$\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}$ Alright, things are getting interesting now. > [!example] Kernel Let's look at $C_\infty([0,1])$, the continuous bounded functions on $[0,1]$ (bounded is redundant). Let $K:[0,1]^2 \to \RR$ be a continuous function. Then, we can define an operator $ Tf(x) = \int_0^1 K(x,y)f(y) \der y $ $K$ is called the ==**kernel**== of $T$. This is linear! We are particularly interested in *continuous* linear operators, defined in the usual manner. By linearity and things, this turns out to be really strong. > [!definition] Operators The ==**operator norm**== on $T:V\to W$ is defined $ \norm{T} = \sup_{v\in V : \norm{v} = 1} \norm{Tv}_W = \sup_{v\in V} \frac{\norm{Tv}_W}{\norm{v}_V} $ An operator is ==**bounded**== if this norm is not infinite. The space of ==**bounded**== linear operators from $V$ to $W$ is denoted $\BB(V,W)$. >[!idea] You might notice that this is analogous to the $\infty$-norm. The analogous $p$-norms are called the [[Schatten Norm]]. > [!problem] A linear operator $T: V\to W$ is continuous iff there exists $C > 0$ such that $\norm{Tv}_W\leq C\norm{v}_V$ for all $v\in V$, i.e. it is bounded. > [!solution]- Suppose $T$ is continuous. Then, the ball $B = \{v : \norm{v} < 1\}$ is by definition open. Thus the preimage $T^{-1}(B)$ is open, and contains $0$. Thus there exists some $\delta > 0$ such that $B_\delta(0) \subseteq T^{-1}(B)$. Thus we can pick $C$ = $1/\delta$, because one can scale any vector so its norm is $1 / \delta - \eps$, and then $Tv$ is at most $1$. > > [!idea] > On the other hand, taking $\delta$ to be the supremum of possible $\delta$, this $C = 1/\delta$ is maximal. > > Suppose now that $T$ is bounded. Over metric spaces, continuity is equivalent to sequential continuity. Thus, pick some sequence $\{v_n\}\to v$ in $V$; we observe that $ \norm{Tv_n - Tv}_W = \norm{T(v_n - v)}_W \leq C\norm{v_n - v}_V $ The RHS vanishes, thus by squeeze theorem the LHS vanishes in the limit, as desired. For example, observe that the above $T$ was bounded by $C = \sup_{x,y\in [0,1]} \abs{K(x,y)}$, and thus automatically continuous. > [!problem] Show that $\BB(V,W)$ is a normed space under the operator norm. > [!solution]- First, we should show that $\BB(V,W)$ is a vector space. This is clear from triangle inequality on $W$, etc. Next, we want the operator norm to be a norm. > > > > - The norm is definite because if $\norm{Tv}$ is identically $0$ then $T$ is identically $0$. > > - Homogeneity follows from $ \norm{\lambda T} = \sup_{v\in V: \norm{v} = 1} \norm{\lambda Tv} = \abs{\lambda} \sup_{v\in V: \norm{v} = 1} \norm{Tv} = \abs{\lambda} \norm{T} $ > > - Triangle inequality follows from $ \norm{T + S} = \sup_{v\in V: \norm{v} = 1} \norm{(T+S)v} \leq \sup_{v\in V: \norm{v} = 1} \norm{Tv} + \norm{Sv} \leq \norm{T} + \norm{S} $ > > > > [!idea] > Observe that all properties followed from the norm on $W$, not $V$. We expect the topological properties of $\BB(T,W)$ to be heavily dependent on $W$. > > Now, the main thing to update your intuition on is that **we only care about bounded (i.e. continuous) linear maps.** This causes a lot of vanilla linear algebra to fail. > [!example] Injective $\centernot\implies$ Invertible In the space $\ell^2$ of square summable sequences, the operator $ Ta = \left\{ \frac{a_1}{1}, \frac{a_2}{2}, \frac{a_3}{3}, \dots \right\} $ is continuous and injective, but not surjective. For instance, it cannot hit the sequence $b = (1, 1/2, 1/3, \dots)\in \ell^2$. Thus $T$ is not invertible. This shows injectivity does not imply invertibility. >[!example] Invertible $\centernot\implies$ what you think it means > If we work over $\ell^\infty$, then the above function is continuous, injective, and surjective. It is a logical fact that this implies the function has an inverse; in our case it is $ T^{-1}a = \left\{ a_1, 2a_2, 3a_3, \dots \right\} $. However this is clearly not bounded. So the definition of an invertible operator is that it must be a homeomorphism as well; its inverse must also be continuous. We do have one saving grace; invertibility works as expected over [[banach|Banach spaces]]. See [[zabreikos]]! > [!problem] An operator $T\in \BB(V,W)$ is ==**isometric**== if it preserves the norm. Show that isometric operators must be injective. > [!solution]- This is trivial.