TITLE: Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$
QUESTION [5 upvotes]: I have been following the thread " Convert Airy's Equation $y''-xy=0$  to Bessel equation $$t^2u''+tu'+(t^2-c^2)u$$ " but I can't join the dots to a solve similar equation $y''+x^2y=0$ so as to obtain a solution of the form $$\sqrt x\left(AJ_\frac{1}{4}+BJ_{-\frac{1}{4}}\right)$$
I actually get an equation that looks this way $$t^2\frac{du}{dt}+t\frac{du}{dt}+(t^2+\frac{5}{64})u$$ The above equation can not yield the desired solution. Please help me to clearly see this. Thank you.

REPLY [5 votes]: Perform the change of variables $y=\sqrt{x}f(x)$. Then, using the product rule, $f$ satisfies:
$$
\frac{x^2}{4}f''+\frac{x}{4}f'+(\frac{x^4}{4}-\frac{1}{16})f=0.
$$
Then make the change of variable $t=x^2/2$ and compute:
$$
\frac{df}{dx}=x\frac{df}{dt},
$$
and
$$\frac{d^2f}{dx^2}=\frac{df}{dt}+x^2\frac{d^2f}{dt^2}.
$$
Thus, the equation above becomes
$$
\frac{x^4}{4}f''+\frac{x^2}{2}f'+(\frac{x^4}{4}-\frac{1}{16})f=0,
$$
where the primes now denote derivatives w.r.t. the variable $t$.
Written in the variable $t$, the equation above becomes
$$
t^2f''+tf'+(t^2-\frac{1}{16})f=0.
$$
The solution of which is
$$
f=A J_{1/4}(t)+B J_{-1/4}(t).$$
Thus since $y=\sqrt{x}f(x^2/2)$, you get the desired solution.


TITLE: 24 hour time modulo
QUESTION [0 upvotes]: I'm trying to work out what time it will be from midnight (00:00) and noon (12:00) given the time difference between an origin and destination of travel.
If the destination is, say, 5 hours ahead of the origin, then I could use the method found at this Math Stackexchange page:
Calculating time using modulus
So for finding time at origin when it is midnight at destination, I would do $(5+00:00) \mod 24$ to get $17:00$, or $5$pm.
But how would I do this for negative time difference?
For example, the time difference between Paris (origin) and Dhaka (destination) is $-04:00$ hours.
If I want Paris time when it's midnight in Dhaka, I'll do $(-4+00:00) \mod 24$, but I'll get the negative value of $-4$, but the answer should be $20:00$ ($8$pm).

REPLY [0 votes]: If you are not keeping track of the date, you can simply use $(24+d+t)$ mod $24$ where $d$ is the time difference and $t$ is the time you are starting with. Your example would become $(24-4+00:00)$ mod $24$, which would be $(20+00:00)$ mod $24$, which would then become $20:00$.
If you are keeping track of the date, you will have to decrement it, as the extra 24 hours added to the time difference will push the date forward one day.


TITLE: Let $f$ continuous, show that $\int_0^a (\int_0^x f(x,y) dy) dx = \int_0^a (\int_y^a f(x,y) dx) dy$
QUESTION [3 upvotes]: $$\int_0^a \int_0^x f(x,y) dy dx = \int_0^a \int_y^a f(x,y) dx dy$$
I have tried to help myself, observing the graph and how the area is covered according to the limits, and managed to see that the area of the two integrals are symmetrical. But I am almost certain that that is not necessary to say that the integrals are equal, and I don't know how I could say it.

REPLY [0 votes]: Hint:
$$LHS:\begin{cases}0\le x\le a\\ 0\le y\le x\end{cases}\Rightarrow \begin{cases}0\le x\le a\\ 0\le y\le x\color{red}{\le a}\end{cases}\Rightarrow RHS:\begin{cases}0\le y\le a\\ y\le x\le a\end{cases}$$


TITLE: Projection matrix onto a subspace parallel to a complementary subspace
QUESTION [3 upvotes]: Given an $n \times k$ matrix $A$ and an $n \times (n-k)$ matrix $B$ such that $\text{span}(A) \oplus \text{span}(B) = \mathbb{R}^n$, how to get the projection matrix on $\text{span}(A)$ parallel to $\text{span}(B)$, in terms of $A$ and $B$?

Edit:
Below it is worked out that $P=\Big(\big(^−^ (^ )^{−1}^  \big)^{−1} ^ − (^ )^{−1}^  \big(^−^(^)^{−1}^
\big)^{-1}^ \Big)$.

Does it look right?
Can we possibly simplify it?
Is the horrendous look the reason that we rarely see it fleshed out in textbooks, etc.?

Edit 2:
I have verified numerically that the formula is correct. Also, with SVD $A = U_A \Sigma_A V_A^T$, $B = U_B \Sigma_B V_B^T$ and $U_{AB} = U_A^T U_B$,
it can be simplified that $P = U_A \Big(I - U_{AB}^T U_{AB} \Big)^{-1} U_A^T - U_A U_{AB} \Big(I - U_{AB}^T U_{AB} \Big)^{-1}U_B^T$.
But I have little intuition on how to understand the expression. If anyone can provide some, that'd be great.

REPLY [3 votes]: Let $U = [A|B]^{-1}$ and denote by $\bar U$ the $k\times n$ upper part of $U$. Then $P = A\bar U$. Indeed, since $\bar U[A|B] = \left[\begin{matrix}I_k&0\end{matrix}\right]$, we have $A\bar Ua_j = Ae_j = a_j$ and $A\bar Ub_i = A(0) = 0$.


TITLE: Can we apply u/v rule to dy/dx?
QUESTION [1 upvotes]: Single derivative of a function y is dy/dx. To find double derivative we write d²y/dx² or d(dy/dx)/dx. So it is correct to write it as [d(dy)dx-d(dx)dy]/(dx)² ?

