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<n

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

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: