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!



Fractional integration lemma
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 $g\_2$.


