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}

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.

Lemma 3.1.1 The sequence $(v_k)$ defined above does not contain a convergent subsequence in $L^2(\Omega)$.

Proof. $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\N}{\mathbb{N}}$ Assume that $v_k$ does contain a convergent subsequence $v_{k_i}$. We consider the sequence of supports $\Omega_{k_{i}}$ of $v_{k_i}$. Now consider the sequence of centres $x_{k_i}^{(centre)}$ of the supports. By the Bolzano-Weierstrass theorem, every bounded sequence in $\R^n$ has a convergent subsequence and so a subsequence $x_|v_{k_i}-v_{\ast}||<\frac{\epsilon}{2}. \end{equation} We know that for any point $x$ lying in the exterior of the support we have $v_{k_i}(x)=0$, then from (\ref{eq:Supp}) we have $$ v_{k_i}(x)=0 \mbox{ for } x\in\Omega\setminus\Omega_{k_i} $$ and therefore (\ref{help0}) becomes \begin{equation} \label{help1} ||v_{\ast}||_{L^2(\Omega\setminus\Omega_{k_i})} = ||v_{k_{i - v_{\ast}||_{L^2(\Omega\setminus\Omega_{k_i})} \leq ||v_{k_{i}} - v_{\ast}||_{L^2(\Omega)} <\frac{\epsilon}{2} \end{equation} for all $\epsilon>0$ and $i>n_1$.

We define the functions $s_{ki}:\Omega\rightarrow \{0,1\}$ such that $$ s_{ki}(x)=\left\{\begin{array}{c} 1 \; \; \mbox{if } x\in supp_{ki},
0 \; \; \mbox{otherwise.} \end{array}\right. $$ Then, noting that $v_{k_{j}}$ and thus also $v_{\ast}$ is positive, we have \begin{eqnarray}

&=&\int_{\Omega}|v_{\ast}(y)|^2|s_{k_i}(y)|dy
&=&\left\langle v_{\ast}^2,s_{k_i} \right\rangle_{L^2(\Omega)}
&\leq& || v_{\ast}^2||_{L^2(\Omega)}\cdot ||s_{k_i}||_{L^2(\Omega)}\label{eq:Argument1}
&=& || v_{\ast}^2 ||_{L^2(\Omega)}\cdot |\Omega_{k_{i}}|. \end{eqnarray} by the Cauchy-Schwarz inequality.

As $i\rightarrow \infty$, the $L$ and $n_L$ in (\ref{def:vk}) tend to infinity and therefore as $i\rightarrow\infty$, the size of $supp_{k_i}$ tends to zero. Then the right hand side of (\ref{eq:Argument1}) tends to zero and $\forall\epsilon >0$, $\exists n_2\in\N$ such that $\forall i>n_2$ $$

$$ Finally, if we take $n=\max\{n_1,n_2\}$ we obtain $$

$$ for all $i>n$.

However we cannot have $v_{k_i}$ with $||v_{k_i}||_{L^2(\Omega)}=1$ tending to zero. Thus $v_k$ does not contain a convergent subsequence. $\Box$