Volterra power-kernel integral ($(V_{\beta }f)(x)={\int }_0^x(x-y{)}^{\beta -1}f(y)dy,\beta >0$)

For nonnegative Lebesgue measurable $f$ on $(0,1)$, the displayed Volterra power-kernel integral is defined for every $0<x<1$ as a nonnegative extended Lebesgue integral. The factor is finite and continuous on the open integration interval $(0,x)$, and an endpoint value can be assigned arbitrarily. The result may initially be infinite; for example $f(y)=1/y$ makes every value infinite. At $\beta =1$, it reduces to the ordinary Volterra operator.

For joint integration, use the integral kernel $(x-y{)}^{\beta -1}$ on $0<y<x<1$ and zero elsewhere. This piecewise prescription avoids the singular diagonal and never raises a negative number to a noninteger power. The L1 bound for a Volterra power-kernel integral gives an integrable output, finite almost everywhere, whenever $f$ is integrable.

For nonnegative integrable $f$, Tonelli's theorem applied on the triangle $0<y<x<1$ gives the exact identityFor signed or complex integrable $f$, first apply the same calculation to $|f|$ and use the triangle inequality for an integral. The defining integral then exists absolutely for almost every $x$, and the same L1 norm bound holds. The constant $1/\beta$ is sharp: use $f_n=n{1}_{(0,1/n)}$, whose L1 norm is one and whose weighted integrals above tend to $1/\beta$.

Finiteness at every point is stronger than this L1 space assertion. For $0<\beta <1$, take $x_0=1/2$, $\gamma =(1+\beta )/2$, and $f(y)=(x_0-y{)}^{-\gamma }$ on $(0,x_0)$, zero otherwise. It is integrable because $\gamma <1$, but $(V_{\beta }f)(x_0)=\infty$ because the power near $y=x_0$ is $\beta -1-\gamma =(\beta -3)/2<-1$. This is compatible with an integrable output after changing values on a null set.