REPLY [0 votes]: You are applying the quotient rule to a symbol not a fraction.
There are cases that we consider dy/dx as a fraction.
For example if $$x=x(t)$$ and $$y=y(t)$$ are parametric functions then $$  dy/dx=\frac {dy/dt}{dx/dt}$$


TITLE: Is $xy=0$ a function?
QUESTION [1 upvotes]: Graphically, it passes the vertical line test. However, I am confused because for $x=0$, there are infinite $y$ values that satisfy $xy=0$. Can someone help me understand this?

REPLY [1 votes]: As a function of one variable it is not (in general) a function, indeed recall that by definition
$$f:A\subseteq\mathbb{R}\to \mathbb{R}\quad\forall x\in A \quad \exists! y \quad y=f(x)$$
and, as noticed by Rushabh Mehta, for $x=0$ we have that $y$ can assume any value.
It would be a function, in implicit form, if we assume for example
$$f:\mathbb{R}\setminus\{0\}\to \mathbb{R}$$
which correspons to the function $y=0$ with domain $x\neq 0$.
Refer also to the related

basic question on definition of function

REPLY [1 votes]: It is not a function (as it was pointed out in a comment), but a relation, like a circle (a set of ordered pairs): $R=\{(x,y)|xy=0\}$.


TITLE: Polynomial ring homomorphism and group ring
QUESTION [1 upvotes]: Suppose that $G$ is a cyclic group of order $n$. I want to show that $R[X]/(X^n-1)\cong R(G)$ where $R(G)$ is a group ring.

Could you give me an idea about it ?
Thank you

REPLY [1 votes]: If $A$ is an $R$-algebra, we have natural bijections
$$\hom_{\mathsf{Alg}(R)}(R[G],A) \cong \hom_{\mathsf{Grp}}(G,A^*) \cong \{a \in A : a^n=1\} \cong \hom_{\mathsf{Alg}(R)}(R[X]/(X^n-1),A),$$
hence, $R[G] \cong R[X]/(X^n-1)$ by Yoneda. If $G=\langle \sigma \rangle$, the isomorphism maps $\sum_i r_i \sigma^i$ to $\sum_i r_i X^i$. If you don't like universal properties and instead have fun with calculating things, you can also check directly that this defines an isomorphism.
The same proof shows: If
$$G = \langle g_1,\dotsc,g_n : R_1(g_1,\dotsc,g_n)=\dotsc=R_p(g_1,\dotsc,g_n)=1 \rangle$$
is a group presentation, then we get an algebra presentation
$$R[G] \cong R \langle X_1^{\pm 1},\dotsc,X_n^{\pm 1} : R_1(X_1,\dotsc,X_n)=\dotsc=R_p(X_1,\dotsc,X_n)=1 \rangle.$$


text:
\begin{document}
\title {$n$-Homomorphisms}
\author{S. Hejazian}
\address{Department of Mathematics, Ferdowsi University, P. O. Box 1159, Mashhad 91775, Iran}
\email{hejazian@math.um.ac.ir}
\author{M. Mirzavaziri}
\address{Department of Mathematics, Ferdowsi University, P. O. Box 1159, Mashhad 91775, Iran}
\email{mirzavaziri@math.um.ac.ir}
\author{M. S. Moslehian}
\address{Department of Mathematics, Ferdowsi University, P. O. Box 1159, Mashhad 91775, Iran}
\email{moslehian@ferdowsi.um.ac.ir} \subjclass{Primary 47B48;
Secondary 16N60, 46L05, 46J10, 16Wxx.}
\keywords{$n$-homomorphism,
semiprime algebra, idempotent, commutator, commutativity,
continuity, $C^*$-algebra, second dual, partial isometry.}
\begin{abstract}
Let $\mathcal A$ and $\mathcal B$ be two (complex) algebras. A
linear map $\varphi:{\mathcal A}\to{\mathcal B}$ is called
$n$-homomorphism if $\varphi(a_{1}\ldots
a_{n})=\varphi(a_{1})\ldots\varphi(a_{n})$ for each
$a_{1},\ldots,a_{n}\in{\mathcal A}.$ In this paper, we
investigate $n$-homomorphisms and their relation to
homomorphisms. We characterize $n$-homomorphisms in terms of
homomorphisms under certain conditions. Some results related to
continuity and commutativity are given as well.
\end{abstract}
\maketitle
\section{Introduction}

Let $\mathcal A$ and $\mathcal B$ be two algebras. A linear
mapping $\varphi:{\mathcal A}\to{\mathcal B}$ is called an
$n$-homomorphism if $\varphi(a_{1}\ldots a_{n})=
\varphi(a_{1})\ldots\varphi(a_{n})$ for each
$a_{1},\ldots,a_{n}\in{\mathcal A}$. A $2$-homomorphism is then a
homomorphism, in the usual sense, between algebras.

For a homomorphism $\varphi:{\mathcal A}\to {\mathcal B}$ we can
see that $\varphi(a_{1}\ldots
a_{n})=\varphi(a_{1})\ldots\varphi(a_{n})$ for each
$a_{1},\ldots,a_{n}\in {\mathcal A}$ and for each $n$. The
problem is to consider the converse: Is every $n$-homomorphism a
homomorphism? The answer is negative in general. As a
counterexample, let $\omega$ be an $n$-root of unity and a
homomorphism $\psi: A \to B$. Then $\varphi : = \omega \psi$ is an
$n$-homomorphism which is not an $m$-homomorphism for any $2\leq
m\leq n-1$.

The paper consists of five sections. The second section is
devoted to examine the relationship between notions of
$n$-homomorphism and homomorphism. The third section concerns
$n$-homomorphisms on algebras generated by idempotents. In the
forth section, we investigate $n$-homomorphisms which preserve
commutativity under some conditions. A study of $n$-homomorphisms
on Banach algebras is given in the fifth section.

Throughout the paper, all Banach algebras are assumed to be over
the complex field ${\C}$.

