On Spectral Families of Hermitian Operators

Kreyszig Functional Analysis

Throughout these notes, we use $H$ to denote an arbitrary complex Hilbert space. We begin with a brief discussion of projection operators. Recall that if $A$ is a closed subspace of $H$, $A^{\perp }$ is also a closed subspace of $H$, and for any $x\in H$ there is a unique representation of the form $x=y+z$where $y\in A$ and $z\in A^{\perp }$ (alternatively, $H=A\oplus A^{\perp }$). The operator that maps every $x\in H$ to its corresponding $y\in A$ we call the projection of $H$ onto $A$. Conversely, we call an operator $T:H\to H$ a projection iff there exists a closed subspace $A\subset H$ such that $T$ is the projection of $H$ onto $A$. Note that $T$ is a projection iff $T(H)$ is closed, $T$ is the identity operator on its range, and $\mathcal{N}(T)=T(H{)}^{\perp }$. In this case, $T$ becomes precisely the projection from $H$ onto the closed subspace $T(H)$.

The following theorem gives a simple criteria for identifying projections:

Theorem 1. An operator $P:H\to H$ is a projection if and only if $P$ is self-adjoint and idempotent ($P^2=P$). Note self-adjoint, by definition, implies boundedness and linearity.

Proof) Let $P$ be a projection onto the closed subspace $A=P(H)$. We can see that $I-P$ is precisely the projection onto the closed subspace $A^{\perp }$ by definition, as for any $x\in H$ and $x=y+z$ where $y\in A$ and $z\in A^{\perp }$, $(I-P)x=y+z-Px=z$. $P$ must be idempotent as $Px=x$ for all $x\in P(H)$, implying $P^{2}x=Px$ for all $x\in H$. Furthermore, if we take an arbitrary pair $x_1,x_2\in H$ with $x_1=y_1+z_1$ and $x_2=y_2+z_2$ where $y_i\in A$ and $z_i\in A^{\perp }$,

$$⟨P{x}_1,x_2⟩=⟨y_1,x_2⟩=⟨y_1,y_2⟩=⟨x_1,y_2⟩=⟨x_1,P{x}_2⟩$$

this implies that $P$ is self-adjoint.

We prove the converse. Assume that $P$ is idempotent and self-adjoint. $P(H)$ is precisely the null space of the operator $I-P$, as $(I-P)x=0$ implies $x=Px\in P(H)$ and $x=Py$ for some $y\in H$ implies $(I-P)x=(I-P)Py=(P-P^2)y=0\Rightarrow x\in \mathcal{N}(I-P)$. Therefore, the range $P(H)$ is closed. $P$ also clearly acts as the identity operator on its range, as $P^{2}x=Px$ for all $x\in H$ implies $Py=y$ for all $y\in P(H)$. The null space of $P$ is $P(H{)}^{\perp }$, as $Px=0$ implies for all $y\in H$,

$$0=⟨Px,y⟩=⟨x,Py⟩\Rightarrow x\in P(H{)}^{\perp }$$

and $x\in P(H{)}^{\perp }$ implies for all $y\in H$, $0=⟨x,Py⟩=⟨Px,y⟩\Rightarrow Px=0$. By definition, $P$ is forced to be the projection of $H$ onto the closed subspace $P(H)$.

Let $P_1:H\to H$ and $P_2:H\to H$ be arbitrary projection operators. The following facts follow quite straightforwardly from Theorem 1 and the definition of projections. (1) For any $x\in H$, $⟨Px,x⟩=\Vert Px{\Vert }^2$ (2) $0\le P\le I$ (3) $P_1{P}_2$ is a projection iff $P_1$ and $P_2$ commute, and in this case $P_1{P}_2$ is a projection onto the closed subspace $P_1(H)\cap P_2(H)$. (4) Two closed subspaces $Y$ and $V$ are orthogonal iff their corresponding projections satisfy $P_Y{P}_V=0$. (5) $P_1+P_2$ is a projection iff $P_1(H)$ and $P_2(H)$ are orthogonal. In this case $P_1+P_2$ is a projection onto the closed subspace $P_1(H)\oplus P_2(H)$. (6) The following conditions are equivalent regarding partial order of projections: i. $P_1{P}_2=P_2{P}_1=P_1$ ii. $P_1(H)\subset P_2(H)$ iii. $\mathcal{N}(P_1)\supset \mathcal{N}(P_2)$ iv. $\Vert P_{1}x\Vert \le \Vert P_{2}x\Vert$ for all $x\in H$ v. $P_1\le P_2$ (7) $P_1-P_2$ is a projection iff $P_2(H)\subset P_1(H)$ and in this case $P_1-P_2$ is a projection onto the closed subspace $P_1(H)\cap P_2(H{)}^{\perp }$.

We prove the following theorem as a nice exercise with projections.

Theorem 2. Let $P_n$ be a monotonically increasing sequence of projections in $H$. Then, $P_n$ is strongly convergent to an operator $P$, where $P$ is a projection with

$$P(H)=\overline{\bigcup _{i=1}^{\infty }{P}_i(H)},\mathcal{N}(P)=\bigcap _{i=1}^{\infty }\mathcal{N}(P_i)$$

Proof) As $P_n$ is bounded above and monotonically increasing, it must be strongly convergent to some $P\in B(H,H)$. By continuity, this $P$ must be idempotent and self-adjoint, implying it is a projection by Theorem 1. For any $P(x)\in P(H)$, the sequence $P_i(x)\to P(x)$, and thus $P(x)\in \overline{\bigcup P_i(H)}$. For any $x\in \bigcup P_i(H)$, there is some $n$ and $y\in H$ such that $P_{n}y=x$. As $P_n$ is idempotent, $P_{n}x=x$; for any $m>n$, $P_m{P}_n=P_n$ and thus $P_{m}x=P_m{P}_{n}x=P_{n}x=x$. Therefore, $P_{i}x\to x$ and $x\in P(H)$. We have shown that $\bigcup P_i(H)\subset P(H)\subset \overline{\bigcup P_i(H)}$. As $P(H)$ is closed, this gives

$$P(H)=\overline{\bigcup _{i=1}^{\infty }{P}_i(H)}$$

By continuity, $P\ge P_n$ for any $n$ and thus $\mathcal{N}(P)\subset \mathcal{N}(P_n)\Rightarrow \mathcal{N}(P)\subset \bigcap \mathcal{N}(P_i)$. Furthermore, if $x\in \bigcap \mathcal{N}(P_i)$, $P_{i}x=0$ for all $i$ and thus $Px=0$, implying $\bigcap \mathcal{N}(P_i)\subset \mathcal{N}(P)$. This gives

$$\mathcal{N}(P)=\bigcap _{i=1}^{\infty }\mathcal{N}(P_i)$$

as desired.

A spectral family $\xi =(E_{\lambda }{)}_{\lambda \in \mathbb{R}}$ is a map from $\mathbb{R}$ to the space of projections $H\to H$ such that: (1) If $\lambda <\mu$, $E_{\lambda }\le E_{\mu }$ (2) For all $x\in H$: ${\lim }_{\lambda \to -\infty }{E}_{\lambda }x=0$ (3) For all $x\in H$: ${\lim }_{\lambda \to \infty }{E}_{\lambda }x=x$ (4) For all $x\in H$: $E_{\lambda +0}x={\lim }_{\mu \to \lambda +}{E}_{\mu }x=E_{\lambda }x$ (2), (3), and (4) can be condensed into the statement that under the strong operator topology, $E_{\lambda }\to I$ as $\lambda \to \infty$, $E_{\lambda }\to 0$ as $\lambda \to -\infty$, and $E$ is continuous from the right. Furthermore we call $\xi$a spectral family on $[a,b]$ if $E_{\lambda }=0$ for $\lambda <a$ and $E_{\lambda }=I$ for $\lambda \ge b$.

