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Spectrum can shrink in a larger Banach algebra

Statement refuted

Let $D = {∣ z ∣ < 1}$ and let

$A (\overline{D}) : = {f : \overline{D} \to C : f continuous on \overline{D}, f holomorphic on D}$

be the disc algebra with the supremum norm, and let $ρ : A (\overline{D}) \to C (T)$ be restriction to the unit circle $T$ . Then $A (\overline{D})$ is a unital commutative complex Banach algebra, $ρ$ is an isometric unital algebra homomorphism, and for the coordinate function $z$ one has

$σ_{A (\overline{D})} (z) = \overline{D}, σ_{C (T)} (ρ (z)) = T,$

so the spectrum strictly shrinks when the element is regarded in the larger algebra $C (T)$ . Here $C (T)$ is the algebra of Continuous functions form a commutative Banach algebra and spectra are taken as in Spectrum and resolvent set in a Banach algebra with the algebra indicated.

Facts & Assumptions

Given: The disc $D$ , its closure $\overline{D}$ , the circle $T$ , the disc algebra $A (\overline{D})$ with the supremum norm, the restriction map $ρ$ , and the coordinate function $z$ .

[L1]

A continuous complex-valued function on an open set is holomorphic if and only if its integral around the boundary of every filled triangle in the set vanishes; uniform limits of continuous functions are continuous (Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions, A uniform limit of continuous complex-valued functions is continuous).

[L2]

Uniformly convergent sequences of continuous functions on a contour may be integrated term by term (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L3]

A continuous function on the closure of a bounded domain that is holomorphic in the domain attains its maximum modulus on the boundary (Boundary maximum modulus principle on a bounded domain).

[L6]

If $f$ is holomorphic on an open set $U$ and a filled triangle lies in $U$ , then the integral of $f$ around its boundary vanishes (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L4]

On the compact Hausdorff space $T$ the algebra $C (T)$ is a unital commutative Banach algebra with spectrum of $g$ equal to $g [T]$ (Continuous functions form a commutative Banach algebra).

[L5]

$A^{\times}$ consists of the elements with a two-sided inverse; $λ \notin σ (a)$ exactly when $λ 1 - a$ is invertible (Unital Banach algebra, Spectrum and resolvent set in a Banach algebra).

Counterexample

technique · direct

1.1

$A (\overline{D})$ is complete: if $(f_{n})$ is uniformly Cauchy on $\overline{D}$ , then it converges uniformly to a continuous $f$ by [L1]; for every filled triangle contained in $D$ , 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 $D$ by [L1] and $A (\overline{D})$ is closed under uniform limits.

L1L2L6

2.1

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

step 1.1L5algebra

3.1

The restriction map is isometric by the maximum modulus principle: $∥ ρ (f) ∥_{\infty} = \sup_{T} ∣ f ∣ = \sup_{\overline{D}} ∣ f ∣ = ∥ f ∥_{\infty}$ for every $f \in A (\overline{D})$ , using [L3] and continuity.

step 2.1L3algebra

3.2

Spectrum in the disc algebra: if $∣ λ ∣ > 1$ then $1 / (λ - z)$ is holomorphic on a neighbourhood of $\overline{D}$ , so $λ - z$ is invertible in $A (\overline{D})$ ; if $∣ λ ∣ \leq 1$ then $λ = z (λ)$ with $λ \in \overline{D}$ , and evaluating the identity $g \cdot (λ - z) = 1$ at $z = λ$ gives $g (λ) \cdot 0 = 1$ , impossible; hence $σ_{A (\overline{D})} (z) = \overline{D}$ .

step 2.1L5algebra

3.3

Spectrum in $C (T)$ : the restriction $ρ (z)$ is the function $ζ \mapsto ζ$ on the circle, whose image is $T$ ; by [L4], $σ_{C (T)} (ρ (z)) = T$ .

step 2.1L4algebra

4.1

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

step 3.1step 3.2step 3.3algebra∎

Remarks

Why the two spectra differ. In $A (\overline{D})$ the inverse of $λ - z$ for $∣ λ ∣ \leq 1$ would have to be a function continuous on the closed disc and holomorphic inside, and no such function exists because the value would have to blow up at the point $λ$ of the closed disc. In $C (T)$ the same element is invertible as soon as $∣ λ ∣ \neq 1$ , because the circle avoids the zero $λ$ . The homomorphism is isometric, so the difference is not a norm effect.
The larger algebra need not be an extension of the element's algebra. The example embeds $A (\overline{D})$ isometrically into $C (T)$ and compares spectra there; the containment $σ_{C (T)} (ρ (z)) \subseteq σ_{A (\overline{D})} (z)$ is the general inclusion for a closed subalgebra with the same unit, as the isometric image $ρ (A (\overline{D}))$ is here, and it is strict here.

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Sources

Theo Bühler and Dietmar A. Salamon, Functional Analysis — §5.1.1 and §5.2.1 (the disc algebra), printed pp. 209–214 and 219–222 (standard reference, not scraped)