q(q^{-1}(W)=W$ so we are done if we show that W is open.
Note that $q^{-1}(W)$ is open in X (because A is open in X). $q^{-1}(W)=\{a\in A: q(a)\in W\}=\{a\in A: p(a)\in W\}=p^{-1}(W)\cap A$.
Suppose that $X, Y$ are given non empty sets. Let $C, D$ be collection of subsets of $X$ and $Y$ respectively. Suppose that there exists sequences $\{C_n\}$ (of elements of $C$) and $\{D_n\}$ (of elements of $D$) such that $\cup_n C_n= X, \cup_n D_n=Y$. Then, $M(C\times D)=M(C)\otimes M(D)$, whe...
Let $V,W$ be finite-dimensional vectors spaces and let $U \subseteq V$ and $T \subseteq W$ be subspaces. Define $S \subseteq Hom(V,W)$ to be $S := \{ \phi : V \to W : \phi(U) \subseteq T\}$.
It's not difficult to show $S$ is a subspace. Now I am trying to determine the dimension. I want to say that $S$ is isomorphic to $Hom(V/U,W/T)$ but I am not entirely sure.
Given $\phi : V \to W$ with $\phi(U) \subseteq T$, it is not difficult to show that $\widehat{\phi} : V/U \to W/T$ defined by $\widehat{\phi}(v+U) = \phi(v) + T$ is well-defined. So, I would like to show that $\phi \mapsto \widehat{\phi}$ is an isomorphism, but I don't see injectivity.
@Jakobian Oh, I see...hmm...Let me think about that.
Okay, that makes sense. I was sort of on the right track. I just need to consider the restriction of $\phi$ to $U \mapsto to T$...Let me write this up now...
So the mapping $S\to \text{Hom}(V/U, W)\times \text{Hom}(U, T)$ would be given by $L\mapsto (\bar{L}, L\restriction_U)$ where $\bar{L}\circ q = L$, $q:V\to V/U$
Choose complements $V=U\oplus U^{\perp}$ and $W=T\oplus T^{\perp}$. Then, $\mathrm{Hom}(V,W)=\mathrm{Hom}(U,T)\oplus\mathrm{Hom}(U^{\perp},T)\oplus\mathrm{Hom}(U,T^{\perp})\oplus\mathrm{Hom}(U^{\perp},T^{\perp})$. $S$ corresponds precisely precisely to the subspace where the $\mathrm{Hom}(U,T^{\perp})$ coordinate is $0$.
There's this exercise from linear regression that shows something like, if correlation coefficient between $X, Y$ is $\pm 1$ then $X = \pm Y$ almost surely or something.
I don't know, it's just some measure of how two random variables relate to each other
@TedShifrin Hello, I have a question about Example 1 on page 189 of "Multivariable Mathematics". When you say "We recognize these as the three ($\mathcal{C}^1$) local inverse functions $\phi_1, \phi_2$ and $\phi_3$ of $g(x)=x^3-3x$,..." shouldn't it be $g(y)=y^3-3y$ since $f(x,y)=y^3-3y-x=0$ ?
It’s irrelevant what letter you use for your independent variable. But if you’re thinking of graphing the inverse by reflecting across the diagonal, then you still use $x$.
Another question. On page 99, in the proof of the Chain Rule, in the line beginning with (**) shouldn't it be $0<||\mathbf{k}||<\eta\Rightarrow ||\mathbf{f(b+k)}-\mathbf{f(b)}-D\mathbf{f(b)k}||\leq\varepsilon ||\mathbf{k}||$ instead of $||\mathbf{k}||<\eta\Rightarrow ||\mathbf{f(b+k)}-\mathbf{f(b)}-D\mathbf{f(b)k}||\leq\varepsilon ||\mathbf{k}||$
since only later we set $\mathbf{k}=\mathbf{g(a+h)}-\mathbf{g(a)}$ from which, since $0<||\mathbf{h}||<\delta_1$ it follows that $0<||\mathbf{k}||$?
If I pick $a=(a_n)\in l^{\infty}$ and define a linear map $M:l^{\infty}\rightarrow l^{\infty}$ by $(x_n)\mapsto (a_nx_n)$. Which conditions do I need to put on $a$ such that $M$ is invertible?
