$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ # Compact Compactness is pretty important. >[!idea]- Recall from point-set topology >The following various equivalent definitions of a compact subset of a *metric space*: > - Sequential compactness: every sequence has a convergent subsequence. > - Completeness and total boundedness: for every $\eps > 0$, there exists a covering by finitely many $\eps$-balls. > - Usual compactness: every open cover has a finite subcover. > > In addition, a subset $C$ is ==**pre-compact**== iff $\bar{C}$ is compact. > - Equivalently, all sequences in $C$ have a subsequence convergent in the whole space. ($\bar{C}$ will be sequentially compact, because one can pick representatives in $C$ which are $\leq \frac1k$ away from $x_k\in \bar{C}$). > - This immediately implies all subsets of a pre-compact space are pre-compact. This is also true in non-metrizable spaces, lol. We'll present two ways to check something is compact on a Hilbert space. First, an elegant basis-free approach: > [!theorem] Finite Dimensional Approximations $K\subset \HH$ is compact iff: > - $K$ is closed (which immediately implies $K$ is complete). > - $K$ is bounded. > - For all $\eps > 0$, there exists a finite-dimensional subspace $W\subset \HH$ such that for all $u\in K$, there exists a $w\in W$ such that $\norm{u - w} < \eps$. This actually isn't too difficult; its just a slight rewording of the standard total boundedness condition to be more amenable to vector spaces. > [!proof]- Statement implies Compact Pick an $\eps$; we're going to fill up $K$ with $\eps$-balls. First off, use the statement to find a finite-dimensional subspace $W$ such that all points of $K$ are within $0.001\eps$ of $W$. I know that $K$ is bounded; $K\subset B_R(0)$. WLOG $R \gg \eps$. The set of $W\cap \overline{B_{2R}(0)}$ is closed and bounded, but it is also a subset of $W\equiv \CC^n$, thus by Heine-Borel it is compact. Thus, it can be covered by finitely many $0.001\eps$-balls. Finally, take all of those balls and increase their radius to $\eps$. We conclude. > [!proof]- Compact implies Statement We show the contrapositive. Flipping all the quantifiers, suppose $ \exists \eps \forall W \exists u\in K \forall w\in W,\qquad \norm{u - w}\geq \eps. $ Well, fix that $\eps$. We'll create a sequence of $Ws and $us. Let $W_0 = \emptyset$. Then, given $W_i$, let $u_i \in K$ be such that for all $w\in W_i$, $\norm{u_i - w} \geq \eps$. Then, let $W_{i+1}$ be the span of $W_i$ and $u_i$. Then, each $W_i$ is finite dimensional, $W_i\subset W_{i+1}$, and we have found a sequence of $u_is such that $\norm{u_i - u_j}\geq \eps$ for all $i\neq j$. This clearly has no converging subsequence, so $K$ is not compact. Here is a second approach: > [!definition] Equi-Small Tails Let $\{e_k\}$ be a countable orthonormal subset. A subset $K\subset H$ is said to have ==**equi-small tails**== with respect to $\{e_k\}$ if for all $\eps > 0$, there is some $N$ such that for all $v\in K$ we have $ \sum_{k > N} \abs{\braket{e_k | v}}^2 < \eps $ Note that $\forall v \exists N$ is already guaranteed by Bessel's inequality; the strength here is that the convergence is uniform. > [!problem] Equi-Small Tails Characterization Let $\HH$ be a *separable* Hilbert space with orthonormal basis $\{e_k\}$. A subset $K\subset H$ is compact iff $K$ is closed, bounded, and has equi-small tails with respect to $\{e_k\}$. ## Finite Rank Operators > [!definition] Finite Rank Operators An operator $T\in \BB(\HH, \HH)$ is said to have ==**finite rank**== if $\im T$ is finite-dimensional. The