# The spectrum of a self-adjoint operator is a compact subset of $\mathbb{R}$

Jordan Bell
April 3, 2014
###### Abstract

In these notes I prove that the spectrum of a bounded linear operator from a Hilbert space to itself is a nonempty compact subset of $\mathbb{C}$, and that if the operator is self-adjoint then the spectrum is contained in $\mathbb{R}$. To show that the spectrum is nonempty I prove various facts about resolvents.

### 1.1 Operator norm

Let $H$ be a Hilbert space with inner product $\left\langle\cdot,\cdot\right\rangle:H\times H\to\mathbb{C}$, and define $I:H\to H$ by $Ix=x$, $x\in H$.. For $v\in H$, let $\left\|v\right\|=\sqrt{\left\langle v,v\right\rangle}$, and if $T:H\to H$ is a bounded linear map, let

 $\left\|T\right\|=\sup_{\left\|v\right\|\leq 1}\left\|Tv\right\|.$

namely, the operator norm of $T$.

### 1.2 Definition of adjoint

The Riesz representation theorem states that if $\phi:H\to\mathbb{C}$ is a bounded linear map then there is a unique $v_{\phi}\in H$ such that

 $\phi(x)=\left\langle x,v_{\phi}\right\rangle$

for all $x\in H$. Let $T:H\to H$ be a bounded linear map, and for $y\in H$, define $\phi_{y}:H\to\mathbb{C}$ by

 $\phi_{y}(x)=\left\langle Tx,y\right\rangle.$

$\phi_{y}:H\to\mathbb{C}$ is a bounded linear map, so by the Riesz representation theorem there is a unique $v_{y}$ such that

 $\phi_{y}(x)=\left\langle x,v_{y}\right\rangle$

for all $x\in H$. Define $T^{*}:H\to H$ by

 $T^{*}y=v_{y}.$

$T^{*}y$ is well-defined because of the uniqueness in the Riesz representation theorem. For all $x,y\in H$,

 $\left\langle x,T^{*}y\right\rangle=\left\langle x,v_{y}\right\rangle=\phi_{y}(% x)=\left\langle Tx,y\right\rangle.$

We call $T^{*}:H\to H$ the adjoint of $T:H\to H$.

### 1.3 Adjoint is linear

For $y_{1},y_{2}\in H$, we have for all $x\in H$ that

 $\displaystyle\left\langle x,T^{*}(y_{1}+y_{2})\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle Tx,y_{1}+y_{2}\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle Tx,y_{1}\right\rangle+\left\langle Tx,y_{2}\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle x,T^{*}y_{1}+T^{*}y_{2}\right\rangle.$

Hence for all $x\in H$,

 $\left\langle x,T^{*}(y_{1}+y_{2})-T^{*}y_{1}-T^{*}y_{2}\right\rangle=0.$

In particular this is true for $x=T^{*}(y_{1}+y_{2})-T^{*}y_{1}-T^{*}y_{2}$, so by the nondegeneracy of $\left\langle\cdot,\cdot\right\rangle$ we get

 $T^{*}(y_{1}+y_{2})-T^{*}y_{1}-T^{*}y_{2}=0.$

We similarly obtain for all $\lambda\in\mathbb{C}$ and all $y\in H$ that

 $T^{*}(\lambda y)-\lambda T^{*}y=0.$

Hence $T^{*}:H\to H$ is a linear map.

### 1.4 Adjoint is bounded

For $x,y\in H$, by the Cauchy-Schwarz inequality we have

 $|\phi_{y}(x)|=|\left\langle x,v_{y}\right\rangle|\leq\left\|x\right\|\left\|v_% {y}\right\|,$

so $\left\|\phi_{y}\right\|\leq\left\|v_{y}\right\|$, i.e. the operator norm of $\phi_{y}$ is less than or equal to the norm of $v_{y}$. If $v_{y}\neq 0$, then $\left\|\frac{v_{y}}{\left\|v_{y}\right\|}\right\|=1$ and

 $\left|\phi_{y}\left(\frac{v_{y}}{\left\|v_{y}\right\|}\right)\right|=\left% \langle\frac{v_{y}}{\left\|v_{y}\right\|},v_{y}\right\rangle=\left\|v_{y}% \right\|.$

It follows that

 $\left\|\phi_{y}\right\|=\left\|v_{y}\right\|.$

Then for $y\in H$, by the Cauchy-Schwarz inequality and because $T$ is bounded we have

