\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{parskip}
\usepackage{amssymb}
\usepackage{fancyhdr}
\usepackage{enumerate}
\usepackage[bookmarksnumbered]{hyperref}
\usepackage[english]{babel}
\usepackage{dashrule}


\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\C}{\mathbb{C}}
\renewcommand{\P}{\mathbb{P}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\half}{\frac{1}{2}}

\usepackage{ textcomp }

\newcommand{\A}{\mathcal{A}}
\renewcommand{\H}{\mathcal{H}}
\newcommand{\1}{\mathbb{1}}
\newcommand{\teq}[1]{\stackrel{#1}{=}}
\newtheorem{lemma}{Lemma}
\newtheorem{theorem}[lemma]{Theorem}
\newtheorem{definition}[lemma]{Definition}

\pagestyle{fancy}

\fancyhead[L]{Hand-in 6 Functional Analysis}
\fancyhead[R]{Thomas van Maaren (9825827)}
\begin{document}
\textbf{Exercise 9.} Let $\mathcal{H}$ be a real Hilbert space. Let $a: \mathcal{H} \times \mathcal{H} \rightarrow \mathbb{R}$ be bilinear and continuous, and let $\Lambda \geq 0$ be such that

$$
|a(x, y)| \leq \Lambda\|x\|\|y\| \quad \forall x, y \in \mathcal{H}
$$

Suppose that $a$ is coercive, i.e. there exists $\lambda>0$ such that

$$
a(x, x) \geq \lambda\|x\|^{2} \quad \forall x \in \mathcal{H}
$$

(a)

\begin{enumerate}[(i)]
	\item  Let $x \in \mathcal{H}$. Show that there exists a unique vector $z \in \mathcal{H}$ such that $a(x, y)=\langle z, y\rangle$ for all $y \in \mathcal{H}$

	\item  Define a map $A: \mathcal{H} \rightarrow \mathcal{H}$ by $x \mapsto z$ (where $z$ is as above), and show that $A \in B(H)$, with $\|A\| \leq \Lambda$.

	\item  Prove that $A$ is injective. (Hint: To do this you can estimate $\langle A x, x\rangle$ using (2)).

	\item  Using Exercise 8, prove that $\operatorname{Im} A$ is closed.

	\item  Show that $(\operatorname{Im} A)^{\perp}=\{0\}$. (Hint: For $x \in(\operatorname{Im} A)^{\perp}$, consider the inner product $\langle A x, x\rangle$.)

	\item  Show that $A$ is surjective.

	\item  Show that $A^{-1} \in B(\mathcal{H})$, and prove $\left\|A^{-1}\right\| \leq \lambda^{-1}$.
\end{enumerate}

\hdashrule{\textwidth}{.4pt}{1pt}

\begin{enumerate}[(i)]
	\item For this exercise we use the Riesz-Fr\'echet Theorem (Theorem 5.2). We know that $a(x,\cdot):\H\to \R$ for every $x\in \H'$, hence $a(x,\cdot) \in \H'$. 
		Because of theorem 5.2 we know there is a unique $z\in \H$ such that $a(x,y) = \langle y, z\rangle = \langle z,y\rangle$.
	\item What is B(H)? Because of (i) $A$ is well-defined. We see that \begin{align}\|Ax\|^2 = \langle Ax, Ax \rangle = a(x,Ax) \leq |a(x,Ax)| \leq \Lambda \|x\|\|Ax\|\label{toBeDivided}\end{align} for
			all $x\in \H$. If $Ax=0$ it is trivial that $\|Ax\| =0 \leq \Lambda \|x\|$ and if $Ax\neq 0$ we can divide both sides of \eqref{toBeDivided} by $\|Ax\|$ and see that
			$$\|Ax\|\leq \Lambda \|x\|$$

			hence $\|A\|\leq \Lambda $
	\item Let $x,x'\in \H$ and assume that $Ax=Ax'$. This means that
		$a(x,y) = \langle Ax, y\rangle = \langle Ax', y\rangle =
		a(x',y)$ for all $y\in \H$. Because of the bi-linearity 
		of $a$ this means that $a(x-x',y) = a(x,y)-a(x',y) = 0$ for all $y\in \H$, hence $\lambda \|x-x'\|^2 \leq a(x-x',x-x') =0$. Because $\Lambda>0$ we know that $\|x-x'\|^2=0$, os $x=x'$.
	\item For all $x\in \H$ we see that \begin{align}\|Ax\|\|x\| \geq \langle Ax , x\rangle = a(x,x) \geq \lambda \|x\|^2\label{toBeDivided2}\end{align} If $x=0$ we see that $\|Ax\| = 0 = c\cdot 0 = c\|x\|$. And otherwise we can divide $\eqref{toBeDivided2}$ by $\|x\|$ and get that $\|Ax\| \geq  \lambda\|x\|$ for all $x\in \H$. Also because $\H$ is a hilbert it must
			be a Banach space and in exercise (ii) we proved that $A\in B(\H,\H)$. Beause of exercise 8 we know that $\text{Im} A$ is closed.
	\item We need to show that if $\langle Ax, y\rangle =0$ for all $x\in \H$  that $y=0$. Note that if $\langle Ax, y\rangle =0$ for all $x\in \H$ that $
		\lambda \|y\|^2 \leq a(y,y) = \langle Ay, y\rangle=0$, hence $\|y\|=0$ and also $y=0$.
	\item Because $A$ is linear $\text{Im} A$ must also be linear. We know from (iv) that $\text{Im} A$ is closed, so $\text{Im} A = \left(\text{Im} A\right)^{\perp \perp}$ by corollary 3.35. We know from (v) that $\text{Im} A ^\perp = \{0\}$, hence $\text{Im} A = \left(\text{Im} A\right)^{\perp \perp}= \{0\}^\perp = \H$, hence $A$ is surjective.
	\item Because (iii) and (vi) we know that $A$ is injective and surjective, hence $A^{-1}$ is well-defined. Know we only need to show that it is bounded. By definition we know
		that $a(A^{-1}z,y)= \langle z,y\rangle$ for any $z,y\in \H$. We then see
		\begin{align}\lambda \|A^{-1} z\|^2 \leq a(A^{-1} z,A^{-1} z) = \langle z, A^{-1} z \rangle \leq \|z\| \|A^{-1} z\|\label{toBeDivided3}\end{align}
			If $A^{-1}z=0$ we see that $\|A^{-1}z\|=0 \leq \lambda \|z\|$ and otherwise we can divide both sides of \eqref{toBeDivided3} by $\lambda \|A^{-1}z\|$ and get
			that $|A^{-1}z\|\leq \lambda^{-1} \|z\|$. 

\end{enumerate}
\end{document}