Let $T:H\to H$ be an arbitrary self-adjoint operator. We define the spectral family $\xi (T)=(E_{\lambda })$ associated with $T$ as follows:

$$B_{\lambda }=(T_{\lambda }^2{)}^{1/2}\text{ (the positive square root of }{T}_{\lambda }^2)$$ $$T_{\lambda }^{+}=\frac{1}{2}(B_{\lambda }+T_{\lambda }),T_{\lambda }^{-}=\frac{1}{2}(B_{\lambda }-T_{\lambda })$$

We let $E_{\lambda }$ be the projection onto $\mathcal{N}(T_{\lambda }^{+})$. Geometrically, $B$ is the operator which preserves the magnitude of scaling for each direction in $T_{\lambda }$ but flips all negatively scaled directions such that they are positively scaled. Therefore, $T_{\lambda }^{+}$ isolates the positive scaling of $T_{\lambda }$ while $T_{\lambda }^{-}$ preserves the negative scaling. $E_{\lambda }$, being the projection onto the null space of $\mathcal{N}(T_{\lambda }^{+})$, is essentially the projection onto all directions that are either annihilated or negatively scaled by $T_{\lambda }$. As the negative scaling behavior of $T_{\lambda }$ increases as $\lambda$ increases, we can intuitively see $E_{\lambda }$ should project onto a monotonically growing space as $\lambda$ increases, and thus compose a valid spectral family. We flesh out this argument in the following theorem.

Theorem 3. $\xi (T)$, as defined above, is a valid spectral family on $[m,M]$, where $m={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $M={\sup }_{\Vert x\Vert =1}⟨Tx,x⟩$.

Proof) Clearly, all operators of the form $B_{\lambda }$, $T_{\lambda }^{+}$, $T_{\lambda }^{-}$, and $T_{\lambda }$ are self-adjoint and commute with each other. $E_{\lambda }$ must also commute with all self-adjoint operators $S$ that commute with $T_{\lambda }^{+}$. This is because for any $x\in H$, $T_{\lambda }^{+}{E}_{\lambda }x=0\Rightarrow S{T}_{\lambda }^{+}{E}_{\lambda }x=T_{\lambda }^{+}S{E}_{\lambda }x=0$, implying $S{E}_{\lambda }(H)\subset \mathcal{N}(T_{\lambda }^{+})$. Therefore, $E_{\lambda }S{E}_{\lambda }=S{E}_{\lambda }\Rightarrow E_{\lambda }S=(S{E}_{\lambda }{)}^{\star }=(E_{\lambda }S{E}_{\lambda }{)}^{\star }=E_{\lambda }S{E}_{\lambda }=S{E}_{\lambda }$. We can therefore safely say all relevant operators ($B$, $T^{+}$, $T^{-}$, $T_{\lambda }$, $E$) commute with one another.

Before we proceed, we must also formalize our intuition that $T^{+}$ is the positive part of $T$ while $T^{-}$ is its negative part (we omit the $\lambda$ subscripts for now). Note that $T=T^{+}-T^{-}$ while $B=T^{+}+T^{-}$. Furthermore, $T^{+}{T}^{-}=T^{-}{T}^{+}=(B^2-T^2)/4=0$. As $T^{+}$ and $T^{-}$ are self-adjoint, this implies $T^{+}(H)\perp T^{-}(H)$. Also, it implies $T^{-}(H)\subset \mathcal{N}(T^{+})$ and therefore $T^{-}=E{T}^{-}=T^{-}E=BE$. We can also see that $E{T}^{+}=T^{+}E=0$. Together, this gives $T^{+}=B-T^{-}=B(I-E)\ge 0$ and $T^{-}=BE\ge 0$.

We first show that if $\lambda <\mu$, $E_{\lambda }\le E_{\mu }$, which is equivalent to $\mathcal{N}(T_{\lambda }^{+})\subset \mathcal{N}(T_{\mu }^{+})$. Note that $T_{\lambda }^{+}\ge T_{\lambda }\ge T_{\mu }$. Multiplying $T_{\mu }^{+}$ gives $T_{\mu }^{+}{T}_{\lambda }^{+}\ge T_{\mu }^{+2}$. Therefore, for any $x\in H$,

$$⟨T_{\lambda }^{+}x,T_{\mu }^{+}x⟩\ge \Vert T_{\mu }^{+}x{\Vert }^2\Rightarrow T_{\lambda }^{+}x=0\text{ implies }{T}_{\mu }^{+}x=0$$

we are done.