 $\displaystyle\left\|T^{*}y\right\|$ $\displaystyle=$ $\displaystyle\left\|v_{y}\right\|$ $\displaystyle=$ $\displaystyle\left\|\phi_{y}\right\|$ $\displaystyle=$ $\displaystyle\sup_{\left\|x\right\|\leq 1}\left\|\phi_{y}(x)\right\|$ $\displaystyle=$ $\displaystyle\sup_{\left\|x\right\|\leq 1}|\left\langle Tx,y\right\rangle|$ $\displaystyle\leq$ $\displaystyle\sup_{\left\|x\right\|\leq 1}\left\|T\right\|\left\|x\right\|% \left\|y\right\|$ $\displaystyle\leq$ $\displaystyle\left\|T\right\|\left\|y\right\|.$

Therefore $T^{*}$ is bounded. Thus if $T:H\to H$ is a bounded linear map then its adjoint $T^{*}:H\to H$ is a bounded linear map.

### 1.5 Adjoint is involution

Because $T^{*}:H\to H$ is a bounded linear map, it has an adjoint $T^{**}:H\to H$, and $T^{**}$ is itself a bounded linear map. For all $x,y\in H$,

 $\displaystyle\left\langle Tx,y\right\rangle$ $\displaystyle=$ $\displaystyle\left\langle x,T^{*}y\right\rangle$ $\displaystyle=$ $\displaystyle\overline{\left\langle T^{*}y,x\right\rangle}$ $\displaystyle=$ $\displaystyle\overline{\left\langle y,T^{**}x\right\rangle}$ $\displaystyle=$ $\displaystyle\left\langle T^{**}x,y\right\rangle.$

Hence for all $x,y\in H$,

 $\left\langle Tx-T^{**}x,y\right\rangle=0.$

This is true in particular for $y=Tx-T^{**}x$, so by the nondegeneracy of $\left\langle\cdot,\cdot\right\rangle$ we obtain

 $Tx-T^{**}x=0,\qquad x\in H.$

Thus for any bounded linear map $T:H\to H$, $T^{**}=T$. In words, if $T$ is a bounded linear map from a Hilbert space to itself, then the adjoint of its adjoint is itself. We have shown already that $\left\|T^{*}\right\|\leq\left\|T\right\|$. Hence also $\left\|T\right\|=\left\|T^{**}\right\|\leq\left\|T^{*}\right\|$, so

 $\left\|T\right\|=\left\|T^{*}\right\|.$

If $T^{*}=T$, we say that $T$ is self-adjoint.

## 2 Bounded linear operators

Let $\mathscr{B}(H)$ be the set of bounded linear maps $H\to H$. With the operator norm, one checks that $\mathscr{B}(H)$ is a Banach space. We define a product on $\mathscr{B}(H)$ by $T_{1}T_{2}=T_{1}\circ T_{2}$, and thus $\mathscr{B}(H)$ is an algebra. We have

 $\left\|T_{1}T_{2}\right\|=\sup_{\left\|x\right\|\leq 1}\left\|T_{1}(T_{2}x)% \right\|\leq\sup_{\left\|x\right\|\leq 1}\left\|T_{1}\right\|\left\|T_{2}x% \right\|=\left\|T_{1}\right\|\sup_{\left\|x\right\|\leq 1}\left\|T_{2}x\right% \|\leq\left\|T_{1}\right\|\left\|T_{2}\right\|,$

and thus $\mathscr{B}(H)$ is a Banach algebra.11 1 The adjoint map $*:\mathscr{B}(H)\to\mathscr{B}(H)$ satisfies, for $\lambda\in\mathbb{C}$ and $T_{1},T_{2}\in\mathscr{B}(H)$, $T^{**}=T,\quad(T_{1}+T_{2})^{*}=T_{1}^{*}+T_{2}^{*},\quad(\lambda T)^{*}=% \overline{\lambda}T^{*},\quad\left\|T^{*}T\right\|=\left\|T\right\|^{2}.$ Thus $\mathscr{B}(H)$ is a $C^{*}$-algebra. $I\in\mathscr{B}(H)$, so we say that $\mathscr{B}(H)$ is unital. Let $\mathscr{B}_{\mathrm{sa}}(H)$ be the set of all $T\in\mathscr{B}(H)$ that are self-adjoint.

