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\begin{document}
	\section{Conjugates}
	\begin{definition}
		Let $F$ be a field. A field automorphism $\overline \cdot : F \to F$ is called a \textbf{conjugate} on $F$ if
		\begin{itemize}
			\item $\overline{\cdot}$ is an involution, i.e. ${\overline \cdot}^{-1} = \overline \cdot$
			\item The fixed points of $\phi$, that is the set
			\[ F^{\bar \cdot} := \{ x \in F \mid \overline x = x \} \]
			is an \textbf{ordered subfield} of $F$, when inheriting addition and multiplication from $F$.
			\item For all $f \in F$, $f \overline{f} \geq 0_F$.
		\end{itemize}
	\end{definition}
	\begin{lemma}
		The usual conjugate on $\C$ given by
		\[ \overline{x + iy} := x - iy \]
		is indeed a conjugate on $\C$.
	\end{lemma}
	\begin{proof}
		It is trivial to show that $\overline{\cdot}$ is a field automorphism. We can show that it is an involution and in particular invertible by (for all $x, y \in \R$)
		\[ \overline{\overline{x + iy}} = \overline{x - iy} = x + iy \]
		therefore, $\overline \cdot \circ \overline \cdot = \id_\C$. Now let $z := x + iy \in \C$ be a fixed point of $\overline \cdot$. Then
		\[ x + iy = x - iy \implies y = 0 \implies z = x \in \C \]
		which clearly is ordered by inheriting the usual order on $\R$, and reversing the embedding $\R \to \C$ (which is invertible onto its image, and we have just shown that fixed points of the conjugate lie in its image). The last property follows from
		\[ (x + iy)(x - iy) = x^2 + y^2 \geq 0 \]
	\end{proof}
	\begin{definition}
		Let $F$ be a field with conjugate, define the set
		\[ F^{\overline{\cdot}}_{\geq 0} := \{ f \in F^{\overline{\cdot}} \mid f \geq 0 \} \]
		We also define the following map, called the \textbf{induced absolute value}
		\[ |\cdot| : F \to F^{\overline{\cdot}}_{\geq 0} \]
		\[ f \mapsto \sqrt{f \overline{f}} \]
		where $\sqrt{\cdot}$ is the inverse of $f \mapsto f \cdot f$ as a map $F^{\overline{\cdot}}_{\geq 0} \to F^{\overline{\cdot}}_{\geq 0}$, provided this is invertible.
	\end{definition}
	\begin{lemma}
		The induced absolute value is an absolute value, in the sense that it is positive definite, multiplicative and satisfies the triangle inequality, if $F$ has infinite characteristic.
	\end{lemma}
	\begin{proof}
		We show the individual properties.
		\begin{itemize}
			\item It suffices to show for positive definiteness that if for $f \in F$ we have $|f| = 0$  then $f = 0$. Well
			\[ |f| = 0\implies \sqrt{f \overline{f}} = 0 \implies f \overline{f} = 0 \implies f = 0 \]
			where the last equality holds because we have that $f = 0$ or $\overline{f} = 0$, which are actually equivalent.
