MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello everyone.

I am trying to establish a fractional integration lemma of this form.

For $\alpha\geq 0$, and $1\leq p,q<\infty$ and $0\leq \frac{1}{q}-\frac{1}{p}=\frac{\alpha}{d}$ or $1\leq p,q\leq\infty$ and $0\leq \frac{1}{q}-\frac{1}{p}<\frac{\alpha}{d}$, 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

$\Vert\Lambda_{t}^{-a}f(t,\cdot)\Vert_{p}\leq C t^{\frac{a}{2}-\frac{3}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\Vert f\Vert_{q}$ 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.

Thanks in advance to anyone who has ever done fractional integration!

share|cite|improve this question
up vote 1 down vote accepted

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.

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 $s=\sqrt{t}$, $$ s^a\| S_{1/s}Z^a(|D|)S_s f\|_p\le C s^{a+3/p-3/q}\|f\|_q. $$

By the scaling property $$\|S_sf\|_p=s^{-n/p}\|f\|_p$$

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 $g_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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