###### Theorem 1.

If $T\in\mathscr{B}(H)$, then $T$ is self-adjoint if and only if $\left\langle Tx,x\right\rangle\in\mathbb{R}$ for all $x\in H$.

###### Proof.

If $T\in\mathscr{B}_{\mathrm{sa}}(H)$, then for all $x\in H$,

 $\left\langle Tx,x\right\rangle=\left\langle x,T^{*}x\right\rangle=\left\langle x% ,Tx\right\rangle=\overline{\left\langle Tx,x\right\rangle},$

so $\left\langle Tx,x\right\rangle\in\mathbb{R}$.

If $T\in\mathscr{B}(H)$ and $\left\langle Tx,x\right\rangle\in\mathbb{R}$ for all $x\in H$, then

 $\left\langle Tx,x\right\rangle=\left\langle x,T^{*}x\right\rangle=\overline{% \left\langle T^{*}x,x\right\rangle}=\left\langle T^{*}x,x\right\rangle,$

so, putting $A=T-T^{*}$, for all $x\in H$ we have

 $\left\langle Ax,x\right\rangle=0.$

Thus, for all $x,y\in H$ we have

 $\left\langle Ax,x\right\rangle=0,\qquad\left\langle Ay,y\right\rangle=0,\qquad% \left\langle A(x+y),x+y\right\rangle=0,$

and combining these three equations,

 $0=\left\langle Ax,x\right\rangle+\left\langle Ax,y\right\rangle+\left\langle Ay% ,x\right\rangle+\left\langle Ay,y\right\rangle=0+\left\langle Ax,y\right% \rangle+\left\langle Ay,x\right\rangle+0.$

But $A^{*}=-A$, so we get

 $\left\langle Ax,y\right\rangle+\left\langle y,-Ax\right\rangle=0,$

hence

 $\left\langle Ax,y\right\rangle-\overline{\left\langle Ax,y\right\rangle}=0.$ (1)

As well, for all $x,y\in H$ we have

 $\left\langle Ax,-iy\right\rangle-\overline{\left\langle Ax,-iy\right\rangle}=0,$

so

 $\left\langle Ax,y\right\rangle+\overline{\left\langle Ax,y\right\rangle}=0.$ (2)

By (1) and (2), for all $x,y\in H$ we have

 $\left\langle Ax,y\right\rangle=0,$

and thus $A=0$, i.e. $T=T^{*}$. ∎

Using the above characterization of bounded self-adjoint operators, we can prove that a limit of bounded self-adjoint operators is itself a bounded self-adjoint operator.

###### Theorem 2.

$\mathscr{B}_{\mathrm{sa}}(H)$ is a closed subset of $\mathscr{B}(H)$.

###### Proof.

If $T_{n}\in\mathscr{B}_{\mathrm{sa}}(H)$ and $T_{n}\to T\in\mathscr{B}(H)$, then for $x\in H$ we have

 $\left\langle Tx,x\right\rangle=\lim_{n\to\infty}\left\langle T_{n}x,x\right% \rangle\in\mathbb{R},$

hence $T\in\mathscr{B}_{\mathrm{sa}}(H)$. ∎

If $T\in\mathscr{B}_{\mathrm{sa}}(H)$ and $\left\langle Tx,x\right\rangle\geq 0$ for all $x\in H$, we say that $T$ is positive. Let $\mathscr{B}_{\mathrm{+}}(H)$ be the set of all positive $T\in\mathscr{B}_{\mathrm{sa}}(H)$. For $S,T\in\mathscr{B}_{\mathrm{sa}}(H)$, if

 $T-S\in\mathscr{B}_{\mathrm{+}}(H)$

we write $S\leq T$. Thus, we can talk about one self-adjoint operator being greater than or equal to another self-adjoint operator. $S\leq T$ is equivalent to

 $\left\langle Sx,x\right\rangle\leq\left\langle Tx,x\right\rangle$

for all $x\in H$.

## 3 A condition for invertibility

###### Theorem 3.

If $T\in\mathscr{B}(H)$ and there is some $\alpha>0$ such that $\alpha I\leq TT^{*}$ and $\alpha I\leq T^{*}T$, then $T^{-1}\in\mathscr{B}(H)$.

###### Proof.

By $\alpha I\leq T^{*}T$, we have for all $x\in H$,

 $\left\|Tx\right\|^{2}=\left\langle Tx,Tx\right\rangle=\left\langle T^{*}Tx,x% \right\rangle\geq\left\langle\alpha x,x\right\rangle=\alpha\left\|x\right\|^{2},$

so $\left\|Tx\right\|\geq\sqrt{\alpha}\left\|x\right\|$. This implies that $T$ is injective. By $\alpha I\leq TT^{*}$, we have for all $x\in H$,

 $\left\|T^{*}x\right\|^{2}=\left\langle T^{*}x,T^{*}x\right\rangle=\left\langle TT% ^{*}x,x\right\rangle\geq\left\langle\alpha x,x\right\rangle=\alpha\left\|x% \right\|^{2},$

so $\left\|T^{*}x\right\|\geq\sqrt{\alpha}\left\|x\right\|$, and hence $T^{*}$ is injective. Let $Tx_{n}\to y\in H$. Then,

 $\left\|Tx_{n}-Tx_{m}\right\|^{2}=\left\|T(x_{n}-x_{m})\right\|^{2}\geq\alpha% \left\|x_{n}-x_{m}\right\|^{2}.$

Since $Tx_{n}$ converges it is a Cauchy sequence, and from the above inequality it follows that $x_{n}$ is a Cauchy sequence, hence there is some $x\in H$ with $x_{n}\to x$. As $T$ is continuous, $y=Tx\in T(H)$, showing that $T(H)$ is a closed subset of $H$. But it is a fact that if $T\in\mathscr{B}(H)$ then the closure of $T(H)$ is equal to $(\ker T^{*})^{\perp}$.22 2 It is straightforward to show that if $v$ is in the closure of $T(H)$ and $w\in\ker T^{*}$ then $\left\langle v,w\right\rangle=0$. It is less straightforward to show the opposite inclusion. Thus, as we have shown that $T^{*}$ is injective,

 $T(H)=(\ker T^{*})^{\perp}=\{0\}^{\perp}=H,$

i.e. $T$ is surjective. Hence $T:H\to H$ is bijective. It is a fact that if $T\in\mathscr{B}(H)$ is bijective then $T^{-1}\in\mathscr{B}(H)$, completing the proof.33 3 $T^{-1}:H\to H$ is linear. The open mapping theorem states that if $X$ and $Y$ are Banach spaces and $S:X\to Y$ is a bounded linear map that is surjective, then $S$ is an open map, i.e., if $U$ is an open subset of $X$ then $S(U)$ is an open subset of $Y$. Here, $T\in\mathscr{B}(H)$ and $T$ is bijective, and so by the open mapping theorem $T$ is open, from which it follows that $T^{-1}:H\to H$ is continuous, and so bounded (a linear map between normed vector spaces is continuous if and only if it is bounded).

## 4 Spectrum

For $T\in\mathscr{B}(H)$, we define the spectrum $\sigma(T)$ of $T$ to be the set of all $\lambda\in\mathbb{C}$ such $T-\lambda I$ is not bijective, and we define the resolvent set of $T$ to be $\rho(T)=\mathbb{C}\setminus\sigma(T)$. To say that $\lambda\in\rho(T)$ is to say that $T-\lambda I$ is a bijection, and if $T-\lambda I$ is a bijection it follows from the open mapping theorem that its inverse function is an element of $\mathscr{B}(H)$: the inverse of a linear bijection is itself linear, but the inverse of a continuous bijection need not itself be continuous, which is where we use the open mapping theorem.

We prove that the spectrum of a bounded self-adjoint operator is real.

###### Theorem 4.

If $T\in\mathscr{B}_{\mathrm{sa}}(H)$, then $\sigma(T)\subseteq\mathbb{R}$.

###### Proof.

If $\lambda\in\mathbb{C}\setminus\mathbb{R}$, $\lambda=a+ib$, $b\neq 0$, and $X=T-\lambda I$, then

 $\displaystyle XX^{*}$ $\displaystyle=$ $\displaystyle(T-\lambda I)(T-\lambda I)^{*}$ $\displaystyle=$ $\displaystyle(T-(a+ib)I)(T-(a-ib)I)$ $\displaystyle=$ $\displaystyle T^{2}-(a-ib)T-(a+ib)T+(a^{2}+b^{2})I$ $\displaystyle=$ $\displaystyle(a^{2}+b^{2})I-2aT+T^{2}$ $\displaystyle=$ $\displaystyle b^{2}I+(aI-T)^{2}$ $\displaystyle=$ $\displaystyle b^{2}I+(aI-T)(aI-T)^{*}$ $\displaystyle\geq$ $\displaystyle b^{2}I.$

$X^{*}X=XX^{*}\geq b^{2}I$ and $b>0$, so by Theorem 3, $X=T-\lambda I$ has an inverse $(T-\lambda I)^{-1}\in\mathscr{B}(H)$, showing $\lambda\not\in\sigma(T)$. ∎

## 5 The spectrum of a bounded linear map is bounded

If $\lambda\in\rho(T)$ then we define $R_{\lambda}=(T-\lambda I)^{-1}\in\mathscr{B}(H)$, called the resolvent of $T$.

###### Theorem 5.

If $T\in\mathscr{B}(H)$ and $|\lambda|>\left\|T\right\|$ then $\lambda\in\rho(T)$.

###### Proof.

Define $R_{\lambda,N}\in\mathscr{B}(H)$ by

 $R_{\lambda,N}=-\frac{1}{\lambda}\sum_{n=0}^{N}\frac{T^{n}}{\lambda^{n}}.$

As $\frac{\left\|T\right\|}{|\lambda|}<1$, the geometric series $\sum_{n=0}^{\infty}\frac{\left\|T\right\|^{n}}{|\lambda|^{n}}$ converges, from which it follows that $R_{\lambda,N}$ is a Cauchy sequence in $\mathscr{B}(H)$ and so converges to some $S_{\lambda}\in\mathscr{B}(H)$. We have

 $\displaystyle\left\|S_{\lambda}(T-\lambda I)-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|S_{\lambda}(T-\lambda I)-R_{\lambda,N}(T-\lambda I)\right\|$ $\displaystyle+\left\|R_{\lambda,N}(T-\lambda I)-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|S_{\lambda}-R_{\lambda,N}\right\|\left\|T-\lambda I\right% \|+\left\|-\frac{T}{\lambda}\sum_{n=0}^{N}\frac{T^{n}}{\lambda^{n}}+\sum_{n=0}% ^{N}\frac{T^{n}}{\lambda^{n}}-I\right\|$ $\displaystyle=$ $\displaystyle\left\|S_{\lambda}-R_{\lambda,N}\right\|\left\|T-\lambda I\right% \|+\left\|-\frac{T^{N+1}}{\lambda^{N+1}}\right\|$ $\displaystyle\leq$ $\displaystyle\left\|S_{\lambda}-R_{\lambda,N}\right\|\left\|T-\lambda I\right% \|+\left(\frac{\left\|T\right\|}{|\lambda|}\right)^{N+1},$

which tends to $0$ as $N\to\infty$. Therefore $S_{\lambda}(T-\lambda I)=I$. And,

 $\displaystyle\left\|(T-\lambda I)S_{\lambda}-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|(T-\lambda I)S_{\lambda}-(T-\lambda I)R_{\lambda,N}\right\|$ $\displaystyle+\left\|(T-\lambda I)R_{\lambda,N}-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|T-\lambda I\right\|\left\|S_{\lambda}-R_{\lambda,N}\right% \|+\left(\frac{\left\|T\right\|}{|\lambda|}\right)^{N+1},$

whence $(T-\lambda I)S_{\lambda}=I$, showing that

 $S_{\lambda}=(T-\lambda I)^{-1}.$

Thus, if $|\lambda|>\left\|T\right\|$ then $\lambda\in\rho(T)$. ∎

The above theorem shows that $\sigma(T)$ is a bounded set: it is contained in the closed disc $|\lambda|\leq\left\|T\right\|$. Moreover, if $|\lambda|>\left\|T\right\|$ then we have an explicit expression for the resolvent $R_{\lambda}$:

 $R_{\lambda}=-\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{T^{n}}{\lambda^{n}}.$

## 6 The spectrum of a bounded linear map is closed

###### Theorem 6.

If $T\in\mathscr{B}(H)$, then $\rho(T)$ is an open subset of $\mathbb{C}$.

