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Let $T_\alpha$ be a singular integral operator defined by $$ T_{\alpha}[f](t):=\int_{0}^{t}\frac{f(t)-f(s)}{(t-s)^{\alpha+1}}ds $$ for continuous functions $f$ on $[0,\infty)$ and $0<\alpha<1$.

Is it known when $T_\alpha$ is well-defined, i.e., its value exists for each $t>0$; for example, necessary conditions for $f$ etc...?

I know that Hadamard's finite value integral can be considered for such integral. However, my feeling is to handle $T_\alpha$ directly.

I'm glad if you tell me weaker condition because I know that $T_\alpha$ is well-defined if $f$ is $\beta$-Holder, where $\alpha<\beta$.

Thank you.

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  • $\begingroup$ I think you might want to look at the Slobodetskij condition. Though, the condition seems to be no more than requiring integrals of the kind you wrote above to be finite. $\endgroup$
    – Igor Khavkine
    Commented Jun 16, 2016 at 6:20

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Assuming the function $f$ of class $C^1$, you find that, using Taylor's formula with integral remainder, $$ f(t)-f(s)=(t-s) f_1(t,s),\quad \text{with $f_1$ continuous},$$ so that $ (T_\alpha f)(t)=\int_0 ^t (t-s)^{-\alpha} f_1(t,s) ds$ makes sense since $\alpha <1$.

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