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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 $0<t<T$.

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  • $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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