Convolution is the operation that "smears" one function against another, producing a weighted average that inherits the best properties of both factors. It is the algebraic operation underlying three pillars of analysis: $L^p$ theory (where [Young's inequality](/theorems/463) controls its [integrability](/page/Integral)), Fourier analysis (where the [Convolution Theorem](/theorems/250) converts it to pointwise multiplication), and PDE (where solutions are built by convolving with fundamental solutions). This page develops the operation itself; for its role in regularisation see the [mollifier page](/page/Standard%20Mollifier), and for its interaction with the Fourier transform see the [Fourier Transform page](/page/Fourier%20Transform).

[motivation] ## Motivation ### What Convolution Does Given two [functions](/page/Function) $f$ and $g$ on $\mathbb{R}^n$, the convolution $(f * g)(x) = \int f(x - y)g(y) \, d\mathcal{L}^n(y)$ evaluates $f$ at all points near $x$, weighted by $g$. If $g$ is a concentrated bump near the origin, $(f * g)(x)$ is a local average of $f$ around $x$ — a smoothing operation. If $g$ is itself rough, the convolution still makes sense (under integrability conditions) and is at least as regular as the smoother of the two factors. ### Why Domains Matter The integral $(f * g)(x) = \int f(x - y)g(y) \, dy$ requires two things: $y$ must lie in the domain of $g$, and $x - y$ must lie in the domain of $f$. If $f$ is supported on a [set](/page/Set) $A$ and $g$ on a set $B$, then the integrand is nonzero only when $y \in B$ and $x - y \in A$, i.e., $x \in A + B$. The convolution's natural domain is therefore the **Minkowski sum** $A + B = \{a + b : a \in A, \, b \in B\}$, which is typically *larger* than either $A$ or $B$ individually. This domain growth is not a technicality — it is the reason convolution with a kernel of width $\varepsilon$ expands the support by $\varepsilon$, and it is the mechanism by which the [Poisson kernel](/theorems/576) on the ball solves the Dirichlet problem. ### The Three Algebraic Structures Convolution gives $L^1(\mathbb{R}^n)$ the structure of a commutative Banach algebra (without identity). The [Fourier transform](/page/Fourier%20Transform) is the Gelfand transform of this algebra, converting convolution to multiplication and making the spectral theory of translation-invariant operators algebraic. In PDE, every linear constant-coefficient equation $P(D)u = f$ on $\mathbb{R}^n$ is solved by $u = E * f$ where $E$ is the fundamental solution satisfying $P(D)E = \delta_0$: the [heat kernel](/page/Heat%20Equation), the [Newtonian potential](/page/Laplace's%20Equation), and the [Duhamel integral](/theorems/55) are all convolutions. [/motivation]

## Definition and Domain

The definition requires care about where the integral makes sense. The integrand involves $f$ evaluated at $x - y$ and $g$ at $y$, so both the integrability of $f$ and $g$ and the geometry of their supports determine the set of $x$ for which the convolution is finite.

[definition: Convolution] Let $f, g: \mathbb{R}^n \to \mathbb{C}$ be measurable functions. The **convolution** of $f$ and $g$ is the function \begin{align*} (f * g)(x) := \int_{\mathbb{R}^n} f(x - y) \, g(y) \, d\mathcal{L}^n(y), \end{align*} defined for those $x \in \mathbb{R}^n$ for which the integral exists (either as a [Lebesgue integral](/page/Lebesgue%20Integral) when the integrand is absolutely integrable, or as an improper integral in suitable senses). [/definition]

The notation is deceptively simple. The integral involves $f$ evaluated at $x - y$ (a *reflected and translated* version of $f$) multiplied by $g(y)$. For the integral to be finite at a given $x$, one needs $y \mapsto f(x-y)g(y)$ to be integrable. This is where the integrability conditions (Young's inequality) and the support conditions (Minkowski sum) enter.

### When Is the Convolution Well-Defined?

The most common sufficient conditions are:

**$L^1 * L^1$:** If $f, g \in L^1(\mathbb{R}^n)$, Fubini-Tonelli gives $\int\!\!\int |f(x-y)||g(y)| \, dy \, dx = \|f\|_{L^1}\|g\|_{L^1} < \infty$, so $f * g$ is defined a.e. and belongs to $L^1$.

