LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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Submultiplicativity of convolution in the L1 norm

Statement

Assume AC. For $f, g \in C_{c} (G)$ on an LCH group $G$ with fixed left Haar measure, $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1},$ the norms being those of Complex Haar L^p spaces and compactly supported functions.

Facts & Assumptions

Given: An LCH group $G$ , a left Haar measure $μ$ , and $f, g \in C_{c} (G)$ complex-valued of compact support.

[F1]

$(f * g) (x) = \int_{G} f (y) g (y^{- 1} x) d μ (y)$ with $f * g \in C_{c} (G)$ under AC (Compactly supported convolution on a group, Convolution preserves compact support and is associative).

[F2]

Assume AC. For a continuous compactly supported real kernel on a product of LCH spaces the partial integrals are continuous and compactly supported and the iterated integrals commute; the complex case obeys the same identity (Compactly supported kernels admit commuting radon integrals).

[F3]

The measure $μ$ is left invariant, that is $\int_{G} H (a x) d μ (x) = \int_{G} H (x) d μ (x)$ for $μ$ -integrable $H$ and $a \in G$ (Left Haar integral and left Haar measure).

[F4]

$∥ f ∥_{1} = \int_{G} ∣ f ∣ d μ$ and $∣ f ∣$ is continuous of compact support when $f$ is (Complex Haar L^p spaces and compactly supported functions).

[F5]

The integral satisfies $∣ \int_{G} F d μ ∣ \leq \int_{G} ∣ F ∣ d μ$ for integrable complex $F$ (The modulus of an integral is bounded by the integral of the modulus).

[A1]

AC is assumed in the choice-function form of the cited definition (The Axiom of Choice).

Proof

technique · direct

1.1F1F5

For $x \in G$ , [F5] applied to the integrable function $y \mapsto f (y) g (y^{- 1} x)$ gives $∣ (f * g) (x) ∣ \leq \int_{G} ∣ f (y) ∣ ∣ g (y^{- 1} x) ∣ d μ (y)$ .

1.2F2F4

The kernel $H (x, y) : = ∣ f (y) ∣ ∣ g (y^{- 1} x) ∣$ is real, nonnegative, continuous, and compactly supported: it is a product of continuous functions, and $H (x, y) \neq 0$ forces $y \in supp f$ and $x \in supp f \cdot supp g$ , a compact set.

1.3F3F4

For each $y \in G$ left invariance [F3] applied to $H_{y} (x) : = ∣ g (y^{- 1} x) ∣$ gives $\int_{G} H (x, y) d μ (x) = ∣ f (y) ∣ \int_{G} ∣ g (y^{- 1} x) ∣ d μ (x) = ∣ f (y) ∣ \int_{G} ∣ g (w) ∣ d μ (w) = ∣ f (y) ∣ ∥ g ∥_{1}$ .

2.1A1F2step 1.2

Integrating step 1.2 over $y$ for each $x$ and then over $x$ , the iterated integral $\int_{G} \int_{G} H (x, y) d μ (y) d μ (x)$ exists and, by [F2] under [A1], equals the reverse iterated integral $\int_{G} \int_{G} H (x, y) d μ (x) d μ (y)$ .

3.1F4step 1.1step 2.1step 1.3

Chaining steps 1.1, 2.1 and 1.3, $∥ f * g ∥_{1} = \int_{G} ∣ (f * g) (x) ∣ d μ (x) \leq \int_{G} \int_{G} H (x, y) d μ (y) d μ (x) = \int_{G} \int_{G} H (x, y) d μ (x) d μ (y) = ∥ g ∥_{1} \int_{G} ∣ f (y) ∣ d μ (y) = ∥ f ∥_{1} ∥ g ∥_{1}$ . ∎

Submultiplicativity of convolution in the L1 norm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources