---- > [!definition] Definition. ([[reflexive space]]) >Let $X$ be a [[Hausdorff space|Hausdorff]] [[locally convex space|locally convex]] [[topological vector space|topological]] [[vector space|vector]] [[topological space|space]]. The [[natural transformation|natural]] [[linear map|map]] into the [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|double dual]][^1] $\begin{align} X &\to X^{ \vee \vee} \\ x& \mapsto \_(x) \end{align}$ is always a [[linear map|linear]] [[continuous]] [[injection]] (usually in fact a [[topological embedding]], e.g. if $X$ is [[norm|normed]] it's an [[Lipschitz continuous|isometry]][^1]). When it is in fact a [[homeomorphism]], we call $X$ **reflexive**. ^definition > [!note] Remark. > A construction due to R.C. James (1951) shows that it is possible for a [[Banach space|Banach]] space to be $\mathsf{Ban_{M}}$-[[isomorphism|isomorphic]] ([[Lipschitz continuous|isometrically isomorphic]]) to its [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|bidual]] and yet non-reflexive. So reflexivity is specifically about the natural map $x \mapsto \_(x)$ implementing the isometric isomorphism. ^note > [!equivalence] Equivalence for [[Banach space|Banach spaces]] > By the [[open mapping theorem|bounded inverse theorem]], a [[Banach space]] $X$ is reflexive iff the natural map $X \to X^{\vee \vee}$ is a [[surjection]]. ^equivalence > [!equivalence] Dual characterization for normed spaces. > Let $X$ be an [[norm|normed vector space]]. Then $X$ is [[reflexive space|reflexive]] iff $X$ is [[complete]] and $X^{*}$ is [[reflexive space|reflexive]]. > > > > [!proof]- Proof. > > > > $\to.$ Suppose $X$ is [[reflexive space|reflexive]]. Then $X$ is isometrically isomorphic to $X^{* *}$ which is always [[complete]], thus $X$ is [[complete]]. Further, $X^{* * *}\cong (\underbrace{ X^{* *} }_{ \cong X })^{*}\cong X^{*}$ witnessing $X^{*}$ to be reflexive. > > > > $\leftarrow.$ Suppose $X$ is [[complete]] and $X^{*}$ is [[reflexive space|reflexive]]. Then $\Xi:=\operatorname{im }J_{X} \subset X^{* *}$ is complete (it's an isometric isomorphism), hence [[closed set|closed]] in $X^{* *}$ as the latter is always [[complete]]. We claim $\Xi^{\circ}=(0)$, which by [[linear functional characterization of linear subspace closure|will imply]] $\overline{Y}=X^{* *}$, finishing. Indeed, let $\xi \in \Xi^{\circ} \subset X^{* * *}$. Since $X^{*}$ is reflexive, $\xi=\_(f) \in X^{* * *}$ for some $f \in X^{*}$. Thus, if $\xi(\_(x))=0$ for all $\_(x) \in \Xi=\operatorname{im } J$, then $f(x)=0$ for all $x \in X$, which of course is just the statement that $f=0$. Thus $\xi=\_(0)=0$. > > > > [!basicexample] > - Any [[dimension|finite-dimensional]] [[norm|normed]] [[vector space]] is reflexive; [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|see here]] >- [[Lp-norm|For]] $1<p<\infty$, $L^{p}(E)$ is reflexive for any [[measurable function|measurable]] $E \subset \mathbb{F}^{n}$, as follows from [[Lp duality]].[^2] >- The [[Sobolev space]] $W^{k,p}(E)$ is reflexive for $1<p<\infty$. (This follows from the above example/properties and the fact that $W^{k,p}(E)$ can be realized as a closed subspace of a finite product of $L^{p}$ spaces) ^basic-example > [!basicnonexample] > > $c_{0}$ is not reflexive. > - Indeed, $(c_{0})^{* *} \cong \ell^{\infty}$ and $\ell^{\infty}$ is not separable. But $c_{0}$ *is* separable. > $\ell^{1}$ is not reflexive. > - We know $(\ell^{1})^{*} \cong \ell^{\infty}$. If $\ell^{1}$ were reflexive, thus [[separable space|separable]] and reflexive, then $(\ell^{1})^{*} \cong \ell^{\infty}$ would be (see [[separable space|normed space is separable if dual is]]). But $\ell^{\infty}$ is not separable. > $\ell^{\infty}$ is not reflexive. > - Indeed, if $\ell^{\infty}$ were reflexive, then the sub-Banach space $c_{0}$ would be (see 3-from-2 property). Or use the equivalence above: if $\ell^{\infty }$ were reflexive, then $(\ell^{\infty})^{*} \cong c_{0}$ would be and $c_{0}$ isn't. > [!equivalence] Equivalence. (Kakutani) > A [[norm|normed space]] $X$ is [[reflexive space|reflexive]] iff $\mathbb{B}_{X}$ is [[weak topology|weakly]] [[compact]].(See [[Kakutani's weak ball characterization of reflexive normed spaces]]) ^equivalence > [!equivalence] Equivalence. ([[weak topology|Weak]] [[precompact|seq/precompactness]]- a Kakutani corollary) > Suppose $X$ is a [[norm|normed space]] [[dual vector space|such that]] $X^{*}$ is [[separable space|separable]].[^10] Then $X$ is reflexive iff every [[bounded set|bounded sequence]] has a [[weak topology|weakly]] [[sequence|convergent]] [[subsequence]]. > > > [!proof]- Proof. > > $\implies$. Assume $X$ is [[reflexive space|reflexive]]. By [[Kakutani's weak ball characterization of reflexive normed spaces|Kakutani]] $\mathbb{B}_{X}$ is weak-compact, and, [[dual separability makes the ball weakly metrizable|since]] $X^{*}$ is [[separable space|separable]], weakly [[metrizable]] as well. Thus by [[for metrizable spaces, compact iff limit point compact iff sequentially compact|for metrizable spaces, compact iff sequentially compact]] $\mathbb{B}_{X}$ is [[sequentially compact]]. Now any bounded sequence fits in the [[sequentially compact]] set $r\mathbb{B}_{X}$ for some $r >0$, thus has a convergent subsequence in $r\mathbb{B}_{X}$. > > > > $\impliedby$. Suppose bounded sequences in $X$, that is sequences in balls $r\mathbb{B}_{X}$, have weakly convergent subsequences in $X$. $r\mathbb{B}_{X}$ weakly [[closed set|closed]] by [[Mazur's Lemma|Mazur]], hence weakly [[the sequence lemma|sequentially closed]], and so sequences in $r\mathbb{B}_{X}$ converge in $r\mathbb{B}_{X}$. Since $X^{*}$ is separable, we know $r\mathbb{B}_{X}$ is weakly [[metrizable]], and so by [[for metrizable spaces, compact iff limit point compact iff sequentially compact|for metrizable spaces, compact iff sequentially compact]] $r\mathbb{B}_{X}$ is weakly compact. Now [[Kakutani's weak ball characterization of reflexive normed spaces|Kakutani]] implies $X$ is [[reflexive space|reflexive]]. > > [!equivalence] > A [[norm|normed space]] $X$ is [[reflexive space|reflexive]] [[weak topology|iff]] $\sigma(X^{*}, X)=\sigma(X^{*}, X^{* *})$ ^equivalence [^9]: Indeed, $\mathbb{B}_{X}$ with the [[weak topology]] is a [[metric space]] by [[dual separability makes the ball weakly metrizable]] so is weakly [[sequentially compact]] [[for metrizable spaces, compact iff limit point compact iff sequentially compact|iff]] weakly [[compact]]. Now, to be a (norm-)bounded sequence in $X$ is to have a ball around you, and this ball we see is weakly sequentially compact and done. [^10]: In fact we don't have to assume $X^{*}$ is separable to get this