Banach Algebras Spectrum and Holomorphic Functional Calculus: Examples

The supremum norm is finite on every $f$: $\mathrm{\mid }f\mathrm{\mid }$ is continuous and real-valued on the nonempty compact $K$, so it attains a maximum by [L1]; the pointwise operations make $C(K,\mathbb{C})$ a commutative associative complex algebra with unit the constant function $1$, and $\mathrm{\parallel }fg{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ holds because $\mathrm{\mid }f(x)g(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$ while $\mathrm{\mid }1(x)\mathrm{\mid }=1$ gives $\mathrm{\parallel }1{\mathrm{\parallel }}_{\mathrm{\infty }}=1$.

Completeness: a $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$-Cauchy sequence $(f_n)$ is uniformly Cauchy, so by [L2] it converges uniformly to a continuous $f$; uniform convergence is convergence in the supremum norm, so $C(K,\mathbb{C})$ is complete.

Spectral inclusion: if $\lambda \mathrm{\notin }f[K]$ then $m:={\inf }_{x\in K}\mathrm{\mid }\lambda -f(x)\mathrm{\mid }>0$: the function $x\mapsto \mathrm{\mid }\lambda -f(x)\mathrm{\mid }$ is continuous on the nonempty compact $K$ and attains its minimum $m$ by [L1], and $m=0$ would mean $\lambda =f(x)$ for some $x$. Hence $1\mathrm{/}(\lambda -f)$ is a bounded continuous function with $\mathrm{\parallel }1\mathrm{/}(\lambda -f){\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}m$, and it is a two-sided inverse of $\lambda 1-f$; so $\lambda \in \rho (f)$.

Spectral equality: conversely, if $\lambda =f(x_0)$ for some $x_0\in K$ and $g$ were an inverse of $\lambda 1-f$, then evaluating the identity $g\cdot (\lambda 1-f)=1$ at $x_0$ would give $g(x_0)\cdot 0=1$, impossible; hence $\lambda \in \sigma (f)$. Combined with [step 2.2], $\sigma (f)=f[K]$.

$\mathcal{B}(X)$ is an associative complex algebra under composition and pointwise linear structure, with unit $1_X$; by [L1] the norm is submultiplicative and $\mathrm{\parallel }{1}_X\mathrm{\parallel }=1$, and by [L2] with $Y=X$ it is complete; hence it is a unital complex Banach algebra.

The space ${\mathbb{C}}^2$ with the maximum norm is a nonzero complex Banach space by [L3], so the argument of [step 1.1] applies to it and $\mathcal{B}({\mathbb{C}}^2)$ is a unital complex Banach algebra; for the explicit operators $U,V$ on that space $\mathrm{\parallel }U(x,y)\mathrm{\parallel }=\mathrm{\parallel }(y,0)\mathrm{\parallel }=\mathrm{\mid }y\mathrm{\mid }\le \mathrm{\parallel }(x,y)\mathrm{\parallel }$ and $\mathrm{\parallel }V(x,y)\mathrm{\parallel }=\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\parallel }(x,y)\mathrm{\parallel }$, so both are bounded, and $UV(x,y)=U(0,x)=(x,0)$ while $VU(x,y)=V(y,0)=(0,y)$, so $UV\mathrm{\ne }VU$ although $UV$ and $VU$ are the two coordinate projections.

Now let $\dim X\ge 2$ and choose linearly independent $u,v\in X$; put $M:=\mathrm{span}\{u,v\}$, a two-dimensional subspace, and let $P$ be a bounded projection with range $M$ by [L4]. The linear maps $S,T:M\to M$ defined by $S(u)=v$, $S(v)=0$ and $T(u)=0$, $T(v)=u$ are bounded by [L5], and they do not commute, since $ST(u)=S(0)=0$ while $TS(u)=T(v)=u\mathrm{\ne }0$. Then $S\circ P$ and $T\circ P$ are bounded operators on $X$ by [L1], and $P\circ S=S$, $P\circ T=T$ because $S$ and $T$ take values in $M$ while $P$ is the identity on $M$; hence $(S\circ P)(T\circ P)=(S\circ T)\circ P$ and $(T\circ P)(S\circ P)=(T\circ S)\circ P$, and these differ because the first sends $u$ to $(S\circ T)(u)=0$ while the second sends $u$ to $(T\circ S)(u)=u\mathrm{\ne }0$. So $\mathcal{B}(X)$ is noncommutative for every $X$ with $\dim X\ge 2$, while [step 2.1] provides the explicit witness on ${\mathbb{C}}^2$; by [step 1.1] the algebra is unital and Banach.

