Spectral theorem for compact self adjoint operators

Nonzero eigenspaces are finite-dimensional. Let $\lambda \in \mathrm{\Sigma }$ and let $C:=\overline{T(\stackrel{‾}{B})}$, which is compact by [A3]. If $x\in E_{\lambda }$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ then $Tx=\lambda x$, so $x={\lambda }^{-1}Tx\in {\lambda }^{-1}T(\bar{B})\subseteq {\lambda }^{-1}C$; thus ${\bar{B}}_{E_{\lambda }}:=\{x\in E_{\lambda }:\mathrm{\parallel }x\mathrm{\parallel }\le 1\}\subseteq {\lambda }^{-1}C$, and ${\lambda }^{-1}C$ is compact as a continuous image of a compact set [A3]. The set $E_{\lambda }$ is closed [A2], so ${\bar{B}}_{E_{\lambda }}$ is closed in $H$ and hence compact as a closed subset of the compact set ${\lambda }^{-1}C$ [A3]; by the closed-unit-ball criterion [A5] applied to the normed space $E_{\lambda }$, the space $E_{\lambda }$ has finite dimension, so its multiplicity is finite.

Only finitely many eigenvalues above each threshold. Fix $\epsilon >0$ and put ${\mathrm{\Sigma }}_{\epsilon }:=\{\lambda \in \mathrm{\Sigma }:\mathrm{\mid }\lambda \mathrm{\mid }\ge \epsilon \}$. The compact set $C:=\overline{T(\stackrel{‾}{B})}$ has a finite cover by open balls of radius $\epsilon \mathrm{/}2$ with centres in $C$, by applying compactness to the cover by all such balls [A3]. For each ball $U$ in that finite cover let $L_U$ consist of those $\lambda \in {\mathrm{\Sigma }}_{\epsilon }$ for which $Te\in U$ for some unit $e\in E_{\lambda }$. If distinct $\lambda ,\nu$ belonged to $L_U$, their witnessing unit eigenvectors $e,f$ would be orthogonal by [A2], and hence $\mathrm{\parallel }Te-Tf{\mathrm{\parallel }}^2=\mathrm{\mid }\lambda {\mathrm{\mid }}^2+\mathrm{\mid }\nu {\mathrm{\mid }}^2\ge 2{\epsilon }^2$, whereas two points of $U$ have distance less than $\epsilon$. Thus each $L_U$ has at most one element. Every $\lambda \in {\mathrm{\Sigma }}_{\epsilon }$ has a unit eigenvector whose image belongs to $C$, so ${\mathrm{\Sigma }}_{\epsilon }={\bigcup }_U{L}_U$ is finite. This argument makes no infinite choice of eigenvectors. For $a\mathrm{\ne }0$, the ball $B(a,\mathrm{\mid }a\mathrm{\mid }\mathrm{/}2)$ meets $\mathrm{\Sigma }$ only inside ${\mathrm{\Sigma }}_{\mathrm{\mid }a\mathrm{\mid }\mathrm{/}2}$, proving local finiteness away from zero.

$M^{\perp }$ is annihilated by $T$. Since $M$ is the closed linear span of the subspaces $E_{\lambda }$, a vector is orthogonal to $M$ exactly when it is orthogonal to every $E_{\lambda }$; hence $M^{\perp }={\bigcap }_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }^{\perp }$ is a closed linear subspace, and it is $T$-invariant because each $E_{\lambda }^{\perp }$ is $T$-invariant [A2]. As a closed subspace of the Hilbert space $H$, $M^{\perp }$ is complete [A3], and the restriction $S:M^{\perp }\to M^{\perp }$ is compact: [A4] gives compact closure in $H$ of each bounded image, that closure lies in the closed subspace $M^{\perp }$, and its subspace topology is unchanged and self-adjoint, since for $u,v\in M^{\perp }$ one has $⟨Tu,v⟩=⟨u,Tv⟩$ in $H$ and both vectors lie again in $M^{\perp }$. If $S\mathrm{\ne }0$, the extremal eigenvalue lemma [A2] provides $\mu =\pm \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\ne }0$ and $w\in M^{\perp }\setminus \{0\}$ with $Sw=\mu w$, hence $Tw=\mu w$ and $w\in E_{\mu }\subseteq M$; then $w\in M\cap M^{\perp }$, so $⟨w,w⟩=0$ and $w=0$ by positive definiteness, a contradiction. Therefore $S=0$, that is $T$ vanishes on $M^{\perp }$.

