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    \fancyhead[LO]{TMA4140: Homework Set 5}
    \fancyhead[RO]{Henry S. Sjøen}
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  \author{Henry S. Sjøen \& Toralf Tokheim}
  \title{
  \textbf{TMA4140 - Homework Set 5}\\
    \color{darkred}{\textbf{RETTES}}
  }
\begin{document}
    \maketitle
    \thispagestyle{empty}
    \pagebreak
    \tableofcontents
    \pagebreak

    \section{Chapter 4.4}
    \subsection{Exercise 21}
    Use the construction in the proof of the Chinese remainder theorem to find all solutions to the system of congruences $x \equiv 1(mod 2)$, $x \equiv 2 (mod 3)$, $x \equiv (mod 5)$, and $x \equiv 4 (mod 11)$.

    We need to find all solutions to the system of congruences.

    First congruence can be written as:
    \begin{equation}
      \begin{split}
        x \equiv  1(mod2)\\
        x = 2m + 1
      \end{split}
    \end{equation}

    Second congruence can be written as:
    \begin{equation}
      \begin{split}
        x \equiv  2(mod3)\\
        x = 3m + 2
      \end{split}
    \end{equation}

    Third congruence can be written as:
    \begin{equation}
      \begin{split}
      x \equiv  2(mod5)\\
      x = 3m + 2
      \end{split}
    \end{equation}

    And the final congruence can be written as:
    \begin{equation}
      \begin{split}
        x \equiv 4(mod11)\\
        x = 11m+4
      \end{split}
    \end{equation}

    Using the Chinese theorem, we can find the L.C.D for $2,3,5$ and $11$.

    $LCD = 2x3x5x11=330$

    Thus, $x = 330k+I$ where $I$ is $1,2,3,4$ for $mod 2,3,5,11$.

    Then $8,13,18,K,323,K$ are numbers divided by 5 where we get remainder 3.

    $8,13,18,K,323,K$ can be divided by 5 to get remainder 3.
    If the number is $232$ it satisfies all the cases for divisor 2.
    $323 = 2(161)+1$.
    Thus the remainder is 1.

    For divisor 3,$323 = 3(107)+2$.
    Thus the remainder is 2.

    For divisor 11,
    $323=11(29)+4$.
    Thus the remainder is 4.

    And the solution of the provided system of congruences is $x = 232 + 330k$.

    \subsection{Exercise 33}
    Use Fermat’s little theorem to find $7^{121} mod 13$.

    \begin{figure}[h]
      \label{theorem:fermat}
      \centering
      Fermat’s theorem\\
      If $p$ is prime and $a$ is an integer not divisible by $p$, then $ap-1\equiv 1 (mod p)$.\\
      % Furthermore, for every integer a we have $ap \equiv a (mod p)$.
    \end{figure}

By Fermat’s theorem,
$7^{12} \equiv 1(mod13)$
So to find $7^{121}(mod13)$ we have to compute
\begin{equation}
  \begin{split}
    7^{121}&=(7^{12})^{10}\cdot 7(mod13)\\
    &\equiv (7^{12})^{10}(mod13)\cdot 7(mod13)\\
    &\equiv 1(mod13)\cdot7(mod13)\\
    &\equiv 7(mod13)
  \end{split}
\end{equation}
Answer is $7(mod13)$

\subsection{Exercise 37a}
    Show that $2^{340} \equiv 1 (mod 11)$ by Fermat’s little theorem and noting that $2^{340} = (2^{10})^{34}$.

    If $p$ is prime and a is an integer not divisible by p, then
    $a^{p-1}\equiv 1(mod p)$.

    So $2^{10}\equiv 1(mod11)$
    \begin{equation}
      \begin{split}
        2^{340}&=(2^{10})^{34}\\
        &\equiv(2^{10})^{34}(mod11)\\
        &\equiv 1(mod11)
      \end{split}
    \end{equation}

    \section{Chapter 4.5}
    \subsection{Exercise 12}
    Find the sequence of pseudorandom numbers generated by the power generator with $p = 11$, $d=2$, and seed $x_0=3$.

    % The power generator is the method of generating pseudorandom numbers

    Let's use the formulae
    \begin{figure}
      \centering
      \begin{equation}
      \begin{split}
        x_{n+1}=x_n^d mod p\\
        x_{n+1}=x_n^2 mod 1 1
      \end{split}
    \end{equation}
    Where $p=11,d=2, x_0=3$
\end{figure}

    So the sequence can be generated as
    \begin{equation}
      \begin{split}
        x_1=3^2mod1 1\\
        \equiv 9 mod 1 1\\
        x_2=9^2mod1 1\\
        \equiv 4 mod 1 1\\
        x_2=4^2mod1 1\\
        \equiv 5 mod 1 1\\
        x_2=5^2mod1 1\\
        \equiv 3 mod 1 1\\
      \end{split}
    \end{equation}
    Then the sequence repeats...
    So the Sequence is $3,9,4,5,3,9,4,5...$

    \section{Chapter 6.1}
    \subsection{Exercise 27}
    A committee is formed consisting of one representative from each of the 50 states in the United States, where the representative from a state is either the governor or one of the two senators from that state. How many ways are there to form this committee?

    The number of different possible committees are $3^{50}=7,178979877e23$

    \subsection{Exercise 44}
    How many ways are there to seat four of a group of ten people around a circular table where two seatings are considered the same when everyone has the same immediate left and immediate right neighbor?

    The number of ways of seating a group of 4 people from the 10 people are
    \begin{equation}
      \begin{split}
        10*9*8*7 =5040
      \end{split}
    \end{equation}

    But our number should be less than this because we don't count all the possible arrangements. We here get 4 of the same seating arrangements that we have remove...
    The required number of ways are therefore
    \begin{equation}
      \begin{split}
        \frac{5040}{4}=1260
      \end{split}
    \end{equation}

    \section{Chapter 6.2}
    \subsection{Exercise 10}
    Let $(x_i,y_i), i = 1,2,3,4,5$,be a set of five distinct points with integer coordinates in the $xy$ plane. Show that the midpoint of the line joining at least one pair of these points has integer coordinates.

    The middle point of two co-ordinates $(x_i,y_i)$ and $(x_j,y_j)$ is $(\frac{x_i+x_j}{2},\frac{y_i+y_j}{2})$. For the co-ordinates mid-point to be an integer,  $(x_i,y_i)$ and $(x_j,y_j)$ needs to be even number and hence should have the same parity.

    Similarly, $y_i$ and $y_j$ have same parit. There are four possible cases of the parity of pair of integers: $(odd,odd),(odd,even),(even,odd),(even,even)$.

    By \textit{pigeon hole principle}, since there are four integer pairs, at least two of them needs to have same parity and hence will have mid-point having integer coordinates.

    \subsection{Exercise 16}
    How many numbers must be selected from the set ${1, 3, 5, 7, 9, 11, 13, 15}$ to guarantee that at least one pair of these numbers add up to 16?
    Best case only two: $1+15=16$ but worst case?

    % TODO: SHOW SOLUTION?
    To completely guarantee that at least one pair of the numbers add up to 16, we need to pick at least 5 numbers from the set.

    \subsection{Exercise 18}
    Suppose that there are nine students in a discrete mathematics class at a small college. We can list all possible (Male, Female)-combinations as shown in \fref{fig:groupings}.
   \begin{figure}[h]
    \label{fig:groupings}
    \begin{equation}
      \begin{split}
        (0,9),(1,8),(2,7),(3,6),(4,5),(5,4)(6,3),(7,2),(8,1),(9,0)
      \end{split}
    \end{equation}
    \centering
    In the order (Male, Female)
  \end{figure}

    \textbf{a) Show that the class must have at least five male students or at least five female students.}\\
    Considering only two kind of genders, Male and Female. If there are 3 male, then the remaning 6 students must be female and vise versa.

    \textbf{b) Show that the class must have at least three male students or at least seven female students.}\\
    Same as for the exercise above. If 2 male, then the remaning 7 must be female. If more than 2 male, then the statement is still valid.

    \section{Chapter 6.3}
    A coin is flipped 10 times where each flip comes up either heads or tails. How many possible outcomes...

    \subsection{Exercise 19a}
    ...are there in total?\\
    $2^{10}=1024$

    \subsection{Exercise 19b}
    ...contain exactly two heads?\\
    $C(10,2)=45$

    \subsection{Exercise 19c}
    ...contain at most three tails?
    \begin{equation}
      \begin{split}
        &C(10,1)+C(10,2)+C(10,3)\\
        &=1+10+45+120\\
        &=176
      \end{split}
    \end{equation}

\end{document}


\documentclass[12pt]{article}

\usepackage{amsthm}
\usepackage{graphicx}
\usepackage{float}

\begin{document}

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\theoremstyle{remark}
\newtheorem*{remark}{Remark}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Maximum Matching - greedy algorithm}

\begin{theorem}
    The maximum matching greedy algorithm for interval graphs is correct.
\end{theorem}

\begin{proof}
    Let $I:= \{i_1, ... i_n\}$; with the intervals sorted by their right-end point.
    The proof is done by induction on $|I|$.\\
    For $|I|=1$, the algorithm gives $M=\emptyset$; which is correct. \\

    \textit{Induction step}
    \vspace{\baselineskip}

    Suppose the algorithm gives the correct answer for $n-1$ intervals. Let $|I| = n$, and $M$ be the matching computed by the greedy algorithm for $I-i_n$. Let $U$ be the set of unmatched intervals. \\
    If $U$ is empty, when treating $I$ the algorithm will give the same answer: $M$.\\
    Let $U$ be non empty. Let $i_k \in U$. If $i_k \cap i_n \ne \emptyset$, the algorithm will return $M+(i_k, i_n)$.
    Indeed, for $i<k$:
    \begin{itemize}
        \item If $i_i$ could be matched with something in $I-i_n$, it would never choose to match it with $i_n$ as it is the worst choice.
        \item If $i_i$ could not be matched in $I-i_n$, then $i_i$ will be left unmatched; because if it was possible to match it with $i_n$; then $i_i$ and $i_k$ would have not be unmatched $I-i_n$.
    \end{itemize}

    \textit{Augmentating path}
    \vspace{\baselineskip}

    Suppose now that $U$ is non empty, and that for some $i_k \in U$ we can find a shortest augmentating path $P: (i_k, j_1, j_1', j_2, j_2', ... , j_m', i_n)$, namely:
    \begin{itemize}
        \item $i_k \cap j_1 \ne \emptyset$ and $j_m'\cap i_n \ne \emptyset$.
        \item $\forall x \in [1, m-1], j_x' \cap j_{x+1} \ne \emptyset$.
        \item $\forall x \in [1, m-1], (j_x$, $j_x') \in M$.

    \end{itemize}

    Until the end of the proof, we will use the notation $r_k, r_x, r_x'$ for the right-end points of respectively $i_k, j_x, j_x'$.
    \vspace{\baselineskip}

    \textit{Decreasing property}
    \vspace{\baselineskip}

    We shall now prove by a second induction that $\forall x \in [0, m-1], r_{x+1}' < r_x'$ and $r_{x+1} < r_x$. We only need the first property to conclude, but we need both to do the proof. \\
    First let's prove this for $j_1, j_1', j_2$ and $j_2'$:

    Observe that if both $j_1, j_1'$ overlap $i_k$, then: $r_1 < r_k$ and $r_1' < r_k$; otherwise, $i_k$ would have matched with either $j_1$ or $j_1'$. It is also possible that $r_1 > i_k$; and in that case $r_1' < r_k$ and $j_1' \cap i_k = \emptyset$. \\
    Suppose for a moment that $r_2 > r_1$:
    \begin{itemize}
        \item case 1: if $r_1' < r_1$, then $j_2 \cap i_k \ne  \emptyset$, so we can cut $j_1, j_1'$ from $P$.
        \item case 2: if $r_1 < r_1'$, then either $j_2$ overlap $i_k$ as in case 1; either $j_1$ and $i_k$ would be matched as in Fig 2.
        \item case 3: if $r_1 > r_k$, then $j_2 \cap i_k \ne  \emptyset$, so we can cut $j_1, j_1'$ from $P$.
    \end{itemize}

    \begin{figure}[H]
        \centering
        \includegraphics[scale= 0.2]{init_1.png}
        \caption{Skipping $j_1, j_1'$ would be shorter.}
        \includegraphics[scale = 0.2]{init_2.png}
        \caption{$j_1$ and $i_k$ would be matched. }
        \includegraphics[scale = 0.2]{init_3.png}
    \end{figure}

    Now we know that $r_2 < r_1$. Let's suppose for a moment that $r_2' > r_1'$. Notice that the smallest element among $\{r_1, r_2, r_1', r_2'\}$ is either $r_1'$ or $r_2$.
    In any case the algorithm would have match the intervals differently.