$M_n(\mathbb{C})$ is an associative complex algebra under matrix multiplication with unit $I$; by [L1] the operator norm is submultiplicative with $\mathrm{\parallel }I\mathrm{\parallel }=1$, and by [L4] the space is complete; hence it is a unital complex Banach algebra.

If $\det (\lambda I-A)\mathrm{\ne }0$ then $B:=\lambda I-A$ has the two-sided inverse $\det (B{)}^{-1}\mathrm{a}\mathrm{d}\mathrm{j}(B)$ by [L2], so $\lambda \mathrm{\notin }\sigma (A)$ by [L1].

Conversely, if $\lambda I-A$ has a two-sided inverse $C$ in the algebra of [step 1.1], then [L3] gives $\det (\lambda I-A)\det (C)=\det (I)=1$, so $\det (\lambda I-A)\mathrm{\ne }0$.

Combining [step 2.1] and [step 2.2] with the characterization of the spectrum in [L1]: $\lambda \in \sigma (A)$ precisely when $\lambda I-A$ is not invertible, which by the two steps happens precisely when $\det (\lambda I-A)=0$.

Put $r=\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$. Each set $N_j=\{\mathrm{\mid }m\mathrm{\mid }>r+1\mathrm{/}j\}$ is null: by the definition of the infimum there is an a.e. bound strictly below $r+1\mathrm{/}j$. A countable union of measurable null sets is null, by disjointifying the union and applying countable additivity. Outside ${\bigcup }_{j\ge 1}{N}_j$, $\mathrm{\mid }m\mathrm{\mid }\le r$. For any measurable $f$ of finite squared integral, the nonnegative integral therefore gives $\int \mathrm{\mid }mf{\mathrm{\mid }}^2\le r^2\int \mathrm{\mid }f{\mathrm{\mid }}^2$. Thus multiplication defines a bounded linear map $M_m[f]=[mf]$, independent of representatives by [L2]. The same reasoning applies to every essentially bounded measurable symbol $h$. If $h,k$ are such symbols, then on every class $M_h{M}_k[f]=[h(kf)]=[(hk)f]=M_{hk}[f]$, and $M_1=I$; a general multiplier need not be the identity. Since the measure is nonzero, some $E_k$ from [L5] has positive measure (otherwise their union is null). Its indicator has nonzero finite $L^2$ norm. Hence $L^2$ is a nonzero complex Banach space by [L1], and [L4] supplies its operator algebra. AC is inherited from [L4] and supplies the Countable Choice required by [L1].

If $\lambda \mathrm{\notin }\mathrm{essran}(m)$, then there is $\epsilon >0$ with $\mu (\{\mathrm{\mid }m-\lambda \mathrm{\mid }<\epsilon \})=0$; hence $\mathrm{\mid }m-\lambda \mathrm{\mid }\ge \epsilon$ almost everywhere and the measurable function $\varphi :=1\mathrm{/}(m-\lambda )$ on $\{\mathrm{\mid }m-\lambda \mathrm{\mid }\ge \epsilon \}$ (arbitrary, say $0$, on the null complement) is essentially bounded by $1\mathrm{/}\epsilon$. By step 1.1 its multiplication operator satisfies $M_{\varphi }(M_m-\lambda )=(M_m-\lambda )M_{\varphi }=1$ on classes. Thus $M_m-\lambda I$ is invertible, and so is its negative $\lambda I-M_m$, the shifted operator in [L3]; therefore $\lambda \in \rho (M_m)$.

If $\lambda \in \mathrm{essran}(m)$, then for every $n\ge 1$ the set $A_n:=\{\mathrm{\mid }m-\lambda \mathrm{\mid }<1\mathrm{/}n\}$ has positive measure; since $X={\bigcup }_k{E}_k$ with $\mu (E_k)<\mathrm{\infty }$, some $A_n\cap E_k$ has positive measure, since otherwise their countable union $A_n$ would be null. Take the least such $k$; this deterministic choice uses no choice principle.