Next, we show that if $\lambda <m$, $E_{\lambda }=0$. Note that if $\lambda <m$, $T_{\lambda }\ge 0$ as for all $x\in H$, $⟨T_{\lambda }x,x⟩=⟨Tx,x⟩-\lambda \Vert x{\Vert }^2>⟨Tx,x⟩-m\Vert x{\Vert }^2\ge 0$. Therefore, $T_{\lambda }=B_{\lambda }$ and $T_{\lambda }^{+}=T_{\lambda }$. The null space of $T_{\lambda }^{+}=T_{\lambda }$ must be empty as $T_{\lambda }x=0\Rightarrow ⟨Tx,x⟩=\lambda \Vert x{\Vert }^2<m\Vert x{\Vert }^2$, a contradiction. We are done. We can similarly see that if $\lambda \ge M$, $E_{\lambda }=I$. This is because $T_{\lambda }\le 0\Rightarrow -T_{\lambda }\ge 0$, $B=-T_{\lambda }\Rightarrow T_{\lambda }^{+}=0$. Therefore, $\mathcal{N}(T_{\lambda }^{+})=H$ and $E_{\lambda }=I$.

The final piece is to show $E_{\lambda +0}=E_{\lambda }$. For any $\mu >\lambda$, consider the difference operator $E(\Delta )=E_{\mu }-E_{\lambda }$. As $E_{\mu }\ge E_{\lambda }$, $E(\Delta )$ is a projection onto $E_{\mu }(H)\cap E_{\lambda }(H{)}^{\perp }$. Therefore, we have $E(\Delta )E_{\mu }=E_{\mu }E(\Delta )=E(\Delta )$, and $(I-E_{\lambda })E(\Delta )=E(\Delta )(I-E_{\lambda })=E(\Delta )$ as $I-E_{\lambda }$ is the projection onto $E_{\lambda }(H{)}^{\perp }$. Therefore,

$$TE(\Delta )=\lambda E(\Delta )+T_{\lambda }E(\Delta )=\lambda E(\Delta )+T_{\lambda }(I-E_{\lambda })E(\Delta )=\lambda E(\Delta )+T_{\lambda }^{+}E(\Delta )\ge \lambda E(\Delta )$$ $$TE(\Delta )=\mu E(\Delta )+T_{\mu }{E}_{\mu }E(\Delta )=\mu E(\Delta )+(-T_{\mu }^{-})E(\Delta )\le \mu E(\Delta )$$

Combined, we get that $\lambda E(\Delta )\le TE(\Delta )\le \mu E(\Delta )$. Note that $E(\Delta )$, as $\mu \to \lambda +$, is a bounded monotonically decreasing sequence of projections. Thus, there exists some projection $P$ to which $E(\Delta )$ strongly converges to. By continuity, $P$ must satisfy $\lambda P=TP$; that is, $T_{\lambda }P=0$ and $P(H)\subset \mathcal{N}(T_{\lambda })$. Note that if $T_{\lambda }x=0$, $T_{\lambda }^{2}x=0\Rightarrow 0=⟨T_{\lambda }^{2}x,x⟩=\Vert B_{\lambda }x{\Vert }^2\Rightarrow B_{\lambda }x=0$, and thus $T_{\lambda }^{+}x=T_{\lambda }^{-}x=0$. Therefore, $P(H)\subset \mathcal{N}(T_{\lambda }^{+})$. However, for any $x\in \mathcal{N}(T_{\lambda }^{+})$ and $\mu >\lambda$, $E_{\lambda }x=E_{\mu }x=x$ and thus $E(\Delta )x=0$. By continuity, this implies $Px=0$ for any $x\in P(H)$. Therefore, $P=0$, and $E(\Delta )$ strongly converges to $0$.