An orthonormal family with closed span $M$. By [step 1.2] each set ${\mathrm{\Sigma }}_{1\mathrm{/}(n+1)}=\{\lambda \in \mathrm{\Sigma }:\mathrm{\mid }\lambda \mathrm{\mid }\ge 1\mathrm{/}(n+1)\}$, $n\in \mathbb{N}$, is finite, and every $\lambda \in \mathrm{\Sigma }$ lies in some ${\mathrm{\Sigma }}_{1\mathrm{/}(n+1)}$ because $\mathrm{\mid }\lambda \mathrm{\mid }>0$ and $1\mathrm{/}(n+1)<\mathrm{\mid }\lambda \mathrm{\mid }$ for a suitable $n$ by [A7]; hence $\mathrm{\Sigma }={\bigcup }_{n\in \mathbb{N}}{\mathrm{\Sigma }}_{1\mathrm{/}(n+1)}$ is at most countable by [A7]. By Countable Choice [A7], choose for every $\lambda \in \mathrm{\Sigma }$ an orthonormal basis $(e_{\lambda ,1},\dots ,e_{\lambda ,d_{\lambda }})$ of the finite-dimensional space $E_{\lambda }$, which exists by [A5], and let $(e_j{)}_{j\in J}$ be the disjoint union of these finite families indexed by the at most countable set $\mathrm{\Sigma }$, so that $J$ is at most countable. Every $e_j$ has norm $1$, and orthonormal bases of orthogonal eigenspaces [A2] make $(e_j{)}_{j\in J}$ an orthonormal family whose closed linear span is $M$ by the definition of $M$.

The support of the operator. Every eigenvector with nonzero eigenvalue is orthogonal to $\ker T$, because for $y\in \ker T$ and $x\in E_{\lambda }$ with $\lambda \mathrm{\ne }0$ one has $\lambda ⟨x,y⟩=⟨Tx,y⟩=⟨x,Ty⟩=0$ by self-adjointness, so $⟨x,y⟩=0$; since $(\ker T{)}^{\perp }$ is closed [A6] and contains each $E_{\lambda }$, it contains $M$, while [step 1.3] and [A6] give $(\ker T{)}^{\perp }\subseteq (M^{\perp }{)}^{\perp }=\bar{M}=M$; hence $M=(\ker T{)}^{\perp }$. Moreover $\mathrm{ran}T\subseteq (\ker T{)}^{\perp }$ by the same computation read with $x$ arbitrary, so $\overline{\mathrm{ran}T}\subseteq (\ker T{)}^{\perp }$, and if $z\perp \mathrm{ran}T$ then $0=⟨z,Tx⟩=⟨Tz,x⟩$ for every $x$, whence $Tz=0$ and $z\in \ker T$; applying [A6] to the linear subspace $\mathrm{ran}T$ gives $\overline{\mathrm{ran}T}=(\mathrm{ran}T{)}^{\perp \perp }\subseteq (\ker T{)}^{\perp }$, while the previous inclusion reverses after taking complements: $(\ker T{)}^{\perp }\subseteq \mathrm{ran}{T}^{\perp \perp }=\overline{\mathrm{ran}T}$.

