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Banach-Stone weighted composition isometries

Example

Assume the Axiom of Choice (The Axiom of Choice). On $C ([0, 1], C)$ with the supremum norm define

$(T f) (t) : = e^{i t} f (1 - t) .$

Then $T$ is a surjective linear isometry of the weighted-composition form $T f = u \cdot (f \circ h)$ with $h (t) = 1 - t$ and $u (t) = e^{i t}$ (Banach-Stone), and $T$ is neither unital nor multiplicative: it is not the identity in disguise. Over the real scalars, $T f (t) : = - f (1 - t)$ is a surjective linear isometry with weight $u \equiv - 1$ , also neither unital nor multiplicative.

Facts & Assumptions

Given: The Axiom of Choice, the homeomorphism $h : [0, 1] \to [0, 1]$ , $h (t) = 1 - t$ , and the continuous unimodular weight $u (t) = e^{i t}$ .

[L1]

For nonempty compact Hausdorff spaces $K, L$ , a homeomorphism $h : L \to K$ and continuous $u : L \to K$ , where $K = R$ or $C$ , with $∣ u ∣ = 1$ , the map $T f : = u \cdot (f \circ h)$ is a surjective linear isometry $C (K) \to C (L)$ , and every surjective linear isometry arises this way (Banach-Stone, The Axiom of Choice).

[L2]

For real $t$ , $e^{i t} = \cos t + i \sin t$ and $∣ e^{i t} ∣ = 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ); the exponential is entire and hence continuous (The complex exponential is entire and its complex derivative is itself).

[L3]

For $0 < t \leq 2$ , $\sin t \geq t - t^{3} / 6 > 0$ (Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3).

Verification

technique · direct

1.1L1L2algebra

$h$ is a homeomorphism with $h^{- 1} = h$ , since $h (h (t)) = t$ , and $u$ is continuous with $∣ u (t) ∣ = 1$ by [L2]; hence by [L1] the map $T f (t) = u (t) f (h (t))$ is a surjective linear isometry.

1.21.1L2L3algebra

$T$ is not unital: $T 1 = u$ , and $u (t) = e^{i t}$ is not the constant function $1$ because at the endpoint $t = 1$ , [L2] and [L3] give $Im u (1) = \sin 1 \geq 5 / 6 > 0$ .

2.11.11.2algebra

$T$ is not multiplicative: $(T 1) (T 1) = u^{2}$ while $T (1 \cdot 1) = u$ , and $u^{2} \neq u$ because $u (t) = e^{i t} \neq 0$ and $u (t) \neq 1$ for some $t$ ; at $t = 1$ , equality $u (1)^{2} = u (1)$ would, by division by the nonzero $u (1)$ , force $u (1) = 1$ , contradicting [step 1.2].

3.1L1algebra∎

In the real case $u \equiv - 1$ has $∣ u ∣ = 1$ and $h = h^{- 1}$ ; directly, $∥ T f ∥_{\infty} = \sup_{t \in [0, 1]} ∣ f (1 - t) ∣ = ∥ f ∥_{\infty}$ and $T^{2} = I$ , so $T$ is a surjective real-linear isometry. Moreover, $T 1 = - 1 \neq 1$ and $T (1 \cdot 1) = - 1 \neq (- 1) (- 1) = 1$ , so $T$ is neither unital nor multiplicative.

Remarks

The weight is the obstruction. By Banach-Stone the weight is forced to be $u = T 1$ ; an isometry of this form is unital exactly when $u \equiv 1$ , and multiplicative exactly when $u \equiv 1$ (or, in the real case, $u \equiv 1$ ).
No star-property is claimed. These maps are isometries of Banach algebras, not $*$ -homomorphisms of C*-algebras; the commutative Gelfand–Naimark theorem concerns the latter.

Depends on

Used by

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Sources

Orr Shalit, Advanced Analysis Notes 14: the isometric structure of C(K) — Examples before Theorem 2, HTML lines 38–54 (standard reference, not scraped)