    \textit{Induction step}
    \vspace{\baselineskip}

    Suppose $x > 1$, $r_x < r_{x+1}$ and $r_x' < r_{x+1}'$. \\
    We will investigate four different cases:
    \begin{itemize}
        \item 1.1 : $r_{x-1} > r_{x-1}'$ and $r_x > r_x'$
        \item 1.2 : $r_{x-1} > r_{x-1}'$ and $r_x'>r_x$
        \item 2.1 : $r_{x-1}' > r_{x-1}$ and $r_x > r_x'$
        \item 2.2 : $r_{x-1}' > r_{x-1}$ and $r_x' > r_x$
    \end{itemize}

    \textit{Case 1.1}
    One of two things are possible:
    \begin{figure}[H]
        \centering
        \includegraphics[scale= 0.2]{induction_case_1.1_0.png}
        \caption{Subcase 1.1.1: $r_x < r_{x-1}'$}
        \includegraphics[scale=0.2]{induction_case_1.1_1.png}
        \caption{Subcase 1.1.2: $r_x > r_{x-1}'$}
    \end{figure}

    For Subcase 1.1.1, as $r_x < r_{x-1}'$, it means that $r_x \in j_{x-1}'$.
    Consequently, if $r_{x+1} > r_x'$, we must have $r_x \in j_{x+1}$. In other words, $j_{x-1}'$ and $j_{x+1}$ share a common point in $r_x$, so we could remove $j_x, j_x'$ from $P$, so we must have $r_{x+1} < r_x$. \\
    Suppose now that $r_{x+1}' > r_x'$: observe that as $r_{x+1} < r_x$, the algorithm has an incentive to match $j_x'$ with $j_{x+1}$; and the only reason they are not matched is because $j_{x+1}$ and $j_{x+1}'$ were matched first in the execution, ie $r_{x+1} < r_{x+1}'$.

    \vspace{\baselineskip}
    For subcase 1.1.2: suppose for a moment that $r_{x+1} > r_{x}$. The exact same argument with $r_{x-1}'$ in place of $r_x$ as in 1.1.1 still holds here, so $r_{x+1} < r_x$.\\
    Suppose now that $r_{x+1}' > r_x'$. Then the algorithm should have matched $j_{x+1}$ and $j_x'$. Thus $r_{x+1}' < r_x'$ for case 1.1.\\

    \textit{Case 1.2}
    \vspace{\baselineskip}

    \begin{figure}[H]
        \centering
        \includegraphics[scale= 0.3]{induction_case_1.2.png}
        \caption{$r_{x-1} > r_{x-1}'$ and $r_x'>r_x$}
    \end{figure}

    Let $l_{x+1}$ be the left end-point of $j_{x+1}$. Suppose $r_{x+1} > r_x$. If $l_{x+1} < r_x$, then $r_x \in j_{x+1}$. As $r_x \in j_{x-1}'$, it means we could remove $j_x, j_x'$ from $P$. \\
    If $l_{x+1} > r_x$, notice we should at least have $l_{x+1} < r_x'$. As $r_x' < r_{x-1}'$, we have $r_x' \in j_{x-1}'$. So $l_{x+1} \in [r_x, r_x'] \subset j_{x-1}'$; meaning we could remove $j_x, j_x'$ from $P$. Thus, $r_{x+1} < r_x$. \\
    As $r_{x+1}'<r_x'$, the algorithm has an incentive to match $j_x'$ with $j_{x+1}$. As for the 1.1.x cases, the only reason for it to not happen is that $j_{x+1}$ is already matched with $j_{x+1}'$, because $r_{x+1}' < r_x'$.

    \vspace{\baselineskip}

    \textit{Case 2.1}
    \begin{figure}[H]
        \centering
        \includegraphics[scale= 0.3]{induction_case_2.1.png}
        \caption{$r_{x-1}' > r_{x-1}$ and $r_x > r_x'$}
    \end{figure}

    If $r_{x+1} > r_x$, then we can skip $j_x, j_x'$ as in case 1.1. So we must have $r_{x+1}<r_x$.  \\
    If $r_{x+1}' > r_x'$, then $j_{x+1}$ would be matched with $j_x'$ as it ends before $j_{x+1}$. Thus $r_{x+1}' < r_x'$. \\

    \textit{Case 2.2}
    \begin{figure}[H]
        \centering
        \includegraphics[scale= 0.3]{induction_case_2.2.png}
        \caption{$r_{x-1}' > r_{x-1}$ and $r_x' > r_x$}
    \end{figure}

    Suppose $r_{x+1} > r_x$. Using similar arguments as in case 1.2, we should conclude that we can remove $j_x, j_x'$ from $P$. So $r_{x+1} < r_x$. \\
    Suppose now $r_{x+1}' > r_x'$. Then $j_{x+1}$ would be matched with $j_x'$ (or $j_x$ if these overlaps).\\

    \textit{Conclusion} \\
    We have showned by induction that $\forall x>1, r_{x+1}' < r_x'$. This is a contradiction with the properties of $P$: there is no way $j_m'$ could possibly overlap $i_n$. \\
    There is no augmentating path; thus, the greedy algorithm is optimal. \\

\end{proof}

\end{document}


% Copyright 2017 Markus J. Pflaum, licensed under CC BY-NC-ND 4.0
% main author:
%   Markus J. Pflaum
%
\section{Seminorms and gauge functionals}
\label{sec:seminorms-gauge-functionals}
%
%
\para Throughout the rest of this chapter the symbol $\fldK$ will always
stand for the field of real numbers $\R$, the field of complex numbers $\C$ or the
division algebra of quaternions $\H$. We assume these division algebras to be equipped
with their standard absolute values $|\cdot |$.
Moreover, vector spaces are assumed to be defined over the ground field $\fldK$
unless mentioned differently and are always assumed to be left vector spaces.

\subsec{Seminorms and induced vector space topologies}
\begin{definition}
  By a \emph{seminorm} on a vector space $\vectorspE$ one
  understands a map $p:\vectorspE \to \R$ with the following properties:
  \begin{axiomlist}[N]
  \setcounter{enumi}{-1}
  \item\label{axiom:positive}
    The  map $p$ is \emph{positive} that is $p(v)\geq 0$ for all $v\in \vectorspE$.
  \item
  \label{axiom:norm-absolute-homogeneity}
     The map $p$ is \emph{absolutely homogeneous} that means
     \[
       p(r v) = |r| \, p(v)  \quad \text{for all $v \in \vectorspE$ and $r\in \fldK$}.
     \]
  \item
  \label{axiom:norm-subadditivity}
     The map $p$ is \emph{subadditive} or in other words satisfies the \emph{triangle inequality}
     \[
       p(v+w) \leq p(v) + p(w) \quad \text{for all $v,w \in \vectorspE$}.
     \]
  \end{axiomlist}
  A seminorm is called a \emph{norm} if in addition the following axiom is satisfied:
  \begin{axiomlist}[N]
  \setcounter{enumi}{2}
  \item \label{axiom:norm-nondegeneracy}
     For all $v\in \vectorspE$ the relation $p(v) = 0$ holds true if and only if $v = 0$.
  \end{axiomlist}
  A vector space $\vectorspE$ equipped with a norm
  $\| \cdot \| : \vectorspE \to \R_{\geq 0}$ is called a \emph{normed vector space}.
  % If $\vectorspE$ is equipped with a seminorm $p: \vectorspE \to \R_{\geq 0}$ we sometimes say that
  % $\vectorspE$ is a \emph{seminormed vector space}.
\end{definition}

\para
 Let us introduce some useful further properties a map $p:\vectorspE \to \R$
 can have.  One calls such a map $p$
 \begin{axiomlist}[]
 \item\label{ite:definition-positive-homogeneity}
   \emph{positively homogeneous} if $p (tv) = t\, p(v)$ for all $t\in \R_{> 0}$ and all $v\in \tvsE$,
 \item\label{ite:definition-sublinearity}
   \emph{sublinear} if $p (tv + sw) \leq  t\, p(v) + s\, p(w)$ for all $t,s\in \R_{\geq 0}$
   and all $v,w\in \tvsE$, and
 \item\label{ite:definition-convexity}
   \emph{convex} if $p (tv + sw) \leq  t\, p(v) + s\, p(w)$ for all $t,s\in \R_{\geq 0}$
   with $t+s=1$ and all $v,w\in \tvsE$.
 \end{axiomlist}

\begin{lemma}\label{thm:equivalent-characterizations-sublinearity}
   For a real-valued map $p:\vectorspE \to \R$ on a vector space $\vectorspE$ the following are equivalent:
   \begin{romanlist}
   \item\label{ite:sublinearity}
     $p$ is sublinear.
   \item\label{ite:positive-homogeneity-convexity}
     $p$ is positively homogeneous and convex.
   \item\label{ite:positive-homogeneity-subadditivity}
     $p$ is positively homogeneous and subadditive.
   \end{romanlist}
 \end{lemma}
 \begin{proof}
   Let $p$ be sublinear. Then $p$ is subadditive by definition. Subadditivity implies
   $p(0) \leq p(0) + p(0)$, hence $p(0)\geq 0$.
   By sublinearity
   \[ p(0) = p(0\cdot 0 + 0\cdot 0) \leq 0 \cdot p(0) +  0\cdot p(0) =  0 \ , \]
   so $p(0)=0$.  We show that $p$ is positively homogeneous.
   Applying sublinearity again one checks for $v\in \vectorspE$ and $t\geq 0$ that
   \[ p(tv)=p(tv+ 0\cdot 0) \leq t p(v) + 0 \cdot p(0)= tp(v) \ , \]
   so $p$ is positively homogeneous and the implication
   \ref{ite:sublinearity} $\implies$ \ref{ite:positive-homogeneity-subadditivity} follows.
 If $p$ is positively homogeneous and subadditive, then for $v,w \in \vectorspE$ and
 $t,s> 0$ with $t +s =1$
 \[
     p(tv +sw) \leq p(tv) + p(sw) \leq t p(v) + s p(w),
 \]
 so $p$ is convex. This gives the implication
 \ref{ite:positive-homogeneity-subadditivity} $\implies$ \ref{ite:positive-homogeneity-convexity}.
 If $p$ is positively homogeneous and convex, then one computes for $v,w \in \vectorspE$ and $t,s\geq 0$
 with $t+s >0$
 \[
   p( tv +sw )  = % (t+s)\, p \left( \frac{1}{t+s} (tv+sw) \right) =
   (t+s)\, p \left( \frac{t}{t+s} v+ \frac{s}{t+s} w \right) \leq
   (t+s) \left( \frac{t}{t+s} p(v) +  \frac{s}{t+s} p(w) \right) = t p(v)+ sp(w) \ .
 \]
 Since $p(0)=\lim\limits_{t\searrow 0} p(t0) = \lim\limits_{t\searrow 0} t\, p(0) = 0$ by positive homogeneity,
 $p$ then has to be sublinear and one obtains the implication
 \ref{ite:positive-homogeneity-convexity}  $\implies$ \ref{ite:sublinearity}.
\end{proof}

