---- > [!definition] Definition. ([[compact operator]]) > A **compact operator** between [[norm|normed spaces]] $X$ and $Y$ is a [[linear map|linear operator]] $T:X\to Y$ that sends [[bounded set|bounded subsets of]] $X$ to [[precompact]] subsets of $Y$. > > The space of compact operators $X\to Y$ is denoted $K(X,Y)$. ^definition > [!NOTE] Slogan. > "Compact operators are the inverses of differential operators". To get one perspective on this, see e.g. [[Fredholm alternative for elliptic PDE]]. > [!equivalence] > > Let $T:X \to Y$ be a [[linear map|linear operator]]. TFAE: > 1. $T$ is a compact operator; > 2. $T(\mathbb{B})$ ([[closed set|closed]] unit ball) is [[precompact]] in $Y$; > 3. For any [[bounded set|bounded sequence]] $(x_{n}) \subset X$, $(Tx_{n})$ has a [[sequence|convergent]] [[subsequence]] > [!basicnonexample] > The [[identity map]] $\id:X\to X$ is [[compact]] iff $\text{dim }X<\infty$ ([[Riesz's Lemma|cf.]]) ^nonexample > [!basicexample] > [[operator norm|finite-rank operators]] are compact; this follows e.g. from the fact that [[precompact|precompact iff bounded in finite dimensions]]. ^basic-example Suppose not. Then there exists $y_{0} \in r\mathbb{B}_{Y}$ such that $y_{0} \notin \overline{T\mathbb{B}_{X}}$, and [[Hahn-Banach Separation theorem|HBS]] produces a linear functional $f \in Y^{*}$ satisfying $f(y_{0}) > \sup_{y \in \overline{T\mathbb{B}_X}} f(y).$ Now $|f(y_{0})|> \sup_{y \in \overline{T\mathbb{B}_{X}}} f(y) \geq \sup_{x \in \mathbb{B}_{X}} \underbrace{ (f \circ T)(x) }_{ = (T^{*}f)(x) } \geq r \|f\|.$ But since $y_{0} \in r\mathbb{B}_{Y}$, also $|f(y_{0})| \leq \|f\| \|y\| \leq r \|f\|$... contradiction. > [!basicproperties] > - $T$ is certainly [[operator norm|bounded]]. In fact, $K(X,Y) \leq B(X,Y)$, and if $Y$ is [[Banach space|Banach]] then so is $K(X,Y)$. > - *([[Ideal]] property)* Compositions in $\mathsf{Norm_{T}}$ (so, [[continuous]]) of the form $W \xrightarrow{R} X \xrightarrow{ T} Y \xrightarrow{S} Z$ > are compact whenever $T$ is. > - [[Schauder's Theorem]] investigates how compactness behaves under [[dual map|dualization]], using [[Arzelà–Ascoli theorem|Arzelà–Ascoli]]. ^properties $K(X,Y) \leq B(X,Y)$ is by looking at [[subsequence|subsequences]]. We want to show that if $Y$ is [[Banach space|Banach]], then so is $K(X,Y)$. Recalling that $B(X,Y)$ is [[Banach space|Banach]] if $Y$ is, it'll suffice to show that $K(X,Y)$ is [[closed set|closed]] in $B(X,Y)$ when $Y$ is [[Banach space|Banach]]. We'll show it's [[the sequence lemma|sequentially closed]]. So suppose $Y$ is [[Banach space|Banach]]. Let $(T_{k}) \to T$ in $K(X,Y)$. Want to show $T \in K(X,Y)$. So let $(x_{n})$ be a [[bounded set|bounded]] [[sequence]] in $X$; have to show $(Tx_{n}) \subset Y$ has a [[sequence|convergent]] [[subsequence]]. Since $T_{k} \in K(X,Y)$, we know $T_{k}(x_{n})$ has a convergent subsequence. Argue diagonally: - For $T_{1}$, choose subsequence $(x_{n}^{(1)})$ so that $(T_{1}x_{n}^{(1)})$ converges - For $T_{2}$, choose further subsequences $(x_{n}^{(2)})$ so that $(T_{2}x_{n}^{(2)})$ converges *and* $(T_{1} x_{n}^{(1)})$ converges - And so on. The claim is that the diagonal subsequence $(x_{n}^{(n)})$ of $(x_{n})$ has $(Tx_{n}^{(n)})$ convergent. By [[complete|completeness]], it suffices to show [[Cauchy sequence|Cauchy]]. Have $\begin{align} \|Tx_{n}^{(n)}-Tx_{m}^{(m)}\| & \leq \|Tx_{n}^{(n)}- T_{k}x_{n}^{(n)}\| + \|T_{k}x_{n}^{(n)}-T_{k}x_{m}^{(m)}\| + \|T_{k}x_{m}^{(m)} - Tx_{m}^{(m)}\| \end{align}$ and all three terms can ultimately be made small: the outer terms because $T_{k} \to T$, and the inner term by the diagonal construction. *Proof of ideal property.* Easy with sequences, $(w_{n}) \subset W$ bounded sequence is sent to a bounded sequence $(Rw_{n} ) \subset X$ since $R$ is a [[operator norm|bounded linear operator]], then $(TRw_{n}) \subset Y$ converges along a [[subsequence]] since $T$ is [[compact operator|compact]], then $(STRw_{n}) \subset Z$ converges also along a subsequence (the same indices) since $S$ is [[continuous]]/[[operator norm|bounded]]. > [!basicexample] > - Any [[operator norm|finite-rank operator]] $T$ is [[compact]]. Indeed, [[precompact]] $\iff$ [[bounded set|bounded]] in finite dimensions and $T \mathbb{B}$ is [[bounded set|bounded]] (since $T$ is [[operator norm|bounded]] which recall is part of the finite-rank definition). > - If $Y$ is [[Banach space|Banach]], then $K(X,Y)$ is closed and so $T_{n} \to T$ in $B(X,Y)$ with $T_{n}$ [[operator norm|finite-rank]] implies $T \in K(X,Y)$. The converse can fail, but often holds. For example, it holds for $\ell^{1 \leq p < \infty}$ [[the dual of the space of null sequences naturally identifies with l1|and]] $c_{0}$. (This is TODO, in onenote/good exercise) ^basic-example Defining $T_{n}:=P_{n}T$, we have $\begin{align} \|T_{n}-T\| &= \|(I-P_{n})T\| \\ &= \sup_{x \in \mathbb{B}_{X}} \|(I-P_{n})Tx\| \\ &= \sup_{y \in T\mathbb{B}} \|(I-P_{n})y\| \\ &\leq \sup_{y \in \overline{T\mathbb{B}}} \|(I-P_{n})y\| \end{align}$ and this goes to zero by the characterization in [[totally bounded]]. > [!proof] > $(1) \iff (3)$ is immediate from the [[metrizable]]-space characterization of [[precompact|precompactness]]: suppose $T$ is compact operator. Any bounded sequence $(x_{n})$ lives in $r\mathbb{B}$ for some $r>0$, so $T(r\mathbb{B})$ is [[precompact]]; by the [[metrizable]]-space characterization of [[precompact|precompactness]] it has a [[sequence|convergent]] [[subsequence]]. Conversely suppose $(3)$. $B$ bounded, WTS $T(B)$ is [[precompact]], enough to show every [[sequence]] $(Tx_{n})$ in $T(B)$ has a convergent subsequence. Such a sequence is by definition the image of a sequence in $B$, which is bounded because $B$ is bounded. > > $(1) \implies (2)$ is obvious. For $(2) \implies (3)$, note that if $T(\mathbb{B})$ is precompact then so is $rT(\mathbb{B})$, hence so is $T(r\mathbb{B})$, for all $r>0$. Thus if $(x_{n})$ is a bounded sequence in $X$, we can find some $r\mathbb{B}$ containing $(x_{n})$, whose image is precompact in $Y$, and so $Tx_{n}$ has a convergent subsequence. > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```