Fractional integrals problem: $g(x)_{|_{[0,t]}}=\frac{(-1)^{\alpha-1}}{\Gamma(1-\alpha)}\int_x^t(y-x)^{-\alpha}h(y)\,dy$ Let's consider $0<\alpha<1/2$ and denote by $W_T^{1-\alpha,\infty}(0,T)$ the space of measurable functions $g:[0,T]\to\Bbb R$ such that
$$
||g||_{1-\alpha,\infty,T}:=\sup_{0<s<t<T}\left[\frac{|g(t)-g(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|g(y)-g(s)|}{(y-s)^{2-\alpha}}\,dy\right]<+\infty\;\;\;.
$$
Moreover, we define the right sided Riemann-Liouville integral of order $1-\alpha$ of a function $f\in L^p(0,t)$, with $1\le p\le\infty$, as
$$
I_{t-}^{1-\alpha}f(x):=\frac{(-1)^{\alpha-1}}{\Gamma(1-\alpha)}\int_x^t(y-x)^{-\alpha}f(y)\,dy\;\;\;
$$
for a.a. $x\in[0,t]$.
My problem is the following: in a paper by Nulart and Rascanu it is stated that, if $g\in W_T^{1-\alpha,\infty}(0,T)$ then its restriction to $[0,t]$ stays in $I_{t-}^{1-\alpha}(L^{\infty}(0,t))$ for all $0<t<T$; so in other words, given such $g$, there exists $h\in L^{\infty}(0,t)$ such that
$$
g(x)|_{[0,t]}=I_{t-}^{1-\alpha}h(x)=\frac{(-1)^{\alpha-1}}{\Gamma(1-\alpha)}\int_x^t(y-x)^{-\alpha}h(y)\,dy\;\;\;.$$
It seems to me, that I DON'T HAVE to search explicitly the $h$ depending on the $g$ but rather I should use a theoretical argument, which proves the existence of such $h$; but I don't know how to do it.
I'm quite lost, can someone shade a light please?
EDIT: Obviously $\Gamma$ is the Euler Gamma function and $(-1)^{\alpha-1}=e^{i\pi(\alpha-1)}$, but these terms are constant, thus this doesn't play any relevant role here.
SECOND EDIT: We can state the "symmetric" claim; the underlying duality could help.
We denote by $W_0^{\alpha,1}(0,T)$ the space of measurable functions $f:[0,T]\to\Bbb R$ such that
$$
||f||_{\alpha,1}:=\int_0^T\frac{|f(s)|}{s^{\alpha}}\,ds+\int_0^T\int_0^s\frac{|f(s)-f(y)|}{(s-y)^{\alpha+1}}\,dyds<+\infty
$$
As before we define the left sided Riemann-Liouville integral of order $\alpha$ of a function $f\in L^p(0,t)$, with $1\le p\le\infty$, as
$$
I_{0+}^{\alpha}f(x):=\frac{1}{\Gamma(\alpha)}\int_0^x(x-y)^{\alpha-1}f(y)\,dy\;\;\;
$$
for a.a. $x\in[0,t]$.
Then if $g\in W_0^{\alpha,1}(0,T)$ then its restriction to $[0,t]$ stays in $I_{0+}^{\alpha}(L^1(0,t))$ for all $0<t<T$.
 A: Just define
\begin{align*}
\eta_t(x)&:=D_{t-}^{1-\alpha}(f-f(t))(x)\\
&=\frac{(-1)^{1-\alpha}}{\Gamma(\alpha)}\left[\frac{f(x)-f(t)}{(t-x)^{1-\alpha}}+(1-\alpha)\int_x^t\frac{f(x)-f(y)}{(y-x)^{2-\alpha}}\,dy\right]\chi_{]0,t[}(x)
\end{align*}
($D_{t-}^{1-\alpha}$ is the Weyl derivative).
From this, a direct computation shows that $||\eta_t||_{\infty}\le C||f||_{1-\alpha,\infty,t}$, thus $f\in W_T^{1-\alpha,\infty}(0,T)\Rightarrow \eta_t\in L^{\infty}(0,t)$.
Finally
$$
I_{t-}^{1-\alpha}\eta_t(x)=I_{t-}^{1-\alpha}D_{t-}^{1-\alpha}(f-f(t))(x)=f(x)-f(t)
$$
from which it's straightforward that
$$
f(x)=\frac{(-1)^{\alpha-1}}{\Gamma(1-\alpha)}\int_x^t(y-x)^{-\alpha}\underbrace{\left[\eta_t(y)+\frac{\Gamma(1-\alpha)}{(-1)^{\alpha-1}}f(t)\frac{1-\alpha}{(t-x)^{1-\alpha}}\right]}_{=:h_t(y)}\,dy
$$
and clearly $h_t\in L^{\infty}(0,t)$, from which we conclude.
EDIT I ask sorry if I made a question, then I answered it and finally I've even accepted it, but honestly it went like this: I was struggling on this problem, I posted it and then I put a bounty. In the meanwhile I continued to think at it. Then I solved and the solutions satisfied me such that I accepted my own answer.
I did all in good faith, really.