For this least $k$ the normalized indicator $f_n:=\mu (A_n\cap E_k{)}^{-1\mathrm{/}2}{\mathbf{1}}_{A_n\cap E_k}$ is a unit vector in $L^2(\mu )$, and $\mathrm{\parallel }(M_m-\lambda )f_n{\mathrm{\parallel }}_2\le (1\mathrm{/}n)\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2=1\mathrm{/}n\to 0$; hence $M_m-\lambda$ is not bounded below, so by [L3] neither it nor its negative is invertible and $\lambda \in \sigma (M_m)$.

Steps [step 2.1] and [step 3.1] together give both inclusions, so $\sigma (M_m)=\mathrm{essran}(m)$; the boundedness assertion is [step 1.1].

Shift identities: $(Sx{)}_n=x_{n-1}$ for $n\ge 1$ and $(Sx{)}_0=0$, so $\mathrm{\parallel }Sx\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }S\mathrm{\parallel }=1$; moreover $S-\lambda$ is injective for every $\lambda$: from $(S-\lambda )x=0$ the recursion $x_{n-1}=\lambda x_n$ gives $x=0$ for $\lambda \mathrm{\ne }0$ and $Sx=0$ gives $x=0$ for $\lambda =0$, since $S$ is injective.

Spectral containment: if $\mathrm{\mid }\lambda \mathrm{\mid }>1$ then $\mathrm{\parallel }S\mathrm{/}\lambda \mathrm{\parallel }<1$ and $S-\lambda =-\lambda (1-S\mathrm{/}\lambda )$ is invertible by the Neumann series; if $\mathrm{\mid }\lambda \mathrm{\mid }<1$ then $\mathrm{\parallel }(S-\lambda )x\mathrm{\parallel }\ge (1-\mathrm{\mid }\lambda \mathrm{\mid })\mathrm{\parallel }x\mathrm{\parallel }$ by [L2], so $S-\lambda$ is bounded below.

Non-density in the open disc: for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ the vector $c:=({\bar{\lambda }}^n{)}_{n\ge 0}$ lies in ${\mathrm{\ell }}^2$ and annihilates the range: for every $x\in {\mathrm{\ell }}^2$, $⟨(S-\lambda )x,c⟩={\sum }_n{x}_n(\overline{c_{n+1}}-\lambda \overline{c_n})={\sum }_n{x}_n({\lambda }^{n+1}-\lambda {\lambda }^n)=0$. So $\mathrm{ran}(S-\lambda )$ is not dense for $\mathrm{\mid }\lambda \mathrm{\mid }<1$, and ${\sigma }_{cp}(S)\supseteq D$.

Approximate eigenvectors on the circle: for $\mathrm{\mid }\lambda \mathrm{\mid }=1$ and $N\ge 1$ put $v_N:=N^{-1\mathrm{/}2}{\sum }_{k<N}{\lambda }^{-k}{e}_k$, a unit vector; since $S{v}_N=N^{-1\mathrm{/}2}{\sum }_{k<N}{\lambda }^{-k}{e}_{k+1}$, the interior terms cancel in $(S-\lambda )v_N$ and only the two boundary terms survive, $(S-\lambda )v_N=N^{-1\mathrm{/}2}({\lambda }^{-(N-1)}{e}_N-\lambda e_0),$ so $\mathrm{\parallel }(S-\lambda )v_N\mathrm{\parallel }=\sqrt{2}{N}^{-1\mathrm{/}2}\le 2N^{-1\mathrm{/}2}\to 0$ and $S-\lambda$ is not bounded below.

Density on the unit circle: if $c\perp \mathrm{ran}(S-\lambda )$ then the same computation gives $c_{n+1}=\bar{\lambda }{c}_n$ for all $n$, that is, $c_n={\bar{\lambda }}^n{c}_0$. For $\mathrm{\mid }\lambda \mathrm{\mid }=1$ this makes $\mathrm{\mid }{c}_n\mathrm{\mid }=\mathrm{\mid }{c}_0\mathrm{\mid }$ for every $n$, so $c\in {\mathrm{\ell }}^2$ forces $c_0=0$ and the orthogonal complement is $\{0\}$: the range is dense there. For $\mathrm{\mid }\lambda \mathrm{\mid }<1$, on the other hand, $\mathrm{\mid }{c}_n\mathrm{\mid }=\mathrm{\mid }\lambda {\mathrm{\mid }}^n\mathrm{\mid }{c}_0\mathrm{\mid }$ decays, the vector $c=({\bar{\lambda }}^n{)}_{n\ge 0}$ is a nonzero element of ${\mathrm{\ell }}^2$ orthogonal to the range, and the range is not dense — the non-density already computed in [step 2.1].