\section{Relationship Between $n$-Homomorphisms and Homomorphisms}

We begin this section with a typical example as an exension of
that of introduced in the first section.

\begin{example} Let ${\mathcal A}$ be a unital algebra, $a_{0}$ be a
central element of ${\mathcal A}$ with $a_{0}^{n}=a_{0}$ for some
natural number $n$ and let $\theta:{\mathcal A}\to {\mathcal A}$
be a homomorphism. Define $\varphi:{\mathcal A}\to {\mathcal A}$
by $\varphi(a)=a_{0}\theta(a)$. Then we have
\begin{eqnarray*}
\varphi(a_{1}\ldots a_{n})&=&a_{0}\theta(a_{1}\ldots a_{n})\\
&=& a_{0}^{n}\theta(a_{1})\ldots\theta(a_{n})\\
&=& a_{0}\theta(a_{1})\ldots a_{0}\theta(a_{n})\\
&=& \varphi(a_{1})\ldots\varphi(a_{n}).
\end{eqnarray*}
Hence $\varphi$ is an $n$-homomorphism. In addition,
$a_{0}=\varphi(1_{\mathcal A})$ whenever $\theta$ is onto.
\end{example}

The above example gives us an $n$-homomorphism as a multiple of a
homomorphism. Indeed, if ${\mathcal A}$ has the identity
$1_{{\mathcal A}}$ then each $n$-homomorphism is of this form,
where $a_{0}=\varphi(1_{{\mathcal A}})$ as the following
proposition shows.

\begin{proposition} Let ${\mathcal A}$ be a unital algebra with identity $1_{{\mathcal A}}$, ${\mathcal B}$
be an algebra and $\varphi:{\mathcal A}\to {\mathcal B}$ be an
$n$-homomorphism. If $\psi:{\mathcal A}\to {\mathcal B}$ is
defined by $\psi(a)=(\varphi(1_{{\mathcal A}}))^{n-2}\varphi(a)$
then $\psi$ is a homomorphism and $\varphi(a)=\varphi(1_{\mathcal
A})\psi(a)$.
\end{proposition}

\begin{proof} We have
$$\varphi(1_{{\mathcal A}})=\varphi(1_{{\mathcal A}}^{n})=(\varphi(1_{{\mathcal A}}))^{n}.$$
and
\begin{eqnarray*}
\psi(ab)&=&(\varphi(1_{{\mathcal A}}))^{n-2}\varphi(ab)\\
&=&(\varphi(1_{{\mathcal A}}))^{n-2}\varphi(a 1_{{\mathcal A}}^{n-2}b)\\
&=&(\varphi(1_{{\mathcal
A}}))^{n-2}\varphi(a)(\varphi(1_{{\mathcal
A}}))^{n-2}\varphi(b)\\&=&\psi(a)\psi(b).
\end{eqnarray*}
It follows from $(\varphi(1_{{\mathcal A}}))^{n-1}\varphi(a)=
\varphi(1_{{\mathcal A}}^{n-1}a)=\varphi(a)$ that
$(\varphi(1_{{\mathcal A}}))^{n-1}$ is an identity for
$\varphi({\mathcal A})$. Thus
\begin{eqnarray*}
\varphi(1_{{\mathcal A}})\psi(a)&=&\varphi(1_{{\mathcal
A}})((\varphi(1_{{\mathcal A}}))^{n-2}\varphi(a))\\
&=&(\varphi(1_{{\mathcal A}}))^{n-1}\varphi(a)\\&=&\varphi(a).
\end{eqnarray*}
\end{proof}

Whence we characterized all $n$-homomorphisms on a unital algebra.
For a non-unital algebra ${\mathcal A}$ we use the unitization
and some other useful constructions. Recall that for an algebra
${\mathcal A}$, the linear space ${\mathcal A}_{1}={\mathcal
A}\oplus {\C}=\{(a,\alpha)\vert a\in {\mathcal A}, \alpha\in
{\C}\}$ equipped with the multiplication
$(a,\alpha)(b,\beta)=(ab+\alpha b+\beta a,\alpha\beta)$,
so-called the unitization of ${\mathcal A}$, is a unital algebra
with identity $(0,1)$ containing ${\mathcal A}$ as a two-sided
ideal.

Now we shall prove that each $n$-homomorphism is a multiple of a
homomorphism under some conditions.

\begin{definition} An algebra ${\mathcal A}$ is called a factorizable algebra
if for each $a\in {\mathcal A}$ there are $b,c\in {\mathcal A}$
such that $a=bc$.
\end{definition}

\begin{theorem} Let ${\mathcal A}$ and ${\mathcal B}$ be two factorizable algebras,
$lan({\mathcal B})=\{b\in {\mathcal B}; b{\mathcal B}=0\}=\{0\}$
and $\varphi:{\mathcal A}\to {\mathcal B}$ an onto
$n$-homomorphism. Then $\ker \varphi$ is a two-sided ideal of
${\mathcal A}$ and there is a unital algebra $\tilde{{\mathcal
B}}\supseteq {\mathcal B}$ and an $x\in \tilde{{\mathcal B}}$ with
$x^{n-1}=1_{\tilde{{\mathcal B}}}$ such that $\psi:{\mathcal A}\to
\tilde{{\mathcal B}}$ defined by $\psi(a)=x^{n-2}\varphi(a)$ is a
homomorphism.
\end{theorem}

\begin{proof} Suppose that $a\in \ker\varphi$ and $u\in {\mathcal A}$.
Since ${\mathcal A}$ is a factorizable algebra there are $u_1,
\ldots u_{n-1}\in {\mathcal A}$ such that $u=u_1\ldots u_{n-1}$.
Hence \[\varphi(au)=\varphi(au_1\ldots
u_{n-1})=\varphi(a)\varphi(u_1)\ldots \varphi(u_{n-1})=0.\]
Therefore $au\in \ker\varphi$. Similarly $ua\in \ker\varphi$.