 \begin{lemma}\label{thm:properties-subadditive-convex-real-valued-maps-vector-space}
   Let $p:\vectorspE \to \R$ be a real-valued map defined on a vector space
   $\vectorspE$ over $\fldK$.
   \begin{romanlist}
   \item\label{ite:implications-positive-homogeneity}
     If $p:\vectorspE \to \R$ is   positively homogeneous, then $p(0)=0$.
   \item\label{ite:implications-subadditivity}
     If $p:\vectorspE \to \R$ is subadditive, then $p(0)\geq 0$ and for all $v,w\in \vectorspE$
     \[
       |p(v)-p(w)| \leq \max \{ p(v-w),p(w-v) \} \ .
     \]
   \item\label{ite:implications-convexity-real-valued-map}
     If $p:\vectorspE \to \R$ is convex, then the sets
     $\ball_{p,\varepsilon} := \{v\in \vectorspE\mid p(v) < \varepsilon\} $ and
     $\closedball_{p,\varepsilon} := \{v\in \vectorspE\mid p(v) \leq \varepsilon\}$ are convex for all $\varepsilon >0$.
   \item\label{ite:implications-sublinearity-real-valued-map}
     If $p$ is sublinear, then $\ball_{p,\varepsilon}$ and
     $\closedball_{p,\varepsilon}$ are convex and
     absorbent for all $\varepsilon >0$.
   \end{romanlist}
 \end{lemma}
 \begin{proof}
   \begin{adromanlist}
   \item
     As already observed, $p(0)=\lim\limits_{t\searrow 0} p(t0) = \lim\limits_{t\searrow 0} t\, p(0) = 0$.
   \item
    Note that by subadditivity
     \[
       p(0)\leq p(0) + p(0), \quad  p(v) -p(w) \leq p(v-w), \quad \text{and} \quad
       p(w) -p(v) \leq p(w-v) \ .
     \]
     This entails \ref{ite:implications-subadditivity}.
   \item
     Let $v,w\in \{v\in \vectorspE\mid p(v) < \varepsilon\}$ and $0\leq t \leq 1$. Then, by convexity of $p$,
     \[
         p\big(tv + (1-t)w\big) \leq t p(v) + (1-t) p(w)< t\varepsilon + (1-t) \varepsilon = \varepsilon \ .
     \]
     Hence $tv + (1-t)w\in  \{v\in \vectorspE\mid p(v) < \varepsilon\} $.
     The proof for $\{v\in \vectorspE\mid p(v) \leq \varepsilon\}$ is analogous.
   \item
     Convexity of the sets $\ball_{p,\varepsilon}$ and  $\closedball_{p,\varepsilon}$ holds by
     \ref{ite:implications-convexity-real-valued-map}.
     Moreover, $\ball_{p,\varepsilon} \subset \closedball_{p,\varepsilon}$ by definition.
     Hence it suffices  by \Cref{thm:real-absorbance-implies-arbsorbance-finite-dimensional-divison-algebra}
     to show that $\ball_{p,\varepsilon} $ is absorbent in the realification $\vectorspE^\R$.
     Since $p$ is positively homogenous by \Cref{thm:equivalent-characterizations-sublinearity} and
     $0\leq p(v) + p(-v)$ for all $v\in \vectorspE$, one concludes that for all $t\in\R$ and $v\in \vectorspE$
     \[
        |p(tv) | \leq |t|\max\{ p(v),p(-v) \} \ .
     \]
     Hence $tv\in \ball_{p,\varepsilon}$ if $0 < t < \frac{\varepsilon}{\max\{ p(v),p(-v) \} +1}$,
     and $\ball_{p,\varepsilon}$ is absorbent in $\vectorspE^\R$.
   \end{adromanlist}
 \end{proof}

\begin{definition}
  If $p :\vectorspE \to \R$ is a seminorm on a vector space $\vectorspE$, we denote for every $v \in \vectorspE$
  and $\varepsilon >0$  by $\ball_{p,\varepsilon} (v)$   the  (\emph{open}) $\varepsilon$-\emph{ball associated with} $p$ and
  \emph{with center} $v$  that is the set
  \[
    \ball_{p,\varepsilon} (v) = \big\{ w \in \vectorspE \bigmid p(w-v) < \varepsilon \big\} \ .
  \]
  The  \emph{closed} $\varepsilon$-\emph{ball associated with} $p$ and \emph{with center} $v$
  is defined as
  \[
    \closedball_{p,\varepsilon} (v) = \big\{ w \in \vectorspE \bigmid p(w-v) \leq \varepsilon \big\} \ .
  \]
  The positive number $\varepsilon$ is called the \emph{radius} of the ball.
  In case the center of the ball is the origin, we often write  $\ball_{p,\varepsilon}$ and $\closedball_{p,\varepsilon}$
  for $\ball_{p,\varepsilon} (0)$ and $\closedball_{p,\varepsilon} (0)$, respectively.
  If in addition the radius equals $1$, then we usually write only
  $\ball_{p}$ and $\closedball_{p}$ and call these sets the \emph{open} respectively the
  \emph{closed unit ball}. More generally, for the particular radius $1$ we denote  the corresponding
  balls by $\ball_p (v)$ and  $\closedball_p (v)$ and call them the \emph{open} respectively
  \emph{closed}  \emph{unit balls with center} $v$.
  When by the context it is clear which seminorm $p$ a ball is associated with
  we often do not mention $p$ explicitely. This is in particular the case
  when the underlying vector space is a normed vector space.

  If $P$ is a finite set or a finite family of seminorms on $\vectorspE$ we define the
  \emph{open} and \emph{closed} $\varepsilon$-\emph{multiballs with center} $v$ by
  \[
    \ball_{P,\varepsilon} (v) =
    \big\{ w \in \vectorspE\bigmid p(w-v) < \varepsilon \text{ for all } p\in P\big\}
  \]
  and
   \[
    \closedball_{P,\varepsilon} (v) = \big\{ w \in \vectorspE \bigmid p(w-v) \leq \varepsilon
    \text{ for all } p\in P \big\} \ ,
  \]
  respectively. As before, we abbreviate
  $\ball_{P,\varepsilon} = \ball_{P,\varepsilon} (0)$ and
  $\closedball_{P,\varepsilon} = \closedball_{P,\varepsilon} (0)$.
\end{definition}

\begin{remark}
  For convenience, we will also use the symbols $\ball_{p,\varepsilon}$ and $\closedball_{p,\varepsilon}$
  to denote the sets $\big\{ v \in \vectorspE \bigmid p(v) < \varepsilon \big\}$
  and $\big\{ v \in \vectorspE \bigmid p(v) \leq \varepsilon \big\}$, respectively, when $p:\vectorspE \to \R$
  is just a real-valued convex map on the vector space $\vectorspE$. Note that for such a $p$ the
  set  $\big\{ v \in \vectorspE \bigmid p(v) < 0 \big\}$ might be non-empty. But as we have shown in
  \Cref{thm:properties-subadditive-convex-real-valued-maps-vector-space}
  the sets $\ball_{p,\varepsilon}$ and $\closedball_{p,\varepsilon} $ associated to a convex $p$ share with the
  the balls associated to a seminorm several nice properties like convexity.
\end{remark}

\begin{proposition}\label{thm:convexity-balls-seminorms}
  Let $\vectorspE$ be a $\fldK$-vector space, and $P$ a finite set of seminorms on $\vectorspE$.
  Then, for every $\varepsilon >0$ and $v\in \vectorspE$, the $\varepsilon$-multiballs
  $\ball_{P,\varepsilon} (v)$ and  $\closedball_{P,\varepsilon} (v)$ are convex.
  The $\varepsilon$-multiballs
  $\ball_{P,\varepsilon}$ and  $\closedball_{P,\varepsilon}$ centered at the origin are absolutely convex and absorbent.
\end{proposition}
\begin{proof}
  Axiom \ref{axiom:norm-absolute-homogeneity} immediately entails that $\ball_{P,\varepsilon}$ and
  $\closedball_{P,\varepsilon}$ are circled. Axiom \ref{axiom:norm-subadditivity}
  together with  \ref{axiom:norm-absolute-homogeneity} entails that the sets
  $\ball_{P,\varepsilon} (v)$ and $\closedball_{P,\varepsilon} (v)$
  are convex. Namely, if $w_1, w_2 \in \ball_{P,\varepsilon} (v)$
  and $t \in [0,1]$, then one has  for all seminorms $p\in P$
  \[
   p\left( tw_1 + (1-t) w_2 - v \right) \leq
   t \,  p \left( w_1  - v \right) + (1-t) \, p \left(  w_2 - v \right) <
   t \, \varepsilon + (1-t) \, \varepsilon = \varepsilon
  \]
  and likewise $p\left( tw_1 + (1-t) w_2 - v \right) \leq \varepsilon$ for all
  $w_1, w_2 \in \closedball_{P,\varepsilon}(v)$ and  $p\in P$.

  Now let $v\in \vectorspE$ and $\varepsilon >0$ be given.
  Put $t_p = \frac{p(v)+1}{\varepsilon}$ for every $p\in P$ and $t_0 = \max \{t_p\mid p\in P\}$.
  Then one has for all $t\in \fldK$ with $|t|\geq t_0$ and for all $p\in P$
  \[
    p\left( \frac{1}{t} v \right) \leq \frac{\varepsilon}{p(v)+1} \, p(v) < \varepsilon \ ,
  \]
  hence $v \in t \, \ball_{P,\varepsilon}$. So $\ball_{P,\varepsilon}$ is absorbing.
  Since $\closedball_{P,\varepsilon}$  contains the absorbing set $\ball_{P,\varepsilon}$,
  it is absorbing as well.
\end{proof}

\begin{propanddef}
Assume to be given a set $Q$ of seminorms on a vector space $\tvsE$.
Let $\powersetfin{Q}$ be the collection of all finite subsets of $Q$.
A base of a topology on $\tvsE$ then is given by
\[
  \base = \big\{  \ball_{P,\varepsilon} (v)  \bigmid
  P  \in \powersetfin{Q} , \: v \in \tvsE , \:  \varepsilon >0 \big\} \ .
\]
The topology $\topology$ generated by $\base$  is called the
topology \emph{generated}, \emph{induced} or \emph{defined} by $Q$. Moreover,
$\topology$ is a locally convex vector space topology on $\tvsE$.
It coincides with the coarsest translation invariant topology on
$\tvsE$ such that each seminorm in $Q$ is continuous.
\end{propanddef}

\begin{proof}
  Consider the set $\base_0$ of all multiballs $\ball_{P,\varepsilon}$ with
  $P  \in \powersetfin{Q}$ and $\varepsilon >0 $ centered at the origin.
  Clearly, $\base_0$ is a filter base since for $P_1,P_2  \in \powersetfin{Q}$ and
  $\varepsilon_1,\varepsilon_2 >0 $ the multiball
  $\ball_{P_1\cup P_2, \min\{ \varepsilon_1,\varepsilon_2\}}$ is contained in
  $\ball_{P_1,\varepsilon_1}\cap \ball_{P_2,\varepsilon_2}$. Moreover it consists of
  absolutely convex and absorbing sets by \Cref{thm:convexity-balls-seminorms}.

  By a similar argument one shows that $\base$ is  base of a topology. Let
  $\ball_{P_1,\varepsilon_1} (v_1), \ball_{P_2,\varepsilon_2} (v_2) \in \base$ and
  $v \in \ball_{P_1,\varepsilon_1}(v_1)\cap \ball_{P_2,\varepsilon_2}(v_2)$.
  Let $\varepsilon$ be the minium of the numbers $\varepsilon_1 - p_1 (v-v_1)$
  and  $\varepsilon_2 - p_2 (v-v_2)$, where $p_1$ runs through the elements of $P_1$ and
  $p_2$ through the ones of $P_2$. Then $\varepsilon >0$ and
  $\ball_{P_1\cup P_2, \varepsilon}(v)
  \subset\ball_{P_1,\varepsilon_1}(v_1)\cap\ball_{P_2,\varepsilon_2}(v_2)$, and $\base$
  is a base for a topology  $\topology$ indeed. By construction,  $\base_0$ then
  is a base for the filter of zero neighborhoods and each element of  $\base_0$ is open in
  $\topology$. Moreover, each closed multiball
  $\closedball_{P,\varepsilon} (v)$ is closed in $\topology$ since the complement
  $\vectorspE\setminus \closedball_{P,\varepsilon} (v)$  contains with $w$ also
  the open multiball $\ball_{P,\delta} (w)$, where $\delta = \min \{ p(v-w) - \varepsilon| p\in P\}$.

  We now prove continuity of addition with respect to  $\topology$. Let $v_1,v_2\in \tvsE$,
  $P \in \powersetfin{Q}$, and $\varepsilon >0$.
  Since the triangle inequality holds for every seminorm in $F$, one has
  \[
    \ball_{P,\frac\varepsilon 2} (v_1) +   \ball_{P,\frac\varepsilon 2} (v_2)
    \subset \ball_{P,\varepsilon}(v_1+v_2) \ ,
  \]
  which entails continuity of addition at each $(v_1,v_2) \in \tvsE \times \tvsE$.
  Next consider multiplication by scalars and let $\lambda \in \fldK$ and $v\in \tvsE$.
  Again let $P= \{ p_1,\ldots, p_n\} \in \powersetfin{Q}$ and $\varepsilon >0$.
  Let $C_1 = \sup \{ p_j (v) \mid 1\leq j \leq n\} + 1$, $C_2 = |\lambda| + 1$
  and put $\delta_1 = \min \{1,\frac{\varepsilon}{2 \, C_1}  \}$
  and $\delta_2 = \frac{\varepsilon}{2 \, C_2}$.
  Then one obtains by absolute homogeneity and subadditivity of each seminorm
  \[
    p_j ( \mu w -\lambda v) \leq | \mu | \, p_j( w-v) + | \mu -\lambda| \, p_j(v) \quad \text{for all }
    \mu \in \fldK  \text{ and } w \in \tvsE,
  \]
  hence
  \[
    \ball_{\delta_1} (\lambda ) \cdot \ball_{P,\delta_2} (v) \subset
    \ball_{P,\varepsilon} (\lambda \cdot v ) \ ,
  \]
  where $\ball_{\delta_1} (\lambda) = \{ \mu \in \fldK \mid |\mu -\lambda| < \delta_1\} $.
  This shows continuity of scalar multiplication at each $(\lambda,v) \in \fldK \times \tvsE$,
  and $\topology$ is a vector space topology.