###### Proof.

If $\lambda\in\rho(T)$, let $|\mu-\lambda|<\left\|R_{\lambda}\right\|^{-1}$, and define $R_{\mu,N}\in\mathscr{B}(H)$ by

 $R_{\mu,N}=R_{\lambda}\sum_{n=0}^{N}(\mu-\lambda)^{n}R_{\lambda}^{n}.$

Because $|\mu-\lambda|<\left\|R_{\lambda}\right\|^{-1}$, $R_{\mu,N}$ is a Cauchy sequence in $\mathscr{B}(H)$ and converges to some $S_{\mu}\in\mathscr{B}(H)$. We have, as $R_{\lambda}=(T-\lambda I)^{-1}$,

 $\displaystyle\left\|S_{\mu}(T-\mu I)-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|S_{\mu}(T-\mu I)-R_{\mu,N}(T-\mu I)\right\|$ $\displaystyle+\left\|R_{\mu,N}(T-\mu I+\lambda I-\lambda I)-I\right\|$ $\displaystyle\leq$ $\displaystyle\left\|S_{\mu}-R_{\mu,N}\right\|\left\|T-\mu I\right\|$ $\displaystyle+\left\|R_{\mu,N}(T-\lambda I)-R_{\mu,N}(\mu-\lambda)-I\right\|$ $\displaystyle=$ $\displaystyle\left\|S_{\mu}-R_{\mu,N}\right\|\left\|T-\mu I\right\|$ $\displaystyle+\left\|\sum_{n=0}^{N}(\mu-\lambda)^{n}R_{\lambda}^{n}-(\mu-% \lambda)R_{\lambda}\sum_{n=0}^{N}(\mu-\lambda)^{n}R_{\lambda}^{n}-I\right\|$ $\displaystyle=$ $\displaystyle\left\|S_{\mu}-R_{\mu,N}\right\|\left\|T-\mu I\right\|+\left\|-(% \mu-\lambda)^{N+1}R_{\lambda}^{N+1}\right\|$ $\displaystyle=$ $\displaystyle\left\|S_{\mu}-R_{\mu,N}\right\|\left\|T-\mu I\right\|+|\mu-% \lambda|^{N+1}\left\|R_{\lambda}\right\|^{N+1},$

which tends to $0$ as $N\to\infty$. Therefore $S_{\mu}(T-\mu I)=I$. One checks likewise that $(T-\mu I)S_{\mu}=I$, and hence that

 $(T-\mu I)^{-1}=S_{\mu},$

showing that $\mu\in\rho(T)$. ∎

As $\sigma(T)$ is bounded and closed, it is a compact set in $\mathbb{C}$. Moreover, if $\lambda\not\in\sigma(T)$ and $|\mu-\lambda|<\left\|R_{\lambda}\right\|^{-1}$, then

 $R_{\mu}=R_{\lambda}\sum_{n=0}^{\infty}(\mu-\lambda)^{n}R_{\lambda}^{n}.$

## 7 The spectrum of a bounded linear map is nonempty

###### Theorem 7.

If $T\in\mathscr{B}(H)$ is self-adjoint, then $\sigma(T)\neq\emptyset$.

###### Proof.

Suppose by contradiction that $\sigma(T)=\emptyset$.44 4 For each $v,w\in H$ we are going to construct a bounded entire function $\mathbb{C}\to\mathbb{C}$ depending on $v$ and $w$, which by Liouville’s theorem must be constant, and it will turn out to be 0. This will lead to a contradiction. If $\lambda,\mu\in\mathbb{C}$, then

 $\displaystyle(T-\lambda I)(R_{\lambda}-R_{\mu})(T-\mu I)$ $\displaystyle=$ $\displaystyle(I-(T-\lambda I)R_{\mu})(T-\mu I)$ $\displaystyle=$ $\displaystyle T-\mu I-(T-\lambda I)$ $\displaystyle=$ $\displaystyle(\lambda-\mu)I,$

so

 $R_{\lambda}-R_{\mu}=(\lambda-\mu)R_{\lambda}R_{\mu},$ (3)

the resolvent identity. Thus

 $\left\|R_{\lambda}-R_{\mu}\right\|\leq|\lambda-\mu|\left\|R_{\lambda}\right\|% \left\|R_{\mu}\right\|,$

and together with $\left\|R_{\mu}\right\|-\left\|R_{\lambda}\right\|\leq\left\|R_{\mu}-R_{\lambda% }\right\|$ we get

 $\left\|R_{\mu}\right\|(1-|\lambda-\mu|\left\|R_{\lambda}\right\|)\leq\left\|R_% {\lambda}\right\|.$

