\newcommand{\cat}[1]{\mathsf{#1}}

\newcommand{\Sets}[0]{{\mathsf{Set}}}
\newcommand{\Set}[0]{{\mathsf{Set}}}
\newcommand{\sets}[0]{{\mathsf{Set}}}
\newcommand{\set}{{\mathsf{Set} }}
\newcommand{\Poset}[0]{\mathsf{Poset}}
\newcommand{\GSets}[0]{{G\dash\mathsf{Set}}}
\newcommand{\Groups}[0]{{\mathsf{Group}}}
\newcommand{\Grp}[0]{{\mathsf{Grp}}}

% Modifiers
\newcommand{\der}[0]{{\mathsf{d}}}
\newcommand{\dg}[0]{{\mathsf{dg}}}
\newcommand{\comm}[0]{{\mathsf{C}}}
\newcommand{\pre}[0]{{\mathsf{pre}}}
\newcommand{\fn}[0]{{\mathsf{fn}}}
\newcommand{\smooth}[0]{{\mathsf{sm}}}
\newcommand{\Aff}[0]{{\mathsf{Aff}}}
\newcommand{\Ab}[0]{{\mathsf{Ab}}}
\newcommand{\Add}[0]{{\mathsf{Add}}}
\newcommand{\Assoc}[0]{\mathsf{Assoc}}
\newcommand{\Ch}[0]{\mathsf{Ch}}
\newcommand{\Coh}[0]{{\mathsf{Coh}}}
\newcommand{\Comm}[0]{\mathsf{Comm}}
\newcommand{\Cor}[0]{\mathsf{Cor}}
\newcommand{\Corr}[0]{\mathsf{Cor}}
\newcommand{\Fin}[0]{{\mathsf{Fin}}}
\newcommand{\Free}[0]{\mathsf{Free}}
\newcommand{\Tors}[0]{\mathsf{Tors}}
\newcommand{\Perf}[0]{\mathsf{Perf}}
\newcommand{\Unital}[0]{\mathsf{Unital}}
\newcommand{\eff}[0]{\mathsf{eff}}
\newcommand{\derivedcat}[1]{\mathbf{D} {#1} }
\newcommand{\bderivedcat}[1]{\mathbf{D}^b {#1} }
\newcommand{\Cx}[0]{\mathsf{Ch}}
\newcommand{\Stable}[0]{\mathsf{Stab}}
\newcommand{\ChainCx}[1]{\mathsf{Ch}\qty{ #1 } }
\newcommand{\Vect}[0]{{ \mathsf{Vect} }}
\newcommand{\kvect}[0]{{ \mathsf{Vect}\slice{k} }}
\newcommand{\loc}[0]{{\mathsf{loc}}}
\newcommand{\locfree}[0]{{\mathsf{locfree}}}
\newcommand{\Bun}{{\mathsf{Bun}}}
\newcommand{\bung}{{\mathsf{Bun}_G}}

% Rings
\newcommand{\Local}[0]{\mathsf{Local}}
\newcommand{\Fieldsover}[1]{{ \mathsf{Fields}_{#1} }}
\newcommand{\Field}[0]{\mathsf{Field}}
\newcommand{\Number}[0]{\mathsf{Number}}
\newcommand{\Numberfield}[0]{\Field\slice{\QQ}}
\newcommand{\NF}[0]{\Numberfield}
\newcommand{\Art}[0]{\mathsf{Art}}
\newcommand{\Global}[0]{\mathsf{Global}}
\newcommand{\Ring}[0]{\mathsf{Ring}}
\newcommand{\Mon}[0]{\mathsf{Mon}}
\newcommand{\CMon}[0]{\mathsf{CMon}}
\newcommand{\CRing}[0]{\mathsf{CRing}}
\newcommand{\DedekindDomain}[0]{\mathsf{DedekindDom}}
\newcommand{\IntDomain}[0]{\mathsf{IntDom}}
\newcommand{\Domain}[0]{\mathsf{Domain}}
\newcommand{\DVR}[0]{\mathsf{DVR}}
\newcommand{\Dedekind}[0]{\mathsf{Dedekind}}

% Modules
\newcommand{\modr}[0]{{\mathsf{Mod}\dash\mathsf{R}}}
\newcommand{\modsleft}[1]{\mathsf{#1}\dash\mathsf{Mod}}
\newcommand{\modsright}[1]{\mathsf{Mod}\dash\mathsf{#1}}
\newcommand{\mods}[1]{{\mathsf{#1}\dash\mathsf{Mod}}}
\newcommand{\stmods}[1]{{\mathsf{#1}\dash\mathsf{stMod}}}
\newcommand{\grmods}[1]{{\mathsf{#1}\dash\mathsf{grMod}}}
\newcommand{\comods}[1]{{\mathsf{#1}\dash\mathsf{coMod}}}
\newcommand{\algs}[1]{{{#1}\dash\mathsf{Alg}}}
\newcommand{\Quat}[0]{{\mathsf{Quat}}}
\newcommand{\torsors}[1]{{\mathsf{#1}\dash\mathsf{Torsors}}}
\newcommand{\torsorsright}[1]{\mathsf{Torsors}\dash\mathsf{#1}}
\newcommand{\torsorsleft}[1]{\mathsf{#1}\dash\mathsf{Torsors}}
\newcommand{\bimod}[2]{({#1}, {#2})\dash\mathsf{biMod}}
\newcommand{\bimods}[2]{({#1}, {#2})\dash\mathsf{biMod}}
\newcommand{\Mod}[0]{{\mathsf{Mod}}}
\newcommand{\Dmod}[0]{{ \mathcal{D}\dash\mathsf{Mod} }}

\newcommand{\zmod}[0]{{\mathbb{Z}\dash\mathsf{Mod}}}
\newcommand{\rmod}[0]{{\mathsf{R}\dash\mathsf{Mod}}}
\newcommand{\amod}[0]{{\mathsf{A}\dash\mathsf{Mod}}}
\newcommand{\kmod}[0]{{\mathsf{k}\dash\mathsf{Mod}}}
\newcommand{\gmod}[0]{{\mathsf{G}\dash\mathsf{Mod}}}
\newcommand{\grMod}[0]{{\mathsf{grMod}}}
\newcommand{\gr}[0]{{\mathsf{gr}\,}}
\newcommand{\mmod}[0]{{\dash\mathsf{Mod}}}
\newcommand{\Rep}[0]{{\mathsf{Rep}}}
\newcommand{\Irr}[0]{{\mathsf{Irr}}}
\newcommand{\Adm}[0]{{\mathsf{Adm}}}
\newcommand{\semisimp}[0]{{\mathsf{ss}}}

% Vector Spaces and Bundles
\newcommand{\VectBundle}[0]{{ \Bun\qty{\GL_r} }}
\newcommand{\VectBundlerk}[1]{{ \Bun\qty{\GL_{#1}} }}
\newcommand{\VectSp}[0]{{ \VectSp }}
\newcommand{\VectBun}[0]{{ \VectBundle }}
\newcommand{\VectBunrk}[1]{{ \VectBundlerk{#1} }}
\newcommand{\Bung}[0]{{ \Bun\qty{G} }}

% Algebras
\newcommand{\Hopf}[0]{\mathsf{Hopf}}
\newcommand{\alg}[0]{\mathsf{Alg}}
\newcommand{\Alg}[0]{{\mathsf{Alg}}}
\newcommand{\scalg}[0]{\mathsf{sCAlg}}
\newcommand{\cAlg}[0]{{\mathsf{cAlg}}}
\newcommand{\calg}[0]{\mathsf{CAlg}}
\newcommand{\liegmod}[0]{{\mathfrak{g}\dash\mathsf{Mod}}}
\newcommand{\liealg}[0]{{\mathsf{Lie}\dash\mathsf{Alg}}}
\newcommand{\Lie}[0]{\mathsf{Lie}}
\newcommand{\kalg}[0]{{\mathsf{Alg}_{/k} }}
\newcommand{\kAlg}[0]{{\mathsf{Alg}_{/k} }}
\newcommand{\kSch}[0]{{\mathsf{Sch}_{/k}}}
\newcommand{\rAlg}[0]{{\mathsf{Alg}_{/R}}}
\newcommand{\ralg}[0]{{\mathsf{Alg}_{/R}}}
\newcommand{\zalg}[0]{{\mathsf{Alg}_{/\ZZ}}}
\newcommand{\CCalg}[0]{{\mathsf{Alg}_{\mathbb{C}} }}
\newcommand{\dga}[0]{{\mathsf{dg\Alg} }}
\newcommand{\cdga}[0]{{ \mathsf{c}\dga }}
\newcommand{\dgla}[0]{{\dg\Lie\Alg }}
\newcommand{\Poly}[0]{{\mathsf{Poly} }}
\newcommand{\Hk}[0]{{\mathsf{Hk} }}

\newcommand{\Grpd}[0]{{\mathsf{Grpd}}}
\newcommand{\inftyGrpd}[0]{{ \underset{\infty}{ \Grpd } }}
\newcommand{\Algebroid}[0]{{\mathsf{Algd}}}

% Schemes and Sheaves
\newcommand{\Loc}[0]{\mathsf{Loc}}
\newcommand{\Locsys}[0]{\mathsf{LocSys}}
\newcommand{\Ringedspace}[0]{\mathsf{RingSp}}
\newcommand{\RingedSpace}[0]{\mathsf{RingSp}}
\newcommand{\LRS}[0]{\Loc\RingedSpace}
\newcommand{\IndCoh}[0]{{\mathsf{IndCoh}}}
\newcommand{\Ind}[0]{{\mathsf{Ind}}}
\newcommand{\Pro}[0]{{\mathsf{Pro}}}
\newcommand{\DCoh}[0]{{\mathsf{DCoh}}}
\newcommand{\QCoh}[0]{{\mathsf{QCoh}}}
\newcommand{\Cov}[0]{{\mathsf{Cov}}}
\newcommand{\sch}[0]{{\mathsf{Sch}}}
\newcommand{\presh}[0]{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}
\newcommand{\prest}[0]{ {\underset{ \mathsf{pre} } {\mathsf{St} } } }
\newcommand{\Descent}[0]{{\mathsf{Descent}}}
\newcommand{\Desc}[0]{{\mathsf{Desc}}}
\newcommand{\FFlat}[0]{{\mathsf{FFlat}}}
\newcommand{\Perv}[0]{\mathsf{Perv}}
\newcommand{\smsch}[0]{{ \smooth\Sch }}
\newcommand{\Sch}[0]{{\mathsf{Sch}}}
\newcommand{\Schf}[0]{{\mathsf{Schf}}}
\newcommand{\Sh}[0]{{\mathsf{Sh}}}
\newcommand{\St}[0]{{\mathsf{St}}}
\newcommand{\Stacks}[0]{{\mathsf{St}}}
\newcommand{\Vark}[0]{{\mathsf{Var}_{/k} }}
\newcommand{\Var}[0]{{\mathsf{Var}}}
\newcommand{\Open}[0]{{\mathsf{Open}}}