statement, cf. [[Eberlein-Smulian Theorem]]. But we don't prove EST in our course while the statement assuming $X^{*}$ is something we've proven. > [!basicproperties] > - *(Necessity of completeness)* Since $X^{\vee \vee}$ is [[Banach space|Banach]] for any [[norm|normed space]] $X$, $X$ being reflexive means it is isometrically isomorphic ($\mathsf{Norm_{M}}$-[[isomorphism|isomorphic]]) to a [[Banach space]], and hence must itself be [[Banach space|Banach]]. One of the equivalences above elaborates. > - *(Reflexivity is a $\mathsf{Ban_{T}}$-invariant, not merely a $\mathsf{Ban_{M}}$-invariant)* If $X,Y$ are $\mathsf{Ban_{T}}$-isomorphic, then $X$ is reflexive $\iff$ $Y$ is reflexive. This follows from the general fact that the diagram > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \usepackage{amsfonts} > \usepackage[mathscr]{euscript} > \usepackage{tikzit} > \usetikzlibrary{shapes.geometric} > > \begin{document} > % TiKZ style file generated by TikZiT. You may edit this file manually, > % but some things (e.g. comments) may be overwritten. To be readable in > % TikZiT, the only non-comment lines must be of the form: > % \tikzstyle{NAME}=[PROPERTY LIST] > > % Node styles > \tikzstyle{red dot}=[fill={rgb,255: red,191; green,0; blue,64}, draw=black, shape=circle] > \tikzstyle{green dot}=[fill={rgb,255: red,0; green,128; blue,128}, draw=black, shape=circle] > \tikzstyle{rectangle}=[fill=white, draw=black, shape=rectangle] > \tikzstyle{triangle}=[fill=white, draw=black, regular polygon sides=3, regular polygon, shape=regular polygon, tikzit shape=regular polygon, shape border rotate=180, minimum size=5mm, inner sep=0pt] > \tikzstyle{bullet}=[fill=black, draw=black, shape=circle, minimum size=2mm, inner sep=0pt] > \tikzstyle{noborder}=[fill=none, draw=none, shape=circle] > \tikzstyle{widerectangle}=[fill=white, draw=black, shape=rectangle, minimum width=16mm, minimum height=4mm, inner sep=1pt, tikzit shape=rectangle] > > % Edge styles > \tikzstyle{new edge style 0}=[-] > > > > > %% paste tikzpicture here > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAA0QBfU9TXfIRQBGclVqMWbdgD1gAHXk0YMAASLlMLt14gM2PASJlh4+s1aIQATR18DgoqNPVzUq9bkaV6pSu1c4jBQAObwRKAAZgBOEAC2SGQgOBBIAEyukpYgACog1Ax0AEYwDAAK-IZCINFYIQAWOHYgMfFIosmpiADMmRZsOV5+at5aza0JiEkp7TxRsZMZnUjdgVxAA > \begin{tikzcd} > X \arrow[d, "T"'] \arrow[r] & X^{\vee \vee} \arrow[d, "T^{\vee \vee}"] \\ > Y \arrow[r] & Y^{\vee \vee} > \end{tikzcd} > %% end paste > > \end{document} > ``` > always commutes (i.e., from the fact that $J$ is [[natural transformation|natural]]!)[^16] > - *(3 from 2)*[^3] Let $X$ be a [[Banach space]] and $Y$ a [[closed set|closed]] [[linear subspace|subspace]] of $X$. Then $X$ is [[reflexive space|reflexive]] if and only if both $Y$ and $X / Y$ are. (Sheet 2.B.7) > - Finite products of Banach spaces preserve reflexivity > - *(Weak embedding)* The map $J:X \to X^{* *}$ is not only a homeomorphism onto its image in the norm topology, but in the weak(star) topology also. By [[there are no weakly continuous functionals past the generating set|this result]](2), we know $J$ is $\sigma(X, X^{*})$-to-$\sigma(X^{* *}, X^{*})$ continuous if and only if $T'J:X^{ ' ' '} \to X^{'}$ sends elements of $X^{*}$ to elements of $X^{*}$. $(T'J)(\xi)=\xi \circ J$, [^3]: I.e. if $2$ out of the $3$ terms in a [[short exact sequence]] have