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Convolution with a mollifier is smooth, and derivatives pass under the integral sign

Statement

Let $f : R^{n} \to C$ be locally integrable, let $φ \in C_{c}^{\infty} (R^{n})$ have mass $1$ , and let $φ_{ε}$ be the associated mollifier. Then for every $ε > 0$ , the convolution

$(f * φ_{ε}) (x) : = \int_{R^{n}} f (y) φ_{ε} (x - y) d y$

is smooth, and for every multi-index $α$ ,

$\partial^{α} (f * φ_{ε}) = f * (\partial^{α} φ_{ε}) .$

Facts & Assumptions

Given: A locally integrable function $f$ , a unit-mass smooth bump, and $ε > 0$ .

[L1]

The mollifier family is defined in The mollifier family generated by a unit-mass smooth bump.

[L2]

Differentiation under the integral sign is available (Differentiation under the integral sign).

[L3]

Multi-index notation and Euclidean smoothness are fixed in $C^{k}$ maps and multi-index derivative notation in Euclidean space.

Proof

technique · direct

1.1

Fix $x_{0} \in R^{n}$ . Because $φ_{ε}$ has compact [L1, L3, given, choose, algebra] support, there are $r > 0$ and a compact set $K$ such that $φ_{ε} (x - y) = 0$ and $\partial^{α} φ_{ε} (x - y) = 0$ whenever $∣ x - x_{0} ∣ < r$ and $y \notin K$ . Local integrability of $f$ therefore makes $∣ f ∣ 1_{K}$ integrable, so $y \mapsto f (y) φ_{ε} (x - y)$ and $y \mapsto f (y) \partial^{α} φ_{ε} (x - y)$ are integrable for $∣ x - x_{0} ∣ < r$ .

L1L3givenchoosealgebra

2.1

Fix a coordinate index $j$ and a point $x$ with $∣ x - x_{0} ∣ < r / 2$ . For [L2, step 1.1, algebra] $∣ t ∣ < r / 2$ , the point $x + t e_{j}$ still satisfies $∣ x + t e_{j} - x_{0} ∣ < r$ , so $G (y, t) : = f (y) φ_{ε} (x + t e_{j} - y)$ is integrable in $y$ . Because $\partial_{j} φ_{ε}$ is continuous with compact support, some constant $C_{j}$ satisfies $∣ \partial_{j} φ_{ε} (x + t e_{j} - y) ∣ \leq C_{j} 1_{K} (y) (∣ t ∣ < r / 2) .$ Hence $∣ \partial_{t} G (y, t) ∣ = ∣ f (y) \partial_{j} φ_{ε} (x + t e_{j} - y) ∣ \leq C_{j} ∣ f (y) ∣ 1_{K} (y),$ and the right-hand side is integrable by step 1.1. Applying [L2] on the interval $(- r / 2, r / 2)$ gives $\partial_{j} (f * φ_{ε}) (x) = \int f (y) \partial_{j} φ_{ε} (x - y) d y = (f * (\partial_{j} φ_{ε})) (x) .$ Since $x_{0}$ was arbitrary, this holds for every $x$ .

L2step 1.1algebra

3.1

Repeating step 2.1 for higher derivatives and using [L3] yields the general [L2, L3, step 2.1, induction] multi-index formula $\partial^{α} (f * φ_{ε}) = f * (\partial^{α} φ_{ε})$ . Hence $f * φ_{ε}$ is smooth.

L2L3step 2.1induction∎

Depends on

Used by

Mollifying 1_[0,1] at two scales Example
C_c^∞(ℝⁿ) is dense in Lᵖ(ℝⁿ) for 1 ≤ p < ∞ Theorem

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Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)