% Homotopy
\newcommand{\CW}[0]{{\mathsf{CW}}}
\newcommand{\sset}[0]{{\mathsf{sSet}}}
\newcommand{\sSet}[0]{{\mathsf{sSet}}}
\newcommand{\ssets}[0]{\mathsf{sSet}}
\newcommand{\hoTop}[0]{{\mathsf{hoTop}}}
\newcommand{\hoType}[0]{{\mathsf{hoType}}}
\newcommand{\ho}[0]{{\mathsf{ho}}}
\newcommand{\SHC}[0]{{\mathsf{SHC}}}
\newcommand{\SH}[0]{{\mathsf{SH}}}
\newcommand{\Spaces}[0]{{\mathsf{Spaces}}}
\newcommand{\GSpaces}[1]{{G\dash\mathsf{Spaces}}}
\newcommand{\Spectra}[0]{{\mathsf{Sp}}}
\newcommand{\Sp}[0]{{\mathsf{Sp}}}
\newcommand{\Top}[0]{{\mathsf{Top}}}
\newcommand{\Bord}[0]{{\mathsf{Bord}}}
\newcommand{\TQFT}[0]{{\mathsf{TQFT}}}
\newcommand{\Kc}[0]{{\mathsf{K^c}}}
\newcommand{\triang}[0]{{\mathsf{triang}}}
\newcommand{\TTC}[0]{{\mathsf{TTC}}}
\newcommand{\dchrmod}{{\derivedcat{\Ch(\rmod)} }}

% Infty Cats
\newcommand{\Finset}[0]{{\mathsf{FinSet}}}
\newcommand{\Cat}[0]{\mathsf{Cat}}
\newcommand{\Fun}[0]{{\mathsf{Fun}}}
\newcommand{\Kan}[0]{{\mathsf{Kan}}}
\newcommand{\Monoid}[0]{\mathsf{Mon}}
\newcommand{\Arrow}[0]{\mathsf{Arrow}}
\newcommand{\quasiCat}[0]{{ \mathsf{quasiCat} } }
\newcommand{\inftycat}[0]{{ \underset{\infty}{ \Cat}  }}
\newcommand{\inftycatn}[1]{{ \underset{(\infty, {#1})}{ \Cat}  }}
\newcommand{\core}[0]{{ \mathsf{core} }}
\newcommand{\Indcat}[0]{ \mathsf{Ind} }

% New?
\newcommand{\Prism}[0]{\mathsf{Prism}}
\newcommand{\Solid}[0]{\mathsf{Solid}}
\newcommand{\WCart}[0]{\mathsf{WCart}}

% Motivic
\newcommand{\Torsor}[1]{{\mathsf{#1}\dash\mathsf{Torsor}}}
\newcommand{\Torsorleft}[1]{{\mathsf{#1}\dash\mathsf{Torsor}}}
\newcommand{\Torsorright}[1]{{\mathsf{Torsor}\dash\mathsf{#1} }}
\newcommand{\Quadform}[0]{{\mathsf{QuadForm}}}
\newcommand{\HI}[0]{{\mathsf{HI}}}
\newcommand{\DM}[0]{{\mathsf{DM}}}
\newcommand{\hoA}[0]{{\mathsf{ho}_*^{\scriptstyle \AA^1}}}
\newcommand\Tw[0]{\mathsf{Tw}}
\newcommand\SB[0]{\mathsf{SB}}
\newcommand\CSA[0]{\mathsf{CSA}}
\newcommand{\CSS}[0]{{ \mathsf{CSS} } }

% Unsorted
\newcommand{\FGL}[0]{\mathsf{FGL}}
\newcommand{\FI}[0]{{\mathsf{FI}}}
\newcommand{\CE}[0]{{\mathsf{CE}}}
\newcommand{\Fuk}[0]{{\mathsf{Fuk}}}
\newcommand{\Lag}[0]{{\mathsf{Lag}}}
\newcommand{\Mfd}[0]{{\mathsf{Mfd}}}
\newcommand{\Riem}[0]{\mathsf{Riem}}
\newcommand{\Wein}[0]{{\mathsf{Wein}}}
\newcommand{\gspaces}[1]{{#1}\dash{\mathsf{Spaces}}}
\newcommand{\deltaring}[0]{{\delta\dash\mathsf{Ring}}}
\newcommand{\terminal}[0]{{ \mathscr{1}_{\scriptscriptstyle \uparrow} }}
\newcommand{\initial}[0]{{ \mathscr \emptyset^{\scriptscriptstyle \downarrow} }}

% Universal guys

\newcommand{\coeq}[0]{\operatorname{coeq}}
\newcommand{\cocoeq}[0]{\operatorname{eq}}
\newcommand{\dgens}[1]{\gens{\gens{ #1 }}}
\newcommand{\ctz}[1]{\, {\converges{{#1} \to\infty}\longrightarrow 0} \, }
\newcommand{\conj}[1]{{\overline{{#1}}}}
\newcommand{\complex}[1]{{ {#1}_{\scriptscriptstyle \bullet}} }
\newcommand{\cocomplex}[1]{ { {#1}^{\scriptscriptstyle \bullet}} }
\newcommand{\bicomplex}[1]{{ {#1}_{\scriptscriptstyle \bullet, \bullet}} }
\newcommand{\cobicomplex}[1]{ { {#1}^{\scriptscriptstyle \bullet, \bullet}} }

\newcommand{\floor}[1]{{\left\lfloor #1 \right\rfloor}}
\newcommand{\ceiling}[1]{{\left\lceil #1 \right\rceil}}
\newcommand{\fourier}[1]{\widehat{#1}}
\newcommand{\embedsvia}[1]{\xhookrightarrow{#1}}
\newcommand{\openimmerse}[0]{\underset{\scriptscriptstyle O}{\hookrightarrow}}
\newcommand{\weakeq}[0]{\underset{\scriptscriptstyle W}{\rightarrow}}

\newcommand{\fromvia}[1]{\xleftarrow{#1}}
\newcommand{\generators}[1]{\left\langle{#1}\right\rangle}
\newcommand{\gens}[1]{\left\langle{#1}\right\rangle}
\newcommand{\globsec}[1]{{{\Gamma}\qty{#1} }}
\newcommand{\Globsec}[1]{{{\Gamma}\qty{#1} }}
\newcommand{\langL}[1]{ {}^{L}{#1} }

\newcommand{\equalsbecause}[1]{\overset{#1}{=}}
\newcommand{\congbecause}[1]{\overset{#1}{\cong}}
\newcommand{\congas}[1]{\underset{#1}{\cong}}
\newcommand{\isoas}[1]{\underset{#1}{\cong}}
\newcommand{\addbase}[1]{{ {}_{\pt} }}
\newcommand{\ideal}[1]{\mathcal{#1}}
\newcommand{\adjoin}[1]{ { \left[ \scriptstyle {#1} \right] } }
\newcommand{\polynomialring}[1]{ { \left[ {#1} \right] } }
\newcommand{\htyclass}[1]{ { \left[ {#1} \right] } }
\newcommand{\qtext}[1]{{\quad \operatorname{#1} \quad}}
\newcommand{\abs}[1]{{\left\lvert {#1} \right\rvert}}
\newcommand{\stack}[1]{\mathclap{\substack{ #1 }}} 

\newcommand{\powerseries}[1]{ { \left[ {#1} \right] } }
\newcommand{\functionfield}[1]{ { \left( {#1} \right) } }
\newcommand{\rff}[1]{ \functionfield{#1} }

\newcommand{\fps}[1]{{\left[\left[ #1 \right]\right]  }}
\newcommand{\formalseries}[1]{ \fps{#1} }
\newcommand{\formalpowerseries}[1]{ \fps{#1} }
\newcommand\fls[1]{{\left(\left( #1 \right)\right)  }}

\newcommand\lshriek[0]{{}_{!}}
\newcommand\pushf[0]{{}^{*}}

\newcommand{\nilrad}[1]{{\sqrt{0_{#1}} }}
\newcommand{\jacobsonrad}[1]{{J ({#1}) }}
\newcommand{\localize}[1]{ \left[ { \scriptstyle { {#1}\inv}  } \right]}
\newcommand{\primelocalize}[1]{ \left[ { \scriptstyle { { ({#1}^c) }\inv}  } \right]}
\newcommand{\plocalize}[1]{\primelocalize{#1}}
\newcommand{\sheafify}[1]{ \left( #1 \right)^{\scriptscriptstyle \mathrm{sh}} }
\newcommand{\complete}[1]{{ {}_{ \hat{#1} } }}
\newcommand{\takecompletion}[1]{{ \overbrace{#1}^{\widehat{\hspace{4em}}}  }}
\newcommand{\pcomplete}[0]{{ {}^{ \wedge }_{p} }}
\newcommand{\kv}[0]{{ k_{\hat{v}} }}
\newcommand{\Lv}[0]{{ L_{\hat{v}} }}

\newcommand{\twistleft}[2]{{ {}^{#1} #2 }}
\newcommand{\twistright}[2]{{ #2 {}^{#1} }}
\newcommand{\liesover}[1]{{ {}_{/ {#1}} }}
\newcommand{\liesabove}[1]{{ {}_{/ {#1}} }}
\newcommand{\slice}[1]{_{/ {#1}} }
\newcommand{\coslice}[1]{_{{#1/}} }
\newcommand{\quotright}[2]{ {}^{#1}\mkern-2mu/\mkern-2mu_{#2} }
\newcommand{\quotleft}[2]{ {}_{#2}\mkern-.5mu\backslash\mkern-2mu^{#1} }
\newcommand{\invert}[1]{{ \left[ { \scriptstyle \frac{1}{#1} } \right] }}
\newcommand{\symb}[2]{{ \qty{ #1 \over #2 } }}
\newcommand{\squares}[1]{{ {#1}_{\scriptscriptstyle \square} }} 
\newcommand{\shift}[2]{{ \Sigma^{\scriptstyle[#2]} #1 }}

\newcommand\cartpower[1]{{ {}^{ \scriptscriptstyle\times^{#1} }  }}
\newcommand\disjointpower[1]{{ {}^{ \scriptscriptstyle\coprod^{#1} }  }}
\newcommand\sumpower[1]{{ {}^{ \scriptscriptstyle\oplus^{#1} }  }}
\newcommand\prodpower[1]{{ {}^{ \scriptscriptstyle\times^{#1} }  }}
\newcommand\tensorpower[2]{{ {}^{ \scriptstyle\otimes_{#1}^{#2} }  }}
\newcommand\tensorpowerk[1]{{ {}^{ \scriptscriptstyle\otimes_{k}^{#1} }  }}
\newcommand\derivedtensorpower[3]{{ {}^{ \scriptstyle {}_{#1} {\otimes_{#2}^{#3}} }  }}
\newcommand\smashpower[1]{{ {}^{ \scriptscriptstyle\smashprod^{#1} }  }}
\newcommand\wedgepower[1]{{ {}^{ \scriptscriptstyle\smashprod^{#1} }  }}
\newcommand\fiberpower[2]{{ {}^{ \scriptscriptstyle\fiberprod{#1}^{#2} }  }}
\newcommand\powers[1]{{ {}^{\cdot #1} }}
\newcommand\skel[1]{{ {}^{ (#1) } }}
\newcommand\transp[1]{{ \, {}^{t}{ \left( #1 \right) } }}