  Since each of the base elements $\ball_{P,\varepsilon}\in \base_0$ is convex,
  Axiom \hyperref[axiom:tvs-local-convexity]{LCVS} holds true as well
  and the topology $\topology$ is locally convex.

  Every seminorm $p\in Q$ is continuous with respect to the topology $\topology$ since for all
  $a<b$ the preimage $p^{-1} (\openint{a,b}) = \ball_{p,b} \setminus \closedball_{p,a}$ is open in
  $\topology$. Now let   $\topology'$ be a translation invariant topology on $\vectorspE$
  for which every seminorm $p\in Q$ is continuous. In that topology $\base_0$ is a set of
  zero neighborhoods. As shown before, every element $B\in \base_0$ is absolutely
  convex, absorbing and satisfies $\frac 12 B + \frac 12 B \subset B$. Hence
  by \Cref{thm:filter-base-absorbing-absolutely-convex-sets-generating-locally-convex-topology}
  the topology $\topology'$ is finer than the locally convex topology generated by  $\base_0$.
  But the latter topology coincides with $\topology$ by construction.
  This shows  the last part of the claim and the proof is finished.
\end{proof}

\subsec{Gauge functionals and induced seminorms}
\para
As we have seen, any vector space with a topology defined by a family of seminorms
on it is a locally convex topological vector space. The converse also holds true. The fundamental notion
needed for the proof of this is the following.

\begin{definition}
  Let  $\vectorspE$  be a vector space and $A \subset \vectorspE$ absorbent. Then the map
  \[
     p_A : \vectorspE \to \R_{\geq 0} , \: v \mapsto p_A (v)= \inf \big\{ t \in \R_{> 0} \bigmid v \in t A \big\}
  \]
  is called the \emph{gauge functional}, the \emph{Minkowski functional} or the \emph{Minkowski gauge}
  of $A$.
\end{definition}

\begin{remark}
  By definition of an absorbent set, $\big\{ t \in \R_{> 0} \bigmid v \in t A \big\}$ is non-empty
  whenever $A\subset \vectorspE$ is absorbent. Hence $p_A$ is well-defined for such $A$.
\end{remark}

\begin{proposition}\label{thm:properties-gauge-functional-absorbent-set}
  The Minkowski gauge $p_A :\vectorspE \to \R_{\geq 0}$ of an absorbent subset $A$
  of a vector space $\vectorspE$  has the following properties.
  \begin{romanlist}
  \item\label{ite:gauge-functional-positively-homogenous}
    The gauge functional is positively homogeneous that is
    $p_A (t v) = t\,  p_A(v)$ for all $t\in \R_{> 0}$ and all $v\in \vectorspE$.
  \item\label{ite:gauge-functional-subadditivity}
     If $A$ is convex, then $p_A$ is subadditive and
  \[
    \ball_p (v)= \bigcup_{0< t < 1} tA \subset A \subset
    \bigcap_{1< t} tA = \closedball_p (v)  \ .
  \]
\item\label{ite:gauge-functional-seminorm}
  If $A$ is absolutely convex, then $p_A$ is a seminorm on $\vectorspE$.
  \end{romanlist}
\end{proposition}

\begin{proof}
 If $t>0$, then $tv \in sA$ for some $s>0$ if and only if $v \in \frac st A$.
 Hence $\big\{ s \in \R_{> 0} \bigmid t v \in s A \big\}$ and $t \big\{ s \in \R_{> 0} \bigmid v \in s A \big\}$
 coincide for all  $t> 0$, so \ref{ite:gauge-functional-positively-homogenous} follows.

 Assume that $A$ is convex. Let $v,w\in \vectorspE$ and $\varepsilon >0$. Then there exist
 $t>p_A(v)$ and $s > p_A(w)$ such that
 $v \in tA$, $w\in sA$, $t < p_A(v) +\frac \varepsilon 2$ and
 $s < p_A(w) +\frac \varepsilon 2$. By convexity of $A$ and
 \Cref{thm:weighted-sum-convex-absolutely-convex-sets},
 $v+w \in tA+ sA = (t+s)A$.
 Hence $p_A(v+w) \leq (t+s) < p_A(v) + p_A (w) + \varepsilon$.
 Since $\varepsilon >0$ was arbitrary, $p_A(v+w) \leq  p_A(v) + p_A (w)$ and $p_A$ is subadditive.
 If $v \in tA$ for some $t$ with $0 < t<1$, then $p_A(v) \leq t < 1$ by definition.
 Conversely, if $p_A(v) < 1$, then there exists a $t>0$ such that $t<1$ and $v \in tA$.
 Hence the equality $\ball_p (v) = \bigcup_{0< t < 1} tA$ follows.
 Since $A$ is absorbing, $0$ is an element of $A$. By convexity of $A$
 one therefore concludes $tA = (1-t) \{ 0 \}  + tA \subset A$ whenever $0< t < 1$.
 For $t>1$ this shows $\frac 1t A \subset A$, hence $A \subset tA$.
 So the relation $\bigcup_{0< t < 1} tA \subset A \subset \bigcap_{1< t} tA$ is proved.
 Now assume that $v \in tA$ for all $t>1$. Then $p_A(v)\leq 1$ by definition.
 If conversely $p_A(v)\leq 1$, then there exists for each $\varepsilon >0$ an $s\geq 0$
 such that $p_A(v) \leq s$, $v \in sA$ and $s < 1 +\varepsilon$.
 Hence, for $t\geq   1 +\varepsilon$ by \Cref{thm:weighted-sum-convex-absolutely-convex-sets}
 and $0\in A$,
 \[
    v \in sA = s A + (t-s) \{0\} \subset s A + (t-s) A = tA \ .
 \]
 Since $\varepsilon >0$ was arbitrary, $v \in tA$ for all $t>1$ follows. So one obtains
 the equality $\bigcap_{1< t} tA =  \closedball_p (v)$,
 and \ref{ite:gauge-functional-subadditivity} is proved.

 To verify \ref{ite:gauge-functional-seminorm} recall that $A$ is circled whenever
 $A$ is absolutely convex. This entails for $r\in \fldK$, $v\in \vectorspE$
 and absolutely convex $A$
 \[
   p_A (rv)= \inf \big\{ t \in \R_{> 0} \bigmid r v \in tA \big\} =
   \inf \big\{ t \in \R_{> 0} \bigmid |r|v \in t A \big\} = p_A (|r|v) = |r| p_A(v) \ ,
 \]
 where for the last equality we have used \ref{ite:gauge-functional-positively-homogenous}.
\end{proof}

\begin{lemma}
  Let  $A$ and $B$ be absorbent subsets of a vector space $\vectorspE$. Then
  the following holds true.
  \begin{romanlist}
  \item
    $p_{tA}(v) =  p_A(t^{-1} v)$ for all $t\in \fldK^\times$ and $v\in \vectorspE$.
  \item\label{ite:comparison-gauge-functionals-subsets}
    If $B\subset A$, then $p_A \leq p_B$.
  \item
    If $A$ is convex, then $v \in tA$ for all $v\in \vectorspE$ and $t > p_A(v)$.
  \item
    If $A$ and $B$ are convex, then the intersection $A\cap B$ is absorbent and convex and
    $p_{A\cap B} = \sup \{ p_A,p_B\}$, where $\sup \{ p_A,p_B\} (v) = \sup \{ p_A (v) ,p_B (v) \}$
    for all  $v\in \vectorspE$.
  \end{romanlist}
\end{lemma}

\begin{proof}
  \begin{adromanlist}
  \item
    If $t\in \fldK$ is invertible, then $v \in tA$ if and only if $t^{-1}v \in A$.
  \item
    Let $v\in \vectorspE$ and $\varepsilon >0$. Then there exists $t$ with
    $p_B(v) \leq t < p_B(v) + \varepsilon$ such that $v \in tB$. By $B\subset A$ this implies
    $v \in tA$, hence $p_A(v) \leq t < p_B(v) + \varepsilon$. Since $\varepsilon >0$
    was arbitrary, the estimate $p_A \leq p_B$ follows.
  \item
    By definition of the Minkowski gauge there exists $s \in \R$ such that $p_A(v)< s <t$ and  $v \in sA$.
    By convexity of $A$ one concludes $\frac st v = \frac st v + \left( 1 - \frac st \right) \cdot 0 \in sA$, hence
    $v \in tA$.
  \item
    The intersection of convex sets is convex, so $A\cap B$ is convex.
    Let $v\in \vectorspE$ and choose $r_A\geq 0$ and $r_B\geq 0$ such that
    $v \in tA$ for all $t\geq r_A$ and $v \in sB $ for all $s\geq r_B$.
    Then $v \in (tA)\cap (tB) = t(A\cap B) $ for all $t\geq \max\{r_A,r_B\}$,
    so $A\cap B$ is absorbent.
    One has $p_{A\cap B} \geq  \sup \{ p_A,p_B\}$ by \ref{ite:comparison-gauge-functionals-subsets}.
    To show the converse inequality assume that  $v \in \vectorspE$ and $t > \sup \{ p_A (v) ,p_B (v)\}$.
    Then $v \in tA \cap tB = t(A\cap B)$, which implies $p_{A\cap B} (v)\leq t$. Hence
    $p_{A\cap B} (v) \leq  \sup \{ p_A (v) ,p_B (v)\}$ since $t > \sup \{ p_A (v) ,p_B (v)\}$
    was arbitrary.
  \end{adromanlist}
\end{proof}

\begin{lemma}
  Let $p:\vectorspE \to\R$ be a sublinear map on a vector space $\vectorspE$ and
  $A\subset \vectorspE$ convex. If
  \[ \ball_p  \subset A \subset \closedball_p \ ,\]
  then the gauge functional $p_A$ coincides with $\sup \{p,0\} $. If $p$ is even
  a seminorm, then $p=p_A$.
\end{lemma}

\begin{proof}
  Let $p:\vectorspE\to\R$ be sublinear. Observe that then  $\ball_p$ is absorbent by
  \Cref{thm:properties-subadditive-convex-real-valued-maps-vector-space}
  \ref{ite:implications-sublinearity-real-valued-map}.
  Hence $A$ must also be absorbent by assumption, so the associated Minkowski gauge $p_A$ is
  positively homogeneous by
  \Cref{thm:properties-gauge-functional-absorbent-set}
  \ref{ite:gauge-functional-positively-homogenous}.

  Assume now that there exists $v\in\vectorspE$ such that $\max\{p(v),0 \} < p_A(v)$. By positive
  homogeneity of $p$ and $p_A$
  one can achive by possibly multiplying $v$ by a positive real number that $\max\{p(v),0 \} < 1 < p_A(x)$.  The first
  inequality entails $v \in  \ball_p$, the second $v\notin \closedball_p $ which is a contradiction.
  Next assume that there exists $v\in\vectorspE$ with $p_A(v) < \max\{p(v),0 \}$. As before one can then achieve that
  $ p_A(v) < 1 < \max\{p(v),0 \}$ for some $v\in\vectorspE$. By the first inequality one concludes $v \in  A$,
  by the second $v \notin A$. This is a contradiction. So the equality $\max\{p(v),0 \} = p_A(v)$ holds for all $v\in \vectorspE$.

