TITLE: Question about $f :\mathbb{R}\rightarrow \mathbb{R}$ defined as $f(x)=|x|^{\frac{3}{2}}$ (TIFR GS $2010$)
QUESTION [1 upvotes]: Question is :

I am not sure how to check for differentiability,
only thing i know is how to see for differentiability of $f(x)=|x|$
$\lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{|x|}{x}$
But then, when $x$ approach $0$ from positive line we have $\lim_ {x\rightarrow 0^+} \frac{|x|}{x}=1$ and  when $x$ approach $0$ from negativeline we have $\lim_ {x\rightarrow 0^-} \frac{|x|}{x}=-1$
So, limit does not exists so the function $f(x)=|x|$ is not differentiable at $0$
But, I am not getting any idea how to proceed in case of $f(x)=|x|^{\frac{3}{2}}$
$lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{|x|^{\frac{3}{2}}}{x}$ I was thinking to write this as
$$\lim_ {x\rightarrow 0} \frac{|x|^{\frac{3}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{|x|^{2-\frac{1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{|x|^2 .|x|^{\frac{-1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{x}{|x|^{\frac{1}{2}}}$$
as $x$ is in square rot in denominator i am unable to see what are its left and right limits for $x=0$
Please help me to proceed further..
Thank you

REPLY [1 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$$
\totald{\pars{\verts{x}^{3/2}}}{x} = {3 \over 2}\,\verts{x}^{1/2}\sgn\pars{x}
$$


TITLE: What is $57^{46}$ divided by 17?
QUESTION [3 upvotes]: $$\frac{57^{46}}{17}$$
What is best and quickest way to evaluate an approximate answer to this by hand?

REPLY [0 votes]: By Euler's theorem $57^{16}\equiv1 \pmod{17}$.
It could be done using this.


TITLE: Error estimate of definite integral of a taylor expanded function
QUESTION [1 upvotes]: If I consider a monotonic decreasing function $f(x)$ in the interval $[0,+\infty[$, and I consider the definite integral $\int_{0}^{+\infty}f(x)\,\mathrm{d}x$. What is the error committed if I compute the definite integral of the taylor expansion of $f(x)$ around $x=0$ truncate at a certain order $n$?
Thanks in advance

REPLY [2 votes]: The mistake is that if you take $T_n$ which is the $n$-th Taylor polynomial for $f$ around $x$, then
$$\int_0^\infty T_n(x) dx$$
is either $\infty$ (if the leading coefficient is positive) or $-\infty$ (if the leading coefficient is negative). So unless $$\int_0^\infty f(x)dx$$ is also infinite, the error commited is infinite.


TITLE: Variation of a function
QUESTION [3 upvotes]: What exactly is the variation of a function ? Is it a distace or an element of some space
The total of a real valued function $f\colon [0,t] \mapsto \Re $ is as below say $\pi = \{0=t_0,t_1,\cdots , t_n=t\} $ then the $p^{th}$ variation of $f$ is
$$ \lim_{\Vert \pi\Vert \to 0}V^p(f)=\sum_{i=1}^n \vert f(t_i)-f(t_{i-1})\vert^p $$
Now say we have $g\colon [0,t] \mapsto \mathcal{X}$ where $\mathcal{x}$ is some linear, normed,Complete space. it seems natural to say variation of $g$ over $[0,t]$ is
$$ Z_n=\sum_{i=1}^n \Vert g(t_i)-g(t_{i-1})\Vert^p $$ where
$\Vert.\Vert$ is the norm of $\mathcal{X}$
If it exists what is the Total variation in this case is it $\lim_{n \to \infty}Z$ ie a value in $[0,\infty]$ or is it some $x$ such that $\Vert Z_n -  x \Vert \to 0 $
say for wiener process $W(s)$ it quadratic  variation over  $[0,t]$ is $t$ does it mean
$$ \lim_{\Vert \pi\Vert \to 0}V^2(W)=\sum_{i=1}^n \vert W(t_i)-W(t_{i-1})\vert^2 $$ in $L^2(\Omega) $ norm or does it mean some $X \in L^2(\Omega)$

REPLY [4 votes]: Consider the case where $W$ is a standard Brownian motion. Given a partition $\pi  = \{ 0 = t_0  < t_1  <  \cdots  < t_n  = t\}$ of the interval $[0,t]$, define
$$
[W]^\pi   = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 }.
$$
Let $|\pi | = \max _i (t_i  - t_{i - 1} )$ denote the mesh of the partition. As is well known, given a sequence of partitions $\pi_n$  with $|\pi_n| \to 0$ as $n \to \infty$, $[W]^{\pi _n }  \to t$ in the $L^2$ norm, that is ${\rm E}\big[([W]^{\pi _n }  - t)^2 \big] \to 0$. Hence, in particular, $[W]^{\pi _n }  \to t$ in probability, and thus, by definition, the quadratic variation of $W$ is given by $[W]_t = t$. The key point here is that the limit $t$ in the right-hand side of $[W]^{\pi _n }  \to t$ plays a role of a random variable, rather than of a constant. Of course, in this particular example, $t$ is also a constant, but in general the limit is a random variable, which is defined using convergence in probability.
EDIT: A much simpler yet instructive example is given by the Poisson process $N$. It is readily seen that the variation of $N$ is given by
$$
\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {|N(t_{i} ) - N(t_{i-1} )|} = N(t),
$$
and the quadratic variation of $N$ is given by
$$
\mathop {\lim }\limits_{|\pi _n | \to 0} \sum\limits_{i = 1}^n {[N(t_{i} ) - N(t_{i-1} )]^2} = N(t).
$$
You may wish to consider different modes of convergence, as an exercise.


TITLE: Zooming formula
QUESTION [3 upvotes]: I am sorry if this is a noob question, I need help with relatively simple math problem and assurance that I understand the problem correctly.
I have a map-like program that zooms in and zooms out if you press the + and - icons (much like google map). Every time you zoom the map size doubles or halves. I know the minimum and maximum size of the map, how do I calculate the number of the doubling steps?
I was thinking of this:
A on power of 2*x equals B, where A is the minimal zoom, B is the maximal zoom a x is the number of steps between them. Am I right?
How do I calculate x form that formula?
Many thanks

REPLY [2 votes]: If I understand the question right, after the step $1$ you doubled $A$; after
the step $2$, you have $2^{2}A$; after step $x$ you have $$2^{x}A=B.$$ Applying logarithms we get
$$\ln \left( 2^{x}A\right) =\ln B\Leftrightarrow \ln \left( 2^{x}\right) +\ln
A=\ln B$$
$$\Leftrightarrow x\ln 2+\ln A=\ln B\Leftrightarrow x=\frac{\ln B-\ln A}{\ln 2}$$
If $x\notin \mathbb{Z}$, then you have to take
$$\left\lceil x\right\rceil =\left\lceil \frac{\ln B-\ln A}{\ln 2}%
\right\rceil $$
or
$$\left\lfloor x\right\rfloor =\left\lfloor \frac{\ln B-\ln
A}{\ln 2}\right\rfloor .$$
i.e. $n$ steps
$$\left\lfloor \frac{\ln B-\ln A}{\ln 2}\right\rfloor \leq n\leq \left\lceil
\frac{\ln B-\ln A}{\ln 2}\right\rceil $$


TITLE: Quantization of strings, string Fock space and transition to QFT
QUESTION [12 upvotes]: I am not an expert of string theory and am quite uncertain about the basic ideas of string theory that I am going to ask about. I would appreciate some hints of more experienced physicists.
What I am trying to understand is how string theory can describe particles as quantum fields as in quantum field theory. This should be possible since, after all, it is said that QFT can be seen as a low energy limit of string theory.
What I would like to do is to compare second quantization in QFT to string theory. In QFT, if the wave function of a quantum mechanical particle can be described as a superposition of a complete set of wave functions $\{\phi_i(x)\}$, then we can define creation (annihilation) $a_i^+$($a_i$) operators which generate symmetrized/antisymmetrized products of corresponding one-particles states. So, by this construction we have for example $|100\dots\rangle = a_1^+|0\rangle = |\phi_1\rangle$ where again $\langle x|\phi_1\rangle= \phi_1(x)$. The commutation relations for $a_i^+$ and $a_i$ follow by construction, this is then called canonical quantization. Finally, changing the basis from $\{|\phi_i\rangle\}$ to $\{|x\rangle\}$ we obtain the field operators $\psi(x) = \sum_i\phi_i(x)a_i$ and $\psi(x)^+ = \sum_i\phi_i^*(x)a_i^+$, such that $\psi(x)^+|0\rangle = |x\rangle = \sum_i\phi_i^*(x)|\phi_i\rangle$. So, to sum up, field operators $\psi(x)^+$ create superposition states with a probability distribution which is equal to a delta function:
$$ \langle x'|x\rangle = \sum_i\psi_i(x')\psi_i^*(x) = \delta(x-x')$$
And again, commutation relations for field operators follow by definition.
In string theory, the coordinates $(\sigma,\tau)$ on the world sheet which parametrize the embedding $X^\mu(\sigma,\tau)$ of the string in spacetime play the role of spacetime coordinates $(x,t)$ in QFT, and the embedded string $X^\mu(\sigma,\tau)$ plays the same role as field operators $\psi(x)$, and with these identifications the quantization is done along the same lines. But with this construction it is not clear to me, what these operators $X^\mu$ really represent. If in QFT $\psi(x)|0\rangle$ was a localized wave function of a one-particle state, what is $X^\mu(\sigma,\tau)|0\rangle$? A wave-function localized on the world-sheet? How can we then identify something that lives on some fictitions two-dimensional parameter space $(\sigma,\tau)$ with particles in 4-dimensional space-time? And what is the Fock space and the states therein in string theory? If someone tells me that something like $\alpha^\mu|0\rangle$ ($\alpha^\mu$ being the modes of $X^\mu$) can for example be seen as a photon state, the only thing that I see in common with the photon is the vector index $\mu$. Photons that I know are bosons described by a probability distribution over the space-time. In this sense, how can I make the identification with $\alpha^\mu|0\rangle$?
I would very much appreciate any help to disentangle these ideas!

REPLY [1 votes]: A single string looks like a single particle to a low energy observer

This is true but at same time when you 'quantizate' a string you condensate on a more fine structure (a string) so you unlock for your quantum particle the 'string function'.
Viceversa, if you apply a 'stringification' of a quantum particle you fall into -1 state, you get a tachyon condensation so you loose imaginary mass that is, instead, very very important.


TITLE: How heat is added to a system? Probably a noob question
QUESTION [1 upvotes]: ok so I am literally just getting into thermodynamics, this is literally my first search, what I am wondering is how fundamentally heat is added to a system. For example say I have a flame heating an object at 120 degrees F and I add a second flame beneath the object that is also burning at 120 degrees F, does the total heat applied to the object equal 240 degrees F or is it still 120 Degrees F from a larger source.

REPLY [2 votes]: No matter how many heat sources you use that are all 120 degrees F, once the object being heated reaches 120 degrees F, it will be at equilibrium with the heat sources and will not get hotter.

REPLY [1 votes]: As pointed out by @Alchimista temperature and heat are two different things. Temperature is a measure of part of the internal microscopic energy of an object. Heat is defined as energy transfer between objects due to a temperature difference between the objects. So to answer the title of your post heat will transfer to a system (object) if the system is exposed to something that has a higher temperature than the system. But once the temperatures are the same heat transfer stops.
Now everything else being the same, the rate of heat transfer to the system in the cases of heat conduction and convection will be proportional to the magnitude of the temperature difference. So if you double the temperature difference between the heat source and the system you will theoretically double the rate of heat transfer.
Generally, depending on the mass of the system and the heat source, and assuming no system phase changes and no system work, as heat transfers to the system the temperature of the system will rise and the temperature of the source will fall until eventually they both reach the same temperature somewhere between their original temperatures. The two objects are then said to be in thermal equilibrium with each other and heat transfer ceases.
Beyond that if you are really interested in learning more take a course and/or get a good textbook on thermodynamics.
Hope this helps as a primer.


TITLE: The Kelvin-Celsius problem
QUESTION [0 upvotes]: Let's suppose we have temperatures
30°C and 35°C. Converting them to Kelvin we have 303.15K and 308.15K.
In the second case, the temperature difference is 5K. While in first case, temperature difference is 5°C, which is convertible to 278.15K.
This clearly is absurd. I have some ideas why it goes wrong, but could someone please provide a clear explanation? Cannot we use conversions on difference?
This seems to be a general problem with units which have additive relationship (Kelvin and Celsius)  rather than a multiplicative one(metre -centimetres).