\newcommand{\inner}[2]{{\left\langle {#1},~{#2} \right\rangle}}
\newcommand{\inp}[2]{{\left\langle {#1},~{#2} \right\rangle}}
\newcommand{\poisbrack}[2]{{\left\{ {#1},~{#2} \right\} }}
\newcommand\tmf{ \mathrm{tmf} }
\newcommand\taf{ \mathrm{taf} }
\newcommand\TAF{ \mathrm{TAF} }
\newcommand\TMF{ \mathrm{TMF} }
\newcommand\String{ \mathrm{String} }

\newcommand{\BO}[0]{{\B \Orth}}
\newcommand{\EO}[0]{{\mathsf{E} \Orth}}
\newcommand{\BSO}[0]{{\B\SO}}
\newcommand{\ESO}[0]{{\mathsf{E}\SO}}
\newcommand{\BG}[0]{{\B G}}
\newcommand{\EG}[0]{{\mathsf{E} G}}
\newcommand{\BP}[0]{{\operatorname{BP}}}
\newcommand{\BU}[0]{\B{\operatorname{U}}}

\newcommand{\MO}[0]{{\operatorname{MO}}}
\newcommand{\MSO}[0]{{\operatorname{MSO}}}
\newcommand{\MSpin}[0]{{\operatorname{MSpin}}}
\newcommand{\MSp}[0]{{\operatorname{MSpin}}}
\newcommand{\MString}[0]{{\operatorname{MString}}}
\newcommand{\MStr}[0]{{\operatorname{MString}}}
\newcommand{\MU}[0]{{\operatorname{MU}}}


\newcommand{\KO}[0]{{\operatorname{KO}}}
\newcommand{\KU}[0]{{\operatorname{KU}}}

\newcommand{\smashprod}[0]{\wedge}

\newcommand{\ku}[0]{{\operatorname{ku}}}
\newcommand{\hofib}[0]{{\operatorname{hofib}}}
\newcommand{\hocofib}[0]{{\operatorname{hocofib}}}
\DeclareMathOperator{\Suspendpinf}{{\Sigma_+^\infty}}


\newcommand{\Loop}[0]{{\Omega}}
\newcommand{\Loopinf}[0]{{\Omega}^\infty}
\newcommand{\Suspend}[0]{{\Sigma}}
\newcommand*\dif{\mathop{}\!\operatorname{d}}
\newcommand*{\horzbar}{\rule[.5ex]{2.5ex}{0.5pt}}
\newcommand*{\vertbar}{\rule[-1ex]{0.5pt}{2.5ex}}
\newcommand\Fix{ \mathrm{Fix} }
\newcommand\CS{ \mathrm{CS} }
\newcommand\FP{ \mathrm{FP} }
\newcommand\places[1]{ \mathrm{Pl}\qty{#1} }
\newcommand\Ell{ \mathrm{Ell} }
\newcommand\homog{ { \mathrm{homog} } }
\newcommand\Kahler[0]{\operatorname{Kähler}}
\newcommand\Prinbun{\mathrm{Bun}^{\mathrm{prin}}}
\newcommand\aug{\fboxsep=-\fboxrule\!\!\!\fbox{\strut}\!\!\!}
\newcommand\compact[0]{\operatorname{cpt}}
\newcommand\hyp[0]{{\operatorname{hyp}}}
\newcommand\jan{\operatorname{Jan}}
\newcommand\curl{\operatorname{curl}}
\newcommand\kbar{ { \bar{k} } }
\newcommand\ksep{ { k\sep } }
\newcommand\mypound{\scalebox{0.8}{\raisebox{0.4ex}{\#}}}
\newcommand\rref{\operatorname{RREF}}
\newcommand\RREF{\operatorname{RREF}}
\newcommand{\Tatesymbol}{\operatorname{TateSymb}}
\newcommand\tilt[0]{ {}^{ \flat } }
\newcommand\vecc[2]{\textcolor{#1}{\textbf{#2}}}
\newcommand{\Af}[0]{{\mathbb{A}}}
\newcommand{\Ag}[0]{{\mathcal{A}_g}}
\newcommand{\Mg}[0]{{\mathcal{M}_g}}
\newcommand{\Ahat}[0]{\hat{ \operatorname{A}}_g }
\newcommand{\Ann}[0]{\operatorname{Ann}}
\newcommand{\sinc}[0]{\operatorname{sinc}}
\newcommand{\Banach}[0]{\mathcal{B}}
\newcommand{\Arg}[0]{\operatorname{Arg}}
\newcommand{\BB}[0]{{\mathbb{B}}}
\newcommand{\Betti}[0]{{\operatorname{Betti}}}
\newcommand{\CC}[0]{{\mathbb{C}}}
\newcommand{\CF}[0]{\operatorname{CF}}
\newcommand{\CH}[0]{{\operatorname{CH}}}
\newcommand{\CP}[0]{{\mathbb{CP}}}
\newcommand{\CY}{{ \text{CY} }}
\newcommand{\Cl}[0]{{ \operatorname{Cl}} }
\newcommand{\Crit}[0]{\operatorname{Crit}}
\newcommand{\DD}[0]{{\mathbb{D}}}
\newcommand{\DSt}[0]{{ \operatorname{DSt}}}
\newcommand{\Def}{\operatorname{Def} }
\newcommand{\Diffeo}[0]{{\operatorname{Diffeo}}}
\newcommand{\Diff}[0]{\operatorname{Diff}}
\newcommand{\Disjoint}[0]{\displaystyle\coprod}
\newcommand{\resprod}[0]{\prod^{\res}}
\newcommand{\restensor}[0]{\bigotimes^{\res}}
\newcommand{\Disk}[0]{{\operatorname{Disk}}}
\newcommand{\Dist}[0]{\operatorname{Dist}}
\newcommand{\EE}[0]{{\mathbb{E}}}
\newcommand{\EKL}[0]{{\mathrm{EKL}}}
\newcommand{\QH}[0]{{\mathrm{QH}}}
\newcommand{\AMGM}[0]{{\mathrm{AMGM}}}
\newcommand{\resultant}[0]{{\mathrm{res}}}
\newcommand{\tame}[0]{{\mathrm{tame}}}
\newcommand{\primetop}[0]{{\scriptscriptstyle \mathrm{prime-to-}p}}
\newcommand{\VHS}[0]{{\mathrm{VHS} }}
\newcommand{\ZVHS}[0]{{ \ZZ\mathrm{VHS} }}
\newcommand{\CR}[0]{{\mathrm{CR}}}
\newcommand{\unram}[0]{{\scriptscriptstyle\mathrm{un}}}
\newcommand{\Emb}[0]{{\operatorname{Emb}}}
\newcommand{\minor}[0]{{\operatorname{minor}}}
\newcommand{\Et}{\text{Ét}}
\newcommand{\trace}{\operatorname{tr}}
\newcommand{\Trace}{\operatorname{Trace}}
\newcommand{\Kl}{\operatorname{Kl}}
\newcommand{\Rel}{\operatorname{Rel}}
\newcommand{\Norm}{\operatorname{Nm}}
\newcommand{\Extpower}[0]{\bigwedge\nolimits}
\newcommand{\Extalgebra}[0]{\bigwedge\nolimits}
\newcommand{\Extalg}[0]{\Extalgebra}
\newcommand{\Extcomplex}[0]{\cocomplex{ \Extalgebra} }
\newcommand{\Extprod}[0]{\bigwedge\nolimits}
\newcommand{\Ext}{\operatorname{Ext} }
\newcommand{\FFbar}[0]{{ \bar{ \mathbb{F}} }}
\newcommand{\FFpn}[0]{{\mathbb{F}_{p^n}}}
\newcommand{\FFp}[0]{{\mathbb{F}_p}}
\newcommand{\FF}[0]{{\mathbb{F}}}
\newcommand{\FS}{{ \text{FS} }}
\newcommand{\Fil}[0]{{\operatorname{Fil}}}
\newcommand{\Flat}[0]{{\operatorname{Flat}}}
\newcommand{\Fpbar}[0]{\bar{\mathbb{F}_p}}
\newcommand{\Fpn}[0]{{\mathbb{F}_{p^n} }}
\newcommand{\Fppf}[0]{\mathrm{\operatorname{Fppf}}}
\newcommand{\Fp}[0]{{\mathbb{F}_p}}
\newcommand{\Frac}[0]{\operatorname{Frac}}
\newcommand{\GF}[0]{{\mathbb{GF}}}
\newcommand{\GG}[0]{{\mathbb{G}}}
\newcommand{\GL}[0]{\operatorname{GL}}
\newcommand{\GW}[0]{{\operatorname{GW}}}
\newcommand{\Gal}[0]{{ \mathsf{Gal}} }
\newcommand{\bigo}[0]{{ \mathsf{O}} }
\newcommand{\Gl}[0]{\operatorname{GL}}
\newcommand{\Gr}[0]{{\operatorname{Gr}}}
\newcommand{\HC}[0]{{\operatorname{HC}}}
\newcommand{\HFK}[0]{\operatorname{HFK}}
\newcommand{\HF}[0]{\operatorname{HF}}
\newcommand{\HHom}{\mathscr{H}\kern-2pt\operatorname{om}}
\newcommand{\HH}[0]{{\mathbb{H}}}
\newcommand{\HP}[0]{{\operatorname{HP}}}
\newcommand{\HT}[0]{{\operatorname{HT}}}
\newcommand{\HZ}[0]{{H\mathbb{Z}}}
\newcommand{\Hilb}[0]{\operatorname{Hilb}}
\newcommand{\Homeo}[0]{{\operatorname{Homeo}}}
\newcommand{\Honda}[0]{\mathrm{\operatorname{Honda}}}
\newcommand{\Hsh}{{ \mathcal{H} }}
\newcommand{\Id}[0]{\operatorname{Id}}
\newcommand{\Intersect}[0]{\displaystyle\bigcap}
\newcommand{\JCF}[0]{\operatorname{JCF}}
\newcommand{\RCF}[0]{\operatorname{RCF}}
\newcommand{\Jac}[0]{\operatorname{Jac}}
\newcommand{\II}[0]{{\mathbb{I}}}
\newcommand{\KK}[0]{{\mathbb{K}}}
\newcommand{\KH}[0]{ \K^{\scriptscriptstyle \mathrm{H}} }
\newcommand{\KMW}[0]{ \K^{\scriptscriptstyle \mathrm{MW}} }
\newcommand{\KMimp}[0]{ \hat{\K}^{\scriptscriptstyle \mathrm{M}} }
\newcommand{\KM}[0]{ \K^{\scriptstyle\mathrm{M}} }
\newcommand{\Kah}[0]{{ \operatorname{Kähler} } }
\newcommand{\LC}[0]{{\mathrm{LC}}}
\newcommand{\LL}[0]{{\mathbb{L}}}
\newcommand{\Log}[0]{\operatorname{Log}}
\newcommand{\MCG}[0]{{\operatorname{MCG}}}
\newcommand{\MM}[0]{{\mathcal{M}}}
\newcommand{\mbar}[0]{\bar{\mathcal{M}}}
\newcommand{\MW}[0]{\operatorname{MW}}
\newcommand{\Mat}[0]{\operatorname{Mat}}
\newcommand{\NN}[0]{{\mathbb{N}}}
\newcommand{\NS}[0]{{\operatorname{NS}}}
\newcommand{\OO}[0]{{\mathcal{O}}}
\newcommand{\OP}[0]{{\mathbb{OP}}}
\newcommand{\OX}[0]{{\mathcal{O}_X}}
\newcommand{\Obs}{\operatorname{Obs} }
\newcommand{\obs}{\operatorname{obs} }
\newcommand{\Ob}[0]{{\operatorname{Ob}}}
\newcommand{\Op}[0]{{\operatorname{Op}}}
\newcommand{\Orb}[0]{{\mathrm{Orb}}}
\newcommand{\Conj}[0]{{\mathrm{Conj}}}
\newcommand{\Orth}[0]{{\operatorname{O}}}
\newcommand{\PD}[0]{\mathrm{PD}}
\newcommand{\PGL}[0]{\operatorname{PGL}}
\newcommand{\GU}[0]{\operatorname{GU}}
\newcommand{\PP}[0]{{\mathbb{P}}}
\newcommand{\PSL}[0]{{\operatorname{PSL}}}
\newcommand{\Pic}[0]{{\operatorname{Pic}}}
\newcommand{\Pin}[0]{{\operatorname{Pin}}}
\newcommand{\Places}[0]{{\operatorname{Places}}}
\newcommand{\Presh}[0]{\presh}
\newcommand{\QHB}[0]{\operatorname{QHB}}
\newcommand{\PHS}[0]{\operatorname{PHS}}
\newcommand{\QHS}[0]{\mathbb{Q}\kern-0.5pt\operatorname{HS}}
\newcommand{\QQpadic}[0]{{ \QQ_p }}
\newcommand{\ZZelladic}[0]{{ \ZZ_\ell }}
\newcommand{\QQ}[0]{{\mathbb{Q}}}
\newcommand{\QQbar}[0]{{ \bar{ \mathbb{Q} } }}
\newcommand{\Quot}[0]{\operatorname{Quot}}
\newcommand{\RP}[0]{{\mathbb{RP}}}
\newcommand{\RR}[0]{{\mathbb{R}}}
\newcommand{\Rat}[0]{\operatorname{Rat}}
\newcommand{\Reg}[0]{\operatorname{Reg}}
\newcommand{\Ric}[0]{\operatorname{Ric}}
\newcommand{\SF}[0]{\operatorname{SF}}
\newcommand{\SL}[0]{{\operatorname{SL}}}
\newcommand{\SNF}[0]{\mathrm{SNF}}
\newcommand{\SO}[0]{{\operatorname{SO}}}
\newcommand{\SP}[0]{{\operatorname{SP}}}
\newcommand{\SU}[0]{{\operatorname{SU}}}
\newcommand{\F}[0]{{\operatorname{F}}}
\newcommand{\Sgn}[0]{{ \Sigma_{g, n} }}
\newcommand{\Sm}[0]{{\operatorname{Sm}}}
\newcommand{\SpSp}[0]{{\mathbb{S}}}
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\newcommand{\Spf}[0]{\operatorname{Spf}}
\newcommand{\Spc}[0]{\operatorname{Spc}}
\newcommand{\spc}[0]{\operatorname{Spc}}
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\newcommand{\Spin}[0]{{\operatorname{Spin}}}
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\newcommand{\coinv}[0]{{\operatorname{coinv}}}
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\newcommand{\Tr}[0]{\operatorname{Tr}}
\newcommand{\Ug}[0]{{\mathcal{U}(\mathfrak{g}) }}
\newcommand{\Uh}[0]{{\mathcal{U}(\mathfrak{h}) }}
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\newcommand{\plocal}[0]{{ \scriptsize {}_{ \localize{p}  } }}
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\newcommand{\Ad}[0]{{ \operatorname{Ad} }}
\newcommand{\afp}[0]{A_{/\FF_p}}
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%\newcommand{\sech}[0]{\operatorname{sech}}
\newcommand{\csch}[0]{\operatorname{csch}}
\newcommand{\arccot}[0]{\operatorname{arccot}}
\newcommand{\arccsc}[0]{\operatorname{arccsc}}
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\newcommand{\dR}[0]{\mathrm{dR}}
\newcommand{\dV}{\,dV}
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\newcommand{\Mgn}[0]{{ \mathcal{M}_{g, n} }}
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\newcommand{\normm}[1]{{\left\lVert \left\lVert {#1} \right\rVert\right\rVert}}
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\newcommand{\op}[0]{^{\operatorname{op}}}
\newcommand{\ord}[0]{{\operatorname{Ord}}}
\newcommand{\order}[0]{{\operatorname{Ord}}}
\newcommand{\oriented}[0]{{ \operatorname{oriented} } }
\newcommand{\orr}[0]{{\operatorname{ or }}}
\newcommand{\padic}[0]{p\dash\text{adic}}
\newcommand{\pic}[0]{{\operatorname{Pic}}}
\newcommand{\pnorm}[2]{{\left\lVert {#1} \right\rVert}_{#2}}
\newcommand{\poly}[0]{\mathrm{poly}}
\newcommand{\prim}[0]{{\operatorname{prim}}}
\newcommand{\projectivelim}[0]{\varprojlim}
\newcommand{\pr}[0]{{\operatorname{pr}}}
\newcommand{\pt}[0]{{\operatorname{pt}}}
\newcommand{\qc}[0]{{\operatorname{qc}}}
\newcommand{\qst}[0]{{\quad \operatorname{such that} \quad}}
\newcommand{\rank}[0]{\operatorname{rank}}
\newcommand{\realpart}[1]{{\mathcal{Re}({#1})}}
\newcommand{\red}[0]{{ \text{red} }}
\newcommand{\reg}[0]{\mathrm{reg}}
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# Tuesday, January 11

