## Boundedness of fractional integration operator

Suppose $f$ is locally integrable, define the fractional integration operator

$\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n-\alpha}}dy$, $0<\alpha

$\mathbf{Thm:}$ Suppose $f\in L^p(\mathbb{R}^n)$ and $0<\alpha, then

$\displaystyle ||I_\alpha(f)||_q\leq C(p,n,\alpha)||f||_p$

where $\displaystyle \frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n}$ and $p>1$. If $p=1$, then $I_\alpha(f)$ is a weak type $(1,\frac{n}{n-\alpha})$.

$\bf{Lemma:}$ Suppose $\psi:\mathbb{R}^n\to\mathbb{R}^n$ is radially decreasing function, then

$\displaystyle \left|\int_{\mathbb{R}^n}f(x)\psi(x-y)dy\right|\leq M(f)(x)||\psi||_1$

when $||\psi||_1<\infty$.

$\textbf{Proof of Theorem:}$ Write

$\displaystyle I_\alpha(f)(x)=\int_{\mathbb{R}^n}\frac{f(x-y)}{|y|^{n-\alpha}}dy=\int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy+\int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy$

By the lemma,

$\displaystyle \int_{|y|\leq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq M(f)(x)\int_{|y|\leq A}\frac{1}{|y|^{n-\alpha}}dy\leq \frac{nw_n}{\alpha}A^\alpha M(f)(x)$

By holder inequality, if $p>1$

$\displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq ||f||_p\left(\int_{|y|\geq A}\frac{1}{|y|^{(n-\alpha)p'}}\right)^\frac{1}{p'}\leq \left(\frac{qw_n}{p'}\right)^\frac{1}{p'}A^{-\frac{n}{q}}||f||_p$

If $p=1$

$\displaystyle \int_{|y|\geq A}\frac{f(x-y)}{|y|^{n-\alpha}}dy\leq A^{-\frac{n}{q}}||f||_1$ with $\displaystyle q=\frac{n}{n-\alpha}$.

Combing the preceeding result,

$\displaystyle |I_\alpha(f)(x)|\leq C(n,p,\alpha)(A^{-\frac{n}{q}}||f||_p+A^\alpha M(f)(x))$ for $\forall\,A>0$

Choose $A=Mf(x)^{-\frac{p}{n}}||f||_p^{\frac{p}{n}}$,

$\displaystyle |I_\alpha(f)(x)|\leq C(n,\alpha,p)M(f)(x)^{\frac{p}{q}}||f||_p^{1-\frac{p}{q}}$

When $p>1$,  maximal function is bounded from $L^p$ to $L^p$,

$\displaystyle ||I_\alpha(f)||_q\leq C(n,\alpha,p)||f||_p^\frac{p}{q}||f||_p^{1-\frac{p}{q}}=C(n,\alpha,p)||f||_p$

When $p=1$, maximal function is weak (1,1),

$\displaystyle |\{x||I_\alpha(f)(x)|>\lambda|\}|\leq \{x|C(n,\alpha)M(f)(x)^{\frac{n-\alpha}{n}}||f||_1^{\frac{\alpha}{n}}>\lambda\}$

$\displaystyle =|\{x|M(f)(x)>\left(\frac{\lambda}{C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}\right)^{\frac{n}{n-\alpha}}\}|$

$\displaystyle \leq 3^n||f||_1\left(\frac {C(n,\alpha)||f||_1^{\frac{\alpha}{n}}}{\lambda} \right)^\frac{n}{n-\alpha}=C'(n,\alpha)\left(\frac{||f||_1}{\lambda}\right)^\frac{n}{n-\alpha}$

$\square$

When the domain is not $\mathbb{R}^n$ but a bounded one, we can obtain the boundedness of $\displaystyle \frac{1}{q}<\frac{1}{p}-\frac{\alpha}{n}$.

$\textbf{Thm: }$ Suppose $f\in L^p(\Omega)$, $p>1$. We have $I_\alpha(f)$ is bounded from $L^p$ to $L^q$  with

$\displaystyle 0\leq \delta=\frac{1}{p}-\frac{1}{q}<\frac{\alpha}{n}$, $1\leq q\leq \infty$

$\displaystyle ||I_\alpha f||_q\leq Cw_n^{1-\alpha/n}|\Omega|^{\alpha/n-\delta}||f||_p$

$\textbf{Remark:}$ Loukas Grafakos: Modern fourier analysis.