  In case $p$ is a seminorm, then $p(v) \geq 0$ for all $v\in \vectorspE$ and the second claim follows by the first.
\end{proof}

\begin{proposition}
  Let $\tvsE$ be a topological vector space, and $p:\tvsE \to \R$ be sublinear. Then the following are equivalent.
  \begin{romanlist}
  \item\label{ite:continuity-sublinear-map-origin}
    The map $p$ is continuous in the origin.
  \item\label{ite:uniform-continuity-sublinear-map}
    The map $p$ is uniformly continuous.
  \item\label{ite:continuity-sublinear-map}
    The map $p$ is continuous.
  \item\label{ite:unit-ball-sublinear-map-zero-neighborhood}
    The unit ball $\ball_p $ is a zero neighborhood.
  \end{romanlist}
\end{proposition}

\begin{proof}
  Let us first show \ref{ite:continuity-sublinear-map-origin}  $\implies$ \ref{ite:uniform-continuity-sublinear-map}.
  To this end fix $\varepsilon >0$. By assumption there exists a zero neighborhood $V\subset \tvsE$ such that
  $|p(v)| < \varepsilon$ for all $v\in V$. By possibly passing to $V \cap (-V)$ one can assume that
  $V$ is symmetric.
  \Cref{thm:properties-subadditive-convex-real-valued-maps-vector-space} \ref{ite:implications-subadditivity}
  now implies
  \[
     | p(v) - p(w)| <  \varepsilon \quad \text{for all } v,w \in V \ .
  \]
  Hence $p$ is uniformly continuous.
  The implications \ref{ite:uniform-continuity-sublinear-map} $\implies$ \ref{ite:continuity-sublinear-map}
  and \ref{ite:continuity-sublinear-map} $\implies$ \ref{ite:unit-ball-sublinear-map-zero-neighborhood} are trivial.
  It remains to prove
  \ref{ite:unit-ball-sublinear-map-zero-neighborhood} $\implies$ \ref{ite:continuity-sublinear-map-origin}.
  Assume that  $\ball_p (0,1)$ is a zero neighborhood.
  Then there exists a symmetric zero neighborhood $V$ contained in  $\ball_p (0,1)$.  Since $p(0)=0$ one concludes by
  \Cref{thm:properties-subadditive-convex-real-valued-maps-vector-space} \ref{ite:implications-subadditivity}
  \[
     | p(v) | <  \max \{ p(v), p(-v )\} < 1   \quad \text{for all } v \in V \ .
  \]
  But this implies $|p(v)| < \varepsilon $ for all $v \in \varepsilon V$ and $\varepsilon >0$,
  so $p$ is continuous at the origin.
\end{proof}

\subsec{Normability}

\begin{definition}
  A topological vector space $\tvsE$ is called \emph{seminormable} if its topology is generated by a single
  seminorm $p:\tvsE\to \R_{\geq 0}$.
  If the topology on  $\tvsE$ coincides with the vector space topology generated by a norm $\|\cdot\|$,
  then one calls $\tvsE$ \emph{normable}.
\end{definition}

\begin{theorem}[Kolmogorov's normability criterion]
  A topological vector space $\tvsE$ is normable if and only if it is a \ref{axiom:t1} space and
  possesses a bounded convex neighborhood of the origin.
\end{theorem}


\documentclass[revision-guide.tex]{subfiles}
%% Current Author: BC
\setcounter{chapter}{6}
\begin{document}
\chapter{Superposition}
\begin{content}
\item phase difference
\item diffraction
\item interference
\item standing waves
\end{content}

\section*{Candidates should be able to:}
\spec{explain and use the concepts of coherence, path difference, superposition and phase}
\spec{understand the origin of phase difference and path difference, and calculate phase differences from path differences}
\spec{understand how the phase of a wave varies with time and position}

These terms are all used when considering more than one wave.

The phase of a wave is related to how far through an oscillation a wave is. This is expressed in radians or degrees, where one complete oscillation corresponds to $360^{\circ}$ or $2\pi$ radians.

The phase difference between two waves is more useful than the phase of one wave. This refers to the fraction of an oscillation by which one wave 'leads' or 'lags' behind another. If the phase difference is $2n\pi$, where $n$ is an integer, then two waves are said to be \emph{in phase} and if the phase difference is $(2n-1)\pi$ then the waves are completely \emph{out of phase}.

Two waves are said to be coherent if they have a constant phase difference. Most often, this is a phase difference of zero, which means that the waves are in phase, but this does not always have to be the case. For interference patterns to occur, coherence is often a necessary condition.

If two waves, from two different sources, meet at a point, the path difference is the difference in distance travelled between the two waves. To calculate the path difference the smaller distance should be taken away from the larger distance. Path difference is normally expressed as a multiple of wavelength, as this then allows the phase difference to be calculated easily.

When two or more waves meet at a point, superposition will occur. This means that the displacements of the individual waves add up to give a resultant displacement. If the two waves are in phase, then constructive interference will occur and if they are out of phase then destructive interference will occur.

\begin{center}
\begin{tabular}{c|c|c}
\textbf{Path difference} & \textbf{Phase difference} & \textbf{Superposition}\\
\hline
$n\lambda$ & $2n\pi$ & constructive interference\\
\hline
$(n+\frac{1}{2})\lambda$ & $(2n-1)\pi$ & destructive interference
\end{tabular}
\end{center}

\spec{determine the resultant amplitude when two waves superpose, making use of phasor diagrams}

When two waves superpose, the resultant displacement at any point is the vector sum of the individual displacements.

\begin{figure}[h!]
\centering
\includegraphics[width=10 cm]{figs/chapt-7/superposition.JPG}
\caption{Graph showing superposition of two waves}
\end{figure}

Phasors are rotating arrows that can be used to describe waves. A phasor arrow rotates anticlockwise and one full oscillation of the wave corresponds to one complete oscillation of the phasor arrow. The length of the phasor arrow corresponds to the amplitude of the wave.

\begin{figure}[h]
\centering
\includegraphics[width=\textwidth]{figs/chapt-7/phasor.JPG}
\caption{The circle on the left shows the rotating phasors that correspond to this sine wave}
\end{figure}

If we have two waves that superpose, their individual phasor arrows at a particular point can be added up as vectors as shown in the diagram below.

\begin{figure}[h]
\centering
\begin{tikzpicture}
    \draw[thick,->] (0,0) -- (1.5,2.6);
    \draw[thick,->] (1.5,2.6) -- +(3,0);
    \draw[very thick, red,->] (0,0) -- (4.5,2.6);
\end{tikzpicture}
\caption{The sum of two phasors placed end to end}
\end{figure}

\spec{explain what is meant by a standing wave, how such a wave can be formed, and identify nodes and antinodes}

A standing wave arises from a combination of reflection and interference. Consider the set up below.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/melde.JPG}
\caption{Vibration generator connected to a horizontal string under tension}
\end{figure}

The vibration generator leads to a progressive wave travelling to the right. This waves reflects off the fixed end and so there are now two waves on the string, travelling in opposite directions.

These two waves superpose and, in some cases (conditions discussed below), a standing wave can form on the string. If these conditions are met, there will be points where the two waves always meet in phase and interfere constructively. These are called \emph{antinodes}. The points where the two waves always meet out of phase and interfere destructively are called \emph{nodes}.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/string.JPG}
\caption{modes of oscillation on a string}
\end{figure}

Unlike a progressive wave, all points on a stationary wave do not have the same amplitude. The amplitude is at a minimum (often zero) at a node and at a maximum at an antinode.

If the length of the string is $L$, and the speed of waves on the string is $v$, we can work out the frequencies of the various modes of oscillation. The lowest frequency (first mode) shown in the diagram is called the fundamental frequency. You can see that half of a wavelength fits on the string. Therefore we can write:
$$\frac{\lambda}{2} = L$$
$$\lambda = 2L$$

Putting this into the wave equation gives an expression for the fundamental frequency:
$$v = f\lambda = f.2L$$
$$f = \frac{v}{2L}$$

The other modes of oscillation can be worked out in similar ways, by looking at the relationship between $L$ and $\lambda$ and substituting into the wave equation. For a given string under a certain tension the speed is constant. The frequency is the frequency of the vibration generator which can be changed to give the different modes of oscillation.

The boundary conditions for this string were that both ends had to be nodes. In other cases where standing waves occur, for example sound waves, the boundary conditions could be different. Closed ends of tubes are always nodes and open ends are always antinodes.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/sound.JPG}
\caption{Standing waves in a tube {credit:Yonsei Phylab}}
\end{figure}

\spec{understand that a complex wave may be regarded as a superposition of sinusoidal waves of appropriate amplitudes, frequencies and phases}

If more than two waves superpose, the resultant wave can get very complicated! This means that \emph{any} waveform can always be broken down into sinusoidal waves. With combinations of sinusoidal waves of various frequencies, amplitudes and phase differences, any waveform can be made.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/sine1.JPG}
\caption{Example showing how 2 waves combine to form a more complex wave (credit:cyberphysics.co.uk)}
\end{figure}

\spec{recall that waves can be diffracted and that substantial diffraction occurs when the size of the gap or obstacle is comparable to the wavelength}

\spec{recall qualitatively the diffraction patterns for a slit, a circular hole and a straight edge}

When waves pass through an opening, or around a barrier, diffraction occurs and the waves can change in direction and spread out. Diffraction is most significant when the size of the gap or obstacle is comparable to the wavelength. For example, sound waves will diffract through open doors as they have wavelengths of similar orders of magnitudes to the size of the door. However, light will not diffract as much through a door as the wavelength of light is many orders of magnitude smaller than the size of the door.

When waves diffract, diffraction patterns will be formed due to interference of the waves. Specific cases will be discussed below. Here are some examples of diffraction patterns:

\textbf{A slit}

When a wave passes through a slit, and the wave is observed a certain distance away from the slit, a diffraction pattern consisting of points of constructive interference and destructive interference will be formed. Visible patterns will occur when the size of the slit is comparable to the wavelength. For example, with visible light and a very narrow slit, the following pattern will be observed.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/slit.JPG}
\caption{Diffraction pattern for a slit (credit:hyperphysics)}
\end{figure}

There is a central maximum which is twice the width of the maxima on either side and the pattern is symmetrical.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/slitphoto.jpg}
\caption{Photograph of single slit diffraction pattern}
\end{figure}

\textbf{A circular hole}

The diffraction pattern for a circular hole is similar to a single slit, except that instead of fringes, the pattern consists of rings.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/hole.JPG}
\caption{Circular hole diffraction pattern {credit:hyperphysics}}
\end{figure}

Although the examples given here are for visible light, remember that \emph{all} waves can diffract.

\textbf{A straight edge}

A similar pattern of fringes is seen, due to constructive and destructive interference. Here the example given is for radio waves.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/edge.jpg}
\caption{Diffraction at a straight edge {credit:University of Alberta}}
\end{figure}

\spec{recognise and use the equation $n\lambda = b sin\theta$ to locate the positions of destructive superposition for single slit diffraction, where $b$ is the width of the slit}

When an electromagnetic wave, such as light, travels through a single slit, it will diffract. Therefore light from the slit reaches many points on a screen placed at some distance from the slit.

Consider a point on the screen. Light will reach this point from all points within the slit. The distances travelled from various points within the slit to the point on the screen will be different, and so there will be a path and phase difference. Therefore, as we move along the screen, there will be points of constructive interference and points of destructive interference.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/singleslit.JPG}
\caption{Single slit diffraction {credit:hyperphysics}}
\end{figure}

It can be shown that for a single slit of width $b$, the points of \textbf{destructive interference} will be at an angle $\theta$ given by the equation
$n\lambda = b sin\theta$
Here, $n$, refers to the order of the minimum being considered.

\spec{recognise and use the Rayleigh criterion $\theta \approx \frac{\lambda}{b}$ for resolving power of a single aperture, where $b$ is the width of the aperture}

As light travels through an instrument, such as the eye, or a telescope, it passes through a gap or aperture and will diffract. The diffraction pattern will depend on the shape of the aperture and the width.

If light from two different objects passes through the aperture, there will be two diffraction patterns that overlap. The Rayleigh criterion tells us the minimum angle between the two objects at which it is still possible to see them as separate. This is when the first diffraction minimum of one pattern coincides with the central maximum of another.

Working in radians, so that the approximation $\sin{\theta} \approx \theta$, then this minimum angle can be approximated by:

$$\theta \approx \frac{\lambda}{b}$$

\spec{describe the superposition pattern for a diffraction grating and for a double slit and use the equation $n\lambda = d \sin\theta$ to calculate the angles of the principal maxima}

\spec{use the equation $\lambda = \frac{ax}{D}$ for double-slit interference using light}

When light, or any other coherent waves, pass through a double slit, a diffraction pattern consisting of evenly spaced fringes is seen. This is due to the waves from each slit interfering with each other.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/double.JPG}
\caption{Double slit diffraction (credit:hyperphysics)}
\end{figure}

You can see that there is also a single slit envelope, which is much wider than the double slit pattern.

If the number of slits is increased, the pattern becomes more defined, and the fringes get narrower. Therefore for a diffraction grating, the pattern consists of sharp, evenly spaced peaks or bright spots.

If we consider two adjacent slits, you can see that for constructive interference to occur the following must be true:

\begin{equation}\label{dsintheta}
d \sin\theta = n\lambda
\end{equation}

$d$ is the distance between adjacent slits and $n$ is the order of the maxima that is being considered.

\begin{figure}[h!]
\centering
\includegraphics[width=10cm]{figs/chapt-7/diffractioneq.JPG}
\caption{Derivation of grating formula}
\end{figure}

Equation \ref{dsintheta} can be re-arranged to give

\begin{equation}
\sin\theta = \frac{n\lambda}{d}
\end{equation}

This formula can be used for any waves, where there is diffraction from more than one slit.

