Monthly Archives: January 2021

Stein’s spherical maximal function

1. Introduction

Maximal function plays a central role in several places in analysis. As an example, we can prove the celebrated Lebesgue differentiation theorem by using Hardy-Littlewood maximal function. Another example for application of maximal function is the nontangential behavior of Poisson integral defined on the half-plane. From these examples, maximal function helps us to understand a pointwise behavior of certain family of functions. See Stein’s monograph (1970) in Chapters 1, 3 for details.

It is well known from classical theory on PDEs that

    \[   u(x,t) = \int_{\mathbb{S}^{n-1}} g(x-ty) \myd{\sigma(y)} \]

solves wave equation

    \[  \partial_{tt}u-\triangle u=0\quad \text{in }\mathbb{R}^3\times (0,\infty),\quad u(x,0)=g(x),\quad \frac{\partial u}{\partial t}(x,0)=0\quad \text{on } \mathbb{R}^3. \]

Here \sigma denotes the normalized surface measure on the unit sphere \mathbb{S}^{n-1}. Such formula is called Kirchoff’s formula. From this, it is natural to define

    \[  A_r(f)(x) = \int_{\mathbb{S}^{n-1}} f(x+ry)\myd{\sigma(y)}, \]

the spherical average of f. Note that by Minkowski’s integral inequality, A_r(f) is well-defined for f\in \Leb{p}(\mathbb{R}^n), 1\leq p\leq \infty. As we mentioned before, to understand the pointwise behavior of spherical average of f, we define an associated maximal operator M_{S} by

    \[   M_{S}(f)(x) =\sup_{r>0}  \left| A_r(f)(x)\right|.\]

We call the operator M_S the spherical maximal operator. In 1976, Stein proved that the spherical maximal operator is bounded from \Leb{p} to itself when n\geq 3 and p>n/(n-1). In the same paper, he proved that the \Leb{p}-boundedness of spherical maximal operator is failed when 1<p\leq n/(n-1) and n\geq 2. Consider

    \[  f(y)=\frac{1}{|y|^{n-1}} (-\log |y|)^{-1}\chi_{|y|\leq 1/2}.\]

Then it is easy to see that f \in \Leb{p}(\mathbb{R}^n) and 1<p\leq n/(n-1) but M_S(f)=\infty. Hence it remains open when n=2. It was resolved by Bourgain (1986) in 1983 that \Leb{p}-boundedness of M_S holds for p>2 when n=2. There are several extensions on this result. We also mentioned that \Leb{p}\Leb{q} mapping property of spherical maximal function was proved by Schlag (1997, 1998) in n=2 and Schlag-Sogge (1997) under more general settings in the connection with local smoothing estimate. We also mention the result of Mockenhaupt-Seeger-Sogge (1992), which is the first paper containing the connection between local smoothing estimate of wave equation and \Leb{p}-boundedness of circular maximal operator. A standard references on this topic are Stein-Wainger (1978) and the monograph of Stein (1993).

The goal of this note is to prove Stein’s original result.

Theorem 1.1. Let n/(n-1)<p<\infty and n\geq 3. Then there exists a constant C(n,p)>0 such that

    \[   \norm{M_S (f)}{\Leb{p}(\mathbb{R}^n)}\leq C\norm{f}{\Leb{p}(\mathbb{R}^n)}\]

for all f\in \Leb{p}(\mathbb{R}^n).

Originally, Stein used some variant of g-functions and its mapping property. Here we follow a quite modern strategy via Littlewood-Paley decomposition as presented in Rubio de Francia (1986). We write

    \[    m(\xi)  =\widehat{d\sigma}(\xi)=\int_{\mathbb{S}^{n-1}} e^{-2\pi i x\cdot \xi}\myd{\sigma(x)}.  \]

Then we decompose

    \[   m(\xi)=\sum_{j=0}^\infty \psi_j(\xi)m(\xi),\]

where \{\psi_j\} is the usual Littlewood-Paley partition of unity. If we write m_j(\xi)=m(\xi)\psi_j(\xi), we first prove that

    \[     \norm{M_j f}{\Leb{2}(\mathbb{R}^n)}\apprle 2^{\frac{2-n}{2}j} \norm{f}{\Leb{2}(\mathbb{R}^n)} \]

and

    \[   \norm{M_j f}{\Leb{1,\infty}(\mathbb{R}^n)}\leq C 2^j \norm{f}{\Leb{1}(\mathbb{R}^n)}\]

for all f\in \mathcal{S}(\mathbb{R}^n) and for all j\geq 1. Here

    \[  M_j f(x) = \sup_{t>0}  |((m_j(t\cdot)\hat{f}(\cdot))^{\vee}|(x) \]

Since M_S f\leq \sum_{j=0}^\infty M_jf and M_0 f\apprle M f, where M denotes the usual Hardy-Littlewood maximal operator, then the desired result follows by an interpolation and the \Leb{p}-boundedness of Hardy-Littlewood maximal operator.