Let $\tilde{{\mathcal
B}}=\{b_\circ+\beta_1x+\ldots+\beta_{n-2}x^{n-2}; b_\circ\in
{\mathcal B}_1, ~{\rm and }~\beta_1, \ldots, \beta_{n-2}\in {\bf
C}\}$ as a subset of the algebra ${\mathcal B}_1[x]$ of all
polynomials in $x$ with coefficients in the unitization
${\mathcal B}_1$ of ${\mathcal B}$. Using the ordinary
multiplication of polynomials, we define a multiplication on
$\tilde{{\mathcal B}}$ by $x^{n-1}=1$ and
$bx=\varphi(a_1)\varphi(a_2)$ where
$b=\varphi(a)=\varphi(a_1a_2)$ and $a=a_1a_2\in {\mathcal A}$. We
show that the multiplication is well-defined.

Let $b=d\in {\mathcal B}$ and $b=\varphi(a)=\varphi(a_1a_2),
d=\varphi(c)=\varphi(c_1c_2)$ with $a=a_1a_2, c=c_1c_2\in
{\mathcal A}$. Then we have $\varphi(a_1a_2)=\varphi(c_1c_2)$. So
$\varphi(a_1a_2)b_2\ldots b_n=\varphi(c_1c_2)b_2\ldots b_n$
for all $b_2\ldots b_n\in {\mathcal B}$.\\
Since $\varphi$ is onto there exist $u_2\ldots u_n\in {\mathcal
A}$ such that $\varphi(u_i)=b_i$. We can then write
\begin{eqnarray*}
&&\varphi(a_1)\varphi(a_2)\varphi(u_2)\ldots\varphi(u_{n-2})\varphi(u_{n-1}u_n)\\
&=&\varphi(a_1a_2u_2\ldots u_{n-1}u_n)\\
&=&\varphi(a_1a_2)\varphi(u_2)\ldots
\varphi(u_{n-1})\varphi(u_n)\\&=&
\varphi(c_1c_2)\varphi(u_2)\ldots
\varphi(u_{n-1})\varphi(u_n)\\
&=&\varphi(c_1c_2u_2\ldots
u_{n-1}u_n)\\&=&\varphi(c_1)\varphi(c_2)\varphi(u_2)
\ldots\varphi(u_{n-2})\varphi(u_{n-1}u_n).
\end{eqnarray*}
This implies that
$\varphi(a_1)\varphi(a_2)b=\varphi(c_1)\varphi(c_2)b$ for each
$b\in {\mathcal B}$, since ${\mathcal B}$ is a factorizable
algebra. Hence
$(\varphi(a_1)\varphi(a_2)-\varphi(c_1)\varphi(c_2)){\mathcal
B}=0$. Since $lan({\mathcal B})=\{0\}$ we conclude that
$\varphi(a_1)\varphi(a_2)=\varphi(c_1)\varphi(c_2)$. In
particular, $\varphi(a)\varphi(b)x^{n-2}=\varphi(ab)$ for all
$a,b\in {\mathcal A}$. Note that associativity of our
multiplication inherits from that of multiplication of
polynomials.

We can inductively prove that
$\varphi(a_1)\ldots\varphi(a_m)x^{n-m}= \varphi(a_1\ldots a_m)$
for all $m\geq 2$.

To show this, suppose that it holds for $m\geq 2$ and $a_{m+1}\in
{\mathcal A}$. Then
\begin{eqnarray*}
&&\varphi(a_1)\ldots\varphi(a_{m-1})\varphi(a_m)\varphi(a_{m+1})x^{n-m-1}\\
&=&\varphi(a_1)\ldots\varphi(a_{m-1})\varphi(a_m)\varphi(a_{m+1})x^{n-(m+1)}x^{n-1}\\
&=&\varphi(a_1)\ldots\varphi(a_{m-1})(\varphi(a_m)\varphi(a_{m+1})x^{n-2})x^{n-m}\\
&=&\varphi(a_1)\ldots\varphi(a_{m-1})\varphi(a_ma_{m+1})x^{n-m}\\&=&\varphi(a_1\ldots
a_{m-1}a_ma_{m+1}).
\end{eqnarray*}

Now define $\tilde{\varphi}:{\mathcal A}_{1}\to \tilde{{\mathcal
B}}$ by $\tilde{\varphi}(a,\alpha)=\varphi(a)+\alpha x$ for each
$(a,\alpha)\in {\mathcal A}_{1}$. Then for each
$(a_{1},\alpha_{1}), \ldots,(a_{n},\alpha_{n})\in {\mathcal
A}_{1}$ we have
\[\tilde{\varphi}(\prod_{i=1}^{n}(a_{i},\alpha_{i}))=
\tilde{\varphi}(\sum \alpha_{j_{1}}\ldots\alpha_{j_{k}}a_{i_{1}}
\ldots a_{i_{l}}),\] where the summation is taken over all
$i_{1},\ldots,i_{l},j_{1}, \ldots,j_{k}$ with $i_{1}<\ldots
<i_{l}, j_{1}<\ldots <j_{k}, 0\leq k,l\leq n,
\{i_{1},\ldots,i_{l}\}\cap\{j_{1},\ldots,j_{k}\}=\emptyset,
\{i_{1},\ldots,i_{l}\}
\cup\{j_{1},\ldots,j_{k}\}=\{1,\ldots,n\}.$ Thus if $\varphi()$
denotes $1\in {\C}$ then we can write

\begin{eqnarray*}
\tilde{\varphi}(\prod_{i=1}^{n}(a_{i},\alpha_{i}))
&=&\sum \alpha_{j_{1}}\ldots\alpha_{j_{k}}\varphi(a_{i_{1}}\ldots a_{i_{l}})\\
&=&\sum \alpha_{j_{1}}\ldots\alpha_{j_{k}}\varphi(a_{i_{1}})
\ldots\varphi(a_{i_{l}})x^{k}\\
&=&\prod_{i=1}^{n}(\varphi(a_{i})+\alpha_{i}x)=\prod_{i=1}^{n}
\tilde{\varphi}(a_{i},\alpha_{i}).
\end{eqnarray*}