\newpage

\begin{figure}[h]
    \centering
    \begin{tikzpicture}
    \draw[very thick] (0,-.5) -- (0,.5);
    \draw[thick] (7,-3) -- (7,3);
    \draw[<->] (0,-1) -- (7,-1) node[midway, below] {$D$};
    \draw[dashed] (0,0)--(7,0);
    \draw[red,thick,->] (0,0) -- (7,2);
    \draw[<->] (7.5,0)--(7.5,2) node[midway, right] {$x$};
    \end{tikzpicture}
    \caption{The double slit pattern}
    \label{doubleslitscreen}
\end{figure}

Now consider a double slit pattern, where the screen is now a distance $D$ away from the slits and the slit separation is $a$ instead of $d$. Bright fringes will be equally spaced on the screen and we can call the fringe separation $x$.

For the first maxima, $n=1$, the equation from before now becomes:
$$\sin\theta = \frac{\lambda}{a}$$
but from the diagram you can see that:
$$\tan\theta = \frac{x}{D}$$

If $\theta$ is small, then the small angle approximation can be used. This is true if $D$ is much larger than $a$.

$$\sin\theta \approx \tan\theta$$
$$\frac{\lambda}{a} = \frac{x}{D}$$
$$\lambda = \frac{ax}{D}$$

\end{document}


\chapter{Statistical Inference}
\label{chp:statisticalinference}

\begin{multicols}{2}[\subsubsection*{Contents of this chapter}]
   \printcontents{}{1}{\setcounter{tocdepth}{2}}
\end{multicols}

\input{./chapters/sections/stats_modelsinference.tex}

\input{./chapters/sections/stats_functionals.tex}

\input{./chapters/sections/stats_bootstrap.tex}

\input{./chapters/sections/stats_parametricinference.tex}

\input{./chapters/sections/stats_directional.tex}

\input{./chapters/sections/stats_features.tex}

\input{./chapters/sections/stats_fattails.tex}

\section{$R^2$ Value}

\section{Regression Diagnostics}

\section{t-Statistics}

\section{AIC and BIC}

\section{Factor Regression}

\section{Factor Models}

\section{How to Combine Estimates with Different Uncertainties}
A series of classic papers exist on this subject.

\section{How to Tackle Very High Dimensional Feature Spaces}
In practice, high-dimensional feature spaces amplify the risk of overfitting. The answer, framed in the most general way, is to introduce limitations on the family of functions that are fit by the estimator of choice. The generic methods are regularization (i.e. penalizing variance), variable selection (i.e. filtering out irrelevant features) and dimensionality reduction through feature transformation (i.e. finding a lower-rank representation of the data). It's not possible to draw neat distinctions between this methods in terms of their effects. For example, regularization often amounts to variable selection through soft (or hard) thresholding. Drop out regularization in neural networks biases the model towards particular types of feature transformations.


%
% OK
%
% revised at Jan. 29th; 2009
% revised at Feb. 29th; 2009
% made the discussion of the complete sets more clear
%
% revised on Mar. 22th, 2009
% to use the environment of theorem and law to improve it
%
% revised at July 26th, 2009
% add the discussion of basis functions as examples
% to understand the Hilbert space
%
% revised at Aug. 1st, 2009
% revise the discussion between the relation of wave functions
% and the Hilbert space.
% now it's seems satisfied.
%
%
%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% important general outline for this chapter:
% 1   Hilbert space is the vector space added
%     the additional inner product rules
% 2   like the vector space, it also has the linear independent
%     relations
% 3   relations between Hilbert space and wave functions
% 4   Bra and Ket

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\chapter{Hilbert space}
%
%
%
\section{Definition of the Hilbert space}
%
% 1  relations between vector space and the Hilbert space
%     Hilbert space is more general, functions can be contained
% 2  rules of the inner product, and its meaning
% 3  the definition of the Hilbert space
% 4  difference between Hilbert space and the common vector space
%

The Hilbert space is a kind of extension to the common vector space.
In the Hilbert space, the element can be the common vectors, or the
functions, or anything which satisfy the definition of Hilbert
space. In the last chapter, the wave functions for the infinite well,
the wave functions for the free particle motion etc. are all the
examples for the Hilbert space. Here we do not intend to give more
examples, they can be found in the common mathematical book or book
related to quantum mechanics. Below we will all use the ``vector'' to
signal the common vectors in the vector space or the elements in the
Hilbert space, its meaning will be judged to be clear by the context.

The Hilbert space has all the operations in the common vector space.
Besides the normal operations, the Hilbert space has some operation
called inner product; that is to pick up two vectors from the space,
we always has a number(real or complex) corresponds to it:
\begin{equation}\label{}
(\Phi,\Psi) = c
\end{equation}
Here the $\Phi$ and $\Psi$ denotes two vectors in the Hilbert space.
In quantum mechanics, the inner product between wave functions is
always expressed as:
\begin{equation}\label{}
\int\Phi^{*}\Psi d\tau
\end{equation}
Where the $\Phi^{*}$ is conjugated with the $\Phi$. In vector space,
that is corresponding to the common dot product between two vectors.

The inner product is highly important in the quantum mechanics. In
the following content, it can see that this is the base for the
``bra'' and ``ket'', on the other hand; the probability character of
the wave functions(also the average eigen values), the orthogonality
etc. are expressed by such inner product between the wave functions
and its conjugated form.

Then we introduce the definition of Hilbert space. If any of the
vectors satisfy the rules below; they are configuring the Hilbert
space (here $a, b$ is used to indicate some arbitrary number and the
capital Greek letter is the vector from the given space):
\begin{eqnarray}
\label{Hilberteq:2}
% \nonumber to remove numbering (before each equation)
  \Phi + \Psi &=& \Psi + \Phi \nonumber \\
  \Phi + (\Psi + \Omega) &=& (\Phi + \Psi) + \Omega \nonumber \\
  \Phi + 0 &=& \Phi  \quad
  \text{we have zero vector in the space} \nonumber \\
  \Phi + \Psi &=& 0  \quad
  \text{for any $\Phi$, such $\Psi$ exists} \nonumber \\
  \Phi 1 &=& \Phi  \quad
  \text{any vector multiply number 1 is still itself} \nonumber \\
  (\Phi a)b &=& \Phi(ab) \nonumber \\
    \Phi (a+b) &=& \Phi a+ \Phi b \nonumber \\
      (\Phi + \Psi)a &=& \Phi a + \Psi a \nonumber \\
  (\Phi,\Psi) &=& (\Psi, \Phi)^{*} \nonumber \\
  (\Phi,\Psi + \Omega) &=&  (\Phi,\Omega) + (\Phi,\Omega)\nonumber \\
  (\Phi,\Psi a) &=& (\Phi,\Psi)a \nonumber \\
  (\Psi,\Psi) & \geq& 0 \qquad
  \text{for any $\Psi$, if it equals to 0, the $\Psi = 0$ }
\end{eqnarray}
Here we note that for the operation between number and the vector,
it should remember that:
\begin{equation}\label{}
\Phi a = a\Phi
\end{equation}
The relations defined in the (\ref{Hilberteq:2}) are some common
extension from the vector space. Besides, it can be easily seen that
the wave functions for a specific system satisfy all the conditions
above. So the function space is a kind of special Hilbert space.

Here in the Hilbert space two vectors are orthogonal if their inner
product is 0. The inner product of $(\Psi,\Psi)$, which can be
abbreviated as $|\Psi|^{2}$; represents the ``length'' of this
vector; in quantum mechanics, it's directly connected to the
probability distribution of the wave functions; therefore it
requires that for any wave function we should have $|\Psi|^{2} = 1$.

In common vector space, we always refer to some finite number of
vectors, however, the Hilbert space can be infinite; and the vectors
can be continuously changed (such as the case in the free particles,
if $\hei{p}$ changes in the continuous way; the wave functions also
changes coordinately); or discontinuously changed (see the case of the
deep infinite well; where the wave functions depends on a natural
number of $n$ inside its expression). However, since that in most of
the cases the discrete case and the continuous case are sharing the
same results, so for simplicity in the following content we only
concentrate on the discussion on the finite number of the vectors in
the Hilbert space.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Linear independent vectors}
\label{LIV_in_Hilbert}
%
% introduce the linear independent vectors in the vector space, two points:
%   1  the definition of the linear independent vectors
%      how to understand it. bridge the orthogonality with
%      the linear independent condition
%   2  Schmidt method
%   3  complete sets introduction, identical sets and dimension
%   4  why we introduce the Hilbert space
%   3  degeneracy phenomenon, related to the eigen values
% same in the Hilbert space, but the element form are more complicated
%

In the Hilbert space, if we have $n$ vectors:
\begin{equation}\label{Hilberteq:1}
\sum^{n}_{i=1}a_{i}\Psi_{i} = 0
\end{equation}
This equation is only satisfied as long as all the $a_{i} = 0
(i=1,2,\cdots, n)$; then it can say that such $n$ vectors are linear
independent.

From this definition, we can get some important character for the
linear independent vectors in the Hilbert space:
\begin{theorem}
if the given $n$ arbitrary vectors are not linear independent, then
these $n$ vectors can not be orthogonal with each other.
\end{theorem}

\begin{proof}
  Suggest that we have some linear combination of these $n$ linear
  dependent arbitrary vectors equals to zero, and suggest that the
  coefficient for the $\Phi$ is not zero:
\begin{align}\label{}
  a_{1}\Phi + \sum_{i=2}^{n}a_{i}\Psi_{i} &= 0 \Rightarrow \nonumber
  \\
  \sum_{i=2}^{n}a_{i}\Psi_{i} &=-a_{1}\Phi
\end{align}
Now to multiply the above equation with $\Phi^{*}$ and make
integration; then we have (here we suggest that $(\Phi, \Phi) = 1$,
but in fact it can be any number but without hurting the proof here):
\begin{equation}\label{Hilberteq:4}
\sum_{i=2}^{n}a_{i}(\Phi,\Psi_{i}) = -a_{1}
\end{equation}
Here if all the vectors are orthogonal with each other, that implies
we have:
\begin{equation}\label{}
\left\{
  \begin{array}{ll}
    (\Phi,\Psi_{i}) = 0, & i=2,3,\cdots, n \\
    (\Psi_{i},\Psi_{j}) = 0, & i \neq j
  \end{array}
\right.
\end{equation}
Therefore it can see that the left side (\ref{Hilberteq:4}) must be
zero due to the orthogonality condition, yet the right side is some
non-zero number of $-a_{1}$; that brings contradiction. Hence there
must be at least two vectors in the Hilbert space that they are not
orthogonal with each other. \qedhere
\end{proof}

This simple derivation reveals some important fact that if the vectors
in the Hilbert space are all orthogonal with each other then they must
be linear independent. However, we note that the reverse proposition
is not correct. For $n$ linear independent vectors in the Hilbert
space they are not needed to be orthogonal with each other. Let's take
the vectors as an example, the vector of $\alpha_{1} = (1, 0, 0)$,
$\alpha_{2} = (0, 1, 0)$ and $\alpha_{3} = (0, 1, 1)$ are three linear
independent vectors; but we do not have $(\alpha_{2}, \alpha_{3}) =
0$.