REPLY [0 votes]: It's only meaningful to describe the temperature of an object if the zeroth law of thermodynamics is true. There's probably no proof of the zeroth law of thermodynamics. See my answer at Is there any proof for the 2nd law of thermodynamics?. Assuming the zeroth law of thermodynamics, just because adding 2 distance given in meters gives the same result is converting them into kilometers then adding them the converting the result back to meters doesn't mean the same goes for the Celcius temperautre scale and the Kelvin temperature scale. For 2 objects of any temperature, English works in such a way that it's correct to say that a temperature is some number of degrees Celsius when the difference in the numbers that describe their temperature Celcius is that number. It's also correct to say their tempertuare difference is a certain number of Kelvins when the difference between the numbers that describe their Kelvin temperatures is that number. Although saying an object has a temperature of 5°C is another way of saying it has a temperature of 278.15 K, saying 2 objects have a temperature difference of 5°C is not another way of saying they have a temperature difference of 278.15 K. Because of the distributivity of multiplication over addition, given 2 distances in meters, you get the same answer if you just add them as if you convert them into kilometers then add them then convert the result back into meters.


TITLE: Every Hilbert space is isometrically isomorphic with $\ell^2$
QUESTION [2 upvotes]: Let $H$ be a hilbert space and let $\{u_\alpha\}_{\alpha \in A}$ be a
  orthornormal basis ($A$ is not supposed to be countable a priori).
  Then there is an isometric isomorphism between $H$ and
  $\ell^2(\mathcal{A})$ given by $x \mapsto \hat{x}$ with $\hat{x}_\alpha = \langle x,u_\alpha\rangle$.

Proof:
Linearity follows by linearity of scalar product: $(\widehat{ax + by})_\alpha = \langle ax + by,u_\alpha\rangle = a\langle x,u_\alpha\rangle + b\langle y,u_\alpha\rangle = a\hat{x}_\alpha + b\hat{y}_\alpha$
From Parseval we have $\|x\| = \sum_{\alpha \in A} |\langle x, u_\alpha\rangle| = \|\hat{x}\|_2$ and then the mapping is an isometry (and hence also injective).
To conclude we need to prove that this isometry is also surjective.
Let then $f$ be in $\ell^2(A)$. Then, by definition of $\ell^2$, $\sum_{\alpha \in A} |f_\alpha|^2 < \infty$, this implies that $|\{\alpha : f_\alpha \not = 0 \}| \le |\mathbb{N}|$ and thus also the sum $\sum_{\alpha \in A} f_\alpha u_\alpha$ has countable many nonzero terms. Then, by Pythagorean theorem (using continuity of inner product) $\|\sum f_\alpha u_\alpha \|^2 = \sum_{\alpha \in A} |f_\alpha|^2 < \infty$ so this series is absolutely convergent and then $z = \sum_{\alpha \in A} f_\alpha u_\alpha$ exists. Clearly $\hat{z} = f$
$\square$
Q1: Is this proof correct?

This proof is worked out from Folland's Real Analysis book. In the book though, Prof. Folland writes:

If $f \in \ell^2(A)$ then $\sum_{\alpha \in A} |f_\alpha|^2 < \infty$
  so the Pythagorean theorem shows that partial sums of the series
  $\sum_{\alpha \in A} f_\alpha u_\alpha$ (of which only countable many
  terms are nonzero) are Cauchy.

I know the definitions but I don't get why Prof. Folland says that Pythagorean theorem asserts that partial sums are Cauchy. I think Pythagorean theorem in this case give us absolute convergence of the sum, not just the fact partial sums are Cauchy. In this case the space is complete so there is no real distinction between Cauchy and convergence but I care about logical order of the proof.
Q2: Why Pythagorean theorem asserts only Cauchy property of partial sums and not absolute convergence as I said?
Thanks in advance

Note: Uncountable sums of non negative terms are defined like this:
$$ \sum_{\alpha \in A} g(\alpha) := \sup \left\{ \sum_{\alpha \in F} g(\alpha) : F \subset A, \ F \ \text{finite} \right\}$$
with $g: A \to [0,\infty)$

REPLY [2 votes]: If $\mathcal{H}$ is a Hilbert space and $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is an orthonormal subset on $\mathcal{H}$, then
$$
     x=\left(x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\right)+\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n} \;\;\; (\dagger)
$$
is an orthogonal decomposition. Therefore,
\begin{align}
          \|x\|^2&=\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2+\|\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2 \\
        &\ge \|\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2 \\
        &= \sum_{n=1}^{N}|(x,e_{\alpha_n})|^2
\end{align}
Therefore, for any $\epsilon > 0$, the following set is finite:
$$
             \Lambda_{x,\epsilon}=\{ \alpha \in \Lambda : |(x,e_{\alpha})| \ge \epsilon \}
$$
Therefore, the following set is empty, finite, or countable infinite.
$$
           \{ \alpha\in \Lambda : (x,e_{\alpha}) \ne 0 \} = \bigcup_{N=1}^{\infty}\Lambda_{x,1/N}.
$$
Let $x\in\mathcal{H}$, and let $\{\alpha_n\}$ be an enumeration of the $\alpha\in\Lambda$ for which $(x,e_{\alpha})\ne 0$. The only interesting case is where this set of indices is countably infinite. In that case, the original orthogonal decomposition $(\dagger)$ gives
$$
         \|x\|^2=\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2+\sum_{n=1}^{N}|(x,e_{\alpha_n})|^2.
$$
Therefore $\|x\|^2=\lim_{N}\sum_{n=1}^{N}|(x,e_{\alpha_n})|^2$ iff
$$
            \lim_{N}\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|=0.
$$
So, Parseval's equality holds for some $x$ with respect to $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ iff $\lim_{N}\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}=x$ converges in the norm of $\mathcal{H}$. And, in that case, any ordering for the sum will give the same limit because the Parseval equality involves an absolutely convergent series of positive numbers and, hence, may be ordered in any way. This is what is meant by the unordered sum converging.


TITLE: Texts on moduli of elliptic curves
QUESTION [5 upvotes]: I want to study FLT (Fermat's Last Theorem), and now I'm studying moduli of elliptic curves.
I've heard that Deligne-Rapoport, Katz-Mazur, Mazur's "Modular curves...", and Katz's "p-adic..." are very good for this topic.
But I don't know what's the difference between these papers.
For me it seems that these treat almost same topics.
So what should I read at first?
(Now I'm reading Katz-Mazur.
It's easy to read even for me, a beginner of arithmetic geometry.
So I think I should read it at first.
And glancing through Mazur, it seems to use many results from other 3 papers.)
And would you recommend other good papers which I read understand for understanding FLT?
Any help will be much appreciated!

REPLY [2 votes]: Not having been mentioned before, I would recommend the two books "Fermat's Last Theorem, Basic Tools" and "Fermat's Last Theorem, The Proof" by Takeshi Saito. https://bookstore.ams.org/mmono-243 and https://bookstore.ams.org/mmono-245.
In particular, moduli of elliptic curves appear in chapter 2.2 but only over $\mathbb{Q}$. Then the second volume starts with chapter 8 "Modular curves over $\mathbb{Z}$" which covers the topics on the moduli spaces of elliptic curves needed for Fermat's Last Theorem.


TITLE: Reedy model structure on sSet
QUESTION [3 upvotes]: According to this question, there is a model structure on $\mathrm{Set}$ in which the cofibrations are the monomorphisms, the fibrations are maps which are either epimorphisms or have empty domain, and the weak equivalences are the maps f:X→Y such that X and Y are both empty or both nonempty.
As $\Delta^{op}$ is a Reedy category. We can consider the model structure induced on $\mathrm{Set}^{\Delta^{op}}$.
How does this model structure compare to the well known model structures of $\mathrm{sSet}$ (Quillen, Joyal)?
There is probably little hope, but can we hope a structure on $\mathrm{Set}$ which induces the Quillen model structure?

REPLY [7 votes]: There are only nine possible model structures on $\mathbf{Set}$. You can easily check that none of them give the right weak equivalences:

Either everything is a weak equivalence,
or $X \to Y$ is a weak equivalence if and only if both $X$ and $Y$ are empty or both $X$ and $Y$ are non-empty,
or only isomorphisms are weak equivalences.


TITLE: Potentially identity elements in an Abelian group
QUESTION [4 upvotes]: I didn't see this problem before. I motivated by the questions

Is every commutative group structure underlying at least one (unitary, commutative) ring structure
A basic question about rings

Suppose A is an Abelian group such that it is possible to define an associative unitary ring structure on A. An element $x\in A$ is called potentially identity if there exists an associative ring $R$ such that $A$ is the additive group of $R$ and $x=1_R$, the identity element of $R$. Determine the set of all potentially identity elements of $A$.

REPLY [1 votes]: Maybe it will be useful for you the notion of potentially invertible elements of a semigroup introduced by E. Shutov [E.S. Lyapin, "Semigroups" , Amer. Math. Soc.].


TITLE: Find function $F(x,y)$ given its total differential
QUESTION [0 upvotes]: If I have a differential $dF = 2xy e^{xy^2} dy + y^2e^{xy^2} dx $, what are the steps to find the original function? Is there a formula?

REPLY [1 votes]: Note that the total differential is for a function, $F$, of two variables is given by
$$dF=\partial_{x}Fdx+\partial_{y}Fdy$$
Now to find the function $F$, we need only to choose $\partial_{x}F$ or $\partial_{y}F$ and integrate appropriately. So
$$F=\int\partial_{x}Fdx=\int y^2e^{xy^2}dx=e^{xy^2}+g(y)$$
Note that we have a function of $y$ instead of a constant of integration since we integrated with respect to $x$.
To solve for $g$, we take the derivative of $F$ with respect to $y$ and compare to the form in the total differential.
So
$$\partial_{y}F=\frac{\partial}{\partial y}(e^{xy^2}+g(y))=2xye^{xy^2}+g'(y)=2xye^{xy^2}$$
Thus
$$g'(y)=0\Rightarrow g(y)=C$$.
Therefore
$$F(x,y)=e^{xy^2}+C$$


TITLE: If $f:X\to Y$ is linear and $y^\ast \circ \ f \in X^\ast$ for all $y^\ast \in Y^\ast$, then $f$ is continuous
QUESTION [2 upvotes]: Let $X,Y$ be normed over $\mathbb{K}$ (which can be $\mathbb{R}$ or $\mathbb{C}$). If $f:X\to Y$ is linear and $y^\ast \circ \ f \in X^\ast$ for all $y^\ast \in Y^\ast$, then $f$ is continuous.
I came up to a solution to this problem which I know is false, but I can't spot where is the wrong argument. The problem is that my solution only need to use one $y^\ast \in Y^\ast$ to prove $f$ is continuous. I want to know what I'm doing wrong and how is the correct argument to solve this problem. If it makes things better you may consider balls instead neighborhoods in all the arguments below.
Here is my "solution": Given any neighborhood $V$ of $0\in Y$, it's enough to show there is a neighborhood $U$ of $0\in X$ such that $f(U) \subset V$. Start considering any neighborhood $A$ of $0\in\mathbb{K}$ and any $y^\ast \in Y^\ast$, we know $(y^\ast)^{-1}(A)$ is a neighborhood of $0\in Y$, for $y^\ast$ is linear and continuous. Therefore the collection $\{\frac{1}{n}(y^\ast)^{-1}(A)\}_{n\in\mathbb{N}}$ is a local base for $Y$. In particular, exists $n\in\mathbb{N}$ such that $\frac{1}{n}(y^\ast)^{-1}(A) \subset V$.
PS: in this context, local base is a collection of neighborhoods of $0\in Y$ such that, for any neighborhood $W$ of $0\in Y$, there is member in this collection contained in $W$.
We have that $y^\ast\circ f$ is also continuous, so there is a neighborhood $U$ of $0\in X$ such that $y^\ast( f(U)) \subset A$, so we get $f(U)\subset (y^\ast)^{-1}(A)$. Now note that
$$f\Big(\frac{1}{n}U\Big) = \frac{1}{n}f(U) \subset \frac{1}{n}(y^\ast)^{-1}(A) \subset V.$$
Therefore $\frac{1}{n}U$ is the neghborhood of $0\in X$ we are looking for, and from this we can conclude $f$ is continuous.