If $|\lambda-\mu|\leq\frac{1}{2}\cdot\left\|R_{\lambda}\right\|^{-1}$, then

 $\left\|R_{\mu}\right\|\leq 2\left\|R_{\lambda}\right\|,$

whence, for $|\lambda-\mu|\leq\frac{1}{2}\cdot\left\|R_{\lambda}\right\|^{-1}$,

 $\left\|R_{\lambda}-R_{\mu}\right\|\leq 2|\lambda-\mu|\left\|R_{\lambda}\right% \|^{2}.$

Therefore, $\lambda\mapsto R_{\lambda}$ is a continuous function $\mathbb{C}\to\mathscr{B}(H)$. From this and (3) it follows that for each $\lambda\in\mathbb{C}$,55 5 There are no complications that appear if we do complex analysis on functions from $\mathbb{C}$ to a complex Banach algebra rather than on functions from $\mathbb{C}$ to $\mathbb{C}$. Thus this statement is that $\lambda\to R_{\lambda}$ is a holomorphic function $\mathbb{C}\to\mathscr{B}(H)$.

 $\lim_{\mu\to\lambda}\frac{R_{\lambda}-R_{\mu}}{\lambda-\mu}=R_{\lambda}^{2}.$

Let $v,w\in H$ and define $f_{v,w}:\mathbb{C}\to\mathbb{C}$ by

 $f_{v,w}(\lambda)=\left\langle R_{\lambda}v,w\right\rangle,\qquad\lambda\in% \mathbb{C}.$

For $\lambda\in\mathbb{C}$,

 $\lim_{\mu\to\lambda}\frac{f_{v,w}(\lambda)-f_{v,w}(\mu)}{\lambda-\mu}=\lim_{% \mu\to\lambda}\left\langle\frac{R_{\lambda}-R_{\mu}}{\lambda-\mu}v,w\right% \rangle=\left\langle R_{\lambda}^{2}v,w\right\rangle.$

Thus $f_{v,w}$ is an entire function. For $|\lambda|>\left\|T\right\|$, $R_{\lambda}=-\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{T^{n}}{\lambda^{n}}$, so, for $r=\frac{\left\|T\right\|}{|\lambda|}$,

 $\displaystyle\left\|R_{\lambda}\right\|$ $\displaystyle=$ $\displaystyle\frac{1}{|\lambda|}\left\|\sum_{n=0}^{\infty}\frac{T^{n}}{\lambda% }\right\|$ $\displaystyle\leq$ $\displaystyle\frac{1}{|\lambda|}\sum_{n=0}^{\infty}r^{n}$ $\displaystyle=$ $\displaystyle\frac{1}{|\lambda|}\frac{1}{1-r}$ $\displaystyle=$ $\displaystyle\frac{1}{|\lambda|}\frac{1}{1-\frac{\left\|T\right\|}{|\lambda|}}$ $\displaystyle=$ $\displaystyle\frac{1}{|\lambda|-\left\|T\right\|}.$

Hence, for $|\lambda|>\left\|T\right\|$,

 $\displaystyle|f_{v,w}(\lambda)|$ $\displaystyle=$ $\displaystyle|\left\langle R_{\lambda}v,w\right\rangle|$ $\displaystyle\leq$ $\displaystyle\left\|R_{\lambda}\right\|\left\|v\right\|\left\|w\right\|$ $\displaystyle\leq$ $\displaystyle\frac{\left\|v\right\|\left\|w\right\|}{|\lambda|-\left\|T\right% \|},$

from which it follows that $f_{v,w}$ is bounded and that $\lim_{|\lambda|\to\infty}f_{v,w}(\lambda)=0$. Therefore by Liouville’s theorem, $f_{v,w}(\lambda)=0$ for all $\lambda$. Let’s recap: for all $v,w\in H$ and for all $\lambda\in\mathbb{C}$, $\left\langle R_{\lambda}v,w\right\rangle=0$. Switching the order of the universal quantifiers, for all $\lambda\in\mathbb{C}$ and for all $v,w\in H$ we have $\left\langle R_{\lambda}v,w\right\rangle=0$, which implies that for all $\lambda\in\mathbb{C}$ we have $R_{\lambda}=0$. But by assumption $R_{\lambda}$ is invertible, so this is a contradiction. Hence $\sigma(T)$ is nonempty. ∎