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 $0<t<T$.

  • $\begingroup$ @SimplyBeautifulArt: that's a good idea. Thanks $\endgroup$ – Joe Mar 16 '17 at 16:31
  • $\begingroup$ Existence theorems of functions in integrals smells like Radon-Nikodym to me. $\endgroup$ – Robert Wolfe Mar 17 '17 at 2:05
  • $\begingroup$ @Bryan: that's a good idea but if I'm right, R-N deals only with existence, it doesn't say nothing about the boundedness (we want $h\in L^{+\infty}$) $\endgroup$ – Joe Mar 17 '17 at 22:45

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.

  • $\begingroup$ It is ok to post an answer to your question, read this math.stackexchange.com/help/self-answer. $\endgroup$ – Zaid Alyafeai Mar 23 '17 at 9:17
  • $\begingroup$ :D Nice job! (+1) $\endgroup$ – Simply Beautiful Art Mar 23 '17 at 13:11
  • $\begingroup$ @ZaidAlyafeai: and...is there a possibility to recover my 350 lost points? All in all, I solved my problem! $\endgroup$ – Joe Mar 23 '17 at 23:00
  • $\begingroup$ @SimplyBeautifulArt: many thanks! :-) $\endgroup$ – Joe Mar 23 '17 at 23:00
  • $\begingroup$ @Joe no, bounties are not refundable. But you shouldn't worry too much :-) $\endgroup$ – Simply Beautiful Art Mar 23 '17 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.