REPLY [1 votes]: If $X$ and $Y$ are Banach spaces continuity follows from the closed graph theorem. In the normed case you have to show that $f$ maps bounded sets to bounded sets and this follows from the weak continuity since the norm topology and the weak topology on $Y$ have the same bounded sets.


text:
\begin{document}
\title{Adding one random real}
\author{Tomek Bartoszy\'{n}ski}
\thanks{First author partially supported by  SBOE grant 92--096}
\address{Department of Mathematics\\
Boise State University\\
Boise, Idaho 83725 U.S.A.}
\email{tomek@@math.idbsu.edu}
\author{Andrzej Ros{\l}anowski}
\thanks{Second author partially supported by KBN grant 1065/P3/93/04}
\address{Department of Mathematics\\
Hebrew University\\
Jerusalem, Israel, and\\
Mathematical Institute\\
Wroclaw University\\
Wroclaw, Poland}
\email{roslanow@@sunrise.huji.ac.il}
\author{Saharon Shelah}
\thanks{Third author partially supported by Basic Research Fund,
Israel Academy of Sciences, publication 490}
\address{Department of Mathematics\\
Hebrew University\\
Jerusalem, Israel}
\email{shelah@@sunrise.huji.ac.il}
\subjclass{03E35}
\maketitle
\begin{abstract}
We study the cardinal invariants of measure and category after adding
one random real. In particular, we show that the number of measure
zero subsets of the plane which are necessary to cover graphs of
all continuous functions maybe large while the covering for measure
is small.
\end{abstract}
\section{Introduction}
Let ${\cal J}$ be an ideal of
subsets of the real line (where real line means $\reals$, $2^\omega$
or $[0,1]$).
Define the following  cardinal invariants:
\begin{enumerate}

\item $\add({\cal J})=\min\{|{\cal A}|:
{\cal A} \subseteq {\cal J} \ \&\ \bigcup {\cal A} \not \in
{\cal J}\}$,
\item $\cov({\cal J})=\min\{|{\cal A}|:
{\cal A} \subseteq {\cal J} \ \&\ \bigcup {\cal A} =\reals\}$,
\item $\unif({\cal J})=\min\{|X|: X \subseteq \reals \ \& \ X \not\in
  {\cal J}\}$,
\item $\cof({\cal J})=\min\{|{\cal A}|: {\cal A} \subseteq {\cal J}
  \&\ \forall A \in {\cal J} \ \exists B \in {\cal A} \ A \subseteq B\}$.
\end{enumerate}

Let ${\cal M}$ and ${\cal N}$ be the ideals of meager and of measure zero
subsets of the real line respectively.
Finally let
${\frak b}$ be the size of the smallest
unbounded family in $\omega^{\omega}$ and
${\frak d}$ the size of the smallest
dominating family in $\omega^{\omega}$ .

The relationship between these cardinals is
described in the following diagram, where arrows
  means $\leq $:

\onesetup  \cov(\N) \unif(\M) \cof(\M) \cof(\N) {\goth b}

\twosetup {\goth d} \add(\N) \add(\M) \cov(\M) \unif(\N)

\ccdiagram

In addition $\add({\cal M}) = \min\{{\frak b}, \cov({\cal M})\}$ and
$\cof({\cal M}) = \max\{{\frak d}, \unif({\cal M})\}$.

The proofs of those inequalities can be found in \cite{Bar84Add}, \cite{fre:cichon}
and \cite{Mil81Som}.
In this paper we show that except for $\cov(\N)$ and $\unif(\N)$
values of these invariants do not change when one random real is added.
Let $\bold B $ be the measure algebra adding one random real.
\begin{theorem}[Pawlikowski, Krawczyk]\label{pawlikowski}
The following holds in $\V^{\bold B}$:
\begin{enumerate}
\item $\add(\N)=\add(\N)^{\bold V}$ and
$\cof(\N)=\cof(\N)^{\V}$,
\item $\cov(\N) \geq \cov(\N)^{\V}$ and
$\unif(\N)\leq\unif(\N)^{\V}$,
\item $\cov(\N) \geq {\frak b}$ and $\unif(\N) \leq {\frak d}$,
\item  ${\frak b} = {\frak b}^{\V}$ and ${\frak d} = {\frak d}^{\V}$,
\item $\cov(\M) \geq \cov(\M)^{\V}$ and
$\unif(\M) \leq \unif(\M)^{\V}$,
\item $\add(\M) \geq \add(\M)^{\V}$ and
$\cof(\M) \leq \cof(\M)^{\V}$.
\end{enumerate}
\end{theorem}
\Proof (1), (2) and (4) is folklore (see \cite{Paw86Sol}).
(3) is due to Krawczyk (see \cite{Paw86Sol} or \cite{Kra83}), (5) is
due to Pawlikowski
(\cite{Paw86Sol}) and (6) follows from (5), (4) and the remarks above.~$\QED$

For a set $H \subseteq \reals\times\reals$ and $x,y \in \reals$ let
$(H)_x = \{y : \<x,y\> \in H$ and let $(H)^y=\{x : \<x,y\> \in H\}$.

We will use the following classical lemma:
\begin{lemma}\label{represent}
  Suppose that $r$ is a random real over $\V$. Then
  \begin{enumerate}
    \item for every $x \in \V[r] \cap \reals$ there exists a Borel
      function $f : \reals \longrightarrow \reals$ such that $x =
      f(r)$,
    \item for every Borel measure zero set $F \in \V[r]$ there exists
      a Borel measure zero set $H \subseteq \reals\times\reals$, $H
      \in \V$ such that $F = (H)_r$.~$\QED$
  \end{enumerate}
\end{lemma}
We will need the following characterization of $\cov(\M)$ and
$\unif(\M)$.

Let $$\cal S = \{S \in ([\omega]^{<\omega})^\omega : \forall n \ |S(n)|
\leq (n+1)^2\}.$$

\begin{theorem}[\cite{Bar87Com}]\label{cohen}
  The following conditions are equivalent:
  \begin{enumerate}
  \item $\cov(\M)\geq\kappa$,
  \item for every family $F \subseteq \omega^\omega$ of size $<\kappa$
    there exists $g \in \omega^\omega$ such that
$$\forall f \in F \ \exists^\infty n \ f(n)=g(n).$$
\item for every family $F \subseteq \omega^\omega$ of size $<\kappa$
    there exists $S \in \cal S$ such that
$$\forall f \in F \ \exists^\infty n \ f(n) \in S(n).$$
  \end{enumerate}
Similarly,
\begin{enumerate}
\item $\unif(\M)\geq\kappa$,
\item for every family $F \subseteq \omega^\omega$ of size $<\kappa$
    there exists $g \in \omega^\omega$ such that
$$\forall f \in F \ \forall^\infty n \ f(n) \neq g(n).$$
\item for every family $F \subseteq \cal S$ of size $<\kappa$
    there exists $f  \in \omega^\omega $ such that
$$\forall S \in F \ \forall^\infty n \ f(n) \not \in S(n).~\QED $$
\end{enumerate}
\end{theorem}

\section{Cohen reals}
In this section we will show that invariants $\cov(\M)$ and
$\unif(\M)$ do not change when random reals are added.
\begin{theorem}
  The following holds in $\V^{\bold B}$:
\begin{enumerate}
\item $\cov(\M) = \cov(\M)^{\V}$ and
$\unif(\M) = \unif(\M)^{\V}$,
\item $\add(\M) = \add(\M)^{\V}$ and
$\cof(\M) = \cof(\M)^{\V}$.
\end{enumerate}
\end{theorem}
\Proof (1) By \ref{pawlikowski}, it is enough to show that in
$\V^{\bold B}$, $\cov(\M) \leq \cov(\M)^{\V}$ and
$\unif(\M) \geq \unif(\M)^{\V}$.

By \ref{cohen}, there exists a family $F \subseteq \omega^\omega$ of
size $\cov(\M)^{\V}$ such that
$$\forall S \in {\cal S} \ \exists f \in F \ \forall^\infty n \ f(n)
\not \in S(n).$$

By \ref{cohen}, to finish the proof it is enough to show that
$$\V^{\bold B} \thinks \forall g  \in \omega^\omega  \ \exists f \in F
\ \forall^\infty n \ f(n) \neq g(n).$$

Let $\dot{g}$ be a $\bold B$-name for an element of $\omega^\omega$.
Define for $n \in \omega$,
$$S(n) = \left\{k \in \omega : \mu\left(\lbv \dot{g}(n)=k \rbv_{\bold
B}\right)>\frac{1}{(n+1)^2} \right\},$$
where $\mu$ is the Lebesgue measure.
It is clear that $|S(n)| <(n+1)^2$ for all $n$. Therefore there exists $f
\in F$ and $N \in \omega$ such that $f(n) \not \in S(n)$ for all $n\geq N$.
We claim that
$$\forces_{\bold B} \forall^\infty n \ f(n) \neq \dot{g}(n).$$
Let $p \in \bold B$. Find $n>N$ such that $\sum_{k=n}^\infty k^{-2} <
\mu(p)$. Then
$$q = p - \bigcup_{k=n}^\infty \lbv \dot{g}(k) = f(k)\rbv_{\bold B}
>0$$
and
$$q \forces_{\bold B} \forall k>n \ f(k) \neq \dot{g}(k).$$

\vspace{0.1in}

To show that
$\unif(\M) \geq \unif(\M)^{\V}$ holds in $\V^{\bold B}$, we
``dualize'' the above argument.

Suppose that $F \subseteq \omega^\omega$ is a family of size
$\unif(\M)$ in $\V^{\bold B}$ such that
$$\V^{\bold B} \thinks \forall g \in \omega^\omega \ \exists f \in F \
\exists^\infty n \ f(n)=g(n).$$
Let $\dot{F}=\{\dot{f} : f \in F\}$ be a set of $\bold B$-names for
elements of $F$. Without loss of generality we can assume that
$\dot{F} \in \V$. For $\dot{f} \in \dot{F}$ let $S_f \in \cal S$ be
defined as
$$S_f(n) = \left\{k \in \omega : \mu\left(\lbv \dot{f}(n)=k \rbv_{\bold
B}\right)>\frac{1}{(n+1)^2} \right\}.$$
As before we show that
$$\forall g \in \omega^\omega \ \exists \dot{f} \in \dot{F} \
\exists^\infty n \ g(n) \in S_f(n),$$
which by \ref{pawlikowski}, finishes the proof.

To show the second part use \ref{pawlikowski} and the fact that
$\add({\cal M}) = \min\{{\frak b}, \cov({\cal M})\}$ and
$\cof({\cal M}) = \max\{{\frak d}, \unif({\cal M})\}$.~$\QED$

Recall that a set $X \subseteq \reals$ has strong measure zero if for
every sequence of positive reals  $\<\varepsilon_n: n \in \omega\>$
there exists a sequence of intervals $\<I_n :n \in \omega\>$ such that
the length of $I_n$ is $\leq \varepsilon_n$ and $X \subseteq
\bigcup_{m} I_m$. Note that, equivalently we can request that
$X \subseteq
\bigcap_{n \in \omega} \bigcup_{m>n} I_m$.
\begin{theorem}\label{strmeazer}
  Suppose that $X \subseteq \reals$ and $X \in \V$. Then  $X$ has
  strong measure zero in $\V$ iff $X$ has
  strong measure zero in $\V^{\bold B}$.
\end{theorem}
\Proof
It is easy to see that for every sequence $\<\varepsilon_n: n \in
\omega\> \in \V^{\bold B}$ there exists a sequence $\<\delta_n : n \in
\omega\> \in \V$ such that $\delta_n \leq \varepsilon_n$ for all $n$.
Therefore, if $X$ has
  strong measure zero in $\V$ then $X$ has
  strong measure zero in $\V^{\bold B}$.

