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Borel functional calculus defines a discontinuous characteristic function

Example

Assume AC. Let $H = L^{2} ([0, 1], λ)$ for Lebesgue measure $λ$ and let $T = M_{x}$ be multiplication by the coordinate, $(T h) (x) = x h (x)$ . Then $T$ is bounded self-adjoint with $σ (T) = [0, 1]$ , and the Borel functional calculus applied to the discontinuous function $1_{[0, 1 / 2]}$ produces the orthogonal projection onto the closed subspace $L^{2} ([0, \frac{1}{2}]) = {h \in L^{2} ([0, 1]) : h = 0 a.e. on (\frac{1}{2}, 1]},$ namely $1_{[0, 1 / 2]} (T) = M_{1_{[0, 1 / 2]}}$ , whose range is that subspace and whose kernel is $L^{2} ((\frac{1}{2}, 1])$ .

Facts & Assumptions

[A1]

The multiplication operator $M_{x}$ on $L^{2} ([0, 1])$ is bounded self-adjoint with $σ (M_{x}) = [0, 1]$ , spectral projections $E (B) = M_{1_{B \cap [0, 1]}}$ and Borel calculus $f (M_{x}) = M_{f \circ x}$ for every bounded Borel $f$ on $[0, 1]$ (Pvm of a multiplication operator, Borel functional calculus for a bounded normal operator).

[A2]

The indicator $1_{[0, 1 / 2]}$ is bounded Borel on $[0, 1]$ and satisfies $1_{[0, 1 / 2]}^{2} = 1_{[0, 1 / 2]} = \overline{1_{[0, 1 / 2]}}$ , so its calculus value is an orthogonal projection equal to the spectral projection $E ([0, 1 / 2])$ (Borel functional calculus for bounded normal operators, Spectral projections and resolution of the identity).

[A3]

Elements of $L^{2}$ are equivalence classes modulo almost-everywhere equality; a class is supported in a Borel set $A$ when it has a representative vanishing almost everywhere off $A$ , and $L^{2} (A)$ denotes this subspace of classes (The space $L^{p} (μ)$ as the quotient by null functions, Orthogonality and the orthogonal complement, Hilbert space).

[A4]

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

Verification

technique · direct

Given: Lebesgue measure on $[0, 1]$ , the operator $T = M_{x}$ and the function $1_{[0, 1 / 2]}$ .

1.1

The function $1_{[0, 1 / 2]}$ is bounded Borel on $[0, 1] = σ (T)$ , with values in ${0, 1}$ and equal to its own square and conjugate, so the calculus attaches to it an orthogonal projection.

A1A2

1.2

By the computation of the calculus for multiplication operators, $1_{[0, 1 / 2]} (T) = M_{1_{[0, 1 / 2]} \circ x} = M_{1_{[0, 1 / 2]}}$ , acting by $(M_{1_{[0, 1 / 2]}} h) (x) = 1_{[0, 1 / 2]} (x) h (x)$ .

2.1

Put $A = [0, 1 / 2]$ and $P = M_{1_{A}}$ . For every $h$ , $P h$ vanishes almost everywhere off $A$ ; conversely, if $g$ is supported in $A$ , then $P g = g$ , so $ran P = L^{2} (A)$ . Also $P h = 0$ exactly when $h$ vanishes almost everywhere on $A$ , so $\ker P = L^{2} ((1 / 2, 1])$ . Both subspaces are closed: if $P h_{n} = h_{n} \to h$ , boundedness and $P^{2} = P$ give $P h = \lim P h_{n} = h$ , while if $P h_{n} = 0$ and $h_{n} \to h$ , then $P h = \lim P h_{n} = 0$ .

step 1.2A1A3

3.1

The spectral projection $E ([0, 1 / 2])$ agrees with this multiplication by the pulled-back indicator, so the discontinuous characteristic function of the Borel set has produced a genuine orthogonal projection of the operator, not merely a continuous-calculus value.

step 1.2step 2.1A2A4

4.1

The Borel calculus of $T = M_{x}$ therefore assigns to the discontinuous function $1_{[0, 1 / 2]}$ the orthogonal projection onto the classes supported in $[0, 1 / 2]$ , with kernel the classes supported in $(1 / 2, 1]$ .

step 3.1A4∎

Depends on

Used by

Continuous functional calculus cannot produce every spectral projection Counterexample

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.6.2–5.7, printed pp.277–296 (standard reference, not scraped)