The rest of this note is organized as follows. In Section 2, we list some facts we used in this note. We prove the main theorem in Section 3.

2. Preliminaries

In this section, we first give some results on Hardy-Littlewood maximal operator.

For f\in \Leb{1}_{\loc}(\mathbb{R}^n), we denote by Mf the Hardy-Littlewood maximal function of f which is defined as

    \[  Mf(x)=\sup_{r>0} \fint_{B_r(x)} |f(y)|\myd{y}. \]

Theorem 2.1. We have

    \[    |\{ x\in \mathbb{R}^n : |Mf(x)|>t \} |\leq \frac{5^n}{t} \norm{f}{\Leb{1}(\mathbb{R}^n)},\]

in other words,

    \[   \norm{Mf}{\Lor{1}{\infty}(\mathbb{R}^n)}\leq 5^n \norm{f}{\Leb{1}(\mathbb{R}^n)}.\]

For any 1<p<\infty, there exists a constant C=C(n,p)>0 such that

    \[  \norm{Mf}{\Leb{p}(\mathbb{R}^n)}\leq C \norm{f}{\Leb{p}(\mathbb{R}^n)}\]

for all f\in \Leb{p}(\mathbb{R}^n).

As an application of Theorem 2.1, we have the following Lebesgue differentiation theorem.

Theorem 2.2. Let f\in\Leb{1}_{\loc}(\mathbb{R}^n). Then

    \[   \lim_{r\rightarrow 0+} \fint_{B_r(x)} f(y)\myd{y}=f(x)\quad \text{a.e. on } x. \]

For the proof of Theorems 2.1 and 2.2, see Stein’s monograph (1970) for details.

Next, we will use some decay estimate on \widehat{d\sigma}, which is defined by

    \[   \widehat{d\sigma}(\xi)=\int_{\mathbb{S}^{n-1}} e^{-2\pi i y\cdot \xi}\myd{\sigma(y)}. \]

It is easy to see that \widehat{d\sigma} is smooth. Moreover, we have the following decay estimate.

Proposition 2.3. We have

    \[    |\nabla^k \widehat{\sigma}(\xi)|\apprle |\xi|^{-(n-1)/2+k} \]

for large |\xi|.

See Wolff’s textbook (2003) or Grafakos’s textbook (2014). A proof presented in Wolff (2003) does not involve Bessel function.

Finally, we will use the Marcinkiewicz interpolation theorem. See Grafakos’s textbook (2014) for the proof.

Theorem 2.4. Let T be a sublinear operator and 1\leq p_0<p_1\leq \infty and let T be a sublinear operator defined on \Leb{p_0}(\mathbb{R}^n)+\Leb{p_1}(\mathbb{R}^n) and taking values in the space of measurable functions on \mathbb{R}^n. Suppose that

    \[    \norm{Tf}{\Leb{p_0,\infty}(\mathbb{R}^n)}\leq A_0 \norm{f}{\Leb{p_0}(\mathbb{R}^n)} \]

and

    \[    \norm{Tf}{\Leb{p_1,\infty}(\mathbb{R}^n)}\leq A_1 \norm{f}{\Leb{p_1}(\mathbb{R}^n)} \]

for all f\in \mathcal{S}(\mathbb{R}^n).

Then for 0<\theta<1, define

    \[   \frac{1}{p}=\frac{1-\theta}{p_0}+\frac{\theta}{p_1}.\]

Then there exists a constant C(n,p_0,p_1,\theta)>0 such that

    \[    \norm{Tf}{\Leb{p}(\mathbb{R}^n)}\leq C A_0^{1-\theta} A_1^{\theta}\norm{f}{\Leb{p}(\mathbb{R}^n)}   \]

for all f\in \mathcal{S}(\mathbb{R}^n).

3. Proof of Theorem 1.1

This section is devoted to a proof of Theorem 1.1. Below we assume that n\geq 3 and f\in \mathcal{S}(\mathbb{R}^n). Set m(\xi) = \widehat{d\sigma}(\xi) and define

    \[   m_j(\xi)=\psi_j(\xi)m(\xi),\quad j\geq 0 \]

where \psi_0\in C_c^\infty(\mathbb{R}^n) satisfies

    \[   \psi_0(\xi) = 1 \quad \text{in } B_1,\quad \psi_0(\xi)=0\quad \text{outside } B_2\]

and

    \[  \psi_j(\xi)=\psi_0(2^{-j} \xi)-\psi_0(2^{-j+1}\xi).\]

Note that if we write

    \[   \Psi(x)=\int_{\mathbb{R}^n} \psi_1(\xi) e^{2\pi i x\cdot \xi}\myd{\xi}, \]

then

    \[  \Psi_j(x)=\int_{\mathbb{R}^n} \psi_1\left(\frac{\xi}{2^j} \right)e^{2\pi i x\cdot\xi}\myd{\xi}=2^{jn}\Psi(2^{j}x).  \]

We decompose our symbol as follows:

    \[   m(\xi)=\sum_{j=0}^\infty \psi_j(\xi)m(\xi).\]

If we define

    \[    M_j f(x)=\sup_{t>0}|(\hat{f}(\xi)m_j(t\xi))^{\vee}(x)|,\]

then

    \[   M_{S}f\leq \sum_{j=0}^\infty M_jf.\]

The following lemma will be used in several places.

