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Talk:Volterra operator

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does not represent antidifferentiation on L^2(0,1)

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An L^2 function doesnt even need to have an antiderivative and even even if almost-everywhere-antiderivatives are allowed, they don't need to equal the Volterra integral (not even almost everywhere). Example for both: f(x)=0, for x<1/2, f(x)=1, for x>1/2. I will adjust the article unless someone has objections. Ninjamin (talk) 17:35, 16 May 2019 (UTC)[reply]

More properties

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Some additional properties of the Volterra operator:

  • For each , is Hölder continuous with exponent . Stated and proved in the StackExchange answer linked in the article.
  • Using Fubini's theorem one can compute repeated applications of as .

Because I don't have a source (other than lecture notes) on the latter statement, I didn't want to add them to the article. Columbus240 (talk) 13:29, 12 January 2022 (UTC)[reply]