Convolution

The convolution of integrable functions on the real line is

A periodic convolution operator with continuous periodic kernel is compact on $L^2[0,2\pi ]$, since its kernel is square integrable. In the normalized Fourier series basis it is diagonal, with multipliers $c_n={\int }_0^{2\pi }K(x)e^{-inx}dx$. These multipliers include the full integral rather than its normalized Fourier coefficient.

When all multipliers are nonzero, the Moore–Penrose inverse of an operator acts by dividing each Fourier coefficient $y_n$ by $c_n$. Its data domain is precisely the square-summability condition displayed. The dense range need not be closed, and writing the formal inverse series does not establish the Picard criterion.

For a continuously differentiable periodic kernel, integration by parts and the Riemann-Lebesgue lemma give $c_n=o(1/|n|)$. Nonperiodic data $e^{\alpha x}$ with $\alpha \notin i\mathbb{Z}$ have Fourier coefficients asymptotic to a nonzero multiple of $1/n$. Their inverse coefficients $y_n/c_n$ therefore fail square summability, and there is no Hilbert-space inverse or least-squares minimizer. Periodic exponential data $\alpha =im$ instead give the single-mode solution $e^{imx}/c_m$.

If $v$ is a compactly supported distribution and $u$ is any distribution, define $⟨u\ast v,\varphi ⟩=⟨u_x,⟨v_y,\varphi (x+y)⟩⟩$. The inner function is smooth and supported in $\mathrm{supp}\varphi -\mathrm{supp}v$, hence is a test function for $u$. Finite-order estimates prove continuity, while the tensor product of distributions proves commutativity after inserting compact cutoffs. For every test function $\psi$, $(u\ast v)\ast \psi =u\ast (v\ast \psi )$; evaluating at zero against reflected test functions proves uniqueness. If both distributions have compact support, $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$ and the convolution theorem gives $\widehat{u\ast v}=\hat{u}\hat{v}$. Without a compact factor or another support/growth condition, arbitrary distributional convolution is not generally defined.

For a distribution and a compactly supported smooth kernel, differentiation with constant coefficients commutes with convolution wherever the kernel support stays inside the domain. The identity follows by testing against the translated kernel and tracking the sign of its derivative. A variable coefficient must remain inside the product before convolution; in general $(bu{)}_{\sigma }\ne b{u}_{\sigma }$.

For any distribution $E\in {\mathcal{D}}^{\prime }$ and any test function $f$, the convolution $(E\ast f)(x)=⟨E_y,f(x-y)⟩$ is a smooth function. On a compact set of $x$ values, the translated test functions and all their derivatives have supports in one fixed compact set, so distributional continuity permits every derivative:For constant-coefficient operators, $P(D)(E\ast f)=(P(D)E)\ast f$. Hence a fundamental solution of a linear differential operator supplies a smooth particular solution for every compactly supported smooth datum. No temperedness of $E$ is needed.

For $f\in L^{\infty }({\mathbb{R}}^n)$ and $g\in L^1({\mathbb{R}}^n)$,is bounded by $\Vert f{\Vert }_{\infty }\Vert g{\Vert }_1$. Continuity follows fromand continuity of translation in $L^1$.

If a Lebesgue-measurable set $A\subset {\mathbb{R}}^n$ has positive measure, then its difference set $A-A$ contains an open neighbourhood of zero. For finite measure, the convolution $1_A\ast 1_{-A}$ is continuous and positive at zero; the general case follows by taking a finite positive-measure subset.

If a measurable set $E\subseteq [0,L]$ has measure $m(E)=L/2+\alpha$, then for $|t|<2\alpha$ the sets $E$ and $E+t$ lie in an interval of length $L+|t|$, soThus $(-2\alpha ,2\alpha )\subseteq E-E$. This quantitative overlap argument is a bounded form of the Steinhaus theorem.

An approximate identity is a family of integrable kernels $K_{\epsilon }$ whose total mass is one and whose mass concentrates near zero as $\epsilon \to 0$. Under standard hypotheses, $f\ast K_{\epsilon }$ converges to $f$ in norm and at almost every Lebesgue point.

In the distributional sense,This kernel is not positive, but testing it against a smooth compactly supported function and using the Dirichlet integral proves convergence.

The Cauchy approximate identity isIt is a probability density of total mass one, and its mass outside every fixed neighbourhood of zero tends to zero as $\epsilon \to 0$. It therefore converges to the Dirac delta function as a distribution.

If $1\le p,q,r\le \infty$ satisfy $1+1/r=1/p+1/q$, then

For sequences $a,b$, their discrete convolution is

For $\hat{f}(\xi )=\int f(x)e^{-2\pi ix\cdot \xi }dx$, Fubini gives $\widehat{f\ast g}=\hat{f}\hat{g}$.

For $F(x)=e^{-x}{1}_{x\ge 0}$, induction in the convolution integral givesIts Fourier transform in the angular-frequency convention is $(1+ik{)}^{-n}$.