:::{.remark}
This course: solving $Lf=g$ for $L$ a linear operator, in analogy to solving $Ax=b$ in matrices.
References:

- Hutson-Pym-Cloud, *Applications of Functional Analysis and Operator Theory*
- Reed-Simon, *Methods of Modern Mathematical Physics*
- Brezis, *Functional Analysis, Sobolev Spaces, and PDEs*

:::

:::{.remark}
The issue when passing to infinite-dimensional vector spaces: the topology matters.
E.g. the closure of the unit ball is closed and bounded and thus compact in finite dimensions, but this may no longer be true in $\RR^\infty$ or $\CC^\infty$.
Recall that a Banach space is a complete normed space, and is further a Hilbert space if the norm is induced by an inner product.
See the textbook for a review of vector spaces, metric spaces, norms, and inner products.
:::

:::{.example title="?"}
Our first example of infinite dimensional vector spaces: sequence spaces $\ell$ with elements $f \da (f_1, f_2, \cdots )$ with each $f_i\in \RR$.
:::

:::{.remark}
Linear subspaces are subspaces that contain zero, as opposed to affine subspaces.
An example is $C_0([0, L]; \RR) \leq C([0, L]; \RR)$, the subspace of bounded continuous functionals on $[0, L]$ which vanish at the endpoints.
For any subset $S \subseteq V$, write $[S]$ or $\spanof S$ for the linear span of $S$: all finite linear combinations of elements in $S$.
:::

:::{.example title="?"}
Let $V = C([-1, 1])$ and $x_1\neq x_2\in [-1, 1]$, and set $M_i \da \ts{f\in V \st f(x_i) = 0}$.
Then $M_i \leq V$ is a linear subspace, and in fact $V = M_1 + M_2$ but $V\neq M_1 \oplus M_2$ since the zero function is in both subspaces.
:::

:::{.remark}
Limits of finite operators are compact.
The classical example: set $(A_N)_{i, i} = {1\over i}$, so $A_N = \diag\qty{{1}, {1\over 2}, {1\over 3}, \cdots, {1\over N}}$.
Then $\spec A_N = \ts{1\over n}_{n\leq N}$, but $A\da \lim_N A_N$ is an operator with $0\in \bar{\spec(A)}$ as an accumulation point.
Exercise: what is $\ker A$? 
Is it nontrivial?
:::

:::{.definition title="Convexity"}
A subset $S \subseteq V$ is **convex** iff 
\[
tf + (1-t)g \in S \qquad \forall f, g\in S,\quad \forall t\in (0, 1]
.\]
Equivalently,
\[
{af+bg\over a+b}\in S \qquad \forall f, g\in S,\quad \forall a,b\geq 0
\]
where not both of $a$ and $b$ are zero.
The **convex hull** of $S$ is the smallest convex set containing $S$.
:::

:::{.remark}
Recall Holder's inequality:
\[
\norm{fg}_1 \leq \norm{f}_p \cdot \norm{g}_q
,\]
Schwarz's inequality
\[
\abs{\inner{f}{g}} \leq \norm{f} \norm{g} \qquad \norm{f} \da \sqrt{\inner{f}{f}}
,\]
and Minkowski's inequality
\[
\norm{f+g}_p \leq \norm{f}_p + \norm{g}_p
.\]

A nice proof of Cauchy-Schwarz:


![](figures/2022-01-11_15-43-17.png)

![](figures/2022-01-11_15-43-24.png)

:::

# Thursday, January 13

:::{.remark}
My notes:

- $K \subseteq \mch$ is **complete** iff $K^\perp = 0$.
- Bessel: for $f\in \mch$ write $f_n \da \inner{f}{e_n} e_n$, then $\norm{(f_n)}_{\ell^2(\CC)} \leq \norm{f}_{\mch}^2$.
- Best estimate: for any other sequence $(c_n) \in \ell^2(\CC), \norm{f - \sum c_n e_n} \geq \norm{f - \sum f_n e_n}$.
- For $\ts{e_n}$ orthonormal, $(c_n) \in \ell^2(\CC) \iff \sum c_n e_n$ converges.
  If the series converges, it can be rearranged.
- Differentiating through an integral:

![](figures/2022-01-13_14-58-19.png)

- Parseval, Plancherel, and Fourier inversion:

![](figures/2022-01-13_15-02-34.png)

:::

:::{.remark}
Last time: any norm yields a metric: $d(f, g) \da \norm{f-g}$.