Point spectrum empty and residual spectrum: injectivity is [step 1.1], so ${\sigma }_p(S)=\mathrm{\varnothing }$; for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ the range is non-dense by [step 2.1], so $\lambda \in {\sigma }_r(S)$ and also $\lambda \in {\sigma }_{cp}(S)$; for $\mathrm{\mid }\lambda \mathrm{\mid }>1$ the operator is invertible by [step 1.2], so those points are outside every spectral set.

Combining: $\sigma (S)=\bar{D}$ because $\mathrm{\mid }\lambda \mathrm{\mid }>1$ gives invertibility [step 1.2] and every $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$ lies in ${\sigma }_c(S)$ or ${\sigma }_r(S)$ by [step 3.2] and [step 2.2]; ${\sigma }_{ap}(S)=\mathbb{T}$ by [step 1.2] (bounded below inside the disc), [step 2.2] (on the circle) and [step 1.2] again (invertible, hence bounded below, outside); ${\sigma }_c(S)=\mathbb{T}$ because those points are injective with dense range [step 1.1, step 3.1] and non-surjective (else invertible); ${\sigma }_r(S)={\sigma }_{cp}(S)=D$ by [step 2.1], [step 3.1] and [step 3.2].

$E_{12}(x,y)=(y,0)$, so $\mathrm{\parallel }{E}_{12}(x,y)\mathrm{\parallel }=\mathrm{\mid }y\mathrm{\mid }\le \mathrm{\parallel }(x,y)\mathrm{\parallel }$ for every $(x,y)\in {\mathbb{C}}^2$ (operator norm on the Euclidean plane), with equality at $(0,1)$; hence $\mathrm{\parallel }{E}_{12}\mathrm{\parallel }=1$.

$E_{12}^2=0$, and $\det (\lambda I-E_{12})={\lambda }^2$, whose only zero is $\lambda =0$; by [L1] the spectrum is $\sigma (E_{12})=\{0\}$.

By [L2] the spectral radius is $r(E_{12})=\max \{\mathrm{\mid }z\mathrm{\mid }:z\in \{0\}\}=0<1=\mathrm{\parallel }{E}_{12}\mathrm{\parallel }$; hence the two quantities differ for this element.

$A(\bar{\mathbb{D}})$ is complete: if $(f_n)$ is uniformly Cauchy on $\bar{\mathbb{D}}$, then it converges uniformly to a continuous $f$ by [L1]; for every filled triangle contained in $\mathbb{D}$, its boundary integral of $f$ is the limit of the corresponding integrals of the holomorphic $f_n$ by [L2], and those integrals vanish by [L6]. Hence $f$ is holomorphic on $\mathbb{D}$ by [L1] and $A(\bar{\mathbb{D}})$ is closed under uniform limits.

Pointwise operations make $A(\bar{\mathbb{D}})$ a commutative complex algebra with unit $1$, and the supremum norm is submultiplicative with $\mathrm{\parallel }1{\mathrm{\parallel }}_{\mathrm{\infty }}=1$; by [step 1.1] the algebra is a unital commutative Banach algebra, and the restriction map $\rho$ is a unital algebra homomorphism.

The restriction map is isometric by the maximum modulus principle: $\mathrm{\parallel }\rho (f){\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{\mathbb{T}}\mathrm{\mid }f\mathrm{\mid }={\sup }_{\bar{\mathbb{D}}}\mathrm{\mid }f\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $f\in A(\bar{\mathbb{D}})$, using [L3] and continuity.

Spectrum in the disc algebra: if $\mathrm{\mid }\lambda \mathrm{\mid }>1$ then $1\mathrm{/}(\lambda -z)$ is holomorphic on a neighbourhood of $\bar{\mathbb{D}}$, so $\lambda -z$ is invertible in $A(\bar{\mathbb{D}})$; if $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$ then $\lambda =z(\lambda )$ with $\lambda \in \bar{\mathbb{D}}$, and evaluating the identity $g\cdot (\lambda -z)=1$ at $z=\lambda$ gives $g(\lambda )\cdot 0=1$, impossible; hence ${\sigma }_{A(\bar{\mathbb{D}})}(z)=\bar{\mathbb{D}}$.