			\item Let $a, b \in F$, then
			\[ |ab| = \sqrt{ab \overline{ab}} = \sqrt{a \overline{a} b \overline{b}} = \sqrt{a \overline{a}} \sqrt{b \overline{b}} = |a| |b| \]
			\item Let $a, b \in F$, then we firstly compute for all $t \in F^{\overline{\cdot}}$
			\begin{align*}
				0 \leq |a - tb|^2 &= (a - tb) \overline{a - tb} \\
				&= a \overline{a} - 2t \operatorname{In}(a \overline{b}) + t^2 b \overline{b}
			\end{align*}
			which shows by polynomial theorems (that just require ordered fields) that
			\[ \operatorname{In}(a \overline{b}) \leq |a| |b| \]
			We now return to
			\begin{align*}
				|a + b| &= \sqrt{(a + b) \overline{a + b}} \\
				&= \sqrt{a \overline{a + b} + b \overline{a + b}} \\
				&= \sqrt{a \overline{a} + b \overline{b} + a \overline{b} + b \overline{a}} \\
				&= \sqrt{a \overline{a} + b \overline{b} + a \overline{b} + \overline{a \overline{b}}} \\
				&= \sqrt{a \overline{a} + b \overline{b} + 2 \operatorname{In}(a \overline{b})} \\
				&\leq \sqrt{a \overline{a} + b \overline{b} + 2\sqrt{a \overline{a}} \sqrt{b \overline{b} } } \\
				&= \sqrt{\left(\sqrt{a \overline{a}} + \sqrt{b \overline{b}}\right)^2} \\
				&= |a| + |b| \qedhere
			\end{align*}
		\end{itemize}
	\end{proof}
	\begin{definition}
		Let $f \in F$, where $F$ is a field with conjugate with $\operatorname{char}(F) \neq 2$ (in the rest of the notes we will further assume this). We define the \textbf{invariant} (under conjugate) part of $f$, $\operatorname{In}(f)$, by
		\[ \operatorname{In}(f) := \frac{f + \overline f}{2} \]
	\end{definition}
	\begin{lemma}
		The invariant part is a fixed point of the conjugate.
	\end{lemma}
	\begin{proof}
		By direct computation for all $f \in F$,
		\[ \overline{\left(\frac{f + \overline f}{2}\right)} = \frac{1}{2} \overline{f + \overline f} = \frac{1}{2} \left(\overline{f} + \overline{\overline f}\right) = \frac{f + \overline{f}}{2} \]
		which shows $\overline {\operatorname{In}(f)} = \operatorname{In}(f)$.
	\end{proof}
	\newpage
	\section{Inner products}
	\begin{definition}
		Let $V$ be an $F$-vector space, where $F$ is a field with a conjugate $\bar \cdot$. A map $\inprod{\cdot, \cdot} : V \times V \to F$ is called an \textbf{inner product} if
		\begin{itemize}
			\item For all $x, y \in V$ we have $\inprod{x, y} = \overline{\inprod{y, x}}$
			\item The map $x \mapsto \inprod{x, y}$ is linear for all $y \in V$ 
			\item For all $x \in V \setminus \{0_V\}$, $\inprod{x, x} > 0$
		\end{itemize}
	\end{definition}
	\begin{remark}
		The last condition is well-defined because
		\[ \inprod{x, x} = \overline{\inprod{x, x}} \]
		which makes $\inprod{x, x}$ be a fixed point of the conjugate, thereby being a member of the ordered subfield $F^{\bar \cdot}$.
	\end{remark}
	\begin{lemma}
		Usual inner products over $\C$-vector spaces are also inner products in this generalized sense.
	\end{lemma}
	\begin{theorem}
		Let $V$ be an $F$-vector space with inner product.
		\begin{itemize}
			\item For all $x \in V$, $\inprod{0_V, x} = \inprod{x, 0_V} = 0_F$
			\item For all $x \in V$, $\inprod{x, x} \geq 0$
			\item For all $x, y \in V$ and $a \in F$, $\inprod{x, ay} = \overline{a} \inprod{x, y}$
			\item For all $x, y \in V$,
			\[ \inprod{x + y, x + y} = \inprod{x, x} + 2\operatorname{In}(\inprod{x, y}) + \inprod{y, y} \]
		\end{itemize}
	\end{theorem}
	\begin{proof}
		All of them are easy exercises to show except perhaps the first and second to last property. Note that
		\[ \inprod{0_V, x} = \inprod{0_F 0_V, x} = 0_F \inprod{0_V, x} = 0_F = 0_F \inprod{x, 0_V} = \inprod{x, \overline{0_F} 0_V} = \inprod{x, 0_V} \]
		where $\overline{0_F} = 0_F$ due to $\overline{\cdot}$ being a field homo.\\
		We also have
		\[ \inprod{x, ay} = \overline{\inprod{ay, x}} = \overline{a\inprod{y, x}} = \overline{a} \overline{\inprod{y, x}} = \overline{a} \inprod{x, y} \qedhere \]
	\end{proof}
	\begin{definition}
		Let $(V, +_V, \cdot_V)$ be an $F$ vector space where $F$ is a field with conjugate. We define the set $\overline{V} := V$, $+_{\overline V} := +_V$ and
		\[ \cdot_{\overline V} : F \times \overline V \to \overline V \]
		\[ \lambda, v \mapsto \overline{\lambda} v \]
		Then $(\overline V, +_{\overline V}, \cdot_{\overline V})$ is an $F$ vector space, called the \textbf{conjugate vector space} of $V$.