I know that a linear operator is invertible if it is bounded, linear and there exists $S:l^{\infty}\rightarrow l^{\infty}$ s.t. $SMx=x$, $MSx=x$ and $||S||<\infty$
@TedShifrin At the end of page 190, when we invoke Theorem 5.1 (Implicit Function Theorem, simple case) to justify the process of implicit differentiation, shouldn't we have the additional hypothesis that $f(\mathbf{a})=0$?
@Jakobian It seems that one of the hypotheses of the theorem is that there is a MADF.
I have no idea what theory this is, but the theorem seems to be trying to prove a statement about MADF. So the structure of the theorem is if a MADF satisfies certain properties, then it must satisfy some other properties, too.
So at least one $\mathcal{B}$ exists by hypothesis.
MADF just means a maximally almost disjoint family, where an almost disjoint family means a family of countable subsets which intersect in finite amount of elements
Presumably, $\mathcal{N}$ itself is an upper bound for any chain of ADFs. (Again, I don't really know what an ADF is, but I don't really need to, either).
I assume that the set of ADFs is ordered by inclusion. So apply Zorn to that.
@TedShifrin I asked this before, but I think maybe you were offline at that time: at the end of page 190, when we invoke Theorem 5.1 (Implicit Function Theorem, simple case) to justify the process of implicit differentiation, shouldn't we have the additional hypothesis that $f(\mathbf{a})=0$?
If I pick $a=(a_n)\in l^{\infty}$ and define a linear map $M:l^{\infty}\rightarrow l^{\infty}$ by $(x_n)\mapsto (a_nx_n)$ is it true that the spectrum of $M$ is $\Bbb{C}\setminus \{0\}$ or how do I find the spectrum?
Ah yeah I think something like this works. For each $x\in X$ take a sequence $N_x$ convergent to $x$. This forms an ADF, let $\mathcal{M}$ be a MADF containing it. Let $M\in \mathcal{M}$ such that $M$ doesn't converge to any $x\in X^*$. For all accumulation points $y\in M' \neq\emptyset$ of $M$ we can find $M_y\subseteq M$ which converges to $y$. Then replace every such $M$ by $\{M_y : y\in M'\}$
I have to prove the following:
Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.
What I have tried to do is to find a sequence of fun...
has any one here published at arxiv? I have some basic question: Can one submit first to see if they get no latex errors or any other error, before doing an official submit? Mine has also some code and supplementary files for the code. I wanted to try first to see if I get any errors and if I need to fix the latex. Is such an option available at arxiv? I looked at the website and did not see anything.
@Koro The answers was a little to stingy with the details but his/her answer is correct. The function $\phi:\mathbb{Q}\times\mathbb{R}\rightarrow\mathbb{R}$ given by $\phi(r, x)= f(r, x)$ is measurable as $\{(r, x): f(r, x)<a\}=\bigcup_{r\in\mathbb{Q}}\{r\}\times(f^{-1}_r(-\infty,a))$. The sequences $f_n(x,y) = f(\lfloor nx\rfloor/n, y)=\phi\circ g_n(x,y)$ where $g_n(x, y)=f(\lfloor nx\rfloor/n, y)$.
@Koro: I meant to say $g_n(x,y) = (\lfloor nx\rfloor/n, y)$.
@AlessandroCodenotti maybe, I was just wondering how to show that it's contained specifically in $\mathcal{N}$
to show that there is a MADF on some infinite set, you want to use Zorn's lemma, but that won't necessarily show it has the desired properties, I don't think
If I define a linear operator $S:l^2\rightarrow l^2$ by $(x_1,x_2,...)\mapsto (0,x_1,x_2,...)$, then how do I compute the operator norm?
I would have done it as follows: Remark that $||Sx||_2^2=\sum_{k=1}^\infty |Sx|^2=\sum_{k=1}^\infty |x_k|^2$ hence $||S||=||x||_2\leq 1$. But now pick $x=(1,0,...)$ then $x\in l^2$ and $||x||_2\leq 1$. In particular $||Sx||_2=1$ hence $||S||=1$.
@copper.hat could you maybe give me a hint how to compute the spectrum of $S$. I could show that $S$ has no eigenvalues, but computing the spectrum doesn't work. So I want to find all $\lambda$ such that $(S-\lambda I)(x)=(-\lambda x_1,x_1-\lambda x_2,x_2-\lambda x_3,...)$ is not invertible but is there a trick?
I have to prove the following:
Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.
What I have tried to do is to find a sequence of fun...