This shows that $\tilde{\varphi}$ is an $n$-homomorphism on
${\mathcal A}_{1}$. Now Proposition 2.3 implies that
$\tilde{\psi}: {\mathcal A}_{1}\to\tilde{{\mathcal B}}$ defined by
$\tilde{\psi}(a,\alpha)=(\tilde{\varphi}(1_{{\mathcal
A}_{1}}))^{n-2}
\tilde{\varphi}(a,\alpha)=(\tilde{\varphi}(0,1))^{n-2}(\varphi(a)+\alpha
x) =x^{n-2}(\varphi(a)+\alpha x)$ is a homomorphism on ${\mathcal
A}_{1}$. Thus $\psi: {\mathcal A}\to \tilde{{\mathcal B}}$
defined by $\psi(a)=x^{n-2}\varphi(a)$ is a homomorphism on
${\mathcal A}$.
\end{proof}

\begin{proposition} Let ${\mathcal A}$ and ${\mathcal B}$ be two algebras,
$\psi:{\mathcal A}\to {\mathcal B}$ be a homomorphism and
$\tilde{{\mathcal B}}$ be a unital algebra with $\tilde{{\mathcal
B}}\supseteq {\mathcal B}$. Then for each central element $x\in
\tilde{{\mathcal B}}$ with $x^{n-1}=1_{\tilde{{\mathcal B}}}$,
the mapping $\varphi:{\mathcal A}\to \tilde{{\mathcal B}}$
defined by $\varphi(a)=x\psi(a)$ is an $n$-homomorphism and
$\ker\varphi$ is a two-sided ideal of ${\mathcal A}$.
\end{proposition}

\begin{proof} We have
\begin{eqnarray*}
\varphi(a_1\ldots a_n)&=&x\psi(a_1\ldots
a_n)\\&=&x^n\psi(a_1)\ldots
\psi(a_n)\\&=&x\psi(a_1)x\psi(a_2)\ldots
x\psi(a_n)\\&=&\varphi(a_1)\ldots \varphi(a_n).
\end{eqnarray*}
Moreover if $a\in \ker\varphi$ and $b\in {\mathcal A}$ then
$\varphi(ab)=x\psi(ab)=x\psi(a)\psi(b)=\varphi(a)\psi(b)=0$ and
similarly $\varphi(ba)=0$. Thus $\ker\varphi$ is a two-sided ideal
of ${\mathcal A}$.
\end{proof}

\begin{example} In general, the kernel of an $n$-homomorphism
may not be an ideal. As an example, take algebra ${\mathcal A}$
of all $3$ by $3$ matrices having $0$ on and below the diagonal.
In this algebra product of any $3$ elements is equal to $0$ so
any linear map from ${\mathcal A}$ into itself is a
$3$-homomorphism but its kernel does not need to be an ideal.
\end{example}

\section{Algebras Generated By Idempotents}

Let ${\mathcal A}$ and ${\mathcal B}$ be two algebras. A linear
map $\varphi:{\mathcal A}\to {\mathcal B}$ is called associative
if $\varphi(a)\varphi(bc)=\varphi(ab)\varphi(c)$ for all $a, b,
c\in {\mathcal A}$. For example every mapping preserving zero
products is associative (see Lemma 2.1 of \cite{CKLW}.

\begin{proposition} Let ${\mathcal A}$ be linearly generated by its
idempotents and $\varphi:{\mathcal A}\to {\mathcal B}$ be an
$n$-homomorphism. Then
$\varphi(a)\varphi(bc)=\varphi(ab)\varphi(c)$ for all $a, b, c\in
{\mathcal A}$.
\end{proposition}

\begin{proof} Let $a, c\in {\mathcal A}$ and $e\in {\mathcal A}$ be an
idempotent. Then
\begin{eqnarray*}
\varphi(a)\varphi(ec)&=&\varphi(a)\varphi(e^{n-1}c)\\
&=&\varphi(a)\varphi(e)\ldots\varphi(e)\varphi(c)\\&=&\varphi(ae^{n-1})\varphi(c)\\
&=&\varphi(ae)\varphi(c).
\end{eqnarray*}
Since $\varphi$ is linear and the linear span of idempotent
elements of ${\mathcal A}$ is ${\mathcal A}$ itself, we obtain
$\varphi(a)\varphi(bc)=\varphi(ab) \varphi(c)$ for all $a, b,
c\in {\mathcal A}$.
\end{proof}

The following lemma states some interesting properties of associative mappings.

\begin{lemma} If ${\mathcal A}$ is unital and $\varphi:{\mathcal A}\to {\mathcal B}$ is associative then\\
{\rm (i)} $\varphi(1)$ belongs to the center of image of $\varphi$;\\
{\rm (ii)} For all $a, b\in {\mathcal A}, \varphi(1)\varphi(ab)=\varphi(a)\varphi(b)$;\\
{\rm (iii)} $\varphi$ preserves commutativity;\\
{\rm (iv)} If ${\mathcal B}$ is unital, and $\varphi(1)$ is
invertible or the algebra $<\varphi({\mathcal A})>$ generated by
$\varphi({\mathcal A})$ has an identity, then
$\psi(a)=\varphi(1)^{-1}\varphi(a)$ is a homomorphism. {\rm ( see
Lemma 2.1 of \cite{CKLW})}
\end{lemma}

\begin{proof} We merely prove (iv), the rest is easily
followed from the definition.

If $\varphi(1)$ is invertible in $<\varphi({\mathcal A})>$ and
$\psi(a)=\varphi(1)^{-1}\varphi(a)$, then
\begin{eqnarray*}
\psi(ab)&=&\varphi(1)^{-1}\varphi(ab)\\&=&
\varphi(1)^{-1}\varphi(1)^{-1}\varphi(a)\varphi(b)\\&=&
\varphi(1)^{-1}\varphi(a)\varphi(1)^{-1}\varphi(b)\\&=&\psi(a)\psi(b).
\end{eqnarray*}
If the algebra $<\varphi({\mathcal A})>$ has an identity, say
$u$, then $u=\displaystyle{\sum_{i=1}^n}\varphi(a_{i_1})
\ldots\varphi(a_{i_n})$. Hence
\[u=uu=(\displaystyle{\sum_{i=1}^n}
\varphi(a_{i_1})\ldots\varphi(a_{i_n}))^2=\varphi(1)v=v\varphi(1).\]
Therefore $\varphi(1)$ is invertible in $<\varphi({\mathcal A})>$.
\end{proof}

\begin{remark} The statements of Lemma 3.2 hold for every
unital $C^*$-algebra of real rank zero or $W^*$-algebra
${\mathcal A}$ and every norm-continuous associative mapping
$\varphi$ from ${\mathcal A}$ into a normed algebra.
\end{remark}

The following result is known for mappings preserving zero
products; cf. Corollary 2.8 of \cite{CKLW}. We however reproved
it for the sake of ellegance.

\begin{proposition} Suppose that $H$ and $K$ are infinite
dimensional complex Hilbert spaces and $\varphi:B(H)\to B(K)$ is
a bijective $n$-homomorphism, then there exist $\alpha\in {\C}$
and an invertible bounded linear operator $S$ from $K$ onto $H$
such that $\varphi(T)=\alpha S^{-1}TS$ for all $T\in B(H)$.
\end{proposition}

\begin{proof} It is known that every bounded linear operator
on an infinite dimensional complex Hilbert space is a sum of at
most five idempotent \cite{P-T}. Hence $B(H)$ is generated by its
idempotents. Since $\varphi(1)\neq 0$ belongs to the center the
factor $B(K)$, it is a scalar $\alpha\in {\C}$. Then Lemma 3.2
follows that $\psi=\alpha^{-1}\varphi$ is an isomorphism from
$B(H)$ onto $B(K)$. Applying Theorem 4 of \cite{ARN}, there
exists an invertible bounded linear operator $S:K\to H$ such that
$\psi(T)=S^{-1}TS$ for all $T\in B(H)$.
\end{proof}

\section{Commutativity}

Recall that an algebra ${\mathcal A}$ is called semiprime if
$a{\mathcal A}a=\{0\}$ implies that $a=0$ for each $a\in {\mathcal
A}$. The known techniques similar to \cite{BRE} yield the
following:

\begin{lemma} Let ${\mathcal A}$ be a semiprime algebra and let $a\in {\mathcal
A}$. If $[[a,x],x]=0$ for all $x\in {\mathcal A}$, then $[a,x]=0$
for all $x\in {\mathcal A}$.
\end{lemma}

\begin{proof} For all $x,y\in {\mathcal A}$,
\[0=[[a,x+y],x+y]=[[a,x]+[a,y], x+y]=[[a,x],y]+[[a,y],x] ~~~(1)\]
Taking $yx$ for $y$ in (1) we have
\begin{eqnarray*}
0&=&[[a,x],yx]+[[a,yx],x]\\&=&[[a,x],y]x+[[a,y]x+
y[a,x],x]\\&=&[[a,x],y]x+[[a,y],x]x+[y,x][a,x].
\end{eqnarray*}
Thus for all $x,y\in {\mathcal A}$,
\[ [y,x][a,x]=0~~~(2).\]
Replacing y by az in (2) we obtain
\begin{eqnarray*}
0&=&[az,x][a,x]\\&=&([a,x]z+a[z,x])[a,x]\\&=&[a,x]z[a,x]+a[z,x][a,x]\\&=&[a,x]z[a,x].
\end{eqnarray*}
Hence $[a,x]{\mathcal A}[a,x]=\{0\}. ~~~(3)$

Since ${\mathcal A}$ is semiprime (3) implies that $[a,x]=0$.
\end{proof}

\begin{theorem} Suppose that ${\mathcal A}$ and ${\mathcal B}$ are two
algebras, ${\mathcal B}$ is semiprime and $\varphi:{\mathcal
A}\to {\mathcal B}$ is a surjective $3$-homomorphism. If
${\mathcal A}$ is commutative, then so is ${\mathcal B}$.
\end{theorem}

\begin{proof} For each $a, b\in {\mathcal A}$, $[[a,b],b]=[0,b]=0$. So
that
\[[[\varphi(a),\varphi(b)],\varphi(b)] = \varphi([[a,b],b]) = 0.\]
Applying lemma 4.1, $[\varphi(a),\varphi(b)]=0$. Hence ${\mathcal
B}$ is commutative.
\end{proof}

\section{$n$-Homomorphisms on Banach Algebras}

Recall that the second dual ${\mathcal A}^{**}$ of a Banach
algebra ${\mathcal A}$ equipped with the first Arens product is a
Banach algebra. The first Arens product is indeed characterized
as the unique extension to ${\mathcal A}^{**} \times {\mathcal
A}^{**}$ of the mapping $(a,b) \mapsto ab$
from ${\mathcal A} \times {\mathcal A}$ into ${\mathcal A}$ with the following properties :\\
(i) for each $G \in {\mathcal A}^{**}$, the mapping $F \mapsto FG$ is weak*-continuous on ${\mathcal A}^{**}$;\\
(ii) for each $a \in {\mathcal A}$, the mapping $G \mapsto aG $ is weak*-continuous on ${\mathcal A}^{**}$.\\
The second Arens product can be defined in a similar way. If the
first and the second Arens products coincide on ${\mathcal
A}^{**}$,
then ${\mathcal A}$ is called regular.\\
We identify ${\mathcal A}$ with its image under the canonical
embedding $i : {\mathcal A} \longrightarrow {\mathcal A}^{**}$.

