TITLE: Different definitions of a relatively compact operator
QUESTION [7 upvotes]: Let $T,K$ be unbounded operators on a Hilbert space $H$.
I've seen the following definition of a relatively compact operator:

(i) The operator $K$ is called relatively compact with respect to $T$, if for some $z$ in the resolvent set of $T$, $KR_T(z)$ is compact, where $R_T(z):=(T-z)^{-1}.$

I've also seen:

(ii) The operator $K$ is called relatively compact with respect to $T$, if for every sequence $(x_n)_{n \in \mathbb{N}}\subseteq H$ such that $(Tx_n)_{n \in \mathbb{N}}$ is bounded, $(Kx_n)_{n \in \mathbb{N}}$ contains a convergent subsequence.

Do definitions (i) and (ii) have something to do with each other, or are they distinct? What is the intuition behind these definitions? Definition (ii) looks like a generalisation of a compact operator, but definition (i) is just weird.
This question has also been posted on Math Overflow.

REPLY [2 votes]: Both definitions aim to make the same statement -- the operator $KT^{-1}$ is compact. But we cannot really to do that, because $T^{-1}$ does not necessarily exist, so in (i) we replace the inverse by the resolvent and in (ii) we write the more usual definition of compactness; if $(Tx_n)$ is bounded then $KT^{-1}(Tx_n)= Kx_n$ should contain a convergent subsequence.
Actually, (i) is equivalent to a version of (ii), where we only allow bounded sequences $(x_n)$. Indeed, to see that (ii) implies (i), pick a bounded sequence $(y_n)$. Then the sequence $x_n:= (T-z)^{-1}y_n$ is bounded as well, and $Tx_n= (T-z)x_n + zx_n = y_n + zx_n$ is bounded. By (ii), $(Kx_n)$ contains a convergent subsequence, but $Kx_n = K(T-z)^{-1}y_n$, so we showed compactness of $K(T-z)^{-1}$.
For the other direction, note that if the sequences $(x_n)$ and $(Tx_n)$ are bounded, then $((T-z)x_n)$ is bounded as well, so by (i) $K(T-z)^{-1} (T-z)x_n = Kx_n$ contains a convergent subsequence.
As stated in your question, the definitions are not equivalent (but (ii) still implies (i)). Note that (i) does not imply (ii), because any compact operator satisfies (i), but not necessarily (ii) (look at $T=0$).
Answer cross-posted from Math Overflow.


text:
\section{A subgradient method and numerical experiments}\label{subsec:algo}

Informed by the analysis in the previous sections, we propose a gradient method for the solution of \eqref{eq:minM} and verify empirically its optimal properties.

Recall that the main properties of the landscape of the minimization problem
\eqref{eq:minM} were the global minimum in a neighborhood of the true latent vector $\xstar$ and a flat region in correspondence of $-\rho_d \xstar$. Moreover
the latter region has larger loss function values than those in the vicinity of $\xstar$.  In order to overcome the non-convexity and avoid this bad region we run gradient descent from a random non-zero initialization and its negation. We then pick the iterate which has smaller final loss value.

\begin{algorithm}[H]
	\caption{Gradient method for the minizimization problem \eqref{eq:minM}}\label{alg:subgrad}
	\begin{algorithmic}[1]
		\State {\bfseries Input:} Weights $W_i$, observation matrix $M$, step size $\alpha > 0$, number of iterations $T$
		\State Choose $\hat{x}_0 \in \R^k \backslash \{0\}$ arbitrary
		\State Let $x_{0}^{(1)} = \hat{x}_0$ and $x_{0}^{(2)} = -\hat{x}_0$
		\For{i = 1, 2}
		\For{$j = 0, 1, \ldots, T-1$}
		\State Compute $v_{{x}_{j}}^{(i)} \in \pa f({x}_{j}^{(i)})$
		\State ${x}_{j+1}^{(i)} \gets {x}_{j}^{(i)} - \alpha v_{{x}_{j}}^{(i)}$;
		\EndFor
		\EndFor
		\If{$f({x}_{T}^{(1)}) < f({x}_{T}^{(2)})$}
		\State {Return:} {$x_{T}^{(1)}, G(x_{T}^{(1)})$}
		\Else
		\State {Return:} {$x_{T}^{(2)}, G(x_{T}^{(2)})$}
		\EndIf
	\end{algorithmic}
\end{algorithm}

Note that it will be highly unlikely that the iterates will be at a non-differentiable point, therefore
in practice we can consider Algorithm \ref{alg:subgrad} with descent direction $v_{x} = \nabla f(x)$.