- Open/closed balls: $B_r(f) \da \ts{x\st \norm{f-x} < r}$.
- Bounded subsets: contained in some ball of finite radius.
- $\diam S = \inf_{r, f} \diam B_r(f)$ is the diameter of the smallest ball containing $S$.
- $d(f, S) \da \inf_{x\in S} \norm{f-x}$.
- $V = \RR^n$ with $\norm{f}_2^2 \da \sum_{k\leq n} f_i^2$, $\bar B_0(1)$ is a metric space but not a vector space.
- For $L^2$, there are unique least squares projections, but uniqueness may fail for $L^1$.
  - Counterexample: take a line $M = \ts{\tv{\alpha, \alpha}}$ in $\RR^2$ of angle $\pi/4$ with respect to the $x\dash$axis and consider $f\da \tv[0, 1]$.
    Then for $g\da (\alpha, \alpha)$, $\norm{f-g} = \abs{1-\alpha} + \abs{\alpha} \geq \abs{1-\alpha + \alpha} = 1$, and the minimizer occurs for *any* $\alpha \in [0, 1]$.
  - Similar issues may happen for $L^\infty$ -- but $L^1, L^\infty$ have sharper tails than $L^2$, so this can be useful e.g. in image problems.
- If limits of sequences $(f_n)$ exist, i.e. $\norm{f_n - f_m}\to 0$, then the limiting function $f_n\to f$ is unique by the triangle inequality.
- Example from last time: $\diag\qty{ 1, {1\over 2}, \cdots, {1\over n}} \to A$ a compact self-adjoint operator with $\spec A = \ts{1\over n}_{n\geq 0}$.
  - What is $\ker A$? Note that $0\in \sigma(A)$, where $\sigma(A)$ is the set where $(A-I\lambda)\inv$ is not defined.
    It turns out $\ker A = \ts{0}$.
- Defining closures of subsets: for $S \subseteq V$, say $f\in \bar{S}$ iff there exists a sequence of not necessarily distinct points $f_n \in S$ with $f_n\to f$.
  - Say $S_1 \subseteq S_2 \subseteq V$ is closed in $S_2$ iff $S_1 = C \intersect S_2$ for some $C$ closed in $V$.
    The closure of $S_1$ in $S_2$ is $\cl_V(S_1) \intersect S_2$.
- A set that is neither open nor closed: $X \da [a, b] \intersect \QQ$, and $\bd X = [a,b] \contains X$ is actually larger.
- Recall the little $\ell^p$ norms: $\norm{(f_n)}_{p} \da \qty{ \sum \abs{f_n}^p }^{1\over p}$ and $\norm{(f_n)}_{\infty} \da \sup_n \abs{f_n}$.
:::

:::{.exercise title="?"}
- Prove Jensen's inequality for concave functions.
- Prove Young's inequality.
- Prove Holder's inequality.
  - Idea: consider $a = \hat{f_n} \da \abs{f_n}/\norm{f_n}^p, b = \hat{g_n} \da \abs{g_n}/\norm{g_n}_q$ and apply Young's after summing over $n$.
- Prove Minkowski's inequality.
  - Idea: use that $(p-1)q=p$ and apply the triangle inequality and then Holder to $\sum \abs{f_n + g_n}^p$.
    Also use that $q\inv = 1-p\inv$, and divide through this inequality at the end.
    Be sure to check the cases $\norm{f+g}_p = 0, \infty$.
:::


# Tuesday, January 18


:::{.remark}
Last time:

- $\norm{f}_{\ell^p} = \qty{\sum \abs{f_n} ^p}^{1\over p}$
- $\norm{f}_{\ell^ \infty} = \sup_{n} \abs{f_n}$.

Today:

- $\ell_p = \ts{f \da (f_n) \st \norm{f}_{\ell^p} < \infty }$.
- Example: set $f^k \da (0, 0, \cdots, 1, 1, \cdots)$ with zeros for the first $k-1$ entires and ones for all remaining entries.
  Then $f^k_i \convergesto{k\to\infty}0$ for each fixed component at index $i$.
  So $f^k \to (0)$ component-wise, but $\norm{f_k}_{\ell^\infty} = 1$ for every $k$, so this doesn't converge in $\ell_\infty$.

- Recall the $\eps\dash\delta$ and limit definitions of continuity.
- Recall the definition of uniform continuity.
- For $\Omega \subseteq \RR^n$, write $C(\Omega)$ for the $\RR\dash$vector space of continuous bounded functionals $f: \Omega\to \CC$ with the norm $\norm{f}_{L^\infty} = \sup_{x\in \Omega} \abs{f(x)}$.
- Define $C^k(\Omega, \CC^m)$ to be functions with $k$ continuous partial derivatives which are bounded, and set $C^\infty(\Omega, \CC^m) = \intersect _{k\geq 0} C^k(\Omega, \CC^m)$. 
  Define a norm $\norm{f}_{C^k} = \sum_{j\leq k} \sup_{x\in \Omega} \abs{f^{(j)} (x)}$.
- Take $g(x) = 2-x^2$ and consider $\BB_{1\over 2}(g)$ in $C[0, 1]$ with $\norm{\wait}_{L^\infty}$.
- Show that convergent implies Cauchy-convergent using the triangle inequality.
- Lipschitz with $\abs{c} < 1$ implies Cauchy.
- Lemma 1.4.2: $\norm{f_{n+k} - f_n} \leq q^n (1-q)^{-1} \norm{f_1 - f_0}$ for all $k\geq 0$.
  Use
  \[
  \norm{f_{n+k} - f_n} = \norm{\sum_{j=1}^k (f_{n+j} - f_{n+j-1} ) } \leq \norm{f_1 - f_0} \sum_{j=1}^k q^{n+j-1} \convergesto{n\to\infty}0
  .\]
- Counterexample, not all normed spaces are complete: take $V = C[-1, 1]$ with $\norm{f}_{L^1} \da \int_{-1}^1 \abs{f(x) }\dx$.
  Define a sequence of functions $(f_n)$:

  <!-- Xournal file: /home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Spring/FunctionalAnalysis/sections/figures/2022-01-18_15-07.xoj -->

  ![](figures/2022-01-18_15-09-38.png)

  Check that $\norm{f_{n+1} - f_n}_{L^1} \leq q \norm{f_n - f_{n-1}}_{L^1}$, and pointwise $f_n \to \chi_{[-1, 0]}$ which is discontinuous and not in $C[-1, 1]$.

- Banach spaces: complete normed vector spaces.

- Series:
  - Convergent: $f \da \lim_N \sum_{n\leq N} f_n \in V$.
  - Absolute convergence: $\sum \norm{f_n} < \infty$.
  - In a Banach space, absolutely convergent series can be rearranged.

- Theorem: A normed space is complete iff absolute convergence $\implies$ convergence.
  Proof:
  - Step 1: show that every Cauchy sequence has a convergent subsequence.
  - Set $a_n \da \sup_{m > n} \norm{f_n - f_m}$, then Cauchy implies $a_n\to 0$ in $\RR$.
  - Get a convergent subsequence $a_{n_j} \leq j^{-2}$.
  - Set $g_j \da f_{n_j} - f_{n_{j+1}}$, then $g\da \sum g_j$ absolutely converges, say to $g$.
  - Note $f_{n_i} - f_{n_{i+1}} = \sum_{j=1}^i g_j$, so the subsequence $(f_{n_j})$ is convergent.
  - Step 2: use this to show that the original sequence $(f_n)$ converges.
  - Set $f = \lim f_n$, then $\norm{f_n - f} \leq \norm{f_n - f_{n_i}} + \norm{f_{n_i} - f}$, using Cauchy for the first $\eps$ and the convergent subsequence for the second.

:::

---

:::{.remark title="Some random notes"}
Some theorems that hold in Hilbert spaces but not necessarily Banach spaces:

![](figures/2022-01-18_14-48-04.png)

![](figures/2022-01-18_14-48-13.png)

![](figures/2022-01-18_14-54-38.png)

Absolute continuity:

![](figures/2022-01-18_14-55-21.png)

![](figures/2022-01-18_14-56-04.png)

$L_p^\loc$:

![](figures/2022-01-18_14-57-54.png)

$\ker L = 0$ may not be sufficient to guarantee bijectivity in infinite dimensions:

![](figures/2022-01-18_15-27-21.png)

Boundedness:

![](figures/2022-01-18_15-28-55.png)

Bounded iff continuous:


![](figures/2022-01-18_15-29-25.png)

:::


# More Banach Spaces (Thursday, January 20)

:::{.remark}
Last time: complete iff absolutely convergent implies convergent.
Today: wrapping up some results on Banach and Hilbert spaces, skipping a review of $L^p$ spaces, and starting on operators next week.
:::

:::{.remark}
Note that $S \da (0, 1]$ is open and not complete, but $\cl_\RR(S) = [0, 1]$ is both closed and complete.
Generalizing:
:::

:::{.lemma title="?"}
A subset $S \subseteq B$ of a Banach space is complete iff $S$ is closed in $B$.
:::

:::{.proof title="?"}
$\impliedby$:
If $S$ is closed, a Cauchy sequence $(f_n)$ in $S$ converges to some $f\in B$.
Since $S$ is closed in $B$, in fact $f\in S$.

$\implies$:
Suppose that for $f\in \cl_B(S)$, there is a Cauchy sequence $f_n\to f$ with $f_n \in S$ and $f\in B$.
Since $S$ is complete, $f\in S$, so $\cl_B(S) \subseteq S$ making $S$ closed.
:::

:::{.theorem title="?"}
For $\Omega \subseteq \RR^n$, the space $X = (C(\Omega; \CC), \norm{\wait}_\infty)$ is a Banach space. 
:::

:::{.warnings}
This is *not* complete with respect to any other $L^p$ norm with $p<\infty$!
:::

:::{.proof title="of theorem"}
Use the lemma -- write $B$ for the space of all bounded (not necessarily continuous) functions on $\Omega$, which is clearly a normed vector space, so it suffice to show

- $X \subseteq B$ is closed,
- $B$ is a Banach space (i.e. complete).

**Step 1**:
we'll show $\cl_B(X) \subseteq X$.
Take $f$ to be a limit point of $X$, then for every $\eps>0$ there is a $g\in X$ with $\norm{f-g} < \eps$.
Apply the triangle inequality:
\[
\norm{f} \leq \norm{f-g+g}\leq \norm{f-g} + \norm{g} = \eps + C < \infty
,\]
so $f\in \cl_B(X) \subseteq B$ since it is bounded.
It remains to show $f$ is continuous.
Use that $\norm{f-g}\infty <\eps$ and continuity of $g$ to get $\abs{g(x) - g(x_0)} < \eps$ for $\abs{x-x_0}<\eps$.
Now
\[
\abs{f(x) - f(x_0)} 
&= \abs{f(x) - g(x) + g(x) -g(x_0) + g(x_0) - f(x_0)} \\
&\leq \abs{f(x) - g(x)} + \abs{g(x) - g(x_0)} + \abs{f(x_0) - g(x_0)} \\
&\leq 3\eps
.\]
So $X$ is closed in $B$.

