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Continuous functional calculus cannot produce every spectral projection

Statement refuted

Assume AC. For $T = M_{x}$ , multiplication by the coordinate on $L^{2} ([0, 1], λ)$ , there is a continuous $f : [0, 1] \to C$ with $f (T) = E ([0, \frac{1}{2}])$ , where $E$ is the spectral projection valued measure of $T$ .

Facts & Assumptions

[A1]

$T = M_{x}$ is bounded self-adjoint with $σ (T) = [0, 1]$ ; its spectral projections act by $E (B) = M_{1_{B}}$ , and $E ([0, 1 / 2])$ is the orthogonal projection onto the classes supported in $[0, 1 / 2]$ (Pvm of a multiplication operator, Borel functional calculus defines a discontinuous characteristic function).

[A2]

The continuous calculus is the restriction of the Borel calculus to continuous functions: for continuous $f$ the operator $f (T)$ of Continuous functional calculus for bounded normal operators satisfies $f (T) = M_{f \circ x}$ (Pvm of a multiplication operator, Borel functional calculus for a bounded normal operator).

[A3]

Multiplication operators in $L^{2}$ : $M_{g} h = 0$ as a class exactly when $g h = 0$ almost everywhere, and two continuous functions on $[0, 1]$ that agree almost everywhere agree everywhere, because the complement of the closed set on which they agree is open and null (The space $L^{p} (μ)$ as the quotient by null functions, Spectrum and resolvent of a bounded operator).

[A4]

AC is the declared choice hypothesis of this page from the construction item onward (The Axiom of Choice).

Counterexample

technique · direct

Given: Lebesgue measure on $[0, 1]$ , the operator $T = M_{x}$ and the spectral projection $P : = E ([0, 1 / 2]) = M_{1_{[0, 1 / 2]}}$ .

1.1

Suppose, for contradiction, that $f \in C ([0, 1])$ satisfies $M_{f \circ x} = f (T) = P = M_{1_{[0, 1 / 2]}}$ .

A1A2

2.1

Evaluating on $h : = 1_{[0, 1 / 2]}$ , one gets $(f \circ x) h = 1_{[0, 1 / 2]} \cdot h = h$ ; subtracting, $(f - 1) h = 0$ almost everywhere, so $(f \circ x - 1) = 0$ almost everywhere on the set ${h \neq 0} = [0, 1 / 2]$ (modulo a null set), hence $f = 1$ on $[0, 1 / 2]$ by continuity.

step 1.1A3

2.2

Evaluating on the nonzeroth class $h^{'} : = 1_{(1 / 2, 1]}$ , one gets $(f \circ x) h^{'} = 1_{[0, 1 / 2]} \cdot h^{'} = 0$ , so $f = 0$ almost everywhere on $(1 / 2, 1]$ and hence $f = 0$ on $(1 / 2, 1]$ by continuity, which forces $f (1 / 2) = 0$ .

step 1.1A3

3.1

The two evaluations give $f (1 / 2) = 1$ and $f (1 / 2) = 0$ simultaneously, a contradiction.

step 2.1step 2.2

4.1

No continuous function on $[0, 1]$ can satisfy $f (T) = E ([0, 1 / 2])$ for $T = M_{x}$ : this particular spectral projection forces incompatible one-sided values at $1 / 2$ and is not a continuous-calculus value of $T$ .

step 3.1A4∎

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.7, printed pp.290–296 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4 and §5, pp.13–20 (standard reference, not scraped)