The spectrum of a self-adjoint operator is a compact subset of $ℝ$

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 $ℂ$, and that if the operator is self-adjoint then the spectrum is contained in $ℝ$. To show that the spectrum is nonempty I prove various facts about resolvents.

1 Adjoints

1.1 Operator norm

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

$$\parallel T\parallel =\underset{\parallel v\parallel \le 1}{sup}\parallel Tv\parallel .$$

namely, the operator norm of $T$.

1.2 Definition of adjoint

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

$$\varphi (x)=⟨x,v_{\varphi }⟩$$

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

$${\varphi }_y(x)=⟨Tx,y⟩.$$

${\varphi }_y:H\to ℂ$ is a bounded linear map, so by the Riesz representation theorem there is a unique $v_y$ such that

$${\varphi }_y(x)=⟨x,v_y⟩$$

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 $x,y\in H$,

$$⟨x,T^{*}y⟩=⟨x,v_y⟩={\varphi }_y(x)=⟨Tx,y⟩.$$

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

$⟨x,T^{*}(y_1+y_2)⟩$ $=$ $⟨Tx,y_1+y_2⟩$

$=$ $⟨Tx,y_1⟩+⟨Tx,y_2⟩$

$=$ $⟨x,T^{*}{y}_1+T^{*}{y}_2⟩.$

Hence for all $x\in H$,

$$⟨x,T^{*}(y_1+y_2)-T^{*}{y}_1-T^{*}{y}_2⟩=0.$$

In particular this is true for $x=T^{*}(y_1+y_2)-T^{*}{y}_1-T^{*}{y}_2$, so by the nondegeneracy of $⟨\cdot ,\cdot ⟩$ we get

$$T^{*}(y_1+y_2)-T^{*}{y}_1-T^{*}{y}_2=0.$$

We similarly obtain for all $\lambda \in ℂ$ 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

$$|{\varphi }_y(x)|=|⟨x,v_y⟩|\le \parallel x\parallel \parallel v_y\parallel ,$$

so $\parallel {\varphi }_y\parallel \le \parallel v_y\parallel$, i.e. the operator norm of ${\varphi }_y$ is less than or equal to the norm of $v_y$. If $v_y\ne 0$, then $\parallel \frac{v_y}{\parallel v_y\parallel }\parallel =1$ and

$$\left|{\varphi }_y\left(\frac{v_y}{\parallel v_y\parallel }\right)\right|=⟨\frac{v_y}{\parallel v_y\parallel },v_y⟩=\parallel v_y\parallel .$$

It follows that

$$\parallel {\varphi }_y\parallel =\parallel v_y\parallel .$$

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

$\parallel T^{*}y\parallel$ $=$ $\parallel v_y\parallel$

$=$ $\parallel {\varphi }_y\parallel$

$=$ $\underset{\parallel x\parallel \le 1}{sup}\parallel {\varphi }_y(x)\parallel$

$=$ $\underset{\parallel x\parallel \le 1}{sup}|⟨Tx,y⟩|$

$\le$ $\underset{\parallel x\parallel \le 1}{sup}\parallel T\parallel \parallel x\parallel \parallel y\parallel$

$\le$ $\parallel T\parallel \parallel y\parallel .$

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 $x,y\in H$,

$⟨Tx,y⟩$ $=$ $⟨x,T^{*}y⟩$

$=$ $\overline{⟨T^{*}y,x⟩}$

$=$ $\overline{⟨y,T^{**}x⟩}$

$=$ $⟨T^{**}x,y⟩.$

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

$$⟨Tx-T^{**}x,y⟩=0.$$

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

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

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 $\parallel T^{*}\parallel \le \parallel T\parallel$. Hence also $\parallel T\parallel =\parallel T^{**}\parallel \le \parallel T^{*}\parallel$, so

$$\parallel T\parallel =\parallel T^{*}\parallel .$$

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

2 Bounded linear operators

Let $ℬ(H)$ be the set of bounded linear maps $H\to H$. With the operator norm, one checks that $ℬ(H)$ is a Banach space. We define a product on $ℬ(H)$ by $T_1{T}_2=T_1\circ T_2$, and thus $ℬ(H)$ is an algebra. We have

$$\parallel T_1{T}_2\parallel =\underset{\parallel x\parallel \le 1}{sup}\parallel T_1(T_{2}x)\parallel \le \underset{\parallel x\parallel \le 1}{sup}\parallel T_1\parallel \parallel T_{2}x\parallel =\parallel T_1\parallel \underset{\parallel x\parallel \le 1}{sup}\parallel T_{2}x\parallel \le \parallel T_1\parallel \parallel T_2\parallel ,$$

and thus $ℬ(H)$ is a Banach algebra. Let ${ℬ}_{\mathrm{sa}}(H)$ be the set of all $T\in ℬ(H)$ that are self-adjoint.

