LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Finite simple analytic families and their exact endpoint norms

Statement

Let $1 \leq p_{0}, p_{1} < \infty$ , $1 \leq q_{0}, q_{1} \leq \infty$ , and $0 < θ < 1$ . Define $\frac{1}{p} = \frac{1 - θ}{p_{0}} + \frac{θ}{p_{1}}, \frac{1}{q} = \frac{1 - θ}{q_{0}} + \frac{θ}{q_{1}}, \frac{1}{r} = 1 - \frac{1}{q}, \frac{1}{r_{ℓ}} = 1 - \frac{1}{q_{ℓ}} .$ For complex finite simple functions $f = \sum_{j} a_{j} 1_{E_{j}}$ and $g = \sum_{k} b_{k} 1_{F_{k}}$ on their respective measure spaces, with disjoint finite-measure fibers, there are coefficientwise entire families $f_{z}, g_{z}$ bounded in coefficient modulus on $0 \leq Re z \leq 1$ , with $f_{θ} = f, g_{θ} = g$ as a.e. classes. After discarding zero coefficients and null fibers, for nonzero classes and $ℓ \in {0, 1}$ and every $t \in R$ , $∥ f_{ℓ + i t} ∥_{p_{ℓ}} = ∥ f ∥_{p}^{p / p_{ℓ}} .$ If $r < \infty$ , then $∥ g_{ℓ + i t} ∥_{r_{ℓ}} = ∥ g ∥_{r}^{r / r_{ℓ}}$ when $r_{ℓ} < \infty$ , and $∥ g_{ℓ + i t} ∥_{\infty} = 1$ when $r_{ℓ} = \infty$ . If $r = \infty$ , necessarily $q_{0} = q_{1} = 1$ ; take $g_{z} = g$ , retaining its infinity norm. Zero classes have identically zero families.

Facts & Assumptions

[F1]

Finite simple classes and their norms use disjoint measurable fibers; zero classes can be represented by zero Complex Lp classes and Euclidean test-function conventions.

[F2]

Conjugate exponents have reciprocal sum one with reciprocal infinity zero Conjugate exponents, including the endpoint conventions.

[F3]

For positive a, exp(log a)=a The natural logarithm as the inverse of the exponential function.

[F4]

For positive a, a to a real power is exp of that power times log a Real powers for positive bases, with the zero-base positive-exponent convention.

[F5]

The complex exponential is entire The complex exponential is entire and its complex derivative is itself.

[F6]

Compositions of complex differentiable maps are complex differentiable The chain rule for complex derivatives.

[F7]

The modulus of exp(x+iy) is exp(x) $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ .

[F8]

Affine combinations and finite sums and products of entire functions are entire Linearity, product, reciprocal, and quotient rules for complex derivatives.

[F9]

The real exponential is increasing, so an affine real exponent between its endpoint values gives a modulus bounded by the endpoint maximum The exponential function is strictly increasing.

Proof

Given: The objects and hypotheses in the statement.

1.1

Discard null fibers and zero coefficients without changing the classes, and define their omitted contributions to be zero for every z. For the remaining coefficients put $α (z) = p ((1 - z) / p_{0} + z / p_{1})$ and $a_{j} (z) = (a_{j} / ∣ a_{j} ∣) \exp (α (z) \log ∣ a_{j} ∣)$ . Positive coefficient moduli have defined logarithms; at z=theta, $α (θ) = 1$ , so $a_{j} (θ) = a_{j}$ . Set $f_{z} = \sum_{j} a_{j} (z) 1_{E_{j}}$ .

F1F2F3F4

2.1

The affine alpha is entire; the chain rule and the entire exponential make every $a_{j} (z)$ entire. For $z = x + i t$ , the exponential modulus formula gives $∣ a_{j} (z) ∣ = \exp (p ((1 - x) / p_{0} + x / p_{1}) \log ∣ a_{j} ∣) = ∣ a_{j} ∣^{p ((1 - x) / p_{0} + x / p_{1})}$ . For $0 \leq x \leq 1$ this is bounded by the larger of the two boundary powers. The finite list of coefficients is therefore bounded throughout the strip. At $x = ℓ$ with $ℓ = 0$ or $1$ , disjointness gives $∥ f_{ℓ + i t} ∥_{p_{ℓ}}^{p_{ℓ}} = \sum_{j} ∣ a_{j} ∣^{p} μ (E_{j}) = ∥ f ∥_{p}^{p}$ , proving the asserted norm formula.

F3F4F5F6F7F8step 1.1F9

3.1

Since $1 / r_{ℓ} = 1 - 1 / q_{ℓ}$ , we have $1 / r = (1 - θ) / r_{0} + θ / r_{1}$ . If r is finite, put $β (z) = r ((1 - z) / r_{0} + z / r_{1})$ and $b_{k} (z) = (b_{k} / ∣ b_{k} ∣) \exp (β (z) \log ∣ b_{k} ∣)$ . The preceding entire-function and modulus calculations apply with r and the b-coefficients, and $β (θ) = 1$ . For finite $r_{ℓ}$ , $∥ g_{ℓ + i t} ∥_{r_{ℓ}}^{r_{ℓ}} = \sum_{k} ∣ b_{k} ∣^{r} ν (F_{k}) = ∥ g ∥_{r}^{r}$ . If $r_{ℓ} = \infty$ , each surviving coefficient has modulus one on that boundary, so the essential maximum is one: a nonzero class has at least one positive-measure surviving fiber.

F1F2F3F4F5F6F7F8step 2.1

4.1

If $r = \infty$ , the positive weights $1 - θ, θ$ and nonnegative reciprocals force $1 / r_{0} = 1 / r_{1} = 0$ , hence $q_{0} = q_{1} = 1$ . The constant family $g_{z} = g$ is entire coefficientwise, bounded, has $g_{θ} = g$ and unchanged infinity norm. Identically zero families handle zero classes on either side, including empty or zero-measure spaces, with no logarithm of zero. Thus every asserted branch is established.

F1F2step 1.1step 3.1∎

Depends on

Used by

Riesz–Thorin estimate on the finite simple core Lemma

Dependency tree · two levels

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Sources

Laugesen Appendix C proof of Theorem C.6, pp.170–172, equations (C.4)–(C.6); Teschl Theorem 15.2, p.415 (standard reference, not scraped)