	\end{definition}
	\begin{lemma}
		An inner product $\inprod{\cdot, \cdot} : V \times V \to F$ is a bilinear map when we view it as a map $V \times \overline{V} \to F$.
	\end{lemma}
	\begin{corollary}
		An equivalent definition of an inner product is a linear map $V \times \overline{V} \to F$ that is symmetric and (in some sense) positive definite. 
	\end{corollary}
	\begin{theorem}
		Let $V$ be an inner product space. We define the \textbf{induced norm}
		\[ \norm{\cdot} : V \to F^{\overline{\cdot}} \] 
		\[ v \mapsto \sqrt{\inprod{v, v}} \]
		where $\sqrt{\cdot}$ is the inverse of $f \mapsto f \cdot f$ as a map $F^{\overline{\cdot}}_{\geq 0} \to F^{\overline{\cdot}}_{\geq 0}$, provided this is invertible. Then $V$ equipped with this induced norm is a normed vector space.
	\end{theorem}
	\begin{proof}
		We show every property of a norm:
		\begin{itemize}
			\item Let $v \in V$ and suppose $\norm{v} = 0$, then $\inprod{v, v} = 0$ which gives $v = 0$
			\item Let $\lambda \in F$, $v \in V$, then
			\[ \norm{\lambda v} = \sqrt{\inprod{\lambda v, \lambda v}} = \sqrt{|\lambda|^2 \inprod{v, v}} = |\lambda| \sqrt{\inprod{v, v}} = |\lambda| \norm{v} \]
			\item Let $v, w \in V$, then note that
			\[ \operatorname{In}(\inprod{v, w}) \leq \norm{v} \norm{w} \]
			which has a similarly spirited proof as before, so that
			\begin{align*}
				\norm{v + w} &= \sqrt{\inprod{v + w, v + w}} \\
				&= \sqrt{\inprod{v, v} + 2 \operatorname{In}(\inprod{v, w}) + \inprod{w, w} } \\
				&\leq \sqrt{\left(\sqrt{\inprod{v, v}} + \sqrt{\inprod{w, w}}\right)^2} \\
				&= \norm{v} + \norm{w} \qedhere
			\end{align*}
		\end{itemize}
	\end{proof}
	\begin{theorem}[Cauchy-Schwarz]
		For all $v, w \in V$, we have
		\[ |\inprod{v, w}| \leq \norm{v} \norm{w} \]
	\end{theorem}
	\begin{proof}
		Wlog suppose $w \neq 0_V$, otherwise the inequality is trivial. Then
		\[ |\inprod{v, w}| = \frac{\inprod{v, w}^2}{|\inprod{v, w}|} = \inprod{v, \frac{\overline{\inprod{v, w}}}{|\inprod{v, w}|} w} = \operatorname{In}\left(\inprod{v, \frac{\overline{\inprod{v, w}}}{|\inprod{v, w}|} w}\right) \leq \norm{v} \norm{w} \]
		due to the inequality we proved before, that
		\[ \operatorname{In}(\inprod{v, w}) \leq \norm{v} \norm{w} \]
	\end{proof}
\end{document}