**Step 2**:
Let $f_n$ be Cauchy in $B$, and note that we have a pointwise bound $\abs{f_n(x) - f_n(x_0)} \leq \norm{f_n - f_m} \to 0$.
So pointwise, $f_n(x)$ is a Cauchy sequence in $\CC$ which is complete, so $f_n(x) \to f(x)$ for some $f: \Omega\to \CC$.
We now want to show $f_n\to f$ in $X$.
Using that $f_n$ is Cauchy in $X$, produce an $N_0$ such that $n, m\geq N_0 \implies \norm{f_n - f_m}< \eps$.
Now
\[
\norm{f - f_n} 
&= \sup_{x\in \Omega} \abs{f(x) - f_n(x)} \\
&\leq \sup_{x\in \Omega} \sup_{m\geq N_0} \abs{f_m(x) - f_n(x)} \\
&= \sup_{m\geq N_0} \sup_{x\in \Omega} \abs{f_m(x) - f_n(x)} \\
&= \sup_{m\geq N_0} \norm{f_m - f_n} \\
&\leq \eps
.\]
Now use the reverse triangle inequality to show $f_n$ is bounded 
\[
\norm{f} - \norm{f_n} \leq \norm{f-f_n} < \eps \implies \norm{f} < \infty
.\]

Now by problem 1.13, every Cauchy sequence is bounded, so $f_n \to f\in B$. 
:::

:::{.remark}
Extending to vector-valued functions: for $\Omega \subseteq \RR^n$, take $\vector x = \tv{x_1, \cdots, x_n}$ and $F = \tv{f_1, \cdots, f_m}: \CC^n\to \CC^m$.
Then there is a Banach space
\[
X = C^(\Omega, \CC^m), \qquad \norm{f}_{C_1(\Omega)} \da \sum_{i\leq m} \sup_{x\in \Omega} \abs{f(x)} + \sum_{i\leq m, j\leq n} \sup_{x\in \Omega} \abs{\dd{f_i}{x_j}(x) }
.\]
Similarly define $L^p(\Omega)$, noting that $\norm{f}_{L^\infty(\Omega)}$ is the *essential supremum*.
:::

:::{.theorem title="?"}
For $p\in (1, \infty)$, the sequence space $X = (\ell^p, \norm{\wait}_{\ell^p})$ is a Banach space.
:::

:::{.definition title="Closed subspaces"}
A **closed subspace** of a Banach space is a linear subspace $M \leq B$ which is norm-closed in $B$.
:::

:::{.example title="?"}
\[
M \da \ker \divergence = \ts{f\in C(\Omega) \st \divergence f = 0} \leq C(\Omega)
\]
is closed, where $\divergence f$ is the divergence of a function $f$.

For any $S \subseteq B$, one can also take the corresponding closed subspace $\bar{[S]} \da \cl_B \spanof_\CC\ts{s\in S}$, i.e. all linear combinations of elements in $S$ and their limits.
This is called the **closed linear span** of $S$.
:::

:::{.exercise title="?"}
Let $B = ( C[a, b], \norm{\wait}_{L^\infty}$ and for $x_0 \in [a, b]$ define
\[
M \da \ts{f\in C[a, b] \st f(x_0) = 0}, \qquad
N \da \ts{f\in C[a, b] \st f(x_0) \leq c}
.\]
Show that these are closed subspaces with no nontrivial open subsets of $B$, since any $f\in M$ can be perturbed to be nonzero at $x_0$ with an arbitrarily small norm difference.
:::

:::{.remark}
Recall that for $S_1 \subseteq S_2 \subseteq B$, $S_1$ is **dense** in $S_2$ iff $\cl_{S_2}(S_1) = S_2$.
Recall Weierstrass' theorem: for $\Omega \subseteq \RR^n$ is closed and bounded and write $\OO \da \RR[x_1, \cdots, x_n]$ for the polynomials in $n$ variables.
Then $\OO \subseteq C(\Omega)$, and $\cl_{C(\Omega)} \OO = C(\Omega)$, i.e. $\OO$ is a dense subspace in the $L^\infty$ norms.
In fact, piecewise linear functions are dense.
:::

:::{.remark}
Norms are equivalent iff $c_1 \norm{f}_a \leq \norm{f}_b\leq c_2 \norm{f}_a$ for some constants $c_i$.
All norms on $\RR^n$ (resp. $\CC^n$) are equivalent.
:::

:::{.example title="?"}
For $a>0, f\in C[0, 1]$, define
\[
\norm{f} \da \sup_{x\in I} \abs{e^{-ax} f(x)}
,\]
which can be thought of as a weighting on the uniform norm which de-emphasizes the tails of functions near the endpoints.
This is equivalent to $\norm{\wait}_\infty$, since
\[
e^a \sup \abs{f} \leq \sup \abs{e^{-ax} f} \leq 1\cdot \sup \abs{f}
.\]
:::

:::{.remark}
Note that a basis for a norm can be used as a basis with respect to an equivalent norm in finite dimensions.
In infinite dimensions this may not hold -- e.g. for Fourier series, $\ts{e_k(x)}_{k\in \ZZ}$ is not a basis for $C[0, 2\pi]$ with the sup norm.
:::

:::{.definition title="Separable Banach spaces"}
An $B\in \Banach$ is **separable** iff $X$ contains a countable dense subset $S = \ts{f_k}_{k\geq 0}$ such that for each $f\in B$ and $\eps>0$, there is an $f_k\in S$ with $\norm{f-f_k} < \eps$.
:::

:::{.example title="?"}
Show that

- $\Omega \subseteq \RR^n$ a bounded subset, $C(\Omega), \norm{\wait}_{\sup}$ is separable.
- $\ell^p$ is separable for $p\in (1,\infty)$.
- $\ell^\infty$ is *not* separable.

:::

## Random Notes

:::{.remark}

![](figures/2022-01-20_14-48-53.png)

![](figures/2022-01-20_14-49-39.png)

:::


# Tuesday, January 25


:::{.remark}
- Inner products:
  - $\inp{f}{f} \geq 0$ and $\inp{f}{f} = 0 \iff f=0$
  - $\inp{f}{g+h} = \inp{f}{g} + \inp{f}{h}$
  - $\inp{f}{g} = \bar{\inp{g}{f}}$ 
  - $\inp{\alpha f}{g} = \alpha \inp{f}{g}$

- A pre-Hilbert space is an inner-product space.
- Example: $\inp f g \da \int_\Omega w(x) f(x) \bar{g(x)}\dx$ for $f,g\in C(\Omega)$, where $w$ is any weighting function.
- Example: $\inp f g = \sum f_i \bar{g_i}$ for $f,g\in \CC^n$.
- Inner products induce norms: $\norm{f} \da \sqrt{\inp f f}$
  - Orthogonality: write $f\perp g$ iff $\inp f g = 0$ and $S^\perp = \ts{f\in \mch \st f\perp s \, \forall s\in S}$.
- Definition of a Hilbert space: a pre-Hilbert space complete with respect to the norm induced by its inner product.
- Recall $C(\Omega)$ with $\inp f g \da \int_\Omega f\bar{g}$ is not complete, thus not a Hilbert space.
- Example: a common optimization problem, $\mathrm{argmin} \norm{x}$ such that $Ax=0$.
:::


:::{.theorem title="Cauchy-Schwarz"}
\[
\abs{\inp{f}{g}}\leq \norm{f} \norm{g}
.\]
:::

:::{.proof title="?"}
Use $\inp f g = \abs f \abs g \cos \theta_{fg}$ where $\abs{ \cos \theta } \leq 1$.
- Assume $g\neq 0$, then STS $\inp{f}{g\over \norm g} \leq \norm f$.
- Use\[
0 
&\leq \norm{f - \inp f g g}^2 \\
&= \inp{f - \inp f g g}{f - \inp f g g} \\
&= \inp f f - \inp f g \inp g f - \bar{\inp f g} \inp f g + \inp f g \bar{\inp f g } \\
&= \norm{f}^2 - \abs{ \inp f g}^2
.\]
:::

:::{.theorem title="Closest approximations"}
Let $f\in M\leq \mch$ be a closed (and thus complete) subspace of a Hilbert space $\mch$.
Then there is a unique element $g$ in $\mch$ closest to $M$ in the norm.
:::

:::{.proof title="?"}
Let $d\da \dist(f, M)$, choose a sequence $g_n \in M$ such that $\norm{f-g_n}\to d$, which is possible since $d = \inf_{g\in M}\norm{f-g}$.
Apply the parallelogram law to write
\[
\norm{g_n - g_m}^2 
&= \norm{(g_n - f) - (g_m - f) }^2 \\
&= 2\norm{g_n - f}^2 + 2\norm{g_m - f}^2 - 4\norm{ {1\over 2} (g_n + g_m) - f}^2 \\
&\leq 2\norm{g_n - f}^2 + 2\norm{g_m - f}^2 - 4d^2 \\
&\leq 2d^2 + 2d^2 - 4d^2 \\
&= 0
,\]
so the $g_n$ are Cauchy.
Here we've used that ${1\over 2}(g_n + g_m) = m \in M$ since $M$ is a subspace, and $\norm{m-f} \geq d$.
Since $M$ is complete, $g_n\to g\in M$ and moreover $\norm{f-g} = d$.
For uniqueness, if $\norm{f-g'} = d$ then
\[
\norm{ f - {1\over 2} (g+g')}^2 = d^2 - \norm{g-g'}^2 < d^2 \qquad \contradiction
.\]

:::

:::{.theorem title="Projection theorem"}
Let $M\leq \mch$ be a closed subspace of a Hilbert space.
Then $M^\perp \leq \mch$ is closed, and $\mch = M \oplus M^\perp$.
In the decomposition $f= g+h$, in fact $g\in M$ is the closest approximation to $f$ in $M$, making this decomposition unique.
:::

:::{.proof title="?"}
STS $\mch = M + M^\perp$ by the exercises.
If $f\in M$, take $f=g+h$ where $g=f$ and $h=0$, so suppose $f\not\in M$.
Let $g = \mathrm{argmin} \dist(f, M) \in M$, and we claim $f-g\in M^\perp$, so $\inp {f-g}{h} = 0$ for any $h\in M$.
For all $h\in M$ and $\alpha > 0$, we have $g + \alpha h\in M$, so
\[
\norm{f - g}^2 
&\leq \norm{f - (g+\alpha h)}^2 \\
&= \norm{f-g}^2 -2\Re\alpha \inp{h}{f-g} + \alpha^2 \norm{h}^2
,\]
so
\[
2\Re \alpha \inp h {f-g} \leq \alpha^2 \norm{h}^2 \implies 2\Re \inp h {f-g} \leq \alpha \norm{h}^2 \convergesto{\alpha\to 0} 0
,\]
so $\Re\inp{h}{f-g} = 0$.
Similarly $\Im\inp{h}{f-g} = 0$.
:::