Spectrum in $C(\mathbb{T})$: the restriction $\rho (z)$ is the function $\zeta \mapsto \zeta$ on the circle, whose image is $\mathbb{T}$; by [L4], ${\sigma }_{C(\mathbb{T})}(\rho (z))=\mathbb{T}$.

Comparing the two computations: ${\sigma }_{A(\bar{\mathbb{D}})}(z)=\bar{\mathbb{D}}⊋\mathbb{T}={\sigma }_{C(\mathbb{T})}(\rho (z))$, so the spectrum of the same element of the smaller algebra (identified with its image under the isometric embedding $\rho$ of [step 3.1]) is strictly larger than in the ambient algebra $C(\mathbb{T})$.

Associativity: expanding both sides of the associativity identity for $((\lambda ,a)(\mu ,b))(\nu ,c)$ and $(\lambda ,a)((\mu ,b)(\nu ,c))$ by bilinearity gives the common value $(\lambda \mu \nu ,\lambda \mu c+\lambda \nu b+\mu \nu a+\lambda (bc)+\mu (ac)+\nu (ab)+(ab)c):$ the left side produces $\lambda \mu c+\nu (\lambda b+\mu a+ab)+(\lambda b+\mu a+ab)c$ and the right side produces $\lambda (\mu c+\nu b+bc)+\mu \nu a+a(\mu c+\nu b+bc)$, and the two agree because scalars may be moved across the product, the multiplication of $A$ is bilinear, and $a(bc)=(ab)c$ by associativity.

The element $(1,0)$ is a two-sided identity: $(1,0)(\mu ,b)=(\mu ,b+0+0)=(\mu ,b)$ and $(\lambda ,a)(1,0)=(\lambda ,0+a+0)=(\lambda ,a)$. Submultiplicativity holds because $\mathrm{\parallel }(\lambda ,a)(\mu ,b)\mathrm{\parallel }=\mathrm{\mid }\lambda \mu \mathrm{\mid }+\mathrm{\parallel }\lambda b+\mu a+ab\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }b\mathrm{\parallel }+\mathrm{\mid }\mu \mathrm{\mid }\mathrm{\parallel }a\mathrm{\parallel }+\mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }=(\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\parallel }a\mathrm{\parallel })(\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\parallel }b\mathrm{\parallel })=\mathrm{\parallel }(\lambda ,a)\mathrm{\parallel }\mathrm{\parallel }(\mu ,b)\mathrm{\parallel }$ by [L1], and $\mathrm{\parallel }(1,0)\mathrm{\parallel }=1$.

Completeness: a sequence $({\lambda }_n,a_n)$ is Cauchy in the sum norm exactly when $({\lambda }_n)$ is Cauchy in $\mathbb{C}$ and $(a_n)$ is Cauchy in $A$ (the two inequalities $\mathrm{\mid }\lambda \mathrm{\mid },\mathrm{\parallel }a\mathrm{\parallel }\le \mathrm{\parallel }(\lambda ,a)\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\parallel }a\mathrm{\parallel }$ compare the norm with the maximum of the coordinate norms); since $\mathbb{C}$ and $A$ are complete by [L1], the coordinates converge and their pair is the limit; so $\tilde{A}$ is a complex Banach algebra.

The map $j(a):=(0,a)$ is isometric and multiplicative: $j(ab)=(0,ab)=(0,a)(0,b)=j(a)j(b)$, and $\mathrm{\parallel }j(a)\mathrm{\parallel }=0+\mathrm{\parallel }a\mathrm{\parallel }$; its image is a two-sided ideal because $(\lambda ,b)(0,a)=(0,\lambda a+ba)$ and $(0,a)(\mu ,b)=(0,\mu a+ab)$, and it is closed as the kernel of the continuous scalar projection $(\lambda ,a)\mapsto \lambda$.

By [L2] applied in $\tilde{A}$, the spectrum of $a\in A$ is the set of $\lambda$ with $(\lambda ,0)-(0,a)=(\lambda ,-a)$ not invertible, which is the convention displayed in the statement.