config: arxiv


TITLE: Infinitesimal Translation Operator
QUESTION [3 upvotes]: In the book "Modern Quantum Mechanics" (by J.J. Sakurai and Jim Napolitano) page 44, infinitesimal Translation operator is given:
$\mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$ after this, book mentioned properties of the operator as being unitary and adding: $\mathscr{J}\left(d \mathbf{x}^{\prime \prime}\right) \mathscr{J}\left(d \mathbf{x}^{\prime}\right)=\mathscr{J}\left(d \mathbf{x}^{\prime}+d \mathbf{x}^{\prime \prime}\right)$ where I have confused is that:
$\mathbf{x} \mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\mathbf{x}\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle=\left(\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$
The left hand side of the equation is so weird to me because middle one is directly driven from the first equation but to find a meaning at the right hand side of the above equation, I thought it might be true because $d \mathbf{x}^{\prime}$  is infinitesimally small but where  is $\mathbf{x}^{\prime}$came from because initially it was $\mathbf{x}$ at the right hand side of the last equation.

REPLY [3 votes]: I'm not sure I completely understood your question, but perhaps it will become clear if I explain what the last equation means. The left hand side of the equation means:

Take a state $|\mathbf{x}'\rangle$
Translate it by some infinitesimal amount $\text{d}\mathbf{x}'$ using the translation operator
Act on it with the position operator, $\mathbf{\hat{x}}$.

By definition, translating the state $|\mathbf{x}'\rangle$ by $\text{d}\mathbf{x}'$ gives you the state $|\mathbf{x}' + \text{d}\mathbf{x}'\rangle$, and also by definition (since it is now a state of definite position with eigenvalue $\mathbf{x}'+ \text{d}\mathbf{x}'$), when you act on this state with the position operator you get $$\mathbf{\hat{x}} |\mathbf{x}' + \text{d}\mathbf{x}'\rangle = (\mathbf{x}' + \text{d}\mathbf{x}')|\mathbf{x}' + \text{d}\mathbf{x}'\rangle.$$
Does this help?


TITLE: How do I show that $\lim_{x\rightarrow\pm\infty}\frac{\ln(x^2+1)}{x} = 0$ without L'Hôspital's rule?
QUESTION [1 upvotes]: Without the use of series of L'Hôspital's rule, therefore with common limits and some intuition.
I wanted to use the common limit saying $\lim_{x\rightarrow0}\dfrac{\ln(1+x)}{x} = 1$, but that doesn't work as the limit there goes to $0$ and not $1$. Is it even possible without making it difficult to compute this limit without L'Hôspitals or series?

REPLY [2 votes]: The limit at $\infty$ is the same as
$$
\lim_{t\to0^+}t\log(1+1/t^2)=
\lim_{t\to0^+}t(\log(1+t^2)-\log(t^2))
$$
Now, $\lim_{t\to0^+}t\log(t^2)=\lim_{t\to0^+}2t\log t =0$ (this should be already known), so you want to look at
$$
\lim_{t\to0^+}t\log(1+t^2)=\lim_{t\to0^+}t^3\frac{\log(1+t^2)}{t^2}
$$
Since
$$
\lim_{t\to0^+}\frac{\log(1+t^2)}{t^2}=
\lim_{u\to0^+}\frac{\log(1+u)}{u}=1
$$
(a well known limit), you finally get
$$
\lim_{x\to\infty}\frac{\log(1+x^2)}{x}=0
$$
Since the function is odd, you also have
$$
\lim_{x\to-\infty}\frac{\log(1+x^2)}{x}=0
$$


TITLE: Solving Forward Equation
QUESTION [1 upvotes]: I've currently started reading 'Lectures on Partial Differential Equations' by Faris.
Page 44 he states the following forward equation:
$$J=a(y)p-\frac{1}{2}\frac{\partial \sigma(y)^2p}{\partial y} = 0$$
I understand how to solve this and obtain the following:
$$I(y) = \int_{y_0}^{y}\frac{2}{\sigma(z)^2}a(z)dz$$
i.e. integrate once and separate the variables. He then states solving for $p$ produces:
$$p(y)=C\frac{2}{\sigma(y)^2}exp(I(y))$$
I don't understand where the $\frac{2}{\sigma(y)^2}$ comes from. Can someone guide me in the right direction please?

REPLY [2 votes]: It's been four days and you've managed to solve your own question. I'll now share it with you.
Note $J=0$ and $\sigma(y)$ is a function on of variable (i.e. there is no time variable).