Lemma 3.1. For any N\geq n+1, we have

    \[ \int_{\mathbb{R}^n} \frac{f(y)}{(1+|x-y|)^N}\myd{y} \apprle Mf(x), \]

where M denotes the standard Hardy-Littlewood maximal operator.

Proof. It suffices to show the case N=n+1. We decompose

    \begin{align*} &\int_{\mathbb{R}^n}  \frac{f(y)}{(1+|x-y|)^{n+1}}\myd{y}\\  &=\int_{|x-y|< 2}   \frac{f(y)}{(1+|x-y|)^{n+1}}\myd{y}  +   \sum_{j=0}^\infty \int_{2^{j}\leq |x-y|<2^{j+1}} \frac{f(y)}{(1+|x-y|)^{n+1}}\myd{y}.  \intertext{Then by definition of Hardy-Littlewood maximal function, we have }  &\apprle \fint_{B_2(x)}|f(y)|\myd{y} + \sum_{j=0}^\infty \frac{1}{2^{j}}  \fint_{B_{2^{j+1}(x)}} |f(y)|\myd{y} \apprle Mf(x).\end{align*}


This proves the desired result.

From this lemma, we immediately get

    \[   M_0f\apprle Mf \]


since \psi_0 \in \mathcal{S}(\mathbb{R}^n).

Lemma 3.2. There exists a constant C=C(n)>0 such that

    \[  \norm{M_jf}{\Leb{2}(\mathbb{R}^n)}\leq C 2^{\frac{2-n}{2}j}\norm{f}{\Leb{2}(\mathbb{R}^n)}  \]

for all \mathcal{S}(\mathbb{R}^n) and j\geq 1.

Proof. For t>0, we write

    \[     A_{j,t}(f)(x)=\int_{\mathbb{R}^n} \hat{f}(\xi)m_j(t\xi)e^{2\pi i x\cdot\xi }\myd{\xi}.\]

If we write

    \[   \tilde{\Psi}_j(x) = \int_{\mathbb{R}^n} m_j(\xi)e^{2\pi i x\cdot \xi }\myd{\xi} \]

then

    \[     \int_{\mathbb{R}^n} m_j(t\xi)e^{2\pi i x\cdot \xi}\myd{\xi} = \int_{\mathbb{R}^n} m_j(t) e^{2\pi i(x/t)\cdot \xi}\myd{\xi}=t^{-n}\Psi_j(x/t). \]

Since m_j(0)=0, it follows that

    \[   \int_{\mathbb{R}^n}\tilde{\Psi}_j \myd{x}=0.\]

Hence by Lebesgue differentiation theorem, we get

    \[    \lim_{t\rightarrow 0+}  A_{j,t}(f)(x)=f(x)\quad \text{a.e. on  } \mathbb{R}^n.   \]

From this it follows from the fundamental theorem of calculus that

(1)   \begin{equation*}  (A_{j,t}(f)(x))^2=2\int_0^t  A_{j,s}(f)(x) s \frac{\partial }{\partial s} A_{j,s}(f)(x)\frac{ds}{s}\end{equation*}


for almost every x. On the other hand, we have

    \[  \frac{\partial}{\partial s} A_{j,s}(f)(x)=\int_{\mathbb{R}^n} \hat{f}(\xi) \left[\xi\cdot \nabla m_j(s\xi)\right] e^{2\pi i x \cdot \xi} \myd{\xi}. \]


For simplicity, we write

    \[    \tilde{A}_{j,s}(f)(x) = \int_{\mathbb{R}^n} [(s\xi)\cdot \nabla m_j(s\xi)]\hat{f}(\xi)e^{2\pi i x\cdot \xi}\myd{\xi}.\]

Then

    \[  s \frac{\partial }{\partial s} A_{j,s}(f)(x) = \tilde{A}_{j,s}(f)(x).\]

By Cauchy-Schwarz inequality, we have

    \begin{align*}&\int_{\mathbb{R}^n} (A_{j,t}(f)(x))^2 \myd{x}\\  &\leq 2\int_{\mathbb{R}^n} \left(\int_0^\infty |A_{j,s}(f)(x)| ^2 \frac{ds}{s}\right)^{1/2} \left( \int_0^\infty |\tilde{A}_{j,s}(f)(x)| ^2 \frac{ds}{s}\right)^{1/2} \myd{x}\\  &\leq 2\norm{G_j(f)}{\Leb{2}(\mathbb{R}^n)}^2\norm{\tilde{G}_j  (f)}{\Leb{2}(\mathbb{R}^n)}^2,\end{align*}


where G_j(f) and \tilde{G}_j(f) are associated square functions of f.

