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If $1 < p < \infty$ and $q$ is conjugate to $p$ , then $f * g$ is continuous and vanishes at infinity

Statement

Assume the Axiom of Countable Choice.

Let $1 < p < \infty$ and let $q$ be the conjugate exponent. If $f \in L^{p} (R^{n})$ and $g \in L^{q} (R^{n})$ , then the convolution $f * g$ has a continuous representative in $C_{0} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, $1 < p < \infty$ , its conjugate exponent $q$ , and functions $f \in L^{p} (R^{n})$ and $g \in L^{q} (R^{n})$ .

[L1]

Translation is continuous in finite $L^{p}$ ( $∥ τ_{h} f - f ∥_{p} \to 0$ in $L^{p} (R^{n})$ as $h \to 0$ , for $1 \leq p < \infty$ ).

[L2]

Young's inequality gives the $L^{\infty}$ bound in the conjugate case (Young's convolution inequality under Countable Choice).

[L3]

$L^{1}$ convolution is defined almost everywhere, and $C_{c} (R^{n})$ is dense in finite $L^{p}$ (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ , $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L4]

The support of a convolution lies in the closure of the support sumset (The support of a convolution lies in the closure of the support sumset).

Proof

technique · direct

1.1L1L2givenalgebra

For every $x, h \in R^{n}$ , [L1, L2, given, algebra] $∣ (f * g) (x - h) - (f * g) (x) ∣ \leq \int ∣ f (x - h - y) - f (x - y) ∣ ∣ g (y) ∣ d y \leq ∥ τ_{h} f - f ∥_{p} ∥ g ∥_{q} .$ By [L1], the right-hand side tends to $0$ as $h \to 0$ , uniformly in $x$ . So the pointwise-defined convolution is uniformly continuous.

2.1L2L3L4step 1.1choosealgebra

By [L3], choose $u, v \in C_{c} (R^{n})$ with [L2, L3, L4, step 1.1, choose, algebra] $∥ f - u ∥_{p} + ∥ g - v ∥_{q}$ arbitrarily small. Applying [L2] twice gives $∥ f * g - u * v ∥_{\infty} \leq ∥ f - u ∥_{p} ∥ g ∥_{q} + ∥ u ∥_{p} ∥ g - v ∥_{q} .$ The function $u * v$ is continuous by step 1.1 and compactly supported by [L4], so $u * v \in C_{0} (R^{n})$ .

3.1step 1.1step 2.1∎

Therefore $f * g$ is a uniform limit of $C_{0}$ functions, hence itself belongs [step 1.1, step 2.1] to $C_{0} (R^{n})$ . Together with step 1.1, this proves that $f * g$ has a continuous representative vanishing at infinity.

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Sources

Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)
Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

If 1<p<∞1<p<∞ and qq is conjugate to pp, then f∗gf∗g is continuous and vanishes at infinity