Fractional integration lemma - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:33:45Z http://mathoverflow.net/feeds/question/60088 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60088/fractional-integration-lemma Fractional integration lemma Nicolas L 2011-03-30T16:56:25Z 2011-04-13T21:22:13Z <p>Hello everyone.</p> <p>I am trying to establish a fractional integration lemma of this form.</p> <p>For $\alpha\geq 0$, and $1\leq p,q&lt;\infty$ and $0\leq \frac{1}{q}-\frac{1}{p}=\frac{\alpha}{d}$ or $1\leq p,q\leq\infty$ and $0\leq \frac{1}{q}-\frac{1}{p}&lt;\frac{\alpha}{d}$, for $f$ a function of $t\in\mathbb{R}$ and $x\in\mathbb{R}^3$, such that $f(t,\cdot)$ is in $L^q(\mathbb{R}^3)$, we have</p> <p>$\Vert\Lambda_{t}^{-a}f(t,\cdot)\Vert_{p}\leq C t^{\frac{a}{2}-\frac{3}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\Vert f\Vert_{q}$ where $\Lambda^{-\alpha}=\frac{1}{|D|^{\alpha}}$ and $\Lambda_{t}^{-\alpha}=\sqrt{t}^{\alpha}Z^{\alpha}\left(\sqrt{t}|D|\right)$ where $Z$ is a smooth function equal to $|\xi|^{-1}$ for $|\xi|\geq 2$ and equal to $1$ for $|\xi|\leq 1$. ($D$ is the Fourier multiplier, i>e $f(D)u=\mathcal{F}^{-1}(f(\xi))\hat{u}$ It must be linked to Riemann Liouville integral operators but I do not know how and how to consider the rescaling in $t$. Normally Fourier should be a simpler way to understand fractionnal integration but here I am quite stuck.</p> <p>Thanks in advance to anyone who has ever done fractional integration!</p> http://mathoverflow.net/questions/60088/fractional-integration-lemma/60125#60125 Answer by Piero D'Ancona for Fractional integration lemma Piero D'Ancona 2011-03-30T21:11:14Z 2011-03-30T21:11:14Z <p>It is not clear to me why in your estimate you put a function f depending on t. The norms are only in x, the operators act only on the x variable, so t is just a parameter and if your estimate is true it must be true for a function f independent of t. Also, I think the sign in front of $3/2$ should be a plus.</p> <p>Anyway, let's prove it for a function depending only on x. Let me call $S_t$ the scaling operator $S_tf(x)=f(tx)$. Then you can write $$Z(s|D|)f= F^{-1}Z(s|\xi|)Ff=F^{-1}S_sZ(|\xi|)S_{1/s}Ff=S_{1/s}Z(|D|)S_sf$$ so the correct inequality can be written, with $s=\sqrt{t}$, $$s^a\| S_{1/s}Z^a(|D|)S_s f\|_p\le C s^{a+3/p-3/q}\|f\|_q.$$</p> <p>By the scaling property $$\|S_sf\|_p=s^{-n/p}\|f\|_p$$</p> <p>and calling $g=S_{s}f$ all powers of $s$ cancel out, and the inequality to prove reduces to $$\|Z^a(|D|)g\|_p\le C\|g\|_q.$$ Now this is easy but let me tell you how to do it. The following inequality is just Sobolev embedding: $$\||D|^{-a}g\|_p\le C\|g\|_q$$ so if you split $g=g_1+g_2$ with $\widehat g_1$ supported on $|\xi|\ge2$, it takes care of the estimate for $g_1$. On the other hand $g_2$ has a compactly supported Fourier transform and the multiplier $Z$ is just 1 so the estimate is trivial for $g_2$.</p>