We will show that

    \[      \norm{G_j(f)}{\Leb{2}(\mathbb{R}^n)}\apprle 2^{-j(n-1)/2}\norm{f}{\Leb{2}(\mathbb{R}^n)} \]

and

    \[ \norm{\tilde{G}_j(f)}{\Leb{2}(\mathbb{R}^n)}\apprle 2^{j\left(1-\frac{n-1}{2}\right)}\norm{f}{\Leb{2}(\mathbb{R}^n)}. \]

Indeed, by Plancherel’s theorem, we get

    \begin{align*}\norm{G_j(f)}{\Leb{2}(\mathbb{R}^n)}^2= \int_{\mathbb{R}^n} \left(\int_{0}^\infty |\psi_j(t\xi)m(t\xi)|^2 |\hat{f}(\xi)|^2  \frac{dt}{t}\right)  \myd{\xi}.\end{align*}

By Proposition 2.3, we have

    \[   |m(\xi)|\apprle \frac{1}{(1+|\xi|)^{(n-1)/2}}. \]

Since \psi_j is supported in the annulus 2^{j-1}\leq |\xi|\leq 2^{j+1}, it follows that

    \begin{align*}\norm{G_j(f)}{\Leb{2}(\mathbb{R}^n)}^2&\apprle \int_{\mathbb{R}^n} \left(\int_{2^{j-1}/|\xi|}^{2^{j+1}/|\xi|} \frac{1}{(1+|t\xi|)^{n-1}}\frac{dt}{t}^2\right) |\hat{f}(\xi)| \myd{\xi} \\&\apprle 2^{(1-j)(n-1)}  \int_{\mathbb{R}^n} \left(\int_{2^{j-1}/|\xi|}^{2^{j+1}/|\xi|} \frac{dt}{t} \right)|\hat{f}(\xi)|^2 \myd{\xi} \\&\apprle 2^{-j(n-1)} \norm{\hat{f}}{\Leb{2}(\mathbb{R}^n)}^2.\end{align*}


Here we used an elementary calculus

    \[   \int_{a/A}^{b/A} \frac{1}{t} \myd{t} = \log(b/a) \]


for 0<a<b and A>0. Hence we get

    \[    \norm{G_j(f)}{\Leb{2}(\mathbb{R}^n)}\apprle 2^{-j(n-1)/2} \norm{f}{\Leb{2}(\mathbb{R}^n)}. \]

To estimate \norm{\tilde{G}_j(f)}{\Leb{2}(\mathbb{R}^n)}, we recall that

    \[    \widehat{{A}_{j,t}(f)}(\xi) = [(t\xi)\cdot \nabla m_j(t\xi)] \hat{f}(\xi). \]

By Proposition 2.3, we have

    \[   |\nabla m(\xi)|\apprle \frac{1}{(1+|\xi|)^{(n-1)/2}}, \]


Hence following the exactly same argument, one can get

    \[   \norm{\tilde{G}_j(f)}{\Leb{2}(\mathbb{R}^n)} \apprle 2^{j (1-(n-1)/2)}\norm{f}{\Leb{2}(\mathbb{R}^n)}. \]


Such estimate is natural since we have to consider one derivative gain. Therefore, we get

    \[   \norm{A_{j,t}(f)}{\Leb{2}(\mathbb{R}^n)}\apprle 2^{j(1/2-(n-1)/2)} \norm{f}{\Leb{2}(\mathbb{R}^n)}, \]


which completes the proof of Lemma 3.2.

Next, we show that each M_j f is of weak type (1,1). To do this, we recall that for any finite Borel measure \mu, its Fourier transform is defined by

    \[   \hat{\mu}(\xi) = \int_{\mathbb{R}^n} e^{-2\pi i x\cdot \xi}\myd{\mu(x)}. \]

We also define

    \[    h(x)=\int_{\mathbb{S}^{n-1}} f(x-y)\myd{\mu(y)} \]


and we denote it by f*d\mu. One can show that f*d\mu is well-defined for f\in \mathcal{S}(\mathbb{R}^n). This concept will be used in the proof of the following lemma.

Lemma 3.3. There exists a constant C=C(n)>0 such that

    \[   \norm{M_j f}{\Leb{1,\infty}(\mathbb{R}^n)}\leq C 2^j \norm{f}{\Leb{1}(\mathbb{R}^n)}\]

for all f\in \mathcal{S}(\mathbb{R}^n) and for all j\geq 1.