However, for any linear independent vectors, there's always has an way
to make them into the ``basis vectors'' (in the function space we
usually call them as ``basis functions''), in which the vectors are
normalized and orthogonal to each other. Here we introduce the Schmidt
method which transform the linear independent vectors of $\alpha_{1},
\alpha_{2}, \cdots, \alpha_{n}$ into the basis vectors of $\beta_{1},
\beta_{2}, \cdots, \beta_{n}$:
\begin{itemize}
  \item First to normalize the $\alpha_{1}$, so that we have:
\begin{equation}\label{}
\beta_{1} = \frac{\alpha_{1}}{\sqrt{|\alpha_{1}|^{2}}}
\end{equation}
  \item Second to make $\beta_{2}^{'}$ as:
\begin{equation}\label{}
\beta_{2}^{'} = \alpha_{2} - a_{12}\beta_{1}
\end{equation}
$a_{12}$ is some undetermined coefficients. Suggest that we have
orthogonal condition between the $\beta_{1}$ and $\beta_{2}^{'}$,
thus we can determine the $a_{12}$ from this condition:
\begin{align}\label{}
(\beta_{1}, \beta_{2}^{'}) &= 0 \Rightarrow \nonumber \\
a_{12} &= \frac{(\alpha_{1},\alpha_{2})}{\sqrt{|\alpha_{1}|^{2}}}
\nonumber \\
a_{12} &= (\beta_{1},\alpha_{2})
\end{align}
Finally we can achieve the $\beta_{2}$ as:
\begin{equation}\label{}
\beta_{2} = \frac{\beta_{2}^{'}}{\sqrt{|\beta_{2}^{'}|^{2}}}
\end{equation}
  \item Similarly we can have $\beta_{3}^{'}$ as:
\begin{equation}\label{}
\beta_{3}^{'} = \alpha_{3} - a_{13}\beta_{1} - a_{23}\beta_{2}
\end{equation}
By the similar process we can get the coefficients of $a_{13}$ and
$a_{23}$; then finally we can have:
\begin{equation}\label{}
\beta_{3} = \frac{\beta_{3}^{'}}{\sqrt{|\beta_{3}^{'}|^{2}}}
\end{equation}
\item For the arbitrary $\beta_{m}$ ($m \leq n$); likely we can
have:
\begin{equation}\label{}
\beta_{m}^{'} = \alpha_{m} - \sum_{i=1}^{m-1}a_{im}\beta_{i}
\end{equation}
Since that each pair of $\beta_{i}$ and $\beta_{j}$ are orthogonal
for different $i$ and $j$; thus we can get $m-1$ functions to
determine the unknown parameter of $a_{im}$. Finally, we can have:
\begin{equation}\label{}
\beta_{m} = \frac{\beta_{m}^{'}}{\sqrt{|\beta_{m}^{'}|^{2}}}
\end{equation}
\end{itemize}

From the above simple derivation, we can introduce the concept of
``complete sets'', which it is identical to the whole space configured
by the given basis vectors. We note that in the given Hilbert space an
arbitrary vector of $\Psi$ can be always expressed by this complete
sets:
\begin{equation}\label{Hilberteq:5}
\Psi = \sum^{n}_{i=1}a_{i}\Psi_{i}
\end{equation}
Hence, all the other vectors in the given Hilbert space can be
expressed via such chosen complete sets.

It's easy to prove that such complete sets is not unique, there are
variety of complete sets can be used to express the same Hilbert
space; however, it can prove that they are identical with each other
(see the chapter of representation for more details). On the other
hand, the dimension of the space is some fixed number among the
identical complete sets (for detailed proof see the
book\cite{XingLinKe}, PP $7$).

The most important character for the Hilbert space is perhaps
reflected by the (\ref{Hilberteq:5}), where it's actually some kind
of abstraction of the superposition principle in the quantum
mechanics. Based on this property, the only thing need to do in the
quantum mechanics is to search a complete set for the Hilbert space
which corresponds to the investigating system, then we can get any
information form this sets.

Finally, let's conclude our derivation into some theorem which is
called ``perfectibility theorem''. That is:
\begin{theorem}
\label{Hilbert:1}
if $\Psi_{i}$ ($i=1,2,\cdots, n$) is a group of
normalized vectors in Hilbert space where they are orthogonal with
each other; then the four propositions below are identical with each
other:
\begin{itemize}
  \item $\Psi_{i}$ is the complete sets where any other vectors can
  be expressed from this sets;
  \item $\Psi = \sum^{n}_{i=1}(\Psi_{i}, \Psi)\Psi_{i}$ is true for
  any arbitrary vectors in the given Hilbert space;
  \item $(\Psi, \Phi) = \sum^{n}_{i=1}(\Psi, \Psi_{i})(\Psi_{i},
  \Phi)$ holds true for any vectors of $\Phi$ and $\Psi$ in the
  given Hilbert space;
  \item $|\Psi|^{2} = \sum^{n}_{i=1}|(\Psi_{i}, \Psi)|^{2}$ holds
  true for any arbitrary vector of $\Psi$.
\end{itemize}
\end{theorem}

The proof for this theorem is direct and simple so here we omitted
it (if necessary, it can be found at book by XinLin
Ke\cite{XingLinKe}, PP. 9).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Axiom for quantum mechanics in terms of Hilbert space}
\label{sec:AFQMITOHS_in_Hilbert}

As what we have been demonstrated, the quantum mechanics can be
considered as some axiomatic system which is logically based on
several axioms; from these axioms we can build the whole methodology
in quantum mechanics so that to tackle down any problems in this
realm.

The first axiom is related to the Hilbert space, that is:
\begin{axiom}\label{axiom1}
\textbf{The system state in quantum mechanics is described by a
vector in
  Hilbert space. Any two vectors which is different only by some
  complex factor is same with each other.}
\end{axiom}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The relationship between wave functions and Hilbert space}
\label{TRBWFAHS_in_Hilbert}
%
% 1 Hilbert space is some abstraction of the wave functions, wave functions
%   are only some functions which composed into a space
% 2 the variables, x, y, z, t, and the group character
% 3 the probability character
% 4 superposition character
% 5  what the Hilbert space composed of?
%
%
%
In the chapter \ref{basic_chapter}, where we have demonstrated that
the wave function is expressed by the form of $\Psi(\bm{r}, t)$ which
is the function of coordinates and time. Hence, here how can we
understand the relation between such wave functions and the Hilbert
space?

The wave functions indicated in the \ref{basic_chapter}, is actually
the solutions to the Schrodinger equation. For one specific
Schrodinger equation, it always gives a group of functions as the
``solution in cluster'', such cluster is labeled by some arguments.
For example, the clustered solution in free particle system is
controlled by the argument of $\hei{p}$ in (\ref{BASICeq:13}), in
infinite well system the clustered solution is controlled by the $n$
in (\ref{ASPIIDW_in_basic}). As the $\hei{p}$ or the $n$ varies, then
the functions is transforming from one to another in the group. That's
the feature of ``solution in cluster''.

Furthermore, for some given one function in the group, it's the
function of the coordinates and time; namely $x, y, z$ in each time of
$t$ (here the spin variables are temporarily omitted for
simplicity). For each group of certain variables of $x, y, z, t$; the
wave function has some value corresponding to it and different
function in the group usually gives different value for the fixing $x,
y, z, t$. Physically, different wave functions is used to
characterize different state of the corresponding system; for example,
in quantum chemistry calculation, different wave functions always
occupy different energy state, so they are used to describe different
energy state of the chemical system.

One of important feature of the wave functions, is that its modulus
characterizes the probability distribution of the system. This
character brings on an important trait in the Hilbert space, that the
``length'' of the vectors in the Hilbert space has no physical
meaning, only the normalized vectors are meaningful, and the
integration of the modulus over whole space equals to $1$:
$(\Psi,\Psi) = 1$. Thus any vectors in the Hilbert space which
different by no more than a constant will be same in portraying the
system state. However, there also exists some functions which can not
normalized, for example; the wave functions for the free
particle. However, such functions also retain important physical
meaning, in the content below we will present more discussion to this
type of wave functions.

Another important feature of the wave functions, is that they
satisfy the superposition principle; which can be directly observed
from the Schrodinger equation. On the other hand, it can see that
the superposition principle is the foundation of the Hilbert space;
since from the (\ref{Hilberteq:5}) any vector in the space can be
expressed as the linear combination of the vectors in the complete
sets. In a sense, the property in (\ref{Hilberteq:5}) in Hilbert
space really depicts the essence behind the superposition of the
wave functions.

Physically, the Hilbert space can be viewed as the abstraction of
the wave functions in the quantum mechanics. The wave functions is
some concrete function (for example, it can be plane waves, gaussian
functions etc.), while in the Hilbert space we abstracted them as
some ``vector'' and discuss their general properties. Therefore, the
Hilbert space can be seen as some extension for the function
space\cite{Coden}. Such abstraction does not only bring us
convenience, but also actually be indispensable. In the
non-relativistic quantum mechanics, as we can see in the following
chapter; the spin phenomenon is introduced as some additional
``hypothesis'' where we do not have correspondence in the
macroscopical world. Hence, for the discussion of the spin we can
not write down the specific form of operators then to get the
concrete form of wave functions, thus the operator is only
abstractly introduced and as well the wave function (see the chapter
of spin). However, such wave functions can be also expressed by the
``vectors'' in the Hilbert space so that we can still discuss its
properties.

Now let's concentrate on another aspect that what kind of space the
Hilbert space including?

Generally to say, nearly all the wave functions are actually composed
into the normal function space. The function space is some group of
functions (such functions can be discretely varied or continuously
varied, for example; in the \ref{ASPIIDW_in_basic} the functions are
discretely varied from each to another by the variable of $n$), and any
two functions satisfy the definition in (\ref{Hilberteq:2}), more
specifically; the functions should be square-integrable:
\begin{equation}
  \int |\Psi|^{2} d\tau = 1
\end{equation}
So the last item in the (\ref{Hilberteq:2}) is more restricted to be
$(\Psi, \Psi) = 1$. In physics, since the functions characterizes
the distribution of the particle in given state, so it requires that
the functions should be continuously differential, uniform etc. Here
the wave functions in the (\ref{ASPIIDW_in_basic}) is one example of
the function space.

However, the function space is only one part of the Hilbert space, the
Hilbert space also contains some ``irregular space'' in which the wave
functions are not square-integrable. For example, the plane wave
functions we have derived in the (\ref{ASCWFFFP_in_basic}). The
integration of such wave function will lead to infinity. However, it
also satisfy the definition in the (\ref{Hilberteq:2}), so it's also a
kind of space belong to Hilbert type.

In the following content, we will give some examples of the Hilbert
space to see how to bridge their the general properties with the
concrete functions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some examples}
\label{sec:SE_in_Hilbert_space}

Now let's give some examples to demonstrate the above discussion for
the Hilbert space. The examples is all related to the function space,
and can be divided into two classes: one is discrete orthogonal basis
functions in common function space, the other is continuous orthogonal
basis functions which is not square integrable. From the discussion
below, finally we will draw some similarity between them\footnote{This
  part of discussion is taken from \cite{Coden}}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Discrete orthogonal basis functions}
\label{sec:DOBF_in_Hilbert_space}

Now consider a group of wave functions, they are corresponding to some
given $\hat{H}$ (for example, the wave functions for infinite well in
the \ref{ASPIIDW_in_basic}). These functions can be generally
expressed as:
\begin{equation}
\label{Hilberteq:11}
  \mu_{1}(\mathbf{r}) \quad  \mu_{2}(\mathbf{r}) \quad  \mu_{3}(\mathbf{r}) \cdots
\mu_{n}(\mathbf{r})
\end{equation}

We assume that such $n$ functions are orthogonal with each other:
\begin{equation}
  \int \mu_{i}(\mathbf{r}) \mu_{j}(\mathbf{r}) d^{3}\mathbf{r} = \delta_{ij}
\end{equation}

Therefore, according to the (\ref{Hilberteq:5}), for any quantum
states of $\Phi(\mathbf{r})$ corresponding to the system, it can be
expressed as:
\begin{equation}
\label{Hilberteq:3}
  \Phi(\mathbf{r}) = \sum_{i=1}^{n}c_{i}\mu_{i}(\mathbf{r})
\end{equation}

$c_{i}$ is given by:
\begin{align}
  (\mu_{i}(\mathbf{r}), \Phi(\mathbf{r})) &=
  \sum_{j=1}^{n}c_{j}(\mu_{i}(\mathbf{r}), \mu_{j}(\mathbf{r})) \nonumber
  \\
  &=\sum_{j=1}^{n}c_{j}\delta_{ij} \nonumber \\
  &=c_{i}
\end{align}

Therefore, we have:
\begin{equation}
  \Phi(\mathbf{r}) = \sum_{i=1}^{n}(\mu_{i}(\mathbf{r}),
  \Phi(\mathbf{r}))\mu_{i}(\mathbf{r})
\end{equation}

For two different quantum states of $\Phi(\mathbf{r})$ and
$\Psi(\mathbf{r})$, where they are expressed as:
\begin{align}
  \Phi(\mathbf{r}) &= \sum_{i=1}^{n}a_{i}\mu_{i}(\mathbf{r}) \nonumber \\
  \Psi(\mathbf{r}) &= \sum_{i=1}^{n}b_{i}\mu_{i}(\mathbf{r})
\end{align}

Their inner product can be given as:
\begin{align}
\label{Hilberteq:13}
  (\Phi(\mathbf{r}), \Psi(\mathbf{r})) &=
  \sum_{i=1}^{n}\sum_{j=1}^{n}a^{*}_{i}b_{j} (\mu_{i}(\mathbf{r}),
  \mu_{j}(\mathbf{r})) \nonumber \\
  &=  \sum_{i=1}^{n}\sum_{j=1}^{n}a^{*}_{i}b_{j}\delta_{ij} \nonumber \\
  &=  \sum_{i=1}^{n}a^{*}_{i}b_{i}
\end{align}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Closure relation}
\label{sec:CR_in_Hilbert}

Now let's set up another relation which is identical to the definition
of complete sets defined in the section of \ref{LIV_in_Hilbert}. This
is called closure relation.