:::{.exercise title="?"}
Show $S^\perp$ is closed for any $S\in \mch$, and in fact $S^\perp = \spanof_\CC(\cl_\mch(S))^\perp$, and if $f\in S \intersect S^\perp$ then $f=0$.
:::



:::{.exercise title="?"}
Prove the parallelogram law
\[
\norm{f-g}^2 + \norm{f+g}^2 = 2\norm{f}^2 + 2\norm{g}^2
.\]
:::


# Thursday, January 27

:::{.remark}
Notes:

- Bessel:
  \[
  \sum_n \abs{ \inp{f}{\phi_n} } \leq \norm{f}^2,\qquad \norm{f} \da \sqrt{\inp{f}{f}}
  .\]
  - Prove using the fact that
  \[
  0\leq \norm{f - \sum_{n\leq m} \inp{f}{\phi_n} \phi_n }^2 = \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2
  .\]

- Best fit: 
\[
\norm{f - \sum_{n\leq m} c_n \phi_n } \geq \norm{f - \sum_{n\leq m} \inp{f}{\phi_n} \phi_n }
,\]
so define projections $P_M(f) \da \sum \inp{f}{\phi_n} \phi_n$ for $\spanof\ts{\phi_n} = M$.
  - Prove using 
  \[
  \norm{f - \sum_{n\leq m} c_n \phi_n } 
  &= \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2 + \sum_{n\leq m} \abs{\inp{f}{\phi_n} - c_n }^2 \\
  &\geq \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2
  .\]
- Hilbert spaces are separable: have countable dense subsets.
  - $\ell^\infty(\ZZ)$ is not separable.
- For $\ts{\phi_n}$ orthonormal and $\ts{c_n}$ scalars, $\sum c_n \phi_n$ is convergent iff $\ts{a_n}\in \ell^2(\ZZ)$, so $\sum\abs{c_n}^2<\infty$.
  If it converges, it can be rearranged, and 
  \[
  \norm{\sum c_n \phi_n}^2 = \norm{\ts{c_n}}_{\ell^2(\ZZ)}^2 = \sum \abs{c_n}^2
  .\]
  - To prove, use that $\norm{ \sum_{i\leq n\leq j}c_n \phi_n}^2 = \sum_{i\leq n\leq j}\abs{c_n}^2$, so the sequence $\ts{S_m}_{m\geq 0}$ where $S_m \da \sum_{n\leq m} c_n\phi_n$ is Cauchy since $\sum\abs{c_n}^2$ converges.
  - If $g = \sum c_n \phi_n$ and $f = \sum c_{m_n} \phi_{m_n}$ is a rearrangement, if $\ts{c_n}\in \ell^2$ then $\norm{f}^2 = \norm{g}^2 = \sum \abs{c_n}^2$.
  Then $\norm{f-g}^2 = \norm{f}^2 + \norm{g}^2 - 2\Re\inp{f}{g}$, but $\inp{f}{g} = \sum \abs{c_n}^2$, so $\norm{f-g} = 0$.
- If $K = \ts{\phi_n} \subseteq \mch$ is a proper subset, so $\spanof\ts{\phi_n}\neq \mch$, write $P_K(f) = \sum \inp{f}{\phi_n}\phi_n$.
  Then $P_K(f) = 0$ if $f\in \cl(K)^\perp$, $P_K(f) = f$ if $f\in \cl(K)$, and if $f\not \in K$, since $\cl(K)\leq \mch$ is closed then there exists a $g\in \cl(K)$ where $\norm{f-g} = \dist(f, K)$.
  Write $f = h + (f-g) \in \cl(K) \oplus \cl(K)^\perp$, so $f = P_K f + (I-P_K)f$.
- Recall complete subspaces of Banach spaces are closed.
- Next theorem: every separable Hilbert space admits an orthonormal basis.
:::

:::{.theorem title="?"}
If $\mch$ is a separable Hilbert spaces and $K = \ts{\phi_n}$ is an orthonormal set, then TFAE

- $K$ is complete, i.e. $K^\perp = 0$ (taking the closure is not needed),
- $\cl{\spanof K} = \mch$,
- $K$ is an orthonormal *basis*,
- For all $f\in \mch$, $\norm{f} = \sum_{n\geq 0} \abs{\inp{f}{\phi_n}}^2$ (Parseval).
:::

:::{.remark}
Note $\inp{f}{\phi_n}$ is the $n$th Fourier coefficient $\hat{f}(\xi) = \sum \inp{f}{\phi_n}\phi_n(\xi)$, and Parseval says $\norm{f}^2 = \norm{\hat{f}}^2$.
:::

:::{.proof title="?"}
$1\implies 2$: 
Let $f\in \mch \sm \cl\spanof(K)$ and project, so $f= g+h$ with $g,h\neq 0$.
But $g\in \cl\spanof(K)$ and $h\in \cl\spanof(K)^\perp = 0$, forcing $K^\perp = \cl(K)^\perp \neq 0$. $\contradiction$

$2\implies 3$:
Follows directly from previous lemma that $f = P_Kf + (I-P_K)f$.

$3\implies 4$:
Write $f\in \mch$ as $f = \sum \inp{f}{\phi_n} \phi_n$ by sending $m\to \infty$ in the previous lemma.

$4\implies 1$:
Toward a contradiction, suppose $f\neq 0\in K^\perp$.
Then $\norm{f} \neq 0$ but $\inp{f}{\phi_n}=0$ for all $n$, contradicting Parseval. $\contradiction$
:::

# Tuesday, February 01


:::{.remark}
Notes:

- Assume $\mch$ is a **separable** Hilbert space: there exists a countable set of vectors $\ts{v_i}_{i\in \ZZ}$ which span a subspace that is dense in $\mch$, so $\cl(\spanof \ts{v_i}) = \mch$.
- Complete subspaces: $M\leq \mch$ with $M^\perp = 0$. 
- Show that for any $S \subseteq \mch$, $S^\perp$ is closed in $\mch$, $S^\perp = \qty{\cl\spanof S}^\perp$, and $f\in S \intersect S^\perp \implies f=0$.
- If $K = \ts{\phi_k}_{k\in \ZZ}$ is an orthonormal set in $\mch$, then TFAE
  - $K \leq \mch$ is a complete subspace, so $K^\perp = 0$, i.e. $\inp{f}{\phi_k} = 0$ for all $k$ implies $f=0$.
  - $\cl\spanof K = \mch$, so every $f\in \mch$ is the limit of a sequence of vectors from $\spanof K$.
  - $K$ is an orthonormal *basis*
  - Parseval: equality in Bessel, i.e. $\norm{f} = \sum \abs{ \inp{f}{\phi_k} }^2$
- Lemma: if $M, N\leq \mch$ with $\dim M < \dim N$, then $M^\perp \intersect N \neq 0$.
  - Without loss of generality assume $\dim N = n + 1$ where $n\da \dim M$, take a basis $\ts{\psi_k}_{k\leq n+1}$ for $N$.
  - Try to find $f\in N$ with $f\perp M$, i.e. coefficients $\ts{b_i}_{i\leq n}$ with $\sum b_i \psi_i \perp \phi_k$ for every $\phi_k$ basis elements of $M$.
  - This is a linear system of $n$ equations in $n+1$ unknowns, so it has a nontrivial solution.
- Theorem, orthonormal bases are stable: if $\ts{\phi_k}$ is an orthonormal basis and $\ts{\psi_k}$ is an orthonormal *system*, if $\sum \norm{\phi_k - \psi_k}^2 < \infty$ then $\ts{\psi_k}$ is a basis.
  - Assume note, then find a $\psi_0 \in \mch$ with $\norm{\psi_0} = 1$ and $\inp{\psi_0}{\psi_j} = 0$ for all $j$.
  - Choose $N\gg 1$ so that $\sum_{k\geq N} \norm{\psi_k - \phi_k} < 1$.
  - Use previous lemma to produce $w\in \spanof\ts{\psi_0, \psi_1,\cdots, \psi_N}$ with $w\neq 0$ and $w\perp \phi_j$ for all $j\leq N$.
  - Note $w\perp \spanof\ts{\psi_n}_{n > N}$.
  - Apply Parseval: 
  \[
  0 
  &< \norm{w}^2 \\
  &= \sum_{n\geq 1} \abs{\inp{w}{\phi_n}}^2 \\
  &= \sum_{n\geq N+1}\abs{\inp{w}{\phi_n}}^2 \\
  &= \sum_{n\geq N+1} \abs{\inp{w}{\phi_n - \psi_n}}^2 \\
  &\leq \norm{w}^2 \sum_{n\geq N+1}\norm{\phi_n - \psi_n}^2 \\
  &< \norm{w}^2 \cdot 1
  ,\]
  where we've used that $\inp{w}{\psi_n} = 0$ for $n\geq N$.
  $\contradiction$
- $\mch$ admits a countable orthonormal basis iff $\mch$ is separable.
  - $\implies$: clear, since the basis is countable, and every element is a limit of partial sums against the basis.
  - $\impliedby$: Gram-Schmidt.
    - $h_1 = \psi_1$ and $\phi_1 = h_1/\norm{h_1}$
    - $h_n = \psi_n - \sum_{1\leq k\leq n-1} \inp{\psi_k}{\phi_k} \phi_k$ and normalize.
- Exercise: a closed subspace of a separable Hilbert space is separable.
- Linearly isometric inner product spaces: $E\sim F$ iff there is a map $A: E\surjects F$ with 
  - $A(au + bv) = aAu + bAv$
  - $\norm{Au}_F = \norm{u}_E$
- Theorem: if $\mch_1, \mch_2$ are infinite dimensional separable Hilbert spaces, then $\mch_1 \sim \mch_2$. 
  Thus for any Hilbert space $\mch$ over $\CC$, $\mch \sim \ell^2(\CC)$.
  - Pick orthonormal bases $\ts{\phi_k} \subseteq \mch_1, \ts{\psi_k} \subseteq \mch_2$.
  - For $u\in \mch_1$, define $Au \da \sum \inp{u}{\phi_k}\psi_k$, which converges -- this will be the linear isometry, and satisfies condition (i).
  - Check $\norm{Au}_{\mch_2}^2 = \sum_{k\geq 1}\abs{\inp{u}{\phi_k}}^2 = \norm{u}_{\mch_2}$, which is condition (ii).
  - Check $A$ is surjective: for $y\in \mch_2$, write $y = \sum_{k} \inp{y}{\psi_k} \psi_k = Av$ for $v\da \sum_k \inp{y}{\psi_k}\phi_k \in \mch_1$.
- Non-separable spaces: look at *almost-periodic functions*.
  - E.g. $\sum_{k\leq n} c_k \exp(i\lambda_k t)$ for $\lambda_k \in \RR$.
:::


# Tuesday, February 08

:::{.remark}
Motivating question: when is an operator equation solvable?
Today: relation between boundedness and continuity for linear operators.
Nonlinear operators next week.

