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)

More properties
Some additional properties of the Volterra operator: 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)
 * 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 $$(V^nf)(x) = \int_0^x f(t)\frac{(x-t)^{n-1}}{(n-1)!}\text{d}t$$.