Proof. We claim that

(2)   \begin{equation*}|M_jf(x)|\apprle 2^{j}\int_{\mathbb{R}^n} \frac{1}{(1+|x-y|)^{n+2}}|f(y)|\myd{y} \end{equation*}

for j\geq 1. If this is true, then by Lemma 3.1, we have

    \[  |M_jf(x)|\apprle 2^{j}Mf(x). \]

Since M maps \Leb{1}(\mathbb{R}^n) to \Leb{1,\infty}(\mathbb{R}^n), it follows that

    \[   \norm{M_jf}{\Leb{1,\infty}(\mathbb{R}^n)}\leq C 2^j \norm{f}{\Leb{1}(\mathbb{R}^n)}.\qedhere \]

It remains to show (2). Recall that

    \[   M_j f(x) = \sup_{t>0} |(m_j(t\cdot)\hat{f}(\cdot))^{\vee}(x)| \]

where

    \[   m_j(\xi)=\psi_j(\xi)m(\xi). \]

If we write

    \[   \Psi(x) = \int_{\mathbb{R}^n} \psi_1(\xi)e^{2\pi i x\cdot \xi}\myd{\xi} \]

then we have

    \begin{align*}   \int_{\mathbb{R}^n} \psi_j(t\xi)m(t\xi)e^{2\pi i x\cdot \xi}\myd{\xi}&=t^{-n}\int_{\mathbb{R}^n} \psi_j(\xi)m(\xi)e^{2\pi i(x/t)\cdot \xi}\myd{\xi} \\  &=t^{-n} 2^{jn}\int_{\mathbb{S}^{n-1}} \Psi(2^j t^{-1}(x-y))\myd{\sigma(y)}\\  &\apprle t^{-n} 2^{jn}\int_{\mathbb{S}^{n-1}}\frac{1}{(1+2^jt^{-1}|x-y|)^{n+2}}\myd{\sigma(y)}.\end{align*}

By a standard argument, it suffices to estimate

    \[   \int_{\mathbb{S}^{n-1}} \frac{2^{jn}}{(1+2^j|x-y|)^{n+2}}\myd{\sigma(y)}.  \]

We decompose our integral as follows. Write

    \[   S_{-1}(x)=\mathbb{S}^{n-1} \cap \{y \in \mathbb{R}^n : 2^j |x-y|\leq 1\} \]

and

    \[  S_{r}(x) = \mathbb{S}^{n-1} \cap \{y \in \mathbb{R}^n : 2^r<2^j |x-y|\leq 2^{r+1}\}. \]

One can easily show that

    \[   \sigma(\mathbb{S}^{n-1}\cap B_R(y))\approx R^{n-1} \]

for any y\in \mathbb{S}^{n-1} and R>0. Note also that if y\in S_r(x), then |x-y|\leq 2^{r+1-j} and so |x|\leq 2^{r+1-j}+1 since y\in \mathbb{S}^{n-1}. From this, we have

    \begin{align*}  & \int_{\mathbb{S}^{n-1}} \frac{2^{jn}}{(1+2^j|x-y|)^{n+2}}\myd{\sigma(y)}\\  &\leq \sum_{r=-1}^j \int_{S_r(x)}  \frac{2^{jn}}{(1+2^j|x-y|)^{n+2}}\myd{\sigma(y)} + \sum_{r=j+1}^\infty \int_{S_r(x)}\frac{2^{jn}}{(1+2^j|x-y|)^{n+2}}\myd{\sigma(y)}\\  &\apprle 2^{nj} \left[\sum_{r=-1}^j \frac{\sigma(S_r(x))\chi_{B_3}(x)}{2^{r(n+2)}} + \sum_{r=j+1}^\infty \frac{\sigma(S_r(x))\chi_{B_{2^{r+1-j}+1}}(x)}{2^{r(n+2)}}  \right]\\  &\apprle 2^{jn}\left[\sum_{r=-1}^j \frac{2^{(r+1-j)(n-1)}\chi_{B_3}(x)}{2^{r(n+2)}} + \sum_{r=j+1}^\infty \frac{\chi_{B_{2^{r+2-j}}}(x)}{2^{r(n+2)}}  \right]\\  &\apprle 2^j \chi_{B_3}(x)+2^{nj} \sum_{r=j+1}^\infty \frac{1}{2^{r(n+2)}} \frac{(1+2^{r+2-j})^{n+1}}{(1+|x|)^{n+1}} \\  &\apprle \frac{2^j}{(1+|x|)^{n+1}} \left[1+\sum_{r=j+1}^\infty \frac{2^{j-r}}{2^{3j}} \right]\\  &\apprle \frac{2^j}{(1+|x|)^{n+1}}.\end{align*}


This proves the desired result.

We are ready to prove the main theorem.

Proof of Theorem 1.1. So far we proved that

    \[   \norm{M_jf}{\Leb{2}(\mathbb{R}^n)}\leq C 2^{\frac{2-n}{2}j} \norm{f}{\Leb{2}(\mathbb{R}^n)} \]

and

    \[  \norm{M_jf}{\Leb{1,\infty}(\mathbb{R}^n)}\leq C 2^j \norm{f}{\Leb{1}(\mathbb{R}^n)} \]

for all j\geq 1. Hence by Marcinkiewicz interpolation theorem, we get

    \[  \norm{M_j f}{\Leb{p}(\mathbb{R}^n)}\leq C 2^{j(1-n/p')} \norm{f}{\Leb{p}(\mathbb{R}^n)}   \]

for all j\geq 1. Since M_0 f\leq C Mf and M_Sf\leq \sum_{j=0}^\infty M_jf, it follows from the \Leb{p}-boundedness of Hardy-Littlewood maximal function that

    \[   \norm{M_Sf}{\Leb{p}(\mathbb{R}^n)}\apprle\norm{f}{\Leb{p}(\mathbb{R}^n)}+\sum_{j=1}^\infty 2^{j(1-n/p')} \norm{f}{\Leb{p}(\mathbb{R}^n)}\apprle \norm{f}{\Leb{p}(\mathbb{R}^n)}. \]

Here we used the restriction p>n' to guarantee that \sum_{j=1}^\infty 2^{j(1-n/p')} converges. This completes the proof of Theorem 1.1.

1.1. Newtonian Potentials

The goal of this note is to study solvability of second-order elliptic and parabolic equations in Hölder spaces. The prototype of elliptic equation is the Poisson equation

    \[   -\triangle u =f. \]

To understand the property of solution u, one of the easiest ways is to use Newtonian potential. In Section 1.1, we derive Newtonian potential and study some basic properties on this object. Also, we study a maximum principle for Poisson equation in Section 1.2. Such property plays a crucial role when we study second-order elliptic and parabolic equations.  Next, we introduce Hölder spaces which illustrate some smoothness of solutions in Section 1.3. After introducing Hölder spaces, in Section 1.4, we estimate a Hessian of Newtonian potential and a Hessian of solution for Poisson equation in Hölder space. This will lead us to think our main object of this note, so called Schauder estimate

To study properties of solution of the Poisson equation, we first seek a solution satisfying

    \[   -\triangle u =0.\]

A natural candidate for such solution is to assume that

    \[  u(x)=\phi(|x|), \]

i.e. radial solution. Write r=|x|. In spherical coordinate, the Laplacian is written as

    \[ \triangle u = \frac{\partial^2}{\partial r^2} u +\frac{n-1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2} \triangle_{\mathbb{S}^{n-1}}u,  \]

where \triangle_{\mathbb{S}^{n-1}} is the spherical Laplacian (or Laplace-Beltrami operator on unit sphere). Since we seek a radial solution, this implies that \phi should satisfy

    \[    \phi''(r)+\frac{n-1}{r} \phi'(r)=0.\]

Suppose that \phi'(r)\neq 0. Then

    \[ \log(|\phi'|)'=\frac{\phi''}{\phi'}=\frac{1-n}{r}  \]

and hence \phi'(r)=\frac{a}{r^{n-1}} for some constant a. Hence if r>0, then

    \[  \phi(r)=\begin{cases}  b\log r +c &\quad (n=2),\\  \frac{b}{r^{n-2}}+c &\quad (n\geq 3),  \end{cases}\]

where b and c are constants. This motivates us to introduce following one.

Definition 1.1. The function 

    \[  \Gamma(x)=\begin{cases}    \frac{1}{2\pi} \log |x| &\quad (n=2),\\    \frac{1}{(2-n)n \omega_n} \frac{1}{|x|^{n=2}}&\quad (n\geq 3),\end{cases}\]


defined for x\in \mathbb{R}^nx\neq 0, is the fundamental solution of Laplacian. Here \omega_n denotes the volume of the unit ball in \mathbb{R}^n

We wil use the following computation:

    \[ D_{i} \frac{1}{|y|^{n-2}} = -\frac{(n-2)y_i}{|y|^n}  \]

and 

    \begin{align*}    D_{j} \frac{y_i}{|y|^{n-2}}&=\frac{\delta_{ij}|y|^2-ny_i y_j}{|y|^{n+2}}.\end{align*}

From this, we see that \Gamma is harmonic in \mathbb{R}^n\setminus \{0\}
For f and g, we define the convolution of f and g by

    \[  (f*g)(x) = \int_{\mathbb{R}^n}  f(x-y)g(y) \myd{y}. \]

Theorem 1.1. Let f\in C_c^\infty(\mathbb{R}^n) and define 

    \[   u(x)=(\Gamma *f)(x).\]


Then 

    \[  u \in C^\infty(\mathbb{R}^n)\quad \text{and}\quad -\triangle u=f\quad \text{in } \mathbb{R}^n. \]

 

Proof. For simplicity, we prove the theorem when n\geq 3. The case n=2 can be similarly proved. We first show that u is well-defined.

Since f\in C_c^\infty(\mathbb{R}^n), there exists R>0 such that \supp f \subset B_R. A change of variable into polar coordinate gives

    \begin{align*}|u(x)|&\leq c_n \int_{\mathbb{R}^n} \frac{1}{|y|^{n-2}}|f(x-y)|\myd{y}\\&\leq c_n \norm{f}{\Leb{\infty}(\mathbb{R}^n)} \int_{B_{R+|x|}} \frac{1}{|y|^{n-2}}\myd{y}\\&=c_n \norm{f}{\Leb{\infty}(\mathbb{R}^n)} \int_0^{R+|x|} \rho \myd{\rho} \\&=c \norm{f}{\Leb{\infty}(\mathbb{R}^n)} (R+|x|)^2<\infty.\end{align*}

Hence u is well-defined.
Fix 1\leq j\leq n. For 0<|h|<R, we have 

    \[   \supp f(x+he_j-\cdot) \subset B_{2R+|x|}\]

since \supp f \subset B_R. So 

    \begin{align*}  \frac{u(x+he_j)-u(x)}{h}&=\int_{\mathbb{R}^n} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y}\\  &=\int_{B_{2R+|x|}} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y}.\end{align*}

By mean value property, it follows that 

    \begin{align*}  &\left|\int_{B_{2R+|x|}} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y} \right|\\  &\leq \int_{B_{2R+|x|}} \Gamma(y) \norm{D_j f}{\Leb{\infty}(\mathbb{R}^n)}\myd{y}\\  &\leq C(R,|x|) \norm{D_j f}{\Leb{\infty}(\mathbb{R}^n)}<\infty.\end{align*}

Since 

    \[   \lim_{h\rightarrow 0}\frac{f(x+he_j-y)-f(x-y)}{h}=D_j f(x-y)\quad \text{pointwise on }  B_{2R+|x|} \]

it follows from dominated convergence theorem that 

    \[    D_ju(x)=\int_{\mathbb{R}^{n}} \Gamma(y)D_j f(x-y)\myd{y}.   \]

By induction, one can show that u\in C^\infty(\mathbb{R}^n). It remains to show that -\triangle u=f
Fix \varepsilon>0. Then

    \begin{align*}D_{ij}u(x)&=\int_{\mathbb{R}^n} \Gamma(x-y) D_{ij}f(y)\myd{y}\\&=\int_{B_\varepsilon} \Gamma(y) D_{ij}f(x-y)\myd{y}+\int_{\mathbb{R}^n\setminus B_\varepsilon}  \Gamma(y) D_{ij}f(x-y)\myd{y}\\&=I_1+I_2.\end{align*}

Then

    \[   I_1 \leq C  \norm{D_{ij}f}{\Leb{\infty}(\mathbb{R}^n)}\int_{B_\varepsilon} \frac{1}{|y|^{n-2}} \myd{y} = C\norm{D_{ij}f}{\Leb{\infty}(\mathbb{R}^n)}\varepsilon^2. \]

Since f has compact support, integration by part gives

    \begin{align*}  I_2&=-\int_{\mathbb{R}^n\setminus B_\varepsilon} D_j\Gamma(y) D_i f(x-y)\myd{y}-\int_{\partial B_\varepsilon} \Gamma(y) D_i f(x-y) \left(\frac{y_j}{|y|}\right)\myd{\sigma(y)}\\  &=I_{21}+I_{22}.\end{align*}

A change of variable gives that

    \begin{align*}  I_{22}&=-\int_{\partial B_\varepsilon} \Gamma(y) D_i f(x-y) \left(\frac{y_j}{|y|}\right)\myd{y}\\  &\leq \frac{C\norm{\nabla f}{\Leb{\infty}(\mathbb{R}^n)}}{\varepsilon^{n-2}} \varepsilon^{n-1}\\  &=C \varepsilon.\end{align*}

Integration by part again gives

    \begin{align*} I_{21} &=\int_{\mathbb{R}^n\setminus B_\varepsilon} D_{ji}\Gamma(y) f(x-y)\myd{y} +\int_{\partial B_\varepsilon} D_j \Gamma(y) f(x-y) \left(\frac{y_i}{|y|} \right)\myd{\sigma(y)}.\end{align*}

From this calculation, we have

(1)   \begin{align*} \triangle u(x) &=\int_{B_\varepsilon} \Gamma(y)\triangle f(x-y)\myd{y}\\&\relphantom{=}+\int_{\mathbb{R}^n\setminus B_\varepsilon} \triangle \Gamma(y)f(x-y)\myd{y}+\int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}\nonumber. \end{align*}

Since \Gamma is harmonic in \mathbb{R}^n\setminus B_\varepsilon, it follows that

    \[  \int_{\mathbb{R}^n\setminus B_\varepsilon} \triangle \Gamma(y)f(x-y)\myd{y}=0. \]

Since

    \[   \nabla \Gamma(y)\cdot \frac{y}{|y|} = \frac{1}{(2-n)n\omega_n} \nabla \left(\frac{1}{|y|^{n-2}}\right) \cdot \frac{y}{|y|} =-\frac{1}{n\omega_n} \frac{1}{|y|^{n-1}} , \]

it follows that

    \[  \int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}=-\frac{1}{n\omega_n \varepsilon^{n-1}}\int_{\partial B_\varepsilon} f(x-y)\myd{\sigma(y)}. \]

  Since f is continuous and \sigma(\partial B_\varepsilon)=n\omega_n \varepsilon^{n-1}, it follows that   

    \[  \lim_{\varepsilon \rightarrow 0+} \int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}=-f(x).\]

  Hence letting \varepsilon \rightarrow 0+ in (1), we conclude that   

    \[  \triangle u(x)=-f(x) \]

  This completes the proof. 