the property, the so too does the third. Other examples: [[topological group action is connected iff group and quotient are]], [[short exact sequence characterization of Noetherian modules]], [[if quotient by a solvable ideal is solvable, then the whole Lie algebra is solvable]], [[detecting isomorphisms with relative homology]], etc... [^16]: The verification of naturality is a matter of bookkeeping: $\begin{align} T^{**}\big( \_(x) \big) (g) &= \big( \_(x) \big) (g \circ T) \\ &= (g \circ T)(x) \\ &= (T^{*}g)(x) \end{align}$while $J_{Y}\big( Tx \big)(g)=\big(\_(Tx)\big)(g)=g (Tx)=(T^{*}g)(x).$ for all $x \in X$, $g \in Y^{*}$. [^4]: Let us check this: let us show $\|\_(x) \|_{X^{\vee \vee}}=\|x\|_{X}$ for all $x \in X$. For $\leq,$ note that $\begin{align} \|\_(x)\|_{X ^{\vee \vee}}&= \sup_{ \|f\|_{X^{\vee}}=1 } \|f(x)\| \\ &= \sup_{\{ f: \sup_{\|y\|=1} \|f(y)\|=1\}} \|f(x)\| \\ & \leq \|x\|. \end{align}$For $\geq$, need to find a single $f$, $\|f\|=1$, such that $\|x\| \leq \|f(x)\|$. To achieve this, define first $f:\langle x \rangle \to \mathbb{F}$, $x \mapsto \|x\|$, and use [[Hahn-Banach Extension Theorem|HBE]] to extend $f$ to a functional $X \to \mathbb{F}$ while preserving its unit-norm. Thus have $X \xhookrightarrow[\mathsf{Ban_{M}}]{}X ^{\vee \vee}$. > [!note] Slogan. > "$X$ is reflexive" means "the bounded linear functionals on its bidual are precisely the evaluations". ^note - [ ] what is the topology on the double dual? will discuss later (but probably not in this note) > [!equivalence] Equivalence for Normed Vector Spaces. > - [[Kakutani's weak ball characterization of reflexive normed spaces]] ^equivalence [^1]: This notation is $\Phi_{f}=\_(f): X^{\vee} \to \mathbb{F}$ given by the evaluation of the argument $\_$ at $f$, that is $\Phi_{f}(\varphi: X \to \mathbb{F})=\varphi(f)$. The reason the embedding is an isometry is because $\|\_(x)\|= \sup_{\|f\|=1} |f(x)|=\|x\|$ in light of the corollary [[linear functional characterization of linear subspace closure|norm-attaining linear functionals]] of [[Hahn-Banach Extension Theorem|HBE]]. [^2]: Verifying this is bookkeeping. Since $p,p' \in (1, \infty)$, we have canonical isomorphisms $\Psi:L^{p} \to (L^{p'})^{\vee}$, $\Psi(f)=\langle f, - \rangle$, and $\Xi:L^{p'} \to (L^{p})^{\vee}$, $\Xi(g)=\langle -,g \rangle$. Write $\pi:=\Xi ^{-1}:(L^{p})^{\vee} \to L^{p'}$, and denote by $\pi^{\vee}$ its [[dual map]] $\begin{align} \pi^{\vee}: (L^{p'})^{\vee}& \to (L^{p})^{\vee \vee } \\ (\varphi: L^{p'} \to \mathbb{F}) & \mapsto \big( \varphi \circ \pi: (L^{p})^{\vee} \to \mathbb{F} \big). \end{align}$ This gives us a map $L^{p} \xrightarrow{\Psi} (L^{p'})^{\vee} \xrightarrow{\pi^{\vee}} (L^{p})^{\vee \vee}.$ We claim $\pi^{\vee} \circ \Psi$ equals the map $f \mapsto \_(f)$. Indeed, let $f \in L^{p}$ and let $\varphi \in (L^{p})^{\vee}$. Then $(\pi^{\vee} \Psi)(f)(\varphi)=\varphi(f)$, as one has $\begin{align} (\pi^{\vee} \Psi)(f) \ (\varphi) &= \pi^{\vee} (\langle f, - \rangle ) \ (\varphi) \\ &= (\langle f, \pi(-) \rangle ) \ (\varphi) \\ &= \langle f, \pi(\varphi) \rangle \\ &= \varphi(f), \end{align}$ where the last step uses that $\pi(\varphi)$ is the unique element $g \in L^{p'}$ satisfying $\langle h, g \rangle=\varphi(h)$ for all $h \in L^{p}$. ---- #### ---- #### 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 > ```