Suppose that $X$ does not have strong measure zero in $\V$ and let
$\<\varepsilon_n : n \in \omega\>$ be a sequence of positive reals
witnessing that.  Suppose that $X$ has strong measure zero in $\V^{\bold
B}$. Let $\<\delta_n : n \in \omega\>$ be a decreasing sequence of positive
reals such that $\delta_{n} <\varepsilon_k$ for all $k\leq n^3$.  Let
$\<\delta_n':n\in\omega\>$ be a decreasing sequence of positive rationals
such that $\delta_{2k}'=\delta_{2k+1}'$ and $\delta_n'<\delta_n$. By the
assumption we can find a sequence of intervals $\<I_n : n \in
\omega\> \in \V^{\bold B}$ such that $X \subseteq
\bigcap_{n \in \omega} \bigcup_{m>n} I_m$ and the length of $I_m$ is less
than $\delta_{2m}'$.
Let $\<I(n,k):k \in \omega\>$ be a partition of $\reals$ into rational intervals
of the length $\delta_n'$. Each interval $I_m$ is covered by $I(2m,k)\cup
I(2m,k+1)$ for some $k=k(m)$. Let $\<\dot{I}_n:n\in\omega\>$ be a ${\bold
B}$-name for the sequence $\<I(2m,k(m)), I(2m,k(m)+1):m\in\omega\>$ (i.e.
$\dot{I}_{2m}$ is a name for $I(2m,k(m))$ and $\dot{I}_{2m+1}$ is that for
$I(2m, k(m)+1)$). Thus
$$\forces_{\bold B}\text{``the length of $\dot{I}_n$ is $\delta_n'$ \&
$X\subseteq\bigcap_{n\in\omega}\bigcup_{m>n}\dot{I}_m$''.}$$
Now define for $n=2m+i$ ($i=0,1$):
$${\cal A}_n = \left\{I(2m,k) : \mu\left(\lbv\dot{I}_n=I(2m,k) \rbv_{\bold
B}\right)>\frac{1}{(n+1)^2} \right\}.$$
Note that $|{\cal A}_n|<(n+1)^2$ (some ${\cal A}_n$'s may be empty). By the
choice of the sequence $\<\delta_n : n \in \omega\>$ if we order
lexicographically the intervals in $\bigcup_{n \in \omega} {\cal A}_n$ in a
sequence $\<J_n : n \in \omega\>$, then the length of $J_n$ will be $\leq
\varepsilon_n$. Let $x \in X$ be such that $x \not \in
\bigcup_{n \in \omega} J_n$. Note that then for each $n\in\omega$
$$\mu(\lbv x\in\dot{I}_n\rbv)\leq\frac{1}{(n+1)^2}$$
Let $p \in \bold B$. Find $n$ such that $\sum_{k=n}^\infty k^{-2} <
\mu(p)$. Then
$$q = p - \bigcup_{k=n}^\infty \lbv x \in \dot{I}_n\rbv_{\bold B}
>0$$
and
$$q \forces_{\bold B} \forall k>n \ x \not\in \dot{I}_n.~\QED$$

The proof of \ref{strmeazer} seems to suggest that a filter $\F\in\V$ on
$\omega$ which cannot be extended to a rapid filter in $\V$ cannot be
extended to a rapid filter in $\V^{\bold B}$. However, this is not the
case. First, let us recall that a non-principal filter $\F$ on $\omega$ is
called rapid if for every increasing function $f\in\omega^\omega$ there
exists $X\in\F$ such that $|X\cap f(n)|\leq n$ for all $n$.
\begin{theorem}
Suppose that $\D$ is a rapid filter on $\omega$. Then there
exists a filter $\F$ such that:
\begin{enumerate}
\item $\F$ cannot be extended to a rapid filter in $\V$,
\item $\V^{\bold B}\models$``$\F$ can be extended to a rapid filter''.
\end{enumerate}
\end{theorem}
\Proof Let $\F$ be the family of all subsets $A$ of $\omega$ such that for
some set $X\in\D$ the sequence
$$\frac{|A\cap [n^2,(n+1)^2)|}{2n+1} \stackrel{n\in
  X}{\longrightarrow} 1.$$
It should be clear that if $A\subseteq B$, $A\in \F$ then
$B\in \F$ and the same set $X\in\D$ witnesses it. Moreover if $A,B\in\F$ is
witnessed by $X_A,X_B\in \D$ then the intersection $X_A\cap X_B\in\D$
witnesses that $A\cap B\in \F$. Consequenlty $\F$ is a non-principal
filter on $\omega$. We claim that $\F$ cannot be extended to a rapid
filter.
Suppose that a set $A\subseteq\omega$ is such that $|A\cap n^3|\leq n$
for $n \in \omega$. Then for each $m\in\omega$ we have
$$\frac{|A\cap [m^2,(m+1)^2)|}{2m+1}\leq \frac{[(m+1)^{2/3}]+1}{2m+1}\leq
\frac{(m+1)^{2/3}+1}{2m+1}$$
and hence
$$\lim_{m\rightarrow\infty}\frac{|A\cap [m^2,(m+1)^2)|}{2m+1}=0.$$
Consequently the complement $\omega\setminus A$ of the set $A$ belongs to
$\F$ and $A$ cannot be in any filter extending $\F$.

\vspace{0.2in}

To prove the assertion $(2)$ we work with the measure algebra on the
space $\prod\limits_{n\in\omega}[n^2,(n+1)^2)$ equipped with
the natural product measure $\mu$ (we use the same symbol as for the
Lebesgue measure since this measure corresponds to Lebesgue measure
under cannonical mapping of underlying space onto the interval
$[0,1]$).
Suppose that
$r\in\prod\limits_{n\in\omega}
[n^2,(n+1)^2)$ is a random real over $\V$ and work in $\V[r]$.
First note
that for a set $A \in \F$ and $X \in \cal D$,
\begin{multline*}
\mu\left(\left\{x \in  \prod\limits_{n\in\omega}[n^2,(n+1)^2):
\forall^\infty m\in X \ x(m)\notin A\right\}\right) \leq \\
\sum_{n=0}^\infty
\prod^\infty\begin{Sb}m=n\\
m \in X
\end{Sb}
 \left(1-\frac{|A\cap [m^2, (m+1)^2)|}{2m+1}\right)=0.
\end{multline*}

In particular, since $r$ is a random real,
$$\forall X \in \cal D \ \forall A \in \F \ A \cap \ran(r\rest X) \neq
\emptyset.$$

Consequently $\F\cup\{\ran(r\rest X): X \in \cal D\}$ generates a
filter $\F^\star$. We are going
to show that it is a rapid filter. Suppose that $f\in\omega^\omega\cap
\V[r]$ is an increasing function. Since random real forcing is
$\omega^\omega$-bounding we can assume that $f \in \V$.  Since $\D$
was a rapid filter in $\V$ we find
a set $X\in D$ such that $|X\cap f(n)|\leq n$ for $n \in \omega$. Look
at the set $A=\{r(n) : n \in X\}$. For
every $n\in\omega$ we have:
$$|A\cap f(n)|\leq  |X\cap f(n)|
\leq n.$$
The theorem is proved.~$\QED$

\section{Random reals}
Theorem \ref{pawlikowski} shows that in $ \V^{\bold B}$,  $\cov(\N) \geq
\max\{\cov(\N)^{\V}, {\frak b}^{\V}\}$. In this section we will show
that it is consistent that $\cov(\N)^{\V^{\bold B}} >
\max\{\cov(\N)^{\V}, {\frak b}^{\V}\}$.

We will need the following notation:
\begin{definition}
  Let $\N_2$ be the ideal of measure zero subsets of
$\reals\times\reals$ and let $\Borel(\reals)$ be the collection of
all Borel mappings from $\reals $ into $\reals$. Define

\begin{multline*}
\cov^\star(\N)=\min\lft2\{|{\cal A}|:  {\cal A} \subseteq \N_2 \ \&\
\forall f \in \Borel(\reals) \ \forall B \in \Borel \setminus \N \ \exists
H \in {\cal A} \\
 \lft1\{x \in B: \<x,f(x)\> \in H\lft1\}\not\in \N\rgt2\}
\end{multline*}
 and
\begin{multline*}
\unif^\star(\N)=\min\lft2\{|X| : X \subseteq \Borel(\reals) \ \& \ \forall
H \in \N_2 \ \forall B \in \Borel \setminus \N \ \exists f \in X \\
 \lft1\{x
\in B : \<x,f(x)\> \not \in H\rgt1\} \not\in \N\rgt2\}.
\end{multline*}
\end{definition}

As a consequence of \ref{represent}, we get:
\begin{lemma}
  $\cov^\star(\N)=\cov(\N)^{\V^{\bold B}}$ and
  $\unif^\star(\N)=\unif(\N)^{\V^{\bold B}}$.~$\QED$
\end{lemma}

The goal of this section is to show that the coefficient
$\cov^\star(\N)$ can be large while both $\frak b$ and $\cov(\N)$ are
small and that $\unif^\star(\N)$ can be small while both $\unif(\N)$
and $\frak d$ are large.

The key to our construction is the following theorem:
\begin{theorem}
  There exists a forcing notion ${\cal P}$, adding generically a
  continuous function $h_G:\reals \longrightarrow \reals$, such that
  \begin{enumerate}
    \item ${\cal P}$ is $\sigma$-centered,
    \item $\forall f \in \omega^\omega \cap \V^{\cal P} \ \exists g
      \in \V \cap \omega^\omega \ \exists^\infty n \ f(n) \leq g(n)$,
    \item for every $H \in \N_2 \cap \V$, $\{x: \<x,h_G(x)\> \in H\}$
      has measure zero.
  \end{enumerate}
\end{theorem}
\Proof
Let $\t$ consists of all pairs $\<\varepsilon,\phi\>$ where $\varepsilon$ is a
rational number in $(0,1)$ and $\phi:2^{<\omega} \times 2^{<\omega}
\longrightarrow [0,1]$ is a
function such that for $s,t \in 2^{<\omega}$:
\begin{enumerate}
\item $\phi(\emptyset,\emptyset)>0$,
\item $\phi(s,t)\leq 2^{-(|s|+|t|)}$,
\item $\phi(s^\frown 0,t)+\phi(s^\frown 1,t)
=\phi(s,t)=\phi(s,t^\frown 0)+\phi(s,t^\frown 1)$.
\end{enumerate}

We define the partial order ${\cal P}$. Conditions are pairs $p=\<h,u\>$ such
that
\begin{enumerate}
\item $u\in [\t]^{<\omega}$
\item $h:2^{\leq m}\longrightarrow 2^{<\omega}$ for some $m=m(p)$,
\item if $s \subseteq t \in 2^{\leq m}$ then $h(s)\subseteq h(t)$,
\item if $\<\varepsilon,\phi\>\in u$ then
$$\sum_{s\in
2^m} 2^{|h(s)|} \phi\lft1(s,h(s)\rgt1) > \varepsilon.$$
\end{enumerate}

The order $\leq$ on ${\cal P}$ is the natural one:
$$(h,u)\geq (h',u')\iff h\supseteq h'\ \&\ u\supseteq u'.$$

\begin{lemma}
\label{density}
Suppose that $p=(h,u)\in{\cal P}$. Then there is $q=(h',u)\in{\cal P}$
such that $q\geq p$, $m(q)>m(p)$ and if $s \in 2^{m(q)}$ then $
\lft1|h'(s)\rgt1|> \lft2|h'\lft1(s\rest m(p)\rgt1)\rgt2|$.
\end{lemma}
\Proof What we have to do is to extend $h$. Note that if we put
$h'(s^\frown i)=h(s)$ (for $s \in 2^{m(p)}$) then $(h',u)$ is a
condition stronger than $p$.  So the only
problem is to extend the ``values'' of $h$.