Rearrange the Forward Equation and use standard differential operators:
$$a(y)p = \frac{1}{2}\frac{d(\sigma(y)^2p)}{dy}$$
Multiply through by 2 and use the product rule:
$$2a(y)p=\frac{d(\sigma(y)^2)}{dy}p+\frac{d(p)}{dy}\sigma(y)^2$$
Now divide through by p and divide through by $\sigma(y)^2$:
$$\frac{2a(y)}{\sigma(y)^2}=\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}+\frac{1}{p}\frac{d(p)}{dy}$$
Use an integration 'trick':
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}dy+\int_{}{}\frac{1}{p}\frac{d(p)}{dy}dy$$
Simplify!
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{d(\sigma(y)^2)}{\sigma(y)^2}+\int_{}{}\frac{d(p)}{p}$$
Perform rhs integration:
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\ln \sigma(y)^2 +\ln P +\ln C$$
Exponential and rearrange:
$$p(y) = C\frac{1}{\sigma(y)^2}exp(\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy)$$

Note:
I am not sure where Mr Faris obtained the '2' in the final solution. Perhaps someone more knowledgable will be able to explain. I believe it is a mistake and the constant of integration should be correct.


TITLE: Can I directly substitute a specific complex number into a formal power series equation?
QUESTION [2 upvotes]: Suppose there is a formal power series equation, such as
$$\sum_{i=0}^\infty a_i x^i=\sum_{i=0}^\infty b_i\left(\sum_{j=1}^\infty c_j x^j\right)^i.$$
If there is a complex number $r$ which makes $\sum c_jr^j$ and $\sum a_i r^i$ convergent, also, the convergent point of $\sum c_jr^j$, $s$, makes $\sum b_i s^i$ convergent, then, does the equation above still hold when I substitute $r$ into $x$?
Thanks for your discussion.

REPLY [2 votes]: The operation involves re-arranging the power series.  Hence this can only be done if the respective sums converge absolutely.  This is satisfied if the power series is evaluated at a value from inside the respective radius of convergence.
If a power series just converges conditionally, then re-arranging might change its value.  This can only happen if the value where the series is evaluated is on the boundary of the radius of convergence.
See also Riemann's rearrangement theorem.
There are also some posts on mathoverflow that deal with value-preserving permutations of conditionally convergent series, like The stabilizer of the conditionally convergent series.

ad comment: Just that a function is analytic at some point doesn't tell you anything about it's power series expansion.  Take $u(z) = \sum_0 z^n$ for example, with radius of convergence $r_u=1$. $u$ is analytic everywhere except $z=1$, but the series does not converge for any $|z|>1$.  And it does not converge for, say, $z=-1$ or $z=\sqrt{-1}$, either.
Also, if a power series expansion of $f$ around 0 converges for $z$ with $|z|=r_f$, then $f$ is countinuous at $z$.  Again, the opposite is not true in general, and continuiuty of $f$ doesn't tell you anything about $r_f$ or about the becaviour of a power series expansion.
However, your requirement that $f$ is analytic in$^1$ $0\leqslant |z|\leqslant r_f$ for $f\in\{F,G,H\}$ means that the radius of convergence of $f$ is greater than $r_f$ and $r_f$ does not denote the radius of convergence of $f$.  The reason is that, when a power series has a finite radius of convergence $r_f\neq0$, then $f$ has a singularity of some sort for at least one $|z|=r_f$.  Example is polylog $\operatorname{Li}_2(z) = \sum_1 z^n/n^2$ that converges for all $|z|=1$ but it has still a singularity: a branch point at $z=1$.
The conclusion is that you are inside$^1$ the radius of convergence of the functions, thus their power series converge absolutely.
Notice that $z$ above denotes a complex number, and if you replace it by a real number, then what's said above is no more true.  For example, $\arctan z =\sum_0 z^{2n+1}/(2n+1)$ is analytic over $\Bbb R$, yet the power series has finite radius of convergence $r_{\arctan} = 1$.  The reason is that $\arctan$ has branch points at $z=\pm\sqrt{-1}$, so you only see whan't going on over $\Bbb C$.

$^1$Assuming with $D(z,r)$ you denote the closed disk of radius $r$ around $z$.
If, however, $D$ denotes the open disk and $r_f$ stands actually for the respective radius of convergence, then analyticity of $f$ in the open disk $0\leqslant |z| < r_f$ does not tell about whether the power series converges (absolutely) for some $|z|=r_f$.


TITLE: convergence of $\int_a^b \frac{1}{x^2} dx$
QUESTION [2 upvotes]: Why is it true that $\int_0^a \frac{dx}{x^2} = \infty$ but $\int_a^\infty \frac{dx}{x^2} < \infty$?
Shouldn't it be symmetric?

