The Kernels of Henkin

In Chapter 19, we constructed two fundamental solutions for ∂ ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1978.tif"/> on ℂ n https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1979.tif"/> . As mentioned in Chapter 18, if K is a fundamental solution for ∂ ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1980.tif"/> and if f ∈ D * ( ℂ n ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1981.tif"/> with ∂ ¯ f ≡ 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1982.tif"/> , then the equation ∂ ¯ u = f https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1983.tif"/> can be solved on ℂ n https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1984.tif"/> by setting u = K(f). Now suppose D is a bounded domain in ℂ n https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1985.tif"/> and suppose f ∈ E * ( D ¯ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1986.tif"/> with ∂ ¯ f = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1987.tif"/> on D. In this case, we cannot directiy apply K to f without first extending f and then multiplying by suitable cutoff function so that f has compact support. This process produces an extended f which is no longer inclosed. Another way to cut off f is to multiply f by the characteristic function on D (denoted χ_D ). If ∂ ¯ f = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/in1988.tif"/> on D, then ∂ ¯ ( χ D f ) = − [ ∂ D ] 0 , 1 ∧ f . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315140445/fbe50be4-2923-45ad-961d-51ed68ffe025/content/equation1089.tif"/>