Take $\delta>0$ such that for every $\<\varepsilon,\phi\>\in u$
$$\delta\cdot\sum_{s \in 2^{m(p)}}2^{1+|h(s)|}< \sum_{s\in
2^m(p)} 2^{|h(s)|} \phi\lft1(s,h(s)\rgt1)-\varepsilon.$$

\begin{lemma}
There are $m'>m(p)$ and $e:2^{m'}\longrightarrow
2$ such that for each $\<\varepsilon,\phi\>\in u$, $s\in 2^{m(p)}$:
$$\frac{1}{2}\phi\lft1(s,h(s)\rgt1)-\delta<\sum_{s\subseteq t
\in 2^{m'}} \phi\lft1(t,h(s)^\frown e(t)\rgt1).\leqno(\otimes)$$
\end{lemma}
\Proof Let $n=|u|$ and let $m'>m(p)$ be such that $2^{-m'}/\delta^2<1/n$.
Fix $s\in 2^{m(p)}$. We are going to find a function $e_s:\{t\in
2^{m'}:s\subseteq t\}\longrightarrow 2$ satisfying the condition
$(\otimes)$ for each $\<\varepsilon,\phi\>\in u$. Consider the space
$\Omega$ of all functions from $\{t\in 2^{m'}: s\subseteq t\}$ to $2$. The
space carries the natural (product) probability measure $P$. For
$\<\varepsilon,\phi\>\in u$ define a random variable
$Y_\phi:\Omega\longrightarrow [0,1]$ by
$$Y_\phi(e)=\sum_{s\subseteq t\in 2^{m'}}\phi\lft1(t,h(s)^\frown e(t)\rgt1).$$
By the Tchebyshev inequality  we know that
$$P\left(\left|Y_\phi-\int
Y_\phi\, d \Omega
\right|\geq\delta\right)\leq\frac{\bold D^2 Y_\phi}{\delta^2}.$$
If we put $X^t_\phi(e)=\phi\lft1(t,h(s)^\frown e(t)\rgt1)$ (for $t\in 2^{m'},
s\subseteq t$) then $X^t_\phi$'s are independent random variables on
$\Omega$ and $Y_\phi=\sum\limits_{s\subseteq t\in 2^{m'}} X^t_\phi$. Now,
\begin{multline*}
\bold D^2
Y_\phi=\int \left(Y_\phi-\int Y_\phi\, d \Omega\right)^2\, d\Omega
=
\int\left(\sum_{s\subseteq t\in
2^{m'}}\left(X^t_\phi-\int X^t_\phi\, d \Omega\right)\right)^2\, d\Omega =\\
\sum_{s\subseteq t\in
2^{m'}}\int
\left(X^t_\phi-\int X^t_\phi \, d\Omega\right)^2\, d\Omega
\end{multline*}
(for the last equality we use the independence of $X^t_\phi$'s). Since
$\left|X^t_\phi -\int X^t_\phi\, d\Omega\right|\leq
2^{-(m'+|h(s)|+1)}$ we get
$$\bold D^2 Y_\phi\leq 2^{m'-m(p)}\cdot 2^{-2m'-2|h(s)|-2}<2^{-m'}.$$
Hence
$$P\left(\left|Y_\phi-\int Y_\phi\, d\Omega \right|\geq\delta\right)<2^{-m'}/\delta^2<1/n$$
and therefore we can find $e_s\in\Omega$ such that for each
$(\varepsilon,\phi)\in u$ we have $\int  Y_\phi - \delta\, d\Omega \leq
Y_\phi(e_s)$. Since
$$\int  Y_\phi\, d\Omega =\sum\limits_{s\subseteq t\in
2^{m'}}\int  X^t_\phi\, d\Omega = \frac{1}{2}\sum\limits_{s\subseteq
t\in 2^{m'}} \phi(t,h(s))= \frac{1}{2}\phi(s,h(s))$$
we get that $e_s$ is as required.~$\QED$

Define $h':2^{\leq m'}\longrightarrow 2^{<\omega}$ by the following
conditions:
\begin{enumerate}
\item $h'\rest 2^{\leq m(p)}=h$,
\item if $s \in 2^{< m'}\setminus 2^{\leq m(p)}$ then $h'(s)=h(s\rest
m(p))$,
\item if $s \in 2^{m'}$ then $h'(s)=h\lft1(s \rest m(p)\rgt1)^\frown e(s)$.
\end{enumerate}
Thus $m(q)=m'$, but we have to prove that $q=(h',u)$ is a condition.

Note that for $\<\varepsilon,\phi\>\in u$ we have then:
\begin{multline*}
\sum_{t \in 2^{m'}} 2^{|h'(t)|}\phi\lft1(t,h'(t)\rgt1)=
\sum_{s\in 2^{m(p)}}\sum_{s \subseteq t\in
2^{m'}} 2^{1+|h(s)|}\cdot \phi(t,h(s)^\frown e(t))>\\
>\sum_{s \in 2^{m(p)}}2^{|h(s)|}\phi\lft1(s,h(s)\rgt1) -
\delta\cdot\sum_{s\in 2^{m(p)}} 2^{1+|h(s)|}>\varepsilon.~\QED
\end{multline*}

Suppose that $G \subseteq {\cal P}$ is generic over $\V$. Let
$\widetilde{h}_G = \bigcup \{h : \<h,u\> \in G\}$ and  for every $x
\in 2^\omega$, let $h_G(x)=\bigcup_{n \in \omega}
    \widetilde{h}_G(x\rest n)$. It follows
immediately from
\ref{density} that

  \begin{lemma}
    $h_G(x): 2^\omega \longrightarrow 2^\omega $ is a continuous
    function in $\V[G]$.~$\QED$
  \end{lemma}

\begin{lemma}
  For every measure zero set $H \subseteq 2^\omega\times2^\omega$
  which is coded in $\V$, the set
$$\{x \in 2^\omega: \<x,h_G(x)\> \not \in H\}$$
has measure one.
\end{lemma}
\Proof
Fix $H$ as above. Suppose that $p=\<h,u\> \in {\cal P}$ and
$\varepsilon
>0$ are given.

It is enough to show that
$\mu\lft2(\{x \in 2^\omega: \<x,h_G(x)\> \not \in H\}\rgt2) >
1-\varepsilon$ holds for every rational $\varepsilon
>0$. Suppose that $p=\<h,u\> \in {\cal P}$ and $m=m(p)$.

Choose a perfect set $F$ disjoint with $H$ of measure so close to one that
$$\sum_{s\in
2^m} 2^{|h(s)|} \mu([s]\times[h(s)] \cap F) > 1-\varepsilon.$$
Define the function $\phi_F:2^{<\omega}\times2^{<\omega}\rightarrow [0,1]$ by
$$\phi_F(s,t)=\mu\lft1([s]\times [t]\cap F\rgt1)$$
and note that $\<1-\varepsilon, \phi_F\>\in\t$. Moreover, $q=\<h, u
\cup\{\<1-\varepsilon, \phi_F\>\}\>$ is a condition. We show that
$$q \forces_{\cal P}\mu\lft2(\{x \in 2^\omega: \<x,h_{\dot{G}}(x)\>
\not \in H\}\rgt2) \geq 1-\varepsilon.$$
Let $F_n = \bigcup \{[s]\times [t]: s,t \in 2^n \ \& \
\lft1([s]\times [t]\rgt1) \cap F \neq \emptyset\}$. Obviously $F =
\bigcap_{n \in \omega} F_n$.
Given $n \in \omega$ there is  $q'=\<h',u'\>$ stronger thatn $q$ such
that $m'=m(q') \geq n$ and $|h'(s)| \geq n $ for $s \in 2^{m'}$.
Since $\<1- \varepsilon, \phi_F\> \in u \subseteq u'$ also $q''=\<h',
u' \cup \{\<1-\varepsilon, \phi_{F_n}\>\} \>$ is a condition stronger
than $p$ and for $s \in 2^{m'}$, $\lft1([s]\times [h'(s)]\rgt1) \cap
F_n \neq \emptyset$ if and only if $[s]\times [h'(s)] \subseteq F_n$.
Hence
$$\mu\left( \bigcup \{[s] : s \in 2^{m'} \ \& \ [s] \times [h'(s)]
\subseteq F_n\}\right) = \sum_{s \in 2^{m'}} 2^{|h'(s)|}
\phi_{F_n}\lft1(s, h'(s)\rgt1) > 1-\varepsilon$$
and
$$q'' \forces_{\cal P} \mu\lft2(\{x: \<x,h_{G}(x)\> \in F_n\}\rgt2)
\geq 1 -\varepsilon.$$
 Using density argument and passing to the limit we get
$$\mu\lft2(\{x: \<x,h_{G}(x)\> \in F\}\rgt2) \geq
1-\varepsilon.~\QED$$

\begin{lemma}\label{nodom}
There exist centered families $\{{\cal P}_i : i \in I\}$, $I$ countable,
such that $\bigcup_{i \in I} {\cal P}_i$ is dense in $ {\cal P}$ and for
every maximal antichain $\{p_n : n \in \omega\}$ in ${\cal P}$ there exists
a natural number $M(i)$ such that for every condition $q \in {\cal P}_i$
there exists $n \leq M(i)$ such that $q $ and $p_n$ are compatible.

In particular, ${\cal P}$ does not add dominating reals.
\end{lemma}
\Proof
For simplicity we will think of the second coordinates of conditions in $\p$
as finite sequences from $\t$.

Let
$$I=\{(N,k,h,\langle\varepsilon_i: i<N\rangle): k,N\in\omega,
h:2^m\rightarrow2^{<\omega}, \varepsilon_j\in(0,1)\cap\rationals \mbox{ for }
j<N.\}\]
For $i=(N,k,h,\langle\varepsilon_i: i<N\rangle)\in I$ let
\begin{multline*}
{\cal P}_i=\lft2\{\<h,\langle\varepsilon_i,\phi_i\rangle_{i<N}\>\in\p: \forall
i<N\ \lft2(\phi_i(\emptyset,\emptyset) \geq 1/k\ \&  \\
  \sum_{s\in
2^m} 2^{|h(s)|} \phi_i\lft1(s,h(s)\rgt1)
\geq\varepsilon_i+1/k\rgt2)\rgt2\}.
\end{multline*}
Clearly each ${\cal P}_i$ is centered (conditions in $\p$ with the same $h$
can be put together) and they cover $\p$.

We want to show that the families $\p_i$ have the required property.  Assume
not. Thus we have a maximal antichain $\langle p_k:k\in\omega\rangle$ in
$\p$ and a sequence $\langle q_n:n\in\omega\rangle\subseteq\p_i$ (for some
$i=(N,k,h,\langle\varepsilon_i: i<N\rangle)$) such that
$q_n\incomp_{\p} p_k$ for $k  \leq n$.

Let $q_n=\<h,\langle\varepsilon_i, \phi^n_i\rangle_{i<N}\>$,
$\phi^n_i(\emptyset,\emptyset)\geq 1/k$, $\sum_{s \in 2^m} 2^{|h(s)|}
\phi^n_i\lft1(s, h(s)\rgt1)\geq
\varepsilon_i+1/k$. Passing to a subsequence we may assume that for each
$i<N$ the sequence $\langle\phi^n_i: n\in\omega\rangle$ is pointwise
converging (note that the space $[0,1]^\omega$ is compact).

Let
$\phi_i:2^{<\omega}\times2^{<\omega}\longrightarrow [0,1]$ be the
limit functions, i.e.
$$\phi_i(s,t)=\lim_{n\to\infty}\phi^n_i(s,t).$$
The functions $\phi_i$ satisfy conditions (1)--(3) of the definition of $\t$
(for the first condition remember that
$\phi^n_i(\emptyset,\emptyset)\geq 1/k$). Moreover
$$\sum_{s\in
2^m} 2^{|h(s)|} \phi_i\lft1(s,h(s)\rgt1)=\lim_{n\to\infty} \sum_{s\in
2^m} 2^{|h(s)|} \phi^n_i\lft1(s,h(s)\rgt1) \geq\varepsilon_i+1/k.$$
Consequently $\<h,\langle\varepsilon_i,\phi_i\rangle_{i<N}\>\in\p$. We find
$k_0\in\omega$ such that the conditions $p_{k_0}$ and
$\<h,\langle\varepsilon_i,\phi_i\rangle_{i<N}\>$ are compatible. Let
$\<h^\star,\langle\varepsilon_i,\phi_i\rangle_{i<N^\star}\>
\geq\<h,\langle\varepsilon_i,
\phi_i\rangle_{i<N}\>, p_{k_0}$ where $h^\star:2^{\leq m^\star}\longrightarrow
2^{<\omega}$, $N^\star>N$. Then we have for $i<N$:
$$\varepsilon_i< \sum_{s\in
2^{m^\star}} 2^{|h^\star(s)|} \phi_i\lft1(s,h^\star(s)\rgt1) =
\lim_{n\to\infty}  \sum_{s\in
2^{m^\star}} 2^{|h^\star(s)|} \phi^n_i\lft1(s,h^\star(s)\rgt1).$$
Consequently for sufficiently large $n$ we will have
$$\varepsilon_i<\sum_{s\in
2^{m^\star}} 2^{|h^\star(s)|} \phi^n_i\lft1(s,h^\star(s)\rgt1).$$
So take $n>k_0$ such that the above holds for each $i<N$. Then
$\<h^\star,\langle\varepsilon_i,\phi^n_i\rangle_{i <N}\>$ is a condition in $\p$.
Since it is stronger than $q_n$ and compatible with
$\<h^\star,\langle\varepsilon_i,\phi_i\rangle_{i <N^\star}\>$ we conclude that
$q_n\mbox{$\compatible_{\p}$} p_{k_0}$, which contradicts the choice of $q_n$
($n>k_0$ !!!).

Let $\{i_n: n \in \omega\}$ be an enumeration of $I$ with infinitely
many repetitions. Suppose that $\forces_{\cal P} \dot{f}\in
\omega^\omega$.