The spectral expansion. Let $x\in H$. By [A6] and [step 2.2] there is a unique decomposition $x=m+n$ with $m\in M$ and $n\in M^{\perp }$, and by [step 1.3] $n\in \ker T$, so $Tn=0$. By [step 2.1] the family $(e_j)$ is complete in $M$, so the Fourier expansion [A6] gives $m={\sum }_{j\in J}⟨m,e_j⟩e_j$ as the limit of the finite-subset net, and the same holds with $⟨x,e_j⟩$ in place of $⟨m,e_j⟩$ because $x-m\in M^{\perp }$. Writing $P_{\lambda }x:={\sum }_{i=1}^{d_{\lambda }}⟨x,e_{\lambda ,i}⟩e_{\lambda ,i}\in E_{\lambda }$, the family $(P_{\lambda }x{)}_{\lambda \in \mathrm{\Sigma }}$ is orthogonal with ${\sum }_{\lambda }\mathrm{\parallel }{P}_{\lambda }x{\mathrm{\parallel }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}^2$ by Bessel [A6], so by [A6] the finite-subset net ${\sum }_{\lambda \in F}{P}_{\lambda }x$ converges to some $m^{\mathrm{\prime }}\in M$; for every $j$ the difference $m-m^{\mathrm{\prime }}$ is orthogonal to $e_j$, hence to $M$, so $m-m^{\mathrm{\prime }}\in M\cap M^{\perp }=\{0\}$ and $m^{\mathrm{\prime }}=m$. Finally $T({\sum }_{\lambda \in F}{P}_{\lambda }x)={\sum }_{\lambda \in F}\lambda P_{\lambda }x$ for finite $F$ because each $P_{\lambda }x\in E_{\lambda }$, and continuity of $T$ [A8] carries the convergent net $(P_{\lambda }x)$ to $Tm$; hence the finite-subset net of $(\lambda P_{\lambda }x)$ converges to $Tm$, and adding $Tn=0$ gives $Tx={\sum }_{\lambda \in \mathrm{\Sigma }}\lambda P_{\lambda }x$.

A spectral gap off the eigenvalue set. Let $\mu \mathrm{\ne }0$ with $\mu \mathrm{\notin }\mathrm{\Sigma }$. The set $A:=\{\lambda \in \mathrm{\Sigma }:\mathrm{\mid }\lambda \mathrm{\mid }\ge \mathrm{\mid }\mu \mathrm{\mid }\mathrm{/}2\}$ is finite by [step 1.2], and $\mu \mathrm{\notin }A$; set $\delta :=\min (\{\mathrm{\mid }\mu -\lambda \mathrm{\mid }:\lambda \in A\}\cup \{\mathrm{\mid }\mu \mathrm{\mid }\mathrm{/}2\})$, a positive real number because $A$ is finite and every displayed distance is positive. For every $\lambda \in \mathrm{\Sigma }$ one has $\mathrm{\mid }\mu -\lambda \mathrm{\mid }\ge \delta$: if $\mathrm{\mid }\lambda \mathrm{\mid }\ge \mathrm{\mid }\mu \mathrm{\mid }\mathrm{/}2$ this is the definition of $\delta$, and if $\mathrm{\mid }\lambda \mathrm{\mid }<\mathrm{\mid }\mu \mathrm{\mid }\mathrm{/}2$ then $\mathrm{\mid }\mu -\lambda \mathrm{\mid }\ge \mathrm{\mid }\mu \mathrm{\mid }-\mathrm{\mid }\lambda \mathrm{\mid }>\mathrm{\mid }\mu \mathrm{\mid }\mathrm{/}2\ge \delta$ by the triangle inequality [A8]. Consequently, for every $z\in H$ the orthogonal family $((\mu -\lambda {)}^{-1}{P}_{\lambda }z{)}_{\lambda \in \mathrm{\Sigma }}$ has ${\sum }_{\lambda }\mathrm{\mid }\mu -\lambda {\mathrm{\mid }}^{-2}\mathrm{\parallel }{P}_{\lambda }z{\mathrm{\parallel }}^2\le {\delta }^{-2}{\sum }_{\lambda }\mathrm{\parallel }{P}_{\lambda }z{\mathrm{\parallel }}^2\le {\delta }^{-2}\mathrm{\parallel }z{\mathrm{\parallel }}^2<+\mathrm{\infty }$ by Bessel [A6], so [A6] makes its finite-subset net converge to a vector of norm $\le {\delta }^{-1}\mathrm{\parallel }z\mathrm{\parallel }$.