\begin{theorem} Suppose that ${\mathcal A}$ and ${\mathcal B}$ are two Banach algebras
and $\varphi:{\mathcal A}\to {\mathcal B}$ is a continuous
$n$-homomorphism. Then the second adjoint $\varphi^{**}:{\mathcal
A}^{**}\to {\mathcal B}^{**}$ of $\varphi$
is also an $n$-homomorphism.\\
If, in addition, ${\mathcal A}$ is Arens regular and has a
bounded approximate identity, then $\phi$ is a certain multiple
of a homomorphism.
\end{theorem}

\begin{proof} Let $F_1, \ldots, F_n\in {\mathcal A}^{**}$.
By Goldstine' theorem (cf. \cite{D-S}), there are nets $(a^1_i),
\ldots, (a^n_j)$ in ${\mathcal A}$ such that \[{\rm weak}^*{\rm
-}\lim _i a^1_i=F_1, \ldots, {\rm weak}^*{\rm -}\lim _j
a^n_j=F_n.\]
Since $\varphi^{**}$ is ${\rm weak}^*{\rm
-}$continuous we have
\begin{eqnarray*}
\varphi^{**}(F_1 \ldots F_n)&=&\varphi^{**} ({\rm weak}^*{\rm
-}\lim _i\ldots {\rm weak}^*{\rm -}\lim _j a^1_i\ldots
a^n_j)\\&=& {\rm weak}^*{\rm -}\lim _i\ldots {\rm weak}^*{\rm
-}\lim _j \varphi^{**}(a^1_i\ldots a^n_j)\\&=& {\rm weak}^*{\rm
-}\lim _i\ldots {\rm weak}^*{\rm -}\lim _j \varphi^(a^1_i\ldots
a^n_j)\\&=&{\rm weak}^*{\rm -}\lim _i \ldots {\rm weak}^*{\rm
-}\lim _j (\varphi(a^1_i)\ldots \varphi(a^n_j))\\&=& {\rm
weak}^*{\rm -}\lim _i\varphi(a^1_i)\ldots {\rm weak}^*{\rm -}\lim
_j \varphi(a^n_j)\\&=& {\rm weak}^*{\rm -}\lim
_i\varphi^{**}(a^1_i) \ldots {\rm weak}^*{\rm -}\lim _j
\varphi^{**}(a^n_j)\\&=& \varphi^{**}(F_1)\ldots\varphi^{**}(F_n).
\end{eqnarray*}

If ${\mathcal A}$ is Arens regular and has a bounded approximate
identity , it follows from and Proposition 28.7 of \cite{B-D} that
${\mathcal A}^{**}$ has an identity. By proposition 2.3 there
exists a homomorphism $\psi:{\mathcal A}^{**}\to {\mathcal
B}^{**}$ such that $\varphi(a)=\varphi^{**}|_{\mathcal
A}(a)=\varphi^{**}(1_{{\mathcal A}^{**}})\psi(a)$ for all $a\in
{\mathcal A}$.
\end{proof}

\begin{remark} A computational proof similar to that of
Theorem 6.1 of \cite{C-Y} may be used in extending
$n$-homomorphisms to the second duals.
\end{remark}

Now suppose that $\varphi$ is a non-zero $3$-homomorphism from a
unital algebra ${\mathcal A}$ to ${\C}$. Then $\varphi(1)= 1$ or
$-1$. Hence either $\varphi$ or $-\varphi$ is a character on
${\mathcal A}$. If ${\mathcal A}$ is a Banach algebra, then
$\varphi$ is automatically continuous; cf. Theorem 16.3 of
\cite{B-D}. It may however happen that a $3$-homomorphism is not
continuous:

\begin{example} Let ${\mathcal A}$ be the algebra of all $3$ by $3$ matrices
having $0$ on and below the diagonal and ${\mathcal B}$ be the
algebra of all ${\mathcal A}$-valued continuous functions from
$[0,1]$ into ${\mathcal A}$ with sup norm. Then ${\mathcal B}$ is
an infinite dimensional Banach algebra ${\mathcal B}$ in which the
product of any three elements is $0$. Since $\mathcal B$ is
infinite dimensional there are linear discontinuous maps (as
discontinuous $3$-homomorphisms) from ${\mathcal B}$ into itself.
\end{example}

\begin{theorem} Let ${\mathcal A}$ be a $W^*$-algebra and ${\mathcal B}$ a $C^*$-algebra.
If $\varphi:{\mathcal A}\to {\mathcal B}$ is a weakly-norm
continuous 3-homomorphism preserving the involution, then $\Vert
\varphi\Vert\leq 1$.
\end{theorem}

\begin{proof} The closed unit ball of ${\mathcal A}$ is compact in weak
topology. By the Krein-Milman theorem this convex set is the
closed convex hull of its extreme points. On the other hand, the
extreme points of the closed unit ball of ${\mathcal A}$ are the
partial isometries $x$ such that $(1-xx^*){\mathcal
A}(1-x^*x)=\{0\}$, cf. Problem 107 of \cite{HAL}, and Theorem
I.10.2 of \cite{TAK}. Since
$\varphi(xx^*x)=\varphi(x)\varphi(x)^*\varphi(x)$, the mapping
$\varphi$ preserves the partial isometries. Since every partial
isometry $x$ has norm $\Vert x\Vert\leq 1$, we conclude that
$$\Vert \varphi(\displaystyle{\sum_{i=1}^n}\lambda_ix_i)\Vert=
\Vert\displaystyle{\sum_{i=1}^n}\lambda_i\varphi(x_i)\Vert\leq
\displaystyle{\sum_{i=1}^n}\lambda_i\Vert\varphi(x_i)\Vert\leq 1$$
where $x_1, \ldots, x_n$ are partial isometries,
$\lambda_1\ldots \lambda_n>0$ and $\displaystyle{\sum_{i=1}^n}\lambda_i=1$.\\
It follows from weak continuity of $\varphi$ that
$\Vert\varphi\Vert\leq 1$.
\end{proof}

{\bf Question.} Is every $*$-preserving $n$-homomorphism between
$C^*$-algebras continuous?

{\bf Aknowledgment.} The authors would like to thank Professor
Krzysztof Jarosz for providing the Examples 2.6 and 5.3.

\bibliographystyle{amsplain}

config: arxiv


TITLE: differential form and cylindrical coordinate
QUESTION [1 upvotes]: Problem.

