Talk:Volterra operator

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

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

Some additional properties of the Volterra operator:

For each $f$ , $Vf$ is Hölder continuous with exponent ${\frac {1}{2}}$ . Stated and proved in the StackExchange answer linked in the article.
Using Fubini's theorem one can compute repeated applications of $V$ as $(V^{n}f)(x)=\int _{0}^{x}f(t){\frac {(x-t)^{n-1}}{(n-1)!}}{\text{d}}t$ .

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]