set of finite rank operators is denoted $\mathcal{R}(\HH)$. Finite-rank operators are very nice, because we can basically just write them down as a finite-dimensional matrix (or an \"infinite matrix with cofinitely many zeros\"). > [!problem] An operator $T\in \BB(\HH, \HH)$ has finite rank iff there exists a finite orthonormal set $\{e_k\}_{k\leq L}$ and complex entries $c_{ij}$ such that $ Tu = \sum_{i,j \leq L} c_{ij} \braket{e_j | u} e_i $ for all $u\in \HH$. > [!problem] Properties of Finite Rank Operators Show that $\mathcal{R}(\HH)$ is a star-closed two-sided ideal in $\BB(\HH)$. This has three components: > > > > - **Star Closure** means that if $T\in \mathcal{R}(\HH)$, then $T^*\in \mathcal{R}(\HH)$. > > - **Two-Sided Ideal** means: If $T\in \mathcal{R}(\HH)$ and $S,R\in \BB(\HH)$, then $STR\in \mathcal{R}(\HH)$. > - **Closure under Addition:** If $T, S\in \mathcal{R}(\HH)$, then $T + S\in \mathcal{R}(\HH)$. > - **Closure under Multiplication:** If $T\in \mathcal{R}(\HH)$ and $S\in \BB(\HH)$, then $ST, TS\in \mathcal{R}(\HH)$. Although $\mathcal{R}(\HH)$ is a subspace, it isn't closed. > [!example] Finite Rank Operators are not Closed Consider the sequence of operators on $\mathcal{R}(\HH)$ called $ T_n a = \left\{ \frac{a_1}{1}, \frac{a_2}{2}, \frac{a_3}{3}, \dots \frac{a_n}{n}, 0, \dots \right\} $ The limit of this sequence is $ T a = \left\{ \frac{a_1}{1}, \frac{a_2}{2}, \frac{a_3}{3}, \dots \right\} $ The range contains all standard basis vectors, so this cannot be finite-dimensional. So it makes sense to ask what this closure is. ## Compact Operators > [!definition] Compact Operators An operator $T\in \BB(\HH, \HH)$ is said to be ==**compact**== if $\overline{T(B_0(1))}$ is compact. The set of compact operators is denoted $\mathcal{K}(\HH)$. >[!idea] >Equivalently, $T$ maps bounded sets to pre-compact sets. The key ideas about compact operators are the following two problems: > [!problem] Closure of Finite Rank Operators Show that $\mathcal{K}(\HH)$ is the closure of $\mathcal{R}(\HH)$. Hint for $\overline{\mathcal{R}(\HH)}\subset \mathcal{K}(\HH)$: similar to [[Arzela-Ascoli]]. > [!solution]- $\mathcal{K}(\HH)\subset \overline{\mathcal{R}(\HH)}$. Completely Trivial > If $T\in \mathcal{K}(\HH)$ then it maps $B_0(1)$ to a compact $K$. For each $k$, find a finite-dimensional $W\subset \mathcal{H}$ such that all $u\in K$ are at most $\frac1k$ away from $W$. Thus $T_k = P_W\circ T\in \mathcal{R}(\HH)$ such that $\norm{T - T_k}\leq \frac1k$. > [!solution]- $\overline{\mathcal{R}(\HH)}\subset \mathcal{K}(\HH)$. Standard Diagonalization Argument > Easy. First off, $T\in \mathcal{R}(\HH)\subset \mathcal{K}(\HH)$ because the image is contained in $B_0(\norm{T})\cap \im T$ which is compact. (This doesn't work in general because $B_0(1)$ might not be compact!) So we just that to show $\mathcal{K}(\HH)$ is closed. > > Pick some $T_1,\dots, T_n\in \mathcal{K}(\HH)$ which converges to $T\in \mathcal{B}(\HH,\HH)$. We want to show that for all $\left(v^0_i\right)_{i\geq 1} = (v_i)_{i\geq 1}$ with $\norm{v_i}\leq 1$, $Tv_i$ admits a convergent subsequence. > > Alright, bet. WLOG $\norm{T - T_i}\leq \frac{1}{2^i}$. I'm going to let $\left(v^{j+1}_i\right)_{i\geq 1}$ be a subsequence of $\left(v^j_i\right)_{i\geq 1}$ such that $\left(T_{j+1}v_i^{j+1}\right)_{i\geq 1}$ converges to $u_{j+1}$ with $\norm{T_{j+1}v_i^{j+1} - u_{j+1}}\leq \frac{1}{2^j}$. Then, $\left(v^i_i\right)_{i\geq 1}$ works. > [!problem] Star-Closed Two-Sided Ideal Show that $\mathcal{K}(\HH)$ is a star-closed two-sided ideal in $\BB(\HH)$.