If $r, \theta, z$ are the cylindrical coordinate functions on $\mathbb
> R^3$ , then $x = r\cos\theta, y = r\sin\theta, z = z$. Compute the
  volume element dx dy dz of $\mathbb R^3$ in cylindrical coordinates.
  (That is, express dx dy dz in terms of the functions $r, \theta, z$,
  and their differentials.)

My solution.
$dx = \cos\theta dr - r\sin\theta d\theta$
$dy = \sin\theta dr + r\cos\theta d\theta$
$dz = dz$
$$ dx\wedge dy\wedge dz = ... = rdrd\theta dz$$
Actually, these steps are from internet or answer page. I have no idea of how to connect $dxdydz$ and $drd\theta dz$. I don't know why $wedge$ operator is done.
Additionally, I want to know how to convert spherical coordinate using 1-form.

REPLY [0 votes]: Begin with coordinate transformation:
$ ds^2 = g_{ab}dx^adx^b $
$ = g_{ab}\frac{\partial x^a}{\partial \zeta^{\alpha}}\frac{\partial x^b}{\partial \zeta^{\beta}} d\zeta^{\alpha}d\zeta^{\beta}$
$ = \gamma_{\alpha\beta} d\zeta^{\alpha}d\zeta^{\beta}$
where
$ \gamma_{\alpha\beta}=g_{ab}\frac{\partial x^a}{\partial \zeta^{\alpha}}\frac{\partial x^b}{\partial \zeta^{\beta}} $
This relates the Cartesian metric to the cylindrical metric:
$ g_{ab}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & 1 & 0 \\
    0       & 0 & 1
\end{pmatrix}$
$ \gamma_{\alpha\beta}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & r^2 & 0 \\
    0       & 0 & 1
\end{pmatrix}$
The volume form is given as:
$dV = \sqrt{|det(\gamma_{\alpha\beta})|}d\zeta^{(0)} \land d\zeta^{(1)} \land d\zeta^{(2)} $
$= rd\zeta^{(0)} \land d\zeta^{(1)} \land d\zeta^{(2)} $
$= rdr \land d\theta \land dz $

note that: $ Adx \land dy \land dz = Adxdydz $ in most shorthand.

Wedge products are used because they allow for one to construct area and volume elements from infinitesimal length elements $ds$. So $dx$ would be an infinitesimal length element, $dx \land dy$ would be an infinitesimal area, and so on.
To convert to spherical coordinates, you would just redefine the second metric as:
$ \gamma_{\alpha\beta}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & r^2 & 0 \\
    0       & 0 & r^2sin^2(\theta)
\end{pmatrix}$
with $ d\zeta^{(2)} = d\phi$


TITLE: Linear Least Squares with $ {L}_{2} $ Norm Regularization / Penalty Term
QUESTION [2 upvotes]: I want to minimize the following cost function
$$J= \displaystyle\min_{\boldsymbol{\rm x}} \left\{\|A\boldsymbol{\rm x}_\lambda-b\|_2^2 + \lambda\|\boldsymbol{\rm x}_\lambda\|_2\right\}$$
It is a bit different from Tikhonov regularization because the penalty term is not squared. As opposed to Tikhonov, which has an analytic solution, I was not able to find such solution for $J$. Could someone give me some hint which method I can use to solve for $\boldsymbol{\rm x}_\lambda$?

REPLY [3 votes]: You can solve this problem easily using the proximal gradient method or an accelerated proximal gradient method such as FISTA. This is a powerful tool that everyone can easily learn to use. The proximal gradient method solves optimization problems of the form
$$
\text{minimize} \quad f(x) + g(x)
$$
where $f$ and $g$ are convex and $f$ is differentiable (with a Lipschitz continuous gradient) and $g$ is "simple" in the sense that its proximal operator can be evaluated efficiently. The proximal operator of $g$ is defined as follows:
$$
\tag{$\spadesuit$}
\text{prox}_{tg}(\hat x) = \arg \, \min_x \quad g(x) + \frac{1}{2t} \| x - \hat x \|_2^2.
$$
The proximal operator of $g$ reduces the value of $g$ without straying too far from $\hat x$. The parameter $t > 0$ can be thought of as a "step size" that controls how far away from $\hat x$ we are allowed to move. Note that $g$ is not required to be differentiable. (But $g$ is required to be lower-semicontinuous, which is a mild condition that is usually satisfied in practice.)
The proximal gradient method iteration is
$$
x^{k+1} = \text{prox}_{tg}(x^k - t \nabla f(x^k)).
$$
What is happening here is that, starting from the current location $x^k$, we take a gradient descent step to reduce the value of $f$, then we apply the prox-operator of $g$ to reduce the value of $g$.
The step size $t > 0$ is required to satisfy $t \leq 1/L$, where $L$ is a Lipschitz constant for $\nabla f$. (There is also a line search version of the proximal gradient method that might converge faster. In the line search version the step size $t$ is chosen adaptively at each iteration.)
Your optimization problem has the form $(\spadesuit)$ where
$$
f(x) = \| Ax - b \|_2^2
$$
and
$$
g(x) = \lambda \|x\|_2.
$$
The gradient of $f$ can be evaluated using calculus:
$$
\nabla f(x) = 2 A^T(Ax - b).
$$
There is a nice formula for the proximal operator of the $\ell_2$-norm:
$$
\text{prox}_{tg}(x) = \text{prox}_{t \lambda \| \cdot \|_2}(x) = x - P_B(x),
$$
where $B$ is the $2$-norm ball of radius $t \lambda$ centered at the origin and $P_B(x)$ is the projection of $x$ onto $B$. (A formula for the proximal operator of a norm can be found for example in Vandenberghe's UCLA 236c notes. See chapter 8 "The proximal mapping", slide 8-3.)
With these formulas, you are ready to implement the proximal gradient method to solve your optimization problem using just a few lines of code. There is also an accelerated version of the proximal gradient method known as FISTA which is equally simple and likely converges faster.