**$L^p * L^q$:** More generally, [Young's Convolution Inequality](/theorems/463) gives the complete $L^p$ picture.

**Compact support:** If one of $f, g$ has compact support and the other is locally integrable, the integral reduces to a bounded region and $f * g$ is defined everywhere.

[example: Domain Growth Under Convolution] Let $f = \mathbb{1}_{[0,1]}$ and $g = \mathbb{1}_{[2,4]}$ on $\mathbb{R}$. The convolution is \begin{align*} (f * g)(x) = \int_{\mathbb{R}} \mathbb{1}_{[0,1]}(x - y) \, \mathbb{1}_{[2,4]}(y) \, d\mathcal{L}^1(y) = \mathcal{L}^1\!\big([2,4] \cap [x-1, x]\big). \end{align*} The intersection $[2,4] \cap [x-1, x]$ is nonempty exactly when $x - 1 \leq 4$ and $x \geq 2$, i.e., $x \in [2, 5]$. The convolution is: $0$ for $x < 2$; $x - 2$ for $2 \leq x \leq 3$; $1$ for $3 \leq x \leq 4$; $5 - x$ for $4 \leq x \leq 5$; $0$ for $x > 5$. The support of $f * g$ is $[2, 5] = [0, 1] + [2, 4]$ — the Minkowski sum of the two supports. Neither input is supported on $[2, 5]$; the convolution's domain has *grown* by adding the widths. Note that $f$ "lives" on $[0,1]$ and $g$ on $[2,4]$, which are disjoint — yet $f * g$ is nonzero on the connecting interval $[2, 5]$. Convolution reaches across gaps. [/example]

## Commutativity and Algebraic Properties

Convolution on $\mathbb{R}^n$ is commutative: $(f * g)(x) = (g * f)(x)$. The proof requires a global change of variables.

[remark: Commutativity Proof] Substitute $z = x - y$ (so $y = x - z$, $dy = dz$): \begin{align*} (f * g)(x) = \int f(x - y) g(y) \, dy = \int f(z) g(x - z) \, dz = (g * f)(x). \end{align*} This substitution is valid on $\mathbb{R}^n$ because Lebesgue measure is translation-invariant. On other domains — for instance, on a non-abelian [group](/page/Group) — the substitution introduces the group inverse and convolution is *not* commutative in general. [/remark]

[example: Commutativity And Asymmetric Supports] Take $f = \mathbb{1}_{[0,1]}$ and $g = \delta$-approximation $\rho_\varepsilon$ supported on $[-\varepsilon, \varepsilon]$. Then $f * g$ is a smooth function supported on $[-\varepsilon, 1 + \varepsilon]$: the mollification of $f$, which smooths the jumps at $0$ and $1$. Computing $g * f$ instead gives the *same* function — but the intuition is different. In $f * g$, we average $f$ against a narrow bump; in $g * f$, we smear the bump $g$ across the support of $f$. Commutativity says these produce the same output, despite the asymmetry between a rough function ($f$) and a smooth one ($g$). [/example]

Beyond commutativity, convolution is associative — $(f * g) * h = f * (g * h)$ — and distributes over addition. These are verified by Fubini's theorem. The Dirac delta $\delta_0$ acts as a (distributional) identity: $f * \delta_0 = f$. However, no function in $L^1(\mathbb{R}^n)$ serves as a convolution identity: the [Riemann-Lebesgue Lemma](/theorems/245) forces $\hat{f}(\xi) \to 0$ for $f \in L^1$, but a convolution identity would need $\hat{e}(\xi) = 1$ for all $\xi$, which is incompatible with decay.

## The Support of a Convolution

The domain example above illustrates a general principle: convolution adds supports. If $f$ is nonzero on a set $A$ and $g$ on a set $B$, the integrand $f(x-y)g(y)$ can be nonzero only when $y \in B$ and $x - y \in A$, forcing $x \in A + B$. This makes precise the intuition that convolution "spreads out" a function by the width of the kernel.

[quotetheorem:588]