Define a function $g \in \V \cap \omega^\omega$ as $g(k) = M(i_k)$,
where $M(i_k)$ is the number obtained by applying the first part of
the lemma to ${\cal P}_{i_k}$ and to the antichain $p_n = \lbv \dot{f}(k)=n\rbv_{\cal P}$, $n
\in \omega$. It is clear that
$$\forces_{\cal P} \exists^\infty n \ \dot{f}(k) \leq g(k).~\QED$$

\begin{theorem}
  It is consistent with $\ZFCa$ that $\cov^\star(\N)> \max\{\cov(\N),
  \frak b\}$ and that $\unif^\star(\N) < \min\{\unif(\N),
  \frak d\}$.
\end{theorem}
\Proof
To construct the first model let ${\cal P}_{\omega_2}$ be the finite
support iteration of ${\cal P}$ of length $\omega_2$.

Let $\V \thinks 2^{\boldsymbol\aleph_0}=\boldsymbol\aleph_1$. It is
clear that $\V^{{\cal P}_{\omega_2}} \thinks
\cov^\star(\N)=\boldsymbol\aleph_2$. Since ${\cal P}$ is
$\sigma$-centered neither ${\cal P}$ nor a finite support iteration
of ${\cal P}$ adds random reals (see \cite{shelahbook} or
\cite{BJbook}). Similarly, property stated in \ref{nodom} implies that
finite support iteration of ${\cal P}$ does not add dominating reals.
Thus $\cov(\N)$ and $\frak b$ are both equal to $\boldsymbol\aleph_1$
in $\V^{{\cal P}_{\omega_2}}$.

The second part of the theorem is proved similarly.  Let $\V \thinks
{\bold { MA}} \ \& \ 2^{\boldsymbol\aleph_0}=\boldsymbol\aleph_2$ and
let ${\cal P}_{\omega_1}$ be the finite support iteration of ${\cal
  P}$ of length $\boldsymbol\aleph_1$.

By ``dualizing'' the above argument we show that
$$\V^{{\cal P}_{\omega_1}} \thinks
\unif^\star(\N)=\boldsymbol\aleph_1 \ \& \ \unif(\N)=\frak
d=\boldsymbol\aleph_2.~\QED$$

\begin{theorem}
Any of the inequalities  $\cov^\star(\N)> {\frak b}$,
$\unif^\star(\N) < {\frak b}$ is consistent with $\ZFCa$.
\end{theorem}
\Proof
For the first model add $\boldsymbol\aleph_2$ random reals (simultanously) to a model
of $\CH$. Then, in the extension we will have ${\frak d}=\boldsymbol\aleph_1$ and
$\cov^\star(\N)=\boldsymbol\aleph_2$ (for the last note that if $r$ is a random real
over $\V$ then the constant function $h(x)=r$ ``omits'' all measure zero
subsets of the plane coded in $\V$). The second model can be obtained by
adding $\boldsymbol\aleph_1$ random reals to a model of ${\bold {MA}} +
2^{\boldsymbol\aleph_0}=\boldsymbol\aleph_2$.~$\QED $

For the sake of completeness of the picture let us mention the following
result which will appear in \cite{470} (the forcing notion applied for it
is a special case of the scheme presented there):
\begin{theorem}
It is consistent with $\ZFCa$  that  $\cov^\star(\N)<\unif(\M)$.
\end{theorem}

\ifx\undefined\bysame
\newcommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\,}
\fi

config: arxiv


TITLE: Torsion of $\mathbb Z ^{2 \times 2}$
QUESTION [3 upvotes]: I am trying to find the torsion of the ring $R= \mathbb Z ^{2 \times 2}$ seen as an $R-$ left module. So I want to see for which matrices $B$, there exists $A \neq 0$ such that $AB=0$. So if $B=\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}$, then I want to find $A=\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}$ with $$\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}=\begin{pmatrix} 0 & 0 \\  0 & 0  \\  \end{pmatrix}$$
From this condition I get these equations:
$$ax+cy=0$$$$bx+dy=0$$$$az+cw=0$$$$bz+dw=0$$
If $B$ is an invertible matrix in $R$, then it is clear that $B$ is not in the torsion. This led me to think that the torsion may be the set of non invertible matrices, or at least is contained in it. I've tried to find the invertible matrices in this ring but I got stuck.
Any suggestions would be appreciated.

REPLY [0 votes]: we go ahead and find an inverteble rational matrix P as a multiple of elementary matrices reducing A reduced rational row-echelon form B: PAx=0. Then we solve for x and find its general integral form. From these we may have to discard eigen vectors of P.


TITLE: Picard group of classifying stack
QUESTION [2 upvotes]: Suppose $S$ is a scheme, and $G$ a smooth $S$-group scheme.
Then there exists an algebraic stack BG called the classifying stack of $G$, defined as the quotient stack $[S/G]$ where $G$ acts trivially on $S$.
I was wondering what is $Pic(BG)$.
Is it true that $Pic(BG)= H^1(k,G)$ when $S=Speck$ the spectrum of a field?
What can we say in general?

REPLY [6 votes]: $\mathrm{Pic}([S/G])$ is the group of line bundles on $S$ together with a $G$-linearization. When $S=\mathrm{Spec}(k)$, any line bundle on $S$ is trivial, and a $G$-linearization is given by a character $G\rightarrow \mathbb{G}_m$, so $\mathrm{Pic}[S/G]=\mathrm{Hom}(G,\mathbb{G}_m)$. No relation with $H^1(k,G)$, which is not a group when $G$ is not abelian.


TITLE: Calculate the Fourier transform of $\log |x| $
QUESTION [22 upvotes]: How can one prove that the Fourier transform of $\log |x|$ is
$$-\pi \mathrm{pf} \frac{1}{|\xi|} +C \delta,$$
where $\mathrm{pf}\frac{1}{|x|} = D(\mathrm{sign}(x)\log|x|)$ (in the sense of distributions) and how can I compute the constant $C$?

REPLY [3 votes]: I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$ that uses a regularization approach.  This way forward is distinct from the methodology I used in THIS ANSWER.
The result herein includes a distributional interpretation of $\frac1{|x|}$.  Finally, we show that the distributional interpretation of $\frac1{|x|}$ is non-unique and that it differs from other interpretations by a multiple of the Dirac Delta distribution.  With that introduction, we now proceed.

PRELIMARIES
Let $\psi(x)=\log(|x|)$ and let $\Psi$ denote its Fourier Transform .  Then, we write
$$\Psi(x)=\mathscr{F}\{\psi\}(x)\tag 1$$
where $(1)$ is interpreted as a Tempered Distribution.  That is, for any $\phi \in \mathbb{S}$, we can write
$$\langle \mathscr{F}\{\psi\}, \phi\rangle =\langle \psi, \mathscr{F}\{\phi\}\rangle$$
Now, let $\psi_\epsilon(k) =e^{-\varepsilon|k|}\log(|k|)$.  Therefore, $\psi(k)=\lim_{\varepsilon\to 0^+}\psi_\varepsilon(k)$ and we see that
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\}, \phi\rangle&=\lim_{\varepsilon\to 0^+}\langle \psi_\varepsilon, \mathscr{F}\{\phi\}\rangle \\\\
&=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\
&=\langle \mathscr{F}\{\psi\}, \phi\rangle
\end{align}$$
Next, we evaluate the Fourier transform of $\psi_\varepsilon$.

EVALUATING THE FOURIER TRANSFORM OF $\displaystyle \psi_\varepsilon$
Denote by $\Psi_\epsilon$, the Fourier transform of $\psi_\varepsilon$.  Then, we have
$$\begin{align}
\Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\
&=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\
&=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\
&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\
&=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2
\end{align}$$
where
$$\begin{align}
\psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\
\psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\
\psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)
\end{align}$$
Next, we will find the distributional limits of $\psi^{(1)}_\varepsilon$, $\psi^{(2)}_\varepsilon$, and $\psi^{(3)}_\varepsilon$.

DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle
 \psi^{(2)}_\varepsilon$, and $\displaystyle  \psi^{(3)}_\varepsilon$
Again, let $\phi\in \mathbb{S}$.  Then,
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\
&=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\
&=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\
&=-2\pi \gamma \phi(0)\tag3
\end{align}$$

$$\begin{align}
\langle \psi^{(2)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2) \right)\phi(x)\,dx\\\\
&=-2\log(\varepsilon)\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx-\int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\phi(\varepsilon x)\,dx\\\\
&= -2\pi \log(\varepsilon)\phi(0)-2\pi \log(2) \phi(0)+o(\varepsilon)
\end{align}\tag4$$

$$\begin{align}
\langle \psi^{(3)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\right)\phi(x)\,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&=-\phi(0)\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) (\phi(x)-\phi(0))\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&= \left(2\pi \log(\varepsilon) +2\pi \log(2)\right)\phi(0)+o(\varepsilon)\\\\
&-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\tag5
\end{align}$$

FINAL RESULTS
Substituting $(3)$, $(4)$, and $(5)$ into $(2)$, we find that
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\},\phi\rangle =-2\pi \gamma \phi(0)-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\\\\
\end{align}$$
from which we assert that in distribution
$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \left(\frac1{|x|}\right)_1}$$
where we interpret $\left(\frac1{|x|}\right)_1$ to mean that for any $\phi\in \mathbb{S}$,
$$\int_{-\infty}^\infty  \left(\frac1{|x|}\right)_1 \phi(x)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$

NOTE:
It was arbitrary to split the integration in $(5)$ into inervals $|x|\le 1$ and $|x|\ge 1$.  Had we chosen instead the intervals $|x|\le \nu$ and $|x|\ge \nu$ for any $\nu>0$, we would have obtained
$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\psi\}(x)=-2\pi (\gamma+\log(\nu)) \delta(x)-\pi \left(\frac1{|x|}\right)_\nu}$$
where we interpret $\left(\frac1{|x|}\right)_\nu$ to mean that for any $\phi\in \mathbb{S}$,
$$\int_{-\infty}^\infty  \left(\frac1{|x|}\right)_\nu \phi(x)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$


TITLE: Degree and fiber of scheme morphisms in $\mathbb P^1$ induced by linear system of divisors on an elliptic curve
QUESTION [2 upvotes]: Consider curves (in the sense of schemes) over $\mathbb C$. Hartshorne, Algebraic Geometry says that, given an elliptic curve $X$, and a closed point $P_0$, the linear system $|2P_0|$ induces a morphism in $\mathbb P^1$ of degree 2.
Now, I know that the induced morphism exists, but why has it degree 2? And, if I took $|3P_0|$, would it bee of degree $3$.
Moreover, I'd like to investigate the fibers of such (a bit abstractly described) morphism. Case $d=2$ involves hyperellipticity and I can deal with it. But Case $d=3$? Is it true, for example, that there are fibers made of at least one rational point and at least one (closed) non rational point? Or are they somehow "separated"?
Thank you in advance.

REPLY [2 votes]: For any projective curve $X$ and an effective Cartier divisor $D$, if there exists a non-constant $f\in |D|$, $f$ induces a rational map $f:X\to\mathbb{P}^1$ and if $X$ is smooth, this is a morphism. Further the inverse image of a point in $\mathbb{P}^1$ is completely contained in $D$ and thus $\deg f\leq\deg D$. So, in your case, one gets the degree of the map given by $2P_0$ has degree at most 2. But, it can not be degree one, since degree one maps are isomorphisms and the two have different genus. If you took $3P_0$, it has now three linearly independent sections and the natural map is not to the line, but the plane and then it will give you an embedding. So, you have to be more specific as to what you mean.
The latter part is unclear. What does rational point mean to you? You are working over the complex numbers.


TITLE: Continuous function and net (Solution Verification)
QUESTION [0 upvotes]: I have started to learning about nets. I need a solution verification or any necessary comments (answer if needed) for the following problem:
Statement: Let $X$ be a topological space and let $f:X\to \mathbb{R}$ be continuous. Show the following:
$(1)$ If a net $\mathscr{S}:D\to X$ converges to $x_0$, then $f\circ \mathscr{S}$ converges to $f(x_0)$ ($(D, \geq)$ denote a directed set).
$(2)$ If $(f\circ \mathscr{S})(d)\leq 1$ for every $d\in D,$ show that $f(x_0) \leq 1.$
proof:
$(1)$ Consider an open interval $(a,b)$ in $\mathbb{R}$ containing $f(x_0)$. Obviously, $f^{-1}((a,b))$ is open in $X$. Since $\mathscr{S}$ converges to $x_0$, there exists $m\in D$ such that $\mathscr{S}(d)\in f^{-1}((a,b))$ for all $d\geq m$. Note that $f\circ \mathscr{S}:D\to \mathbb{R}$ is also a net in $\mathbb{R}$. And $f\circ \mathscr{S}(d)\in (a,b)$ for all $d\geq m$. Since the open interval was chosen arbitrarily, the proof follows.
$(2)$ If possible, suppose $f(x_0)>1$. Find disjoint pair of open sets $U,~V$ in $\mathbb{R}$ such that $f(x_0)\in U$, each member of $U$ is greater than $1$, and $1\in V$. Now, $f^{-1}(U)$ is open in $X$ containing $x_0$ since $\mathscr{S}$ converges to $x_0$, there exists $m\in D$ such that $\mathscr{S}(d)\in f^{-1}(U)$ for all $d\geq m$. Then $f\circ \mathscr{S}\cap U\neq \emptyset.$ This is contradiction since $f\circ \mathscr{S}(d) \leq 1$ for every $d\in D$.
Any help will be appreciated. Thanks in advance.

REPLY [1 votes]: $(1)$ is fine.
For $(2)$ it's slightly easier: suppose $f(x_0) > 1$. Then $U := (1,\infty)$ is an open set of $\Bbb R$ that contains $f(x_0)$. Now apply $(1)$ so there exists $m \in D$ such that for all $d \ge m$ we have $f(\mathscr{S}(d)) \in U$ so $f(\mathscr{S}(d)) > 1$. The same contradiction ensues. Introducing $V$ too is unnecessary.


TITLE: Shifting Eigenvalues
QUESTION [1 upvotes]: Let $\mathbf{A}$ be a hermitian positive semi-definite matrix and $\mathbf{B}$ be a hermitian positive definite matrix. Then I am interested in the eigenvalues of matrix $\mathbf{C}(t)=\mathbf{A}-t\mathbf{B}$ where $t$ is a real parameter. Note that when $t=0$, $\mathbf{C}$ is a positive semi-definite matrix. As $t$ increases, the eigenvalues of $\mathbf{C}$ moves towards the left on the real line and crosses the origin. At one point, the last positive eigenvalue of $\mathbf{C}$ will also cross origin at which point, $\mathbf{C}$ will become negative semi-definite. I want the smallest $t$ for which this happens. In other words, I want to solve
\begin{align}
\max_{t}~t \\
\lambda_{max}(\mathbf{C}(t))\geq 0
\end{align}
where $\lambda_{max}(.)$ denotes the largest eigenvalue of the argument.

REPLY [1 votes]: $\det(C(t))$ is a polynomial in $t$, of degree $n$ (where your matrices are $n \times n$).  You want the greatest positive root of this polynomial.
You could also call this the greatest eigenvalue of $B^{-1/2} A B^{-1/2}$.


TITLE: Finding the Radius of Convergence of Series.
QUESTION [0 upvotes]: Find the radius of convergence of the series:
  $$\sum_{n=0}^{\infty} 5^{(-1)^n}z^n$$

This was an old homework question.  I tried finding the radius of convergence by using the ratio test, but the professor wrote on my homework that the ratio test does not work.  I, therefore, am not sure how to or what method to use to find the radius of convergence.  The answer should be $1$ according to the back of the book.  Any help, suggestions, and pointers are welcomed.  I have attached the question as a picture.  Thank you.

REPLY [1 votes]: You can split the summation in
$$5(1+z^2+z^4+z^6+\cdots)+\frac15(z+z^3+z^5+z^7+\cdots).$$ This is allowed as all terms are positive.
The radius of convergence of the sum will be the smallest of the two radii.

When it converges, the value of the sum is
$$\frac5{1-z^2}+\frac z{5(1-z^2)}.$$

REPLY [0 votes]: For $|z|=1$, the sequence $(5^{(-1)^n}z^n)$ is neither convergent to $0$ nor unbounded, so radius of convergence is $1$.
Another way to look at it : if $|z|<1$, $\left|5^{(-1)^n}z^n\right|\le 5|z|^n$ which is the term of a convergent geometric series, so radius is greater than $1$, but if $|z|>1$, $\left|5^{(-1)^n}z^n\right|\le \frac15|z|^n$ which is the term of a divergent geometric series.


TITLE: Prove algebraically that $n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$
QUESTION [5 upvotes]: Prove algebraically that
$n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$ where $n$ and $k$ are integers.
I start with the fact that $(x+1)^n=\sum_{k=0}^{n}\dbinom{n}{k}x^k$ and differentiate both sides of the equation twice, with respect to $x$.
The second derivative (with respect to $x$) of the above equation yields: $n(n-1)(x+1)^{n-2}=\sum_{k=0}^{n}k^2-k\dbinom{n}{k}x^{k-2}$.
Now, I am not sure how to proceed. I know I'm supposed to set $x = 1$. But even then, I am missing the $(n+1)$ on the left hand side and there is still a $-k$ on the right hand side.
Any help would be appreciated. Thank you.

REPLY [3 votes]: Using your idea of differentiating the binomial formula & then setting $x=1$ ... gives\begin{eqnarray*}
n2^{n-1} &=& \sum_{k=0}^{n} k \binom{n}{k} \\
n(n-1)2^{n-2}&=& \sum_{k=0}^{n} k(k-1) \binom{n}{k} \\
\end{eqnarray*}
now add these & you will have your result.


TITLE: Every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments
QUESTION [10 upvotes]: Let $E$ be an open set in $\mathbb{R}$.
Fix $x\in E$.
I have proved that statement is true when $\{y\in \mathbb{R}|(x,y)\subset E\}$ is bounded above and $\{z\in \mathbb{R}|(z,x)\subset E\}$ is bounded below.
If at least one of those above are not bounded, $E$ must be equal to one of;
1.$\mathbb{R}$
2.$\{r\in \mathbb{R}|r<k\}$for some $k\in \mathbb{R}$
3.$\{r\in \mathbb{R}|k<r\}$for some $k\in \mathbb{R}$
Are these sets can be the union of an at most countable collection of disjoint 'segments'?

REPLY [3 votes]: Let $E$ be an open subset of $\mathbb{R}$.  We can assume without loss of generality that E is nonempty.  Consider a real number $x \in E$.  In the context of this problem, a segment $(a_1,b_1)$ means the set of all $p \in \mathbb{R}$ such that $a_1 \lt p \lt b_1$.  $ $Let $y = inf \lbrace a \in E \mid (a,x) \subseteq E \rbrace$ and $w = sup \lbrace a \in E \mid (x,a) \subseteq E \rbrace$.  Then $E$ is the union of all such segments $(y,w)$ for all points $x \in E$.  Since each of these segments is open, it remains to show that they are at most countable. First, note that the rationals are known to be dense in the reals.  Also, the rationals are a countable set.  Thus, let $\lbrace r_n \rbrace$ be an ordering of the rational numbers.  Now order the segments $(y,w)$ such that $(y_1,w_1)$ precedes $(y_2,w_2)$ if and only if the least $i$ such that $r_i \in (y_1,w_1)$ is less than the least $j$ such that $r_j \in (y_2,w_2)$.  $r_i$ and $r_j$ exist because the rationals are dense.  This is an ordering of the segements whose union is $E$, showing that the segments are at most countable.


TITLE: Decide if there exist $a$ and $b \in \mathbb{Z}$ such that $a^2=2b^2$.
QUESTION [2 upvotes]: Decide if there exist $a$ and $b \in \mathbb{Z}$ such that $a^2=2b^2$.
$a,b \neq 0$
We have to solve this kinds of problems using the order of a prime function:
$v_p(a) \in \mathbb{Z}$ which tells the exponent of a prime $p$ in the factorization of an integer $a$.
I managed to find that $a$ must be $even$ using
$$
2b^2|a^2 \longrightarrow v_2(a^2)\geq 1, v_2(a^2)=2v_2(a)
$$
$$
2v_2(a)\geq1 \longrightarrow v_2(a)\geq \frac 1 2 \longrightarrow v_2(a)\geq 1
$$
But don't know what to do next...

REPLY [1 votes]: Hint $\ $ Compare parity: $\ v_2(a^2) = 2 v_2(a)\,$ is even, but $\,v_2(2b^2) \,=\, 1+2\, v_2(b)\,$ is odd.


TITLE: 2.17.32 - Mary L Boas.
QUESTION [0 upvotes]: Use a series you know to show that $\sum_{n=0}^\infty\frac{(1+i\pi)^n}{n!}=-e.$

It does look like $e^x$ at first, but i'm not sure how to proceed. Any tips?

REPLY [0 votes]: A hint:
$$
\ln(-e) = \ln(e) +\ln(-1) = 1 + \ln(-1)
$$
You can use Euler equation to find that
$$
\mathrm{e}^{i\pi}
$$
So you have the argument in the summand.


TITLE: Irreducible polynomials over $GF(4)$.
QUESTION [2 upvotes]: I'm working with $GF(4)$ and I'm looking for irreducible polynomials of different degrees over that field. So $GF(4) = \{0,1,\alpha, \alpha + 1\}$ where $\alpha^{2} + \alpha + 1 = 0$, and first I'm looking for an irreducible polynomial of degree 2.
I looked at $\alpha x^2 + x + 1$, and it seems to be irreducible, but in the suggested solution to the problem they have used $x^2 + \alpha x + 1$, which is clearly irreducible. They suggests monic polynomials for the rest of the problems as well, and I'm wondering if they need to be monic?

REPLY [3 votes]: For question of irreducibility (primality) in a domain (and in particular in a UFD), one usually works up to multiplication be an invertible factor. In some cases rather than working with association classes of elements, it is possible and common to impose and additional condition that will select a unique representative from each class; notably in $\Bbb Z$ it is common to only consider positive numbers, and in $K[X]$ it is common to consider only monic polynomials. This is why only monic irreducible polynomials are considered. Their scalar multiples are irreducible as well, but not really of independent interest, just like the irreducible integer $-17$ is not considered independently of the prime$~17$.
In practice this simplifies the classification of the irreducibles considerably. In the current case we can eliminate from the $16$ monic polynomials of degree$~2$, the $X^2+bX+c$ for $b,c\in GF(4)$, the $10$ distinct products of two monic polynomials $X+a$ of degree$~1$ for $a\in GF(4)$. The products involving the irreducible $X$ are the cases with $c=0$, the squares $(X+a)^2=X^2+a^2$ are the cases with $b=0$, and the product $(X+\alpha)(X+\alpha+1)$ is the polynomial $X^2+X+1$ that was irreducible over $GF(2)$ with roots $\alpha,\alpha+1$ in$~GF(4)$. The two remaining reducible polynomials are $(X+1)(X+\alpha)=X^2+(\alpha+1)X+\alpha$ and $(X+1)(X+\alpha+1)=X^2+\alpha X+\alpha+1$. All remaining monic polynomials of degree$~2$ (there are $6$ of them) are irreducible.


TITLE: Confusion about whether a supremum of a set can be a limit point if it belongs to the set.
QUESTION [0 upvotes]: Let $A\subseteq \mathbb{R}$ be a bounded above set of real numbers and $l=SupA$ then which of the following is true?

$l$ is a limit point of A.
$l\in A$
$l\notin A$
Either $l\in A$ or $l$ is the limit point of $A$.

Options (1), (2) and (3) can be easily discarded by taking a finite set, an open set and a closed set. For (4) when $l\notin A$, $l\in (l-\epsilon, l+\epsilon)$. Now since $l$ is the least upper bound of $A$ therefore for any $x\in A$, $x>l-\epsilon$ and $x<l+\epsilon$ $\implies$ $x\in(l-\epsilon,l+\epsilon)$ $\implies$ $l$ is a limit point of $A$. Now my doubt starts where $l$ does not belong to $A$, however we can prove it by taking a finite set as a counter example and there are also many infinite sets which can be taken here as a counter example, but is it possible that $l\in A$ and $l$ is also the limit point of $A$. How can we prove it in general? Also we can always find some $x\in A$ in an infinite set (I am looking for an example) which is in the open interval  $(l-\epsilon,l)$ and hence in $(l-\epsilon, l+ \epsilon)$. So I am just looking for a general proof here. Please help.

REPLY [2 votes]: At first we should add an extra condition: $A\neq\varnothing$.
It is immediate that $4$ is true.
For every $\epsilon>0$ point $l-\epsilon$ is not an upperbound of $A$ simply because $l$ is by definition the least upper bound of $A$.
So for every $\epsilon>0$ we can find an element $a\in A$ that satisfies $a>l-\epsilon$.
Does this mean that $l$ is a limit point of $A$?...
Yes if $l\notin A$ (so that $a\neq l$), but not necessarily if $l\in A$.
Examples:

$A=(0,1)$ so that $l=1\notin A$ and consequently $l$ is a limit point of $A$.
$A=(0,1]$ where $l=1\in A$ is a limit point of $A$.
$A=(0,1)\cup\{2\}$ where $l=2\in A$ is not a limit point of $A$.


