If is a compactly supported distribution and is any distribution, define . The inner function is smooth and supported in , hence is a test function for . Finite-order estimates prove continuity, while the tensor product of distributions proves commutativity after inserting compact cutoffs. For every test function , ; evaluating at zero against reflected test functions proves uniqueness. If both distributions have compact support, and the convolution theorem gives . Without a compact factor or another support/growth condition, arbitrary distributional convolution is not generally defined.
If at least one of is a compactly supported distribution, thenThe sums are Minkowski sums. The first rule follows from the pairing : a test function away from the support sum gives zero. For the second rule, localize the output to a small compact set avoiding the sum of the singular supports. The relevant pairs of input points form a compact set because one input support is compact. Cover it by finitely many product neighborhoods on each of which at least one input distribution is smooth; otherwise the output would meet the sum of the singular supports. Smooth cutoffs in these neighborhoods decompose the localized pairing into finitely many convolutions with a smooth compactly supported factor. Each is smooth by smoothing convolution with a test function. This proves the second rule locally.
The same argument works when addition is a proper map on the product of the two supports. Some condition defining the convolution is necessary; arbitrary pairs of distributions cannot always be convolved.
Target: Cauchy kernel fundamental solution. Parent: Fundamental solution of a linear differential operator. Intended file: distribution-theory.bigb. Articles by others on the same topic
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