This is a page for various notes and quick tests …

Many times I had someone on the other part of the line and 
wanted to talk about mathematics ... so you need some space 
where you can quickly type some ideas in LaTeX and the other
person with internet connection can read it ... 

Embedding PHP works, writing “test” via php: <php> echo “test”; </php>

Use $x$ for points in $\mathbb{R}^3$, then \begin{equation} \Omega = \Big\{ x \in \bigcup_{i=1}^{K} C_i \Big\} \end{equation}

probably better to write \begin{equation} \Omega = \bigcup_{C \in \cal{C}} C \end{equation} where \begin{equation} \cal{C} = \Big\{ L_{j} …\Big\} \end{equation}

Consider a simple situation like

and now two slants, given by \begin{eqnarray} s_1 & = & x_1 + x_4
s_2 & = & x_2 + x_4 \end{eqnarray} leading to a matrix equation \begin{equation} \left( \begin{array}{ccc} 1 & 0 & 0 & 1
0 & 1 & 0 & 1 \end{array} \right) \circ \left( \begin{array}{c} x_1
x_2
x_3
x_4 \end{array} \right) = \left( \begin{array}{c} s_1
s_2 \end{array} \right) \end{equation} We might include some weighting function for taking care of the lenght of the path of the ray through the area denoted by $x_j$, $j=1,…,4$, then we obtain a weighted system of the form \begin{eqnarray} s_1 & = & w_{1,1} x_1 + w_{1,4} x_4
s_2 & = & w_{2,2} x_2 + w_{2,4} x_4 \end{eqnarray}

A ray starting at $b \in \mathbb{R}^3$ with direction $\nu \in \mathbb{R}^3$ is given by \begin{equation} L = \{ x = b + \eta \cdot \nu| \; \eta > 0 \}. \end{equation} An integral along $L$ is written as \begin{equation} f(b,\nu) := \int_{L} \varphi(y) ds(y), \end{equation} where $ds(y)$ denotes the standard Eucliedean measure along $L$. Alternatively, when $||\nu|| = 1$, you might write \begin{equation} f(b,\nu) = \int_{0}^{\infty} \varphi(b + \eta \nu) d\eta \end{equation} We use the notation \begin{equation} (R\varphi)(b,\nu) := f(b,\nu), \;\; b \in \mathcal{B}, \; \nu \in \mathcal{N} \end{equation} where $\mathcal{B}$ and $\mathcal{N}$ denote the set of our base stations and the set of slant directions measured by the stations over some time interval. In discretized form, the operator $R$ maps $\vec{\varphi} \in \R^n$ into $\vec{f} \in \R^m$. It is a linear operator and, thus, we obtain a matrix operator from $\R^n$ into $\R^m$. Thus, GPS tomography boiles down to solving a linear finite dimensional system \begin{equation} R \vec{\varphi} = \vec{f} \end{equation}

We first select a subsequence of $\mathbb{N}$ such that a point in the support $\chi_i$ of $v_{k_{i}}$ has limiting point $x_{\ast} \in \Omega$. This is possible since any bounded sequence in $\mathbb{R}$ has a convergent subsequence. We call it $k_{i}$ again.

Given $\epsilon>0$ there is $i>0$ such that $$

$$ So we know that on the exterior of the support $\Omega_{k_{i}}$ of $v_{k_{i}}$ we have $$

$$ For a function $v_{\ast} \in L^2(\Omega)$ we have that \begin{eqnarray}

\end{eqnarray} Now, we obtain \begin{eqnarray}

+ ||v_{\ast}||_{L^2(\Omega_{k_{i}})}^2 \nonumber
& \rightarrow & 0, \;\; i \rightarrow \infty. \end{eqnarray}

Simple question: is $\frac{1}{x-y}$ analytic in $x$ for fixed $y$?

Answer 1) Yes, since the sum and product of analytic functions are analytic, and the quotient of analytic functions are analytic where the denominator is non-zero.

Answer 2) We use the series expansion $$ \frac{1}{x-y} = (-1) \sum_{n=1}^{\infty} \frac{x^n}{y^{n+1}} $$ for $|x|<|y|$, which I took from Wolfram. For $|x|<\rho \cdot |y|$ with $\rho<1$ we have $$

$$ such that $$

$$ is absolutely convergent by the geometric series. Thus, $\frac{1}{x-y}$ is analytic on $|x|<|y|$. For $|x|>|y|$ we use $$ \frac{1}{x-y} = \sum_{n=1}^{\infty} \frac{y^n}{x^{n+1}} $$ analogously.

Let $g$ be a function in $L^2(\Omega)$ and let $\Omega_i$ be subsets of $\Omega$ with $|\Omega_i| \rightarrow 0$ for $i \rightarrow \infty$. Then we have \begin{equation} \label{eq1} \int_{\Omega_i} |g|^2 dy \rightarrow 0, \;\; i \rightarrow \infty. \end{equation}

Proof. Given $\epsilon$ there is a bounded function $f \in C^{\infty}(\Omega)$ such that \begin{equation} \label{eq2}

\end{equation} Further, for a function $f \in C^{\infty}(\Omega)$ given $\epsilon$ we find $i_0>0$ such that \begin{equation} \label{eq3} \int_{\Omega_{i}} |f(y)|^2 dy \leq \frac{\epsilon}{4} \end{equation} for all $i\geq i_0$. Now, given $\epsilon$ we first choose $f \in C^{\infty}$ such that (\ref{eq2}) is satisfied. Then, we choose $i_0$ such that (\ref{eq3}) is true. This yields \begin{eqnarray} \int_{\Omega_i} |g|^2 dy & = & \int_{\Omega_i} |g-f + f|^2 dy
& \leq & 2 (\int_{\Omega_i} |g-f|^2 dy + \int_{\Omega_i} |f|^2 dy )
& \leq & 2(\frac{\epsilon}{4} + \frac{\epsilon}{4}) = \epsilon \end{eqnarray} for $i \geq i_0$. $\Box$

$A: X \rightarrow Y$ compact, $\varphi_n$ bounded sequence in $X$, then $\psi_n := A\varphi_n$ has a convergent subsequence, which we relable to use index $n$ again.

$\psi_n \rightarrow \psi_{\ast} \in Y$. If $\psi_{\ast} \in R(A)$, then there exists $\varphi_{\ast} \in X$ such that $A \varphi_{\ast} = \psi_{\ast}$. Defining $\tilde{\varphi}_{n}:= \varphi_n - \varphi_{\ast}$ we obtain a sequence such that $$ A \tilde{\varphi}_{n} \rightarrow 0, \;\; n \rightarrow \infty. $$