REPLY [4 votes]: By definition:
$$\begin{align}&\int_0^a\frac{dx}{x^2}=\left.\lim_{\epsilon\to 0^+}\left(-\frac1x\right)\right|^a_\epsilon=-\frac1a+\lim_{\epsilon\to0^+}\frac1\epsilon=\infty\\{}\\
&\int_a^\infty\frac{dx}{x^2}=\left.\lim_{b\to\infty}\left(-\frac1x\right)\right|_a^b=-\lim_{b\to\infty}\frac1b+\frac1a=\frac1a\end{align}$$
Thus, the second improper integral converges, whereas the first one diverges.

REPLY [1 votes]: If you work at the problem, you'll see that you would have to evaluate $\frac{1}{0}$. Anyway, not sure what you mean by symmetric, but think about what $$\int_0^a\frac{1}{x^2}dx$$ is actually saying. It's the area under the curve of $\frac{1}{x^2}$, and the height of the curve goes to infinity as $x \to 0$. It makes sense that you would have infinite area under a curve of infinite height.


TITLE: Adjoint of an adjoint of a matrix
QUESTION [2 upvotes]: Can you please help me on this question?
$\DeclareMathOperator{\adj}{adj}$
$A$ is a real $n \times n$ matrix; show that:
$\adj(\adj(A)) = (\det A)^{n-2}A$
I don't know which of the expressions below might help
$$
\adj(A)A = \det(A)I\\
(\adj(A))_{ij} = (-1)^{i+j}\det(A(i|j))
$$
Editor's note: adjoint here refers to the classical adjoint.

REPLY [1 votes]: We know the property adj(A).A = |A|I
Now consider
$$adj(adj(A))*adj(A)=|adj A|I$$
Post multiply this with A
$$adj(adj(A))*adj(A)*A=|A|^{n-1}.I.A$$
$$adj(adj(A))*|A|=|A|^{n-1}.A$$
$$adj(adj(A))=|A|^{n-2}.A$$


TITLE: $h(\partial \bar{D_r})$ is a convex curve iff $arg(z)+arg(h'(z))+\pi/2$ is an increasing function of $arg(z)$
QUESTION [0 upvotes]: Let $h:V\to V$ be a holomorphic automorphism fixing $0$ of $V$ open in $\mathbb{C}$.
Let $\bar{D_r}\subset V$
$h(\partial \bar{D_r})$ is a convex curve iff $\operatorname{arg}(z)+\operatorname{arg}(h'(z))+\pi/2$ is an increasing function of $\operatorname{arg}(z)$
The derivative of this function with respect to $\operatorname{arg}(z)$ is given by $\operatorname{Re}(\frac{z h''}{h'})+1$.
What does  $\operatorname{arg}(z)+\operatorname{arg}(h'(z))+\pi/2$ mean geometrically?
How can I differentiate this function with respect to $\operatorname{arg}(z)$?
How should I prove above statement?
Any hints or references are appreciated. Thank you

REPLY [1 votes]: Let
$$\partial D_r:\quad t\mapsto z(t)=re^{it}\qquad(0\leq t\leq2\pi)$$
be the standard parametric representation of $\partial D_r$. Then
$$\gamma:\quad t\mapsto p(t):=h\bigl(re^{it}\bigr)$$
is a parametric representation of the image curve $\gamma:=h(\partial D_r)$. One computes
$$p'(t)=h'(re^{it})ir e^{it},\quad p''(t)=-h''(re^{it})r^2 e^{2it}-h'(re^{it})re^{it}\ .$$
Now for the geometry. The tangent direction of the image curve at the point  $p(t)$ is given by
$$\theta(t)={\rm arg}\bigl(p'(t)\bigr)={\rm arg}\bigl(h'(z)\bigr)+{\rm arg}(z)+{\pi\over2}\>,\qquad z:=re^{it}\ .\tag{1}$$
Here we have used that ${\rm arg}$ converts products into sums. If we want $\gamma$ to be strictly convex then we want the function $t\mapsto\theta(t)$ strictly increasing with $t$, hence $\theta'(t)>0$. Write the first part of $(1)$ in the form
$$\theta(t)={\rm arg}\bigl(p'(t)\bigr)={\rm Im}\bigl(\log p'(t)\bigr)\ .$$
Then we obtain
$$\theta'(t)={\rm Im}\left(p''(t)\over p'(t)\right)={\rm Im}\left(-{h''(re^{it})r e^{it}\over ih'(re^{it})}-{1\over i}\right)={\rm Re}\left({zh''(z)\over h'(z)}+1\right)\ .$$