Manuscript Psy02-A@1 · Psy02-A.tex · proposition · d92ce402

Proposition: A concrete compact-test allowance

Carrier specification and finite realizations › Spacetime tests are not one-particle vectors · lines 365–404 · proof included or linked

Psy02-A@1#prop:compact-test, Psy02-A@1#eq:test-bound

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The map $κ_{m} : C_{c}^{\infty} (R^{4}; R) \to L^{2} (R^{3})$ is well defined and satisfies

\norm κ_{m} (f) \leq \frac{\norm f _{L^{1}} + \norm Δ _{x} f _{L^{1}}}{2 2 π} .

The same inequality bounds $\norm κ_{m} (f) - κ_{m} (g)$ with $f - g$ on the right.

If $f = (\partial_{t}^{2} - Δ_{x} + m^{2}) q$ for compact smooth $q$ , then $κ_{m} (f) = 0$ .

Proof. Integration by parts gives

(1 + \abs k^{2}) \abs f (ω, k) \leq \norm f_{L^{1}} + \norm Δ_{x} f_{L^{1}} .

Since $ω_{m} (k) \geq \abs k$ ,

\norm κ_{m} (f)^{2} \leq \frac{( \norm f _{L^{1}} + \norm Δ _{x} f _{L^{1}} ) ^{2}}{2 ( 2 π ) ^{3}} \int_{R^{3}} \frac{d ^{3} k}{\abs k ( 1 + \abs k ^{2} ) ^{2}} .

The integral is $2 π$ , proving (eq:test-bound). Linearity gives the difference estimate. The Fourier multiplier of the displayed wave operator vanishes on $ω = ω_{m} (k)$ , proving the last assertion.

Displayed formulas in this passage

Psy02-A@1#eq:test-bound · line 369
$\norm κ_{m} (f) \leq \frac{\norm f _{L^{1}} + \norm Δ _{x} f _{L^{1}}}{2 2 π} .$
unlabeled · line 385
$(1 + \abs k^{2}) \abs f (ω, k) \leq \norm f_{L^{1}} + \norm Δ_{x} f_{L^{1}} .$
unlabeled · line 391
$\norm κ_{m} (f)^{2} \leq \frac{( \norm f _{L^{1}} + \norm Δ _{x} f _{L^{1}} ) ^{2}}{2 ( 2 π ) ^{3}} \int_{R^{3}} \frac{d ^{3} k}{\abs k ( 1 + \abs k ^{2} ) ^{2}} .$

Proposition: A concrete compact-test allowance

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