Theorem 1.

If $T\mathrm{\in }\mathrm{B}\mathit{}\mathrm{(}H\mathrm{)}$, then $T$ is self-adjoint if and only if $\mathrm{⟨}T\mathit{}x\mathrm{,}x\mathrm{⟩}\mathrm{\in }\mathrm{R}$ for all $x\mathrm{\in }H$.

Proof.

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

$$⟨Tx,x⟩=⟨x,T^{*}x⟩=⟨x,Tx⟩=\overline{⟨Tx,x⟩},$$

so $⟨Tx,x⟩\in ℝ$.

If $T\in ℬ(H)$ and $⟨Tx,x⟩\in ℝ$ for all $x\in H$, then

$$⟨Tx,x⟩=⟨x,T^{*}x⟩=\overline{⟨T^{*}x,x⟩}=⟨T^{*}x,x⟩,$$

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

$$⟨Ax,x⟩=0.$$

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

$$⟨Ax,x⟩=0,⟨Ay,y⟩=0,⟨A(x+y),x+y⟩=0,$$

and combining these three equations,

$$0=⟨Ax,x⟩+⟨Ax,y⟩+⟨Ay,x⟩+⟨Ay,y⟩=0+⟨Ax,y⟩+⟨Ay,x⟩+0.$$

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

$$⟨Ax,y⟩+⟨y,-Ax⟩=0,$$

hence

$$⟨Ax,y⟩-\overline{⟨Ax,y⟩}=0.$$ (1)

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

$$⟨Ax,-iy⟩-\overline{⟨Ax,-iy⟩}=0,$$

so

$$⟨Ax,y⟩+\overline{⟨Ax,y⟩}=0.$$ (2)

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

$$⟨Ax,y⟩=0,$$

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.

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

Proof.

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

$$⟨Tx,x⟩=\underset{n\to \mathrm{\infty }}{lim}⟨T_{n}x,x⟩\in ℝ,$$

hence $T\in {ℬ}_{\mathrm{sa}}(H)$. ∎

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

$$T-S\in {ℬ}_{+}(H)$$

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

$$⟨Sx,x⟩\le ⟨Tx,x⟩$$

for all $x\in H$.

3 A condition for invertibility

Theorem 3.

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

Proof.

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

$${\parallel Tx\parallel }^2=⟨Tx,Tx⟩=⟨T^{*}Tx,x⟩\ge ⟨\alpha x,x⟩=\alpha {\parallel x\parallel }^2,$$

so $\parallel Tx\parallel \ge \sqrt{\alpha }\parallel x\parallel$. This implies that $T$ is injective. By $\alpha I\le T{T}^{*}$, we have for all $x\in H$,

$${\parallel T^{*}x\parallel }^2=⟨T^{*}x,T^{*}x⟩=⟨T{T}^{*}x,x⟩\ge ⟨\alpha x,x⟩=\alpha {\parallel x\parallel }^2,$$

so $\parallel T^{*}x\parallel \ge \sqrt{\alpha }\parallel x\parallel$, and hence $T^{*}$ is injective. Let $T{x}_n\to y\in H$. Then,

$${\parallel T{x}_n-T{x}_m\parallel }^2={\parallel T(x_n-x_m)\parallel }^2\ge \alpha {\parallel x_n-x_m\parallel }^2.$$

Since $T{x}_n$ 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 $T\in ℬ(H)$ then the closure of $T(H)$ is equal to ${(\ker T^{*})}^{⟂}$. Thus, as we have shown that $T^{*}$ is injective,

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

i.e. $T$ is surjective. Hence $T:H\to H$ is bijective. It is a fact that if $T\in ℬ(H)$ is bijective then $T^{-1}\in ℬ(H)$, completing the proof. ∎

4 Spectrum

For $T\in ℬ(H)$, we define the spectrum $\sigma (T)$ of $T$ to be the set of all $\lambda \in ℂ$ such $T-\lambda I$ is not bijective, and we define the resolvent set of $T$ to be $\rho (T)=ℂ\setminus \sigma (T)$. 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 $ℬ(H)$: 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 $T\mathrm{\in }{\mathrm{B}}_{\mathrm{sa}}\mathit{}\mathrm{(}H\mathrm{)}$, then $\sigma \mathit{}\mathrm{(}T\mathrm{)}\mathrm{\subseteq }\mathrm{R}$.