- A map of vector spaces $V\to W$ is a linear map defined on some domain $D(A) \subseteq V$, where $D(A)$ need not equal $V$.
  - Notation: $A(f) = Af = g$.
  - $Af \subseteq W$ is the image of $f$, and $R(A) \da \ts{Af\st f\in D(A)} \subseteq W$ is the range.
  Preimages of $S \subseteq W$: $A\inv(S) = \ts{f\st f\in D(A) \text{ and } Af\in S}$.
- We distinguish operators with different domains, e.g. $Af \da f'$ can be the formula for distinct operators $A_1, A_2$ where $D(A_1) = C^\infty[0, 1] \subseteq C^0[0, 1]$ or $D(A_2) = C^1[0, 1] \subseteq C^0[0, 1]$, so $A_1\neq A_2$.
  - I.e. $A_1 = A_2 \iff D(A_1) = D(A_2)$ and $A_1 f = A_2 f$ for all $f\in D(A_1)=D(A_2)$.
  - If $A_1 f = A_2 f$ with $D(A_1) \subseteq D(A_2)$, say $A_2$ is an extension of $A_1$.
  The extension is proper iff $D(A_1)\neq D(A_2)$.
- Example operator: the Laplace equation $\laplacian f= g$ where $\laplacian = \del_{xx} + \del_{yy}$.
  We can take domains $g\in C[0, 1], L^2[0, 1], H^2[0, 1] = \ts{f\in L^2(0, 1) \st \del_{xx} f, \del_{yy} f\in L^2(0, 1) }$.
  - Why domains matter: boundary conditions affect what eigenfunctions you get. Examples where $A_1\neq A_2$:
  - Dirichlet boundary conditions: $\laplacian f = g, \ro{f}{\bd \Omega} = 0$.
    The relevant solution spaces is $D(A_1) = \ts{\phi\in C^2[0,1]^2 \st \ro \phi {\bd \Omega} = 0}$ for $A_1 \phi \da \laplacian \phi$.
  - Neumann boundary conditions: $\laplacian f = g, \ro{ \dd{f}{\vector n}}{\bd \Omega} = 0$, i.e. there is no flux across the boundary.
    The relevant solution space is $D(A_2) = \ts{\psi \in C^2[0,1]^2\st \ro{ \dd{\psi}{\vector n} }{\bd\Omega} = 0 }$ for $A_2 \psi \da \laplacian \psi$.

- Injectivity: for $A: V\to W$, for every $g\in R(A)$ there is exactly one $f\in D(A)$ with $Af=g$.
  Equivalently for linear operators, $Af = 0 \implies f=0$.
- Surjectivity: $R(A) = W$.
- Example: $A\da x\mapsto \sin(x)$ regarded as a function $A:\RR\to \RR$ is neither surjective nor injective: $R(A) = [-1, 1] \subsetneq \RR$, and $\sin(\pi \ZZ) = 0$.
- Linearity: for $Lf = g$, $L$ is linear if $L(af + bg) = aLf + bLg$.

:::

:::{.exercise title="?"}
Show that the following are equivalent conditions for continuity of $A: V\to W$ at $f_0\in D(A)$:

- $\norm{Af - Af_0}_W < \eps$ for all $f\in D(A)$ with $\norm{f-f_0}_V < \delta$
- For every sequence $\ts{f_k} \to f_0$, $Af_k \to Af_0$.
- Preimages of open sets in $W$ are again open in $V$.

:::


# Tuesday, February 15

:::{.remark}
Last time:

- Continuous operators are bounded:
  - If $\norm{Lf_n} = 1$ and $\norm{f_n} \to 0$, check $\lim (Lf_n) = L(\lim f_n) = L0 = 0$.
  - Take norms to contract $\norm{Lf_n} = 1$.
:::

:::{.theorem title="3.4.4"}
If $L: B\to C$ with dense image (so $\cl_B(D(L)) = B$), if $L$ is continuous on $D(L)$ then it has a unique extension $\tilde L$ to all of $B$, so $D(\tilde L) = B$, with $\norm{L} = \norm{\tilde L}$.
:::

:::{.proof title="of theorem"}
In steps:

- Defining the extension:
  - For $f\in B$, pick $f_n \to f$ with $f_n \in D(L)$ using density.
  - Convergent implies Cauchy, so estimate:
  \[
  \norm{Lf_n - Lf_m} = \norm{L(f_n - f_m)} \leq \norm{L} \norm{f_n - f_m} \to 0
  .\]
  - Thus $Lf_n$ is Cauchy, by completeness $Lf_n \to g$ for some $g$.

- Preservation of norm:
  - Define the extension as $\tilde L f \da g$, by continuity it is independent of the sequence $\ts{f_k}\to f$.
  - Check that $\tilde L$ is a bounded operator:
  \[
  \norm{\tilde L f} \da \norm{g} = \norm{\lim L f_n} = \lim \norm{Lf_n} \leq \lim \norm{L} \norm{f_n} = \norm{L} \norm{f} \\
  \implies \norm{\tilde L} \leq \norm{L}
  .\]
  - Since $\norm{A}_{\op}$ is defined in terms of sups over test functions in $D(A)$ for any operator $A$ and here $D(\tilde L) \contains D(L)$ is a larger set, we have $\norm{\tilde L} \geq \norm{L}$ by definition, yielding $\norm{\tilde L} = \norm{L}$.

- Uniqueness of the extension:
  - Take $\tilde L_1, \tilde L_2$ extending $L$, then
  \[
  \tilde L_1 f = \lim \tilde L_1 f_n = \lim L f_n = \lim \tilde L_2 f_n = \tilde L_2 f
  .\]
  - Use linearity:
  \[
  \tilde L_1 f - \tilde L_2 f = (\tilde L_1 - \tilde L_2)f = 0 \implies \tilde L_1 - \tilde L_2 = 0
  .\]





:::

:::{.example title="?"}
Let $\mcl \in L(\CC^n, \CC^n)$ be defined in coordinates by $(\mcl f)_i \da \sum_{1\leq j\leq n} \alpha_{ij} f_j$ for $1\leq i\leq n$.
Take $\norm{\wait}_{\ell^\infty}$ and check
\[
\norm{Lf}_\infty 
&\da \sup_i \abs{\sum \alpha_{ij} f_j} \\
&\leq \qty{ \sup_i \sum \abs{\alpha_{ij} } } \sup_j \abs{f_j} \\
&\da m \norm{f}_{\ell^\infty}
.\]
So $\norm{L} \leq m$, where $m$ is the largest row sum.
Is there an $f$ for which equality holds?
In this case, we'd need
\[
\norm{Lf}_{\ell^\infty} \geq m \norm{f}_{\ell^\infty}
.\]
Identify the row $k$ so that $m = \sum_{1\leq j\leq n} \abs{\alpha_{kj}}$.
Set $f$ to be a unit vector with coefficients $(f)_j = \bar{\alpha_{kj}} / \abs{\alpha_{kj}}$.
Then
\[
\norm{Lf}_\infty
&= \sup_i \abs{ \sum_j \alpha_{ij} f_j } \\
&\geq \abs{\sum_j \alpha_{kj} f_j } \\
&= \abs{\sum_j \alpha_{kj} {\bar \alpha_{kj} / \abs{\alpha_{k j}} } } \\
&= \sum_j \abs{\alpha_{kj}} \\
&= m \norm{f}_{\ell^\infty}
.\]

So the answer is yes in this case.
Does this also work for $\norm{\wait}_{\ell^p}$ with $p\in (1, \infty)$?
Recall Holder's inequality:
\[
\abs{\sum \alpha_{ij} f_j } 
&\leq \qty{ \sum \abs{\alpha_{ij}}^q }^{1\over q} \qty{\sum \abs{f_j}^p }^{1\over p} \\
&= \qty{ \sum \abs{\alpha_{ij}}^q }^{1\over q} \norm{f}_{\ell^p}
.\]
Check that
\[
\norm{Lf}_{\ell^p}^p
&= \sum_i \abs{(Lf)_i }^p \\
&= \sum_i \abs{\sum_j \alpha_{ij} f_j }^p \\
&\leq \sum_i \qty{ \sum_j \abs{\alpha_{ij} } }^{p\over q} \norm{f}_{\ell^p}^p
,\]
where we've applied Holder in the last line.
Thus
\[
\norm{L} \leq \qty{ \sum_i \qty{ \sum_j \abs{ \alpha_{ij}} }^{p\over q} }^{1\over p}
.\]

:::{.exercise title="?"}
Is there an $f$ that attains this bound in the $\ell^p$ case?
:::

:::

:::{.remark}
For $\mcl \in L(\CC^\infty, \CC^\infty)$ defined by $(Lf)_i = \sum_{j\geq 11} \alpha_{ij} f_j$ for $j\geq 1$, one needs a notion of convergence of the coordinates $\alpha_{ij}$ in order for $\mcl$ to be bounded.
A sufficient condition is $m\da \sup_i \sum_{j\geq 1} \abs{\alpha_{ij}} < \infty$.
:::

:::{.definition title="?"}
Some notation:

\[
\normm{\alpha}_1 &\da \sup_j \sum_i \abs{ \alpha_{ij}} \\
\normm{\alpha}_p &\da \qty{ \sum_i \qty{ \sum_j \abs{\alpha_{ij} }^{q} }^{p\over q} }^{1\over p} \\
\normm{\alpha}_\infty &\da \sup_i \sum_j \abs{ \alpha_{ij}}
.\]
:::

:::{.remark}
Note that if $\mcl: \ell^p\to \ell^p$, then $\norm{L} \leq \normm{\alpha}_p$ for $p\in [1, \infty)$ and for $p=\infty$ this is an equality.
:::

:::{.example title="Kernels"}
Consider $C[a, b]$ with $\norm{\wait}_\infty$ and $k\in C^0( [a,b]\cartpower{2}, \CC)$.
Define
\[
K: C[a, b] &\to C[a, b] \\
f &\mapsto \int_a^b k(x, y) f(y) \dy
.\]
What is $\norm{K}$?
Estimate
\[
\norm{Kf} 
&\leq \sup_{y\in [a, b]}\abs{f(y)}  \sup_{x\in [a, b]} \int_a^b \abs{k(x, y)} \dy \\
&\leq \norm{f}_\infty \norm{k}_\infty
,\]
so $\norm{K} \leq \norm{k}_\infty$.

Define
\[
\normm{k}_1 &\da \sup_y \int \abs{k} \dx \\
\normm{k}_p &\da \qty{\int \qty{ \int \abs{k}^q \dy }^{p\over q} \dx }^{1\over p} \\
\normm{k}_\infty &\da \sup_x \int \abs{k} \dy
.\]

:::