Let $R\_\alpha f(t) = \int\_0^t (t-s)^{-\alpha} f(s)\,ds$ the fractional integration operator. If $f \in L\_q(0,1)$ for some $q>2$ then $R\_{1/2} f$ is (even Hölder) continuous on $[0,1]$. 


My question is what happens in the endpoint: does $R\_{1/2}$ map $L\_2(0,1)$ to $C([0,1])$?


thank you very much, Bernhard



Is the fractional integral of order 1/2 of an L_2 function continuous
The answer is no as the following example shows: Let 
 $$
 F(t)= \begin{cases} a\_nt^{-\frac 12+b\_n},& t\in [t\_{n+1},t\_n),\quad n\geq 1,\\
 0,& \text{otherwise,}
 \end{cases}
 $$
where $0< t\_{n+1}< t\_n<1$, $n\geq 1$. Then $F\in L^2(0,1)$ provided
 $$
 \sum\_{n=1}^\infty \frac{a\_n^2}{2b\_n}(t\_n^{2b\_n}-t\_{n+1}^{2b\_n}) < \infty
 $$
 and $\int\_0^1 t^{-\frac 12} F(t)\, dt = +\infty$ provided
 $$
 \sum\_{n=1}^\infty \frac{a\_n}{b\_n}(t\_n^{b\_n}-t\_{n+1}^{b\_n})= \infty.
 $$
 Now we choose $a\_n= \frac 1{n^2} e^n$, $b\_n=\frac 1n$, $t\_n=e^{-n^2}$, and take $f(t)= F(1-t)$, $t\in [0,1]$ and we conclude that $f\in L^2(0,1)$ but $R\_{1/2}f\notin C([0,1])$.