Proof.

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

$X{X}^{*}$ $=$ $(T-\lambda I){(T-\lambda I)}^{*}$

$=$ $(T-(a+ib)I)(T-(a-ib)I)$

$=$ $T^2-(a-ib)T-(a+ib)T+(a^2+b^2)I$

$=$ $(a^2+b^2)I-2aT+T^2$

$=$ $b^{2}I+{(aI-T)}^2$

$=$ $b^{2}I+(aI-T){(aI-T)}^{*}$

$\ge$ $b^{2}I.$

$X^{*}X=X{X}^{*}\ge b^{2}I$ and $b>0$, so by Theorem 3, $X=T-\lambda I$ has an inverse ${(T-\lambda I)}^{-1}\in ℬ(H)$, showing $\lambda \notin \sigma (T)$. ∎

5 The spectrum of a bounded linear map is bounded

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

Theorem 5.

If $T\mathrm{\in }\mathrm{B}\mathit{}\mathrm{(}H\mathrm{)}$ and $\mathrm{|}\lambda \mathrm{|}\mathrm{>}\mathrm{\parallel }T\mathrm{\parallel }$ then $\lambda \mathrm{\in }\rho \mathit{}\mathrm{(}T\mathrm{)}$.

Proof.

Define $R_{\lambda ,N}\in ℬ(H)$ by

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

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

$\parallel S_{\lambda }(T-\lambda I)-I\parallel$ $\le$ $\parallel S_{\lambda }(T-\lambda I)-R_{\lambda ,N}(T-\lambda I)\parallel$

$+\parallel R_{\lambda ,N}(T-\lambda I)-I\parallel$

$\le$ $\parallel S_{\lambda }-R_{\lambda ,N}\parallel \parallel T-\lambda I\parallel +\parallel -{\displaystyle \frac{T}{\lambda }}{\displaystyle \sum _{n=0}^N}{\displaystyle \frac{T^n}{{\lambda }^n}}+{\displaystyle \sum _{n=0}^N}{\displaystyle \frac{T^n}{{\lambda }^n}}-I\parallel$

$=$ $\parallel S_{\lambda }-R_{\lambda ,N}\parallel \parallel T-\lambda I\parallel +\parallel -{\displaystyle \frac{T^{N+1}}{{\lambda }^{N+1}}}\parallel$

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

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

$\parallel (T-\lambda I)S_{\lambda }-I\parallel$ $\le$ $\parallel (T-\lambda I)S_{\lambda }-(T-\lambda I)R_{\lambda ,N}\parallel$

$+\parallel (T-\lambda I)R_{\lambda ,N}-I\parallel$

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

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

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

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

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

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

6 The spectrum of a bounded linear map is closed

Theorem 6.

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

Proof.

If $\lambda \in \rho (T)$, let , and define $R_{\mu ,N}\in ℬ(H)$ by

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

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

$\parallel S_{\mu }(T-\mu I)-I\parallel$ $\le$ $\parallel S_{\mu }(T-\mu I)-R_{\mu ,N}(T-\mu I)\parallel$

$+\parallel R_{\mu ,N}(T-\mu I+\lambda I-\lambda I)-I\parallel$

$\le$ $\parallel S_{\mu }-R_{\mu ,N}\parallel \parallel T-\mu I\parallel$

$+\parallel R_{\mu ,N}(T-\lambda I)-R_{\mu ,N}(\mu -\lambda )-I\parallel$

$=$ $\parallel S_{\mu }-R_{\mu ,N}\parallel \parallel T-\mu I\parallel$

$+\parallel {\displaystyle \sum _{n=0}^N}{(\mu -\lambda )}^n{R}_{\lambda }^n-(\mu -\lambda )R_{\lambda }{\displaystyle \sum _{n=0}^N}{(\mu -\lambda )}^n{R}_{\lambda }^n-I\parallel$

$=$ $\parallel S_{\mu }-R_{\mu ,N}\parallel \parallel T-\mu I\parallel +\parallel -{(\mu -\lambda )}^{N+1}{R}_{\lambda }^{N+1}\parallel$

$=$ $\parallel S_{\mu }-R_{\mu ,N}\parallel \parallel T-\mu I\parallel +{|\mu -\lambda |}^{N+1}{\parallel R_{\lambda }\parallel }^{N+1},$

which tends to $0$ as $N\to \mathrm{\infty }$. 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 $ℂ$. Moreover, if $\lambda \notin \sigma (T)$ and , then

$$R_{\mu }=R_{\lambda }\sum _{n=0}^{\mathrm{\infty }}{(\mu -\lambda )}^n{R}_{\lambda }^n.$$

7 The spectrum of a bounded linear map is nonempty

Theorem 7.