The candidate inverse. Fix $\mu \mathrm{\ne }0$ with $\mu \mathrm{\notin }\mathrm{\Sigma }$ and, for $z\in H$, write $z=m(z)+n(z)$ with $m(z)\in M$, $n(z)\in M^{\perp }$ as in [step 3.1]. The limit of the finite-subset net in [step 3.2] is unique: if the same net converges to $y$ and to $y^{\mathrm{\prime }}$, then for any $\epsilon >0$ there are finite sets $F_0,F_1\subseteq \mathrm{\Sigma }$ such that every $F\supseteq F_0$ has $\mathrm{\parallel }{a}_F-y\mathrm{\parallel }<\epsilon \mathrm{/}2$ and every $F\supseteq F_1$ has $\mathrm{\parallel }{a}_F-y^{\mathrm{\prime }}\mathrm{\parallel }<\epsilon \mathrm{/}2$; at $F=F_0\cup F_1$ the triangle inequality gives $\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }<\epsilon$, and hence $y=y^{\mathrm{\prime }}$. We may therefore define $Az:={\sum }_{\lambda \in \mathrm{\Sigma }}(\mu -\lambda {)}^{-1}{P}_{\lambda }z+{\mu }^{-1}n(z),$ where the first term is that unique limit. The map $A$ is linear because each $P_{\lambda }$ is linear and limits respect linear combinations, and $\mathrm{\parallel }Az{\mathrm{\parallel }}^2\le {\delta }^{-2}\mathrm{\parallel }m(z){\mathrm{\parallel }}^2+\mathrm{\mid }\mu {\mathrm{\mid }}^{-2}\mathrm{\parallel }n(z){\mathrm{\parallel }}^2\le \max ({\delta }^{-1},\mathrm{\mid }\mu {\mathrm{\mid }}^{-1}{)}^2\mathrm{\parallel }z{\mathrm{\parallel }}^2$ by orthogonality of the decomposition [A6], so $A$ is a bounded linear operator on $H$.

The inverse identities and the spectrum. Work now over $\mathbb{C}$ and fix $\mu \mathrm{\ne }0$ outside $\mathrm{\Sigma }$. For finite $F\subseteq \mathrm{\Sigma }$, put $a_F(z):={\sum }_{\lambda \in F}(\mu -\lambda {)}^{-1}{P}_{\lambda }z$. Since $T$ acts as $\lambda I$ on $E_{\lambda }$, one has $(\mu I-T)a_F(z)={\sum }_{\lambda \in F}{P}_{\lambda }z$. By [step 3.2] and continuity of $\mu I-T$, taking limits gives $(\mu I-T){\lim }_F{a}_F(z)=m(z)$. Also $(\mu I-T)({\mu }^{-1}n(z))=n(z)$ because $Tn(z)=0$. Thus $(\mu I-T)Az=z$. For the other identity, self-adjointness and the finite formula for $P_{\lambda }$ give $P_{\lambda }Tw=\lambda P_{\lambda }w$: for each basis vector $e\in E_{\lambda }$, $⟨Tw,e⟩=⟨w,Te⟩=\lambda ⟨w,e⟩$, since $\lambda$ is real. Moreover $Tw\in M$ by [step 2.2], so $n((\mu I-T)w)=\mu n(w)$ by uniqueness of the orthogonal decomposition. Consequently $A(\mu I-T)w={\lim }_F{\sum }_{\lambda \in F}{P}_{\lambda }w+n(w)=m(w)+n(w)=w$. This proves that $A$ is a bounded two-sided inverse, so $\mu \in \rho (T)$ [A9]. Conversely a nonzero eigenvector makes $\lambda I-T$ noninjective, so every $\lambda \in \mathrm{\Sigma }$ lies in $\sigma (T)$. Therefore $\sigma (T)\cap \{\mu \mathrm{\ne }0\}=\mathrm{\Sigma }$.

Conclusion. Claim 1 is the combination of [step 1.1], [step 1.2], [step 2.1] and the reality of eigenvalues in [A2]; claim 2 is [step 2.2] together with the decomposition of [step 3.1]; claim 3 is [step 3.1]; claim 4 is [step 5.1]. At no point was a Hilbert basis of $\ker T$ selected: the chosen vectors all lie in the eigenspaces $E_{\lambda }$ with $\lambda \mathrm{\ne }0$, which are contained in $(\ker T{)}^{\perp }$ by [step 3.1].