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>

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}

\left|\int_{\Omega} |v_{\ast}(y)|^2 |\chi_{k_{i}}(y)|^2 dy \right| \nonumber
& \leq & ||v_{\ast}||_{L^2(\Omega)}^2 \cdot || \chi_{k_{i}} ||_{L^2(\Omega)}^2. \end{eqnarray} Now, we obtain \begin{eqnarray}

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

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.