TITLE: lebesgue measurability sufficient and necessary conditions
QUESTION [2 upvotes]: I was reading a proof ("my question is also about the steps in the proof itself which I will type out here also") regarding a necessary and sufficient condition for measurability. In particular the theorem says:
A subset A $\subset$ $\mathbb{R}^{n}$ is Lebesgue measurable if and only if for every $\epsilon > 0$, there is an open set $G \supset A$ such that $\mu ^{*}(G \setminus A) < \epsilon$.
Proof:
First we assume that A is measurable and show that it satisfies the condition given in the theorem.
Suppose that $\mu(A) < \infty$ and let $\epsilon >0$. From (2.12) there is an open set $G \supset A$ such that $\mu(G) < \mu^{*}(A) + \epsilon$. Then since A is measurable,
$\mu^{*}(G\setminus A) = \mu^{*}(G)- \mu^{*}(G\cap A) = \mu(G)-\mu^{*}(A) < \epsilon$

My question is specifically regarding the second last step of the expression which is this:
$\mu(G)-\mu^{*}(A)$
I am not sure how the $\mu^{*}(G)$ becomes $\mu(G)$ because I was reading that if G is a rectangle in $R^n$, then yes we have $\mu^{*}(G)$ would equal $\mu(G)$. But in this case, the conditions for the proof only states that G is an open set that contains A (it did not say G is a rectangle, all it says is G is an open set in the statement of the theorem). So I am not sure how the author can use $\mu^{*}(G)$ = $\mu(G)$ in the proof.
Could someone kindly give me some hints as to what the author has used or which part am I missing? The symbol $\mu^{*} (G)$ means the outer-measure of the set $G$. The symbol without the star (i.e. $\mu(G)$) means measure (i.e. it means the usual sense of volume for a box, or the usual sense of area for a 2D-rectangle, or the length for a 1D line). For example, $\mu(G)$ = 27, if G is a rectangle of sides 3 and 9, and in this case $\mu^{*}(G)= \mu(G)$ because we have G being a rectangle. But the theorem only says G is an open set, so I am not sure how the author proves the theorem in particular the steps in the second last statement.
Thank you for your help

REPLY [1 votes]: $\mu(G) = \mu^*(G)$ by definition of $\mu$.  See Definition 2.10 in your notes. This only makes sense where $\mu$ is defined, which is on measurable sets.  All Borel sets, including open sets, are measurable (theorem 2.21).
Possible confusion arises, I speculate, because "$\mu$" is used for the volume of "rectangles" before defining $\mu$ as Lebesgue measure.  So the context of following Definition 2.10 is important.  The stuff about proving that $\mu^*(R)=\mu(R)$ for rectangles ensures consistency of the notation later, but it is overloaded/abused and takes getting used to.  Get used to knowing that $\mu = \mu^*$ wherever the former is defined.


TITLE: Canonical map from coproduct
QUESTION [0 upvotes]: I'm trying to learn some category theory, and I can't figure out what "canonical map" means here. If $\mathcal{C}$ is some category and $X\sqcup X$ is the coproduct of some object $X\in\mathcal{C}$ with itself, what is the canonical map $X\sqcup X\to X$?
My only guess is that, since $X\sqcup X$ is the coproduct, there's a unique map $f: X\sqcup X\to X$ such that $f\circ\iota = 1_X$ where $1_X: X\to X$ is the identity and $\iota: X\to X\sqcup X$ is the map that comes along with $X\sqcup X$ being the coproduct. But, I'm unsure if this is what is meant by the canonical map.

REPLY [2 votes]: Note that there are two maps $\iota_1: X\to X\sqcup X$  and $\iota_2: X\to X\sqcup X$, which are the inclusions of $X$ as each of the two factors of $X \sqcup X$.
With this in mind, the map $f: X\sqcup X\to X$ such that $f \circ \iota_n = 1_X$ for both inclusions is a very canonical map. It's often called the co-diagonal, because it's dual to the diagonal map into a product $\Delta :Y \to Y \times Y$


text:
\section{Dynamic Modified-CS and LS-CS}
\label{algos}
Here, we briefly review dynamic modified-CS and LS-CS.

\subsection{Dynamic modified-CS algorithm}
\label{modcsalgo}
Modified-CS was first proposed in \cite{isitmodcs} for exact reconstruction from noiseless measurements when part of the support is known. It solved $\min_\beta \|\beta_{T^c}\|_1  \ s.t. \ y  = A \beta$. For noisy measurements, we can relax the data constraint in many possible ways.
In this work, we stick to the following, because its error bounds have the simplest form and this makes the corresponding stability result less messy.
\bea
\min_\beta \|\beta_{T^c}\|_1  \ s.t. \ \|y - A \beta\|^2 \le \eps^2
\label{modcs}
\eea
Denote its output by $\xhat_{modcs}$.
In this work, whenever we refer to CS, we refer to the following
\bea
\min_\beta \|\beta\|_1  \ s.t. \ \|y - A \beta\|^2 \le \eps^2
\label{simplecs}
\eea

We summarize below the dynamic modified-CS algorithm \cite{icipmodcs} for time sequences.
At $t=0$, we use large enough number of measurements, $n_0 > n$, and do CS.

\noindent Set $\Nhat_{-1}=\phi$ (empty set). For $t \ge 0$ do,
\ben
\item If $t=0$, set $A := A_{n_0}$, else set $A := A_n$.
\label{step1}

\item {\em Modified-CS.} Solve (\ref{modcs}) with $T = \Nhat_{t-1}$ and $y = y_t$. Denote its output by $\xhat_{t,modcs}$.
\label{modcsstep}

\item {\em Detections / LS.} Compute $\tT_\dett$ and LS estimate using it:
\label{addls}
\bea
\tT_\dett \se T \cup \{i \in T^c: |(\xhat_{t,modcs})_i| > \alpha_{add} \} \nn \\

(\xhat_{t,\dett})_{\tT_\dett} \se {A_{\tT_\dett}}^\dag y_t, \ \ (\xhat_{t,\dett})_{\tT_\dett^c} = 0
\eea

\item {\em Deletions / LS.} Compute $\tT$ and LS estimate using it:
\label{delete}
\bea
\tT \se  \tT_{\dett} \setminus  \{i \in \tT_\dett: |(\xhat_{t,\dett})_i| \le \alpha_{del} \} \nn \\

(\xhat_{t})_{\tT} \se {A_{\tT}}^\dag  y_t, \ \ (\xhat_{t})_{\tT^c} = 0
\label{finalls}
\eea

\item Output $\xhat_t$. Set $\Nhat_t = \tT$. Feedback $\Nhat_t$.

\een
Notice that one could also replace the addition and deletion steps above by a single step that computes $\Nhat_t = \{i: |(\xhat_{t,modcs})_i| > \alpha_* \}$. This would be sufficient when the noise is small and $|\Delta_e|$ is small. But when either of these does not hold, the bias in $\xhat_{t,modcs}$ creates the following problem. Along $T^c$, the solution will be biased towards zero, while along $T$ it may be biased away from zero (since there is no constraint on $(\beta)_T$). The set $T$ contains $\Delta_e$ which needs to be deleted. Since the estimates along $\Delta_e$ may be biased away from zero, one will need a higher threshold to delete them. But that would make detecting additions more difficult especially since the estimates along $\Delta \subseteq T^c$ are biased towards zero.

By adapting the approach of \cite{candes_rip}, the error of the modified-CS step can be bounded as a function of $|T|=|N|+|\Delta_e|-|\Delta|$ and $|\Delta|$ \cite{arxiv}. We state a modified version of \cite{arxiv}'s result.

\begin{theorem}[modified-CS error bound \cite{arxiv}]
If  $\|w\| \le \eps$ and $\delta(\max(3|\Delta|,|N|+|\Delta|+|\Delta_e|)) < \sqrt{2}-1$, then

\bea
\|x_t - \xhat_{t,modcs}\| \sle C_1(\max(3|\Delta|,|N|+|\Delta|+|\Delta_e|)) \eps, \ \text{where} \nn \\
C_1(S) \sdefn \frac{4 \sqrt{1+\delta(S)}}{1 - (\sqrt{2} +1) \delta(S)}
\label{defC1s}
\eea
\label{modcsbnd}
\end{theorem}

\subsection{LS-CS (dynamic CS-residual) algorithm}
LS-CS uses partial knowledge of support in a different way than modified-CS.
The LS-CS algorithm \cite{just_lscs} is the same as the dynamic modified-CS algorithm but with step \ref{modcsstep} replaced by

\bi
\item {\em CS-residual step. }
\bi
\item Use $T:=\Nhat_{t-1}$ to compute the initial LS estimate, $\xhat_{t,\text{init}}$, and the LS residual, $\tty$, using
\label{initls}
\bea
(\xhat_{t,\text{init}})_{T} \se {A_{T}}^\dag y_t, \ \ (\xhat_{t,\text{init}})_{T^c} = 0 \nn \\
\tty \se  y_t - A \xhat_{t,\text{init}}
\label{deftty0}
\eea
\item Do CS on the LS residual, i.e.  solve (\ref{simplecs}) with $y= \tty$ and denote its output by $\betahat_t$. Compute
\bea
\xhat_{t,\CSres} : = \betahat_t + \xhat_{t,\text{init}}.
\label{xhatcsres}
\eea
\ei
\ei

The CS-residual step error can be bounded as follows. The proof follows in exactly the same way as that given in \cite{just_lscs} where CS is done using Dantzig selector instead of (\ref{simplecs}). We use (\ref{simplecs}) here to keep the comparison with modified-CS easier.

\begin{theorem}[CS-residual error bound \cite{just_lscs}]
If  $\|w\| \le \eps$, $\delta_{2|\Delta|} < \sqrt{2}-1$ and $\delta_{|T|} < 1/2$,
\label{lscsbnd}
\bea
&& \|x_t - \xhat_{t,\CSres}\| \le C'(|T|,|\Delta|) \eps +  \theta_{|T|,|\Delta|} C''(|T|,|\Delta|) \|x_\Delta\| \nn \\
&& C'(|T|,|\Delta|)  \defn C_1(2|\Delta|) + \sqrt{2} C_2(2|\Delta|) \sqrt{\frac{|T|}{|\Delta|}} \nn \\
&& C''(|T|,|\Delta|) \defn 2 C_2(2|\Delta|) \sqrt{{|T|}}, \ \text{where} \nn \\
&& \text{$C_1(S)$ is defined in (\ref{defC1s}),~}  C_2(S) \defn 2\frac{1 + (\sqrt{2}-1) \delta(S)}{1 - (\sqrt{2}+1) \delta(S)}
\eea
\end{theorem}


TITLE: Theory without a concrete example.
QUESTION [2 upvotes]: I heard in class that there is a branch of mathematics that has been studied for some decades, but still has no "concrete example" of the theory.
My professor refused to speak out the name of this "theory" for some reason.
What kind of math might it be?

REPLY [0 votes]: Depending on what you mean by "concrete example" there could be many instances of such theories.  For example, if by "concrete" you mean "not relying on the axiom of choice", then an example of a non-concrete theory would be the theory of $\sigma$-additive Lebesgue measures on $\mathbb R$; see this 2017 publication in Real Analysis Exchange for details.


TITLE: Correction factor for the variance
QUESTION [1 upvotes]: In an exercise they asked me: "Why could we use the following correction factor? $\text{varianceX} = \frac{n-1}{n}*\text{varianceY}$
What I said was basically, because the unbiased sample variance have a factor of $\frac{n}{n-1}$, we could multiply the variance by $\frac{n-1}{n}$ to cancel the correction of Bessel and get the biased variance.
But what would be the utility of calculating a biased variance?, maybe I'm wrong about something...

REPLY [1 votes]: Well, after reading a lot on wikipedia I think this is the answer:
In calculating the expected value of the sample variance a factor equal to $\frac{n-1}{n}$ is obtained, which underestimates the expected variance so usually when calculating the sample variance we multiply it by the factor $\frac{n}{n-1}$ which is generally known as the variance unbiased sample variance or corrected sample variance.
The problem with this is that to correct the bias produces a large MSE, so one can choose a scaling factor that behave better than the variance of the corrected sample. This is always scaling down, choosing a 'a' greater than n-1, such that:
$S^2_a = \frac{n-1}{a}S^2_{n-1}$
In my case $a = n$.