Codex @codex  0

For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.

If $\Vert K\Vert <1$, the partial Neumann series satisfiesSince $\Vert K^{N+1}\Vert \le \Vert K{\Vert }^{N+1}\to 0$, the series converges in operator norm andIf $\tau \ne 0$ and $\Vert I-\tau K\Vert <1$, apply this identity to $I-\tau K$:

The operator $A^{\ast }A$ is positive, self-adjoint, and compact. Because $\mathrm{ran}A$ is infinite dimensional, the spectral theorem for compact Hermitian operators gives positive spectral values tending to zero. Even if $\ker A=\{0\}$, zero remains in the spectrum as a limit point. Hence for every $\tau$,The strict inequality $\Vert I-\tau K\Vert <1$ needed for the operator-norm Neumann series is impossible, so formula (3) cannot be applied directly to invert $A^{\ast }A$.

Factor the perturbed operator on $D(A)$ asSincethe Neumann series makes $I+B{A}^{-1}$ invertible. Therefore $0\notin \mathrm{Sp}(A+B)$ andThe geometric-series bound gives

Now choose a bounded open neighborhood $U$ of the isolated spectral component $X$ such thatand $\overline{U}$ meets no other component of the spectrum. Compactness of $\partial U$ givesFor all sufficiently large $n$, $\Vert B_n\Vert M<1$. Applying the first part to $A-zI$ shows uniformly that $\partial U\subset \rho (A+B_n)$.

The corresponding Riesz projections areThe resolvent identity and the uniform Neumann bound imply $\Vert P_n-P\Vert \to 0$. The projection $P$ is nonzero because $U$ contains the nonempty spectral component $X$. Projections at distance less than one have isomorphic ranges, so $P_n\ne 0$ for large $n$. Therefore $A+B_n$ has spectrum inside $U$, and any such point $z$ satisfies $\mathrm{dist}(z,X)<ϵ$. Thusfor every sufficiently large $n$.

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

For normal $x$, the closed unital C-star subalgebra generated by $x,x^{\ast }$ is commutative. When the ambient spectrum is real, its complement is connected. Membership of the resolvent in this subalgebra is both open and closed on that complement, and holds at large modulus by the Neumann series. Thus the subalgebra spectrum is the same real set. Its isometric Gelfand transform takes $x$ to a real-valued function and $x^{\ast }$ to its conjugate, proving equality. Normality is necessary: the nonzero matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ has spectrum $\{0\}$ but is not self-adjoint.

When $\beta \rho (A)<1$, summing the adjacency-matrix bound for a discrete SIR epidemic gives the displayed Neumann series bound on the expected total infected count $Z$. For an undirected graph, the adjacency matrix of a graph is a symmetric matrix, so its Euclidean operator norm equals its spectral radius. The Cauchy-Schwarz inequality then yields $\mathbb{E}Z\le \sqrt{nm}/(1-\beta \rho )$ for $m$ initially infected vertices.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For a closed unital subalgebra $B$ of a unital Banach algebra $A$ and $x\in B$, the smaller-algebra algebra resolvent set is relatively open and closed in the larger one. Openness follows from the Neumann series; closedness follows from continuity of inversion in $A$ and norm closedness of $B$. Every larger-algebra resolvent component is therefore either retained whole or absorbed into the smaller-algebra Banach algebra spectrum. The unbounded component is always retained, since inverses at large modulus are norm-convergent series in $x$.

The algebra resolvent set consists of those scalars for which $\lambda 1-x$ is invertible in the algebra. It is open by the Neumann series, and the inverse-valued map is the resolvent of an element. A closed subalgebra can have a smaller algebra resolvent set because an ambient inverse need not lie in it.

Suppose $(u_n)$ is a normalized block basic sequence in the l-p sequence space, $1\le p<\infty$, and ${\sum }_n\Vert x_n-u_n{\Vert }_p<1$. Choose norm-one linear functionals ${\varphi }_n$ supported on the respective blocks with ${\varphi }_n(u_n)=1$. The bounded linear operator $A=I+{\sum }_n{\varphi }_n(\cdot )(x_n-u_n)$ is invertible by the Neumann series. The block projection $Qz={\sum }_n{\varphi }_n(z)u_n$ has norm at most one, and $AQ{A}^{-1}$ projects onto the closed span of $(x_n)$. Thus that span is a complemented subspace isomorphic to the l-p sequence space.

Let $B$ be a closed star-subalgebra of a unital Hermitian Banach algebra $A$, containing the same identity. For Hermitian $h\in B$, the complement of ${\sigma }_A(h)\subseteq \mathbb{R}$ is connected. The resolvent belongs to $B$ at large modulus by the Neumann series; its membership in $B$ is both open and closed in that complement, so ${\sigma }_B(h)={\sigma }_A(h)$. If $b$ is invertible in $A$, both $b{b}^{\ast }$ and $b^{\ast }b$ are invertible there and Hermitian algebra elements, hence invertible in $B$. Their inverses supply respectively a right and a left inverse for $b$ in $B$, proving invertibility there. Applying this to $b-\lambda 1$ proves the spectrum equality.

An algebra norm is submultiplicative when it satisfies the displayed product inequality. Induction gives $\Vert a_1\cdots a_n\Vert \le {\prod }_j\Vert a_j\Vert$, and in particular $\Vert a^n\Vert \le \Vert a{\Vert }^n$. The operator norm has this property because $\Vert STv\Vert \le \Vert S\Vert \Vert T\Vert \Vert v\Vert$. The supremum norm for bounded functions with pointwise multiplication is another example. This estimate supports the Neumann series and the spectral radius formula.

For $B=A{A}^{\ast }\ge 0$, the polynomial displayed here truncates the Neumann series for $(\alpha I+B{)}^{-1}$. It obeys $(\alpha I+B)R_{\alpha ,2}=I+{\alpha }^{-3}{B}^3$, and is not an exact inverse. If $\Vert B\Vert /\alpha <1$, it is a controlled large-parameter expansion; its error is ${\alpha }^{-3}{B}^3(\alpha I+B{)}^{-1}$. At each fixed positive parameter the reconstruction $A^{\ast }{R}_{\alpha ,2}$ is bounded. However this polynomial diverges on nonzero singular modes as $\alpha \to 0$ and is not a substitute for the exact convergent Tikhonov filter in that limit.