If $T\mathrm{\in }\mathrm{B}\mathit{}\mathrm{(}H\mathrm{)}$ is self-adjoint, then $\sigma \mathit{}\mathrm{(}T\mathrm{)}\mathrm{\ne }\mathrm{\varnothing }$.

Proof.

Suppose by contradiction that $\sigma (T)=\mathrm{\varnothing }$. If $\lambda ,\mu \in ℂ$, then

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

$=$ $T-\mu I-(T-\lambda I)$

$=$ $(\lambda -\mu )I,$

so

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

the resolvent identity. Thus

$$\parallel R_{\lambda }-R_{\mu }\parallel \le |\lambda -\mu |\parallel R_{\lambda }\parallel \parallel R_{\mu }\parallel ,$$

and together with $\parallel R_{\mu }\parallel -\parallel R_{\lambda }\parallel \le \parallel R_{\mu }-R_{\lambda }\parallel$ we get

$$\parallel R_{\mu }\parallel (1-|\lambda -\mu |\parallel R_{\lambda }\parallel )\le \parallel R_{\lambda }\parallel .$$

If $|\lambda -\mu |\le \frac{1}{2}\cdot {\parallel R_{\lambda }\parallel }^{-1}$, then

$$\parallel R_{\mu }\parallel \le 2\parallel R_{\lambda }\parallel ,$$

whence, for $|\lambda -\mu |\le \frac{1}{2}\cdot {\parallel R_{\lambda }\parallel }^{-1}$,

$$\parallel R_{\lambda }-R_{\mu }\parallel \le 2|\lambda -\mu |{\parallel R_{\lambda }\parallel }^2.$$

Therefore, $\lambda \mapsto R_{\lambda }$ is a continuous function $ℂ\to ℬ(H)$. From this and (3) it follows that for each $\lambda \in ℂ$,

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

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

$$f_{v,w}(\lambda )=⟨R_{\lambda }v,w⟩,\lambda \in ℂ.$$

For $\lambda \in ℂ$,

$$\underset{\mu \to \lambda }{lim}\frac{f_{v,w}(\lambda )-f_{v,w}(\mu )}{\lambda -\mu }=\underset{\mu \to \lambda }{lim}⟨\frac{R_{\lambda }-R_{\mu }}{\lambda -\mu }v,w⟩=⟨R_{\lambda }^{2}v,w⟩.$$

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

$\parallel R_{\lambda }\parallel$ $=$ ${\displaystyle \frac{1}{|\lambda |}}\parallel {\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{T^n}{\lambda }}\parallel$

$\le$ ${\displaystyle \frac{1}{|\lambda |}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{r}^n$

$=$ ${\displaystyle \frac{1}{|\lambda |}}{\displaystyle \frac{1}{1-r}}$

$=$ ${\displaystyle \frac{1}{|\lambda |}}{\displaystyle \frac{1}{1-\frac{\parallel T\parallel }{|\lambda |}}}$

$=$ ${\displaystyle \frac{1}{|\lambda |-\parallel T\parallel }}.$

Hence, for $|\lambda |>\parallel T\parallel$,

$|f_{v,w}(\lambda )|$ $=$ $|⟨R_{\lambda }v,w⟩|$

$\le$ $\parallel R_{\lambda }\parallel \parallel v\parallel \parallel w\parallel$

$\le$ ${\displaystyle \frac{\parallel v\parallel \parallel w\parallel }{|\lambda |-\parallel T\parallel }},$

from which it follows that $f_{v,w}$ is bounded and that ${lim}_{|\lambda |\to \mathrm{\infty }}{f}_{v,w}(\lambda )=0$. 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 $\lambda \in ℂ$, $⟨R_{\lambda }v,w⟩=0$. Switching the order of the universal quantifiers, for all $\lambda \in ℂ$ and for all $v,w\in H$ we have $⟨R_{\lambda }v,w⟩=0$, which implies that for all $\lambda \in ℂ$ we have $R_{\lambda }=0$. But by assumption $R_{\lambda }$ is invertible, so this is a contradiction. Hence $\sigma (T)$ is nonempty. ∎