From the (\ref{Hilberteq:3}), we can write:
\begin{align}
\label{Hilberteq:15}
  \Phi(\mathbf{r}) &= \sum_{i=1}^{n}c_{i}\mu_{i}(\mathbf{r}) \nonumber
  \\
&= \sum_{i=1}^{n}(\mu_{i}(\mathbf{r}),
  \Phi(\mathbf{r}))\mu_{i}(\mathbf{r}) \nonumber \\
&= \sum_{i=1}^{n}\left[ \int
\mu_{i}(\mathbf{r}^{'})\Phi(\mathbf{r}^{'})d^{3}\mathbf{r}^{'}
\right]\mu_{i}(\mathbf{r}) \quad
\underrightarrow{\text{interchange $\sum$ and $\int$}} \nonumber \\
&= \int
\left[ \sum_{i=1}^{n}
\mu_{i}(\mathbf{r}^{'})\mu_{i}(\mathbf{r})\right]
\Phi(\mathbf{r}^{'})d^{3}\mathbf{r}^{'} \nonumber \\
&= \int F(\mathbf{r}, \mathbf{r}^{'})\Phi(\mathbf{r}^{'})d^{3}\mathbf{r}^{'}
\end{align}

Hence we have:
\begin{equation}
  \label{Hilberteq:6}
  F(\mathbf{r}, \mathbf{r}^{'}) = \sum_{i=1}^{n}
\mu_{i}(\mathbf{r}^{'})\mu_{i}(\mathbf{r}) = \delta (\mathbf{r}
-\mathbf{r}^{'})
\end{equation}

The relation in the (\ref{Hilberteq:6}) is the closure relation. On
the other hand, we can prove that from the (\ref{Hilberteq:6}) we can
also derive the (\ref{Hilberteq:3})\cite{Coden}, so that to prove it's
identical to the definition of complete sets.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Plane wave functions as continuous basis functions}
\label{sec:PWF_in_Hilbert}

In section of (\ref{sec:HTEAAWFFFP_in_basic}), we have introduced the
free particle wave function as well as the wave packets. From the
(\ref{BASICeq:19}) and (\ref{BASICeq:20}), they can be generally
expressed as:
\begin{align}
  \label{Hilberteq:7}
\Phi(x) &=
\frac{1}{\sqrt[2]{2\pi}}\int_{-\infty}^{+\infty}\Psi(p)e^{ipx}dp
\nonumber \\
\Psi(p) &=
\frac{1}{\sqrt[2]{2\pi}}\int_{-\infty}^{+\infty}\Phi(x)e^{-ipx}dx
\end{align}
Here we have omit the irrelevant time part and restrict to the X
direction for simplicity.

If we write the plane wave function of $\nu_{p}(x)$ as:
\begin{equation}
  \label{Hilberteq:8}
  \nu_{p}(x) = \frac{1}{\sqrt[2]{2\pi}}e^{ipx}
\end{equation}

Then the (\ref{Hilberteq:7}) can be converted to:
\begin{align}
  \label{Hilberteq:9}
\Phi(x) &= \int_{-\infty}^{+\infty}\Psi(p)\nu_{p}(x)dp
\nonumber \\
\Psi(p) &= \int_{-\infty}^{+\infty}\Phi(x)\nu_{p}^{*}(x)dx = (v_{p}(x), \Phi(x))
\end{align}

As $p$ continuously varied, since $|\nu_{p}(x)|^{2} = \frac{1}{2\pi}$,
thus $\nu_{p}(x)$ is not square integrable. However, we can use the
delta function to express their integration that:
\begin{equation}
  \label{Hilberteq:16}
  \int \nu_{p}^{*}(x)\nu_{p}(x) dx = \delta(x)
\end{equation}
The delta function is detailed discussed in the appendix.

Now let's make some comparison between the (\ref{Hilberteq:11}) and
(\ref{Hilberteq:8}), we can see many similarities between the
$\nu_{p}(x)$ and the discrete basis functions.:
\begin{align}
  \label{Hilberteq:10}
p &\Leftrightarrow i \nonumber \\
x &\Leftrightarrow \bm{r} \nonumber \\
\nu_{p}(x) &\Leftrightarrow \mu_{i}(\bm{r}) \nonumber \\
\Phi(x) &\Leftrightarrow \Phi(\bm{r})
\end{align}

Then we can see the relation in the (\ref{Hilberteq:3}) just
equivalent to the (\ref{Hilberteq:9}). So the $c_{i}$ in the
(\ref{Hilberteq:3}) equivalent to the $\Psi(p)$ in the
(\ref{Hilberteq:9}).

Then if we have two wave packets, $\Phi_{1}(x)$ and $\Phi_{2}(x)$,
then we can evaluate their inner product as:
\begin{align}
  \label{Hilberteq:12}
  (\Phi_{1}(x), \Phi_{2}(x)) &= \int_{-\infty}^{+\infty}
  \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}
  \Psi^{*}_{1}(p)\nu^{*}_{p}(x)\Psi_{2}(p^{'})\nu_{p^{'}}(x)dpdp^{'}dx
  \nonumber \\
  &= \int_{-\infty}^{+\infty}
  \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}
  \Psi^{*}_{1}(p), \Psi_{2}(p^{'})\delta(p-p^{'})dpdp^{'}
  \nonumber \\
  &= \int_{-\infty}^{+\infty}\Psi^{*}_{1}(p), \Psi_{2}(p)dp
\end{align}
So finally the result reveals that it's equivalent to the
(\ref{Hilberteq:13}). Here we have imposed the orthogonality of the
continuous basis functions.

Finally, let's reveal how to express the closure relation in the
(\ref{sec:CR_in_Hilbert}), which is due to the orthogonality for the
plane wave functions:
\begin{align}
  \label{Hilberteq:14}
\int \nu_{p}^{*}(x)\nu_{p^{'}}(x)dx = \delta(p - p^{'}) \nonumber \\
\int \nu_{p}^{*}(x)\nu_{p}(x^{'})dp = \delta(x - x^{'})
\end{align}
This is equivalent to the closure relation in the
(\ref{sec:CR_in_Hilbert}).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Subspace}
%
% some general discussion for the subspace
%
%
Sometimes we have encounter such circumstance that in the given
complete sets, there's a small group of vectors which is also
composed into some Hilbert space. This is called ``subspace''
compared with the original Hilbert space.

In quantum mechanics, such subspace is very common. For example, in
the following content we will introduce the eigen function (see the
chapter of ``General Discussion to Operator''):
\begin{equation}\label{}
\hat{A}\Psi_{i} = a_{i}\Psi_{i}
\end{equation}
$\hat{A}$ is some operator. Here we may meet such case that for a
group of $\Psi_{ij}$ ($j=1,2, \cdots, m$) they all give the eigen
value of $a_{i}$. Such $m$ wave functions just compose into some
subspace inside the complete set $\Psi_{ij}$ ($i=1,2,\cdots, n$).

It's easy to prove that if the subspace is in $m$ dimension, then we
can find another $n-m$ vectors to form the whole complete sets for
$\Psi_{ij}$ ($i=1,2,\cdots, n$) together with the $\Psi_{ij}$
($j=1,2, \cdots, m$). The demonstration is direct and
straightforward (see \cite{XingLinKe}, PP. 12). therefore, no matter
whether the subspace exists, the dimension of the complete sets are
fixed.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Bra and Ket}
%
%  1  why we introduce the bra and ket
%  2  definition
%  3  one to one correspondent relation
%  4  different with complex number
%
In the above definition, it can see that the inner product is related
to the multiplication order between two vectors in Hilbert space. That
is: $(\Phi,\Psi) = (\Psi, \Phi)^{*}$.  Furthermore, it's easy to prove
that $(a\Phi,\Psi) = a^{*}(\Phi,\Psi)$, which is conjugated with
$(\Phi,a\Psi) = (\Phi,\Psi)a$. Thus even for the same vector it's
different between the left position and the right position in the
inner product.

Moreover, in the study of quantum mechanics it's found that to
distinguish the left position and the right position is necessary,
so we introduce the ``bra'' and ``ket'' to represent the left
position and right position, respectively. Mathematically the bra is
written as $\langle\Psi|$, and the ket is written as $|\Psi\rangle$.
They are independent vector space, but coordinate with each other.
It was introduced by Dirac, so this symbolic representation is
called Dirac symbol.

Simply to say, the bra and ket are some dual space, but they are
independent with each other. However, how to understand the relation
between the Hilbert space and the space of ``bra'' and ``ket''? In
short, the bra and ket can be viewed as some extention to the Hilbert
space, just to decompose the original single space into some dual
space for the sake of quantum mechanics. Therefore, in bra and ket the
elements should satisfy the same rules in the Hilbert space:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  |\Psi\rangle + |\Phi\rangle&=& |\Phi\rangle + |\Psi\rangle \nonumber \\
  |\chi\rangle + (|\Psi\rangle + |\Phi\rangle) &=&
  (|\chi\rangle + |\Psi\rangle) + |\Phi\rangle \nonumber \\
  |\Psi\rangle + |O\rangle &=& |\Psi\rangle \nonumber \\
  |\Psi\rangle1 &=& |\Psi\rangle \nonumber\\
  |\Psi\rangle + |\Phi\rangle &=& |O\rangle \nonumber\\
  |\Psi\rangle (a+b) &=& |\Psi\rangle a + |\Psi\rangle b \nonumber\\
  (|\Psi\rangle a)b &=& |\Psi\rangle (ab) \nonumber\\
  (|\Psi\rangle + |\Phi\rangle)a &=& |\Psi\rangle a + |\Phi\rangle a
\end{eqnarray}
It's note that the space of bra satisfy the same rules.

For the bra and ket, it's the inner product that connect them
together, that the inner product is defined as the multiplication
between bra and ket: $\langle\Phi|\Psi\rangle$. hence, the bra and ket
has to comply with the rules related to the inner product:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  \langle\Phi|\Psi\rangle &=& \langle\Psi|\Phi\rangle^{*}      \nonumber\\
  \langle\Omega|(|\Phi\rangle+|\Psi\rangle) &=&
  \langle\Omega|\Phi\rangle + \langle\Omega|\Psi\rangle \nonumber \\
  (\langle\Phi|+\langle\Psi|)|\Omega\rangle &=&
  \langle\Phi|\Omega\rangle + \langle\Psi|\Omega\rangle \nonumber \\
  \langle\Phi|(|\Psi\rangle a) &=& \langle\Phi|\Psi\rangle a \nonumber \\
  (a\langle\Phi|)|\Psi\rangle  &=& a\langle\Phi|\Psi\rangle  \nonumber \\
  \langle\Psi|\Psi\rangle&\geq& 0 \qquad \text{for any vector in space }
\end{eqnarray}

From the definition above, it's easy to see that the bra space is
one to one mapping with the ket space. For an arbitrary \kett{\Psi},
there hast have an \brat{\Psi}; makes $\langle\Psi|\Psi\rangle = 1$,
and only one \brat{\Psi} satisfies this correspondence. Furthermore,
the operations on the \kett{\Psi} is same to \brat{\Psi}. The only
correspondence between \kett{\Psi} and \brat{\Psi} is, the
$c\ket{\Psi}$ is corresponding to $c^{*}\bra{\Psi}$ but not
$c\bra{\Psi}$.

Such correspondence between bra and ket, makes us to call them as
``conjugated vectors''. They are similar to the conjugated complex
number such as $a+ib$ and $a-ib$, however; there's essential
distinction between them( to read the book by Dirac \cite{Dirac} for
more further details, PP 17).

For the conjugated vectors, it's impossible to differentiate the
real part and the imaginary part in the \kett{\Psi} and \brat{\Psi}.
For example, if we have $x=a+ib$, $y=a-ib$; thus $a= \frac{x+y}{2}$
and $b = \frac{x-y}{2i}$. Whereas for the \kett{\Psi} and
\brat{\Psi}, they are in different space, and independent with each
other; \kett{\Psi} $+$ \brat{\Psi} has no meaning, only their
multiplication is meaningful. Thus we called them ``conjugated
vectors''.

Because of the close relation between \kett{\Psi} and \brat{\Psi},
we introduce transposing operation to transform them to each other:
$\ket{\Psi}^{*} = \bra{\Psi}$. Since the bra and ket symbol is
obviously different with the complex number, we still use the
conjugated symbol.

Finally we note that usually we always use the \kett{\Psi} to
represent the wave function.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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