Definition 1.2. Let f\in C_c^\infty(\mathbb{R}^n). The function \mathcal{N}f defined by  

    \[   \mathcal{N}f(x)=(\Gamma *f)(x) \]

  is called the Newtonian potential of f.

Remark. Here we obtain another calculation related to some fundamental studies on Poisson equation. Following the above argument, we have

    \begin{align*}D_{ij}\mathcal{N}f(x)&=\int_{\mathbb{R}^n} \Gamma(y)D_{ij}f(x-y)\myd{y}\\&=-\int_{\mathbb{R}^n} D_j \Gamma (y) D_i f(x-y)\myd{y}.\end{align*}

For simplicity, we set x=0. Choose a radial function \zeta \in C^\infty_c(\mathbb{R}^n) such that \zeta=1 near at the origin. Then

    \begin{align*}  D_{ij}\mathcal{N}f(0)=\int_{\mathbb{R}^n} D_i \Gamma(y)D_j\left(f(-y)-f(0)\zeta(y)\right)\myd{y}+f(0)C_{ij},\end{align*}

where

    \[  C_{ij}=\int_{\mathbb{R}^n} D_i \Gamma(y) D_j \zeta(y)\myd{y}.  \]

We write

    \[   K_{ij}(y)=-D_{ij}\Gamma(y) =\frac{1}{n\omega_n} \left(\frac{n y_i y_j-\delta_{ij}|y|^2}{|y|^{n+2}} \right).   \]

Note that

    \[   |K_{ij}(y)|\leq \frac{C}{|y|^n}\quad \text{and}\quad |f(y)-f(0)\zeta(y)|=|f(y)-f(0)+f(0)(1-\zeta(y))|\leq C|y|.  \]

Since f(y)-f(0)\zeta(y) has compact support, integration by part gives

    \[  D_{ij}\mathcal{N}f(0)=\lim_{r\rightarrow 0+}\int_{|y|>r} K_{ij}(y) \left(f(-y)-f(0)\zeta(y)\right)\myd{y}+f(0)C_{ij}. \]

For any r>0, we have

    \[    \int_{|y|> r} K_{ij}(y)\zeta(y)\myd{y}=0. \]

A change of variable gives

    \begin{align*}\int_{|y|> r} K_{ij}(y)\zeta(y)\myd{y}&=\int_r^\infty \int_{\mathbb{S}^{n-1}} K_{ij}(\rho\omega)\myd{\sigma(\omega)} \zeta(\rho) \rho^{n-1}\myd{\rho}.\end{align*}

Hence it suffices to show that

    \[   \int_{\mathbb{S}^{n-1}} K_{ij}(\omega)\myd{\sigma(\omega)}=0.\]

If i\neq j, the integral is zero since K_{ij} is antisymmetric. If i=j, then a change of variable and harmonicity of \Gamma give

    \[     n\int_{\mathbb{S}^{n-1}}K_{ii}(\omega)\myd{\sigma(\omega)}=\int_{\mathbb{S}^{n-1}} K_{11}(\omega)\myd{\sigma(\omega)}+\cdots+\int_{\mathbb{S}^{n-1}} K_{nn}(\omega)\myd{\sigma(\omega)} =0 \]

for 1\leq i\leq n. Integration by part gives

    \begin{align*}  C_{ij}&=\int_{B_\varepsilon} D_i \Gamma(y) D_j \zeta (y)\myd{y}+\int_{\mathbb{R}^n\setminus B_\varepsilon} D_i \Gamma(y)D_j \zeta(y)\myd{y}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\int_{\partial B_\varepsilon} D_i \Gamma(y)\zeta(y)\frac{y_j}{|y|}\myd{\sigma(y)}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\frac{1}{n\omega_n}\int_{\partial B_\varepsilon} \frac{y_i y_j}{|y|^{n+1}} \zeta(y)\myd{\sigma(y)}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\frac{1}{n\omega_n \varepsilon^{n+1}} \int_{\partial B_\varepsilon} y_i y_j \zeta(y)\myd{\sigma(y)}.\end{align*}

If i\neq j, the quantity C_{ij} is zero due to antisymmetric. If i=j, then a change of coordinate and harmonicity of \Gamma give

    \[n C_{11}=C_{11}+C_{22}+\cdots + C_{nn}=-\frac{1}{n\omega_n \varepsilon^{n-1}}\int_{\partial B_\varepsilon} \zeta(y)\myd{\sigma(y)}.  \]

Letting \varepsilon \rightarrow 0+, we conclude that

    \[ C_{ii}=- \frac{1}{n}\quad \text{for all } 1\leq i\leq n.  \]

This shows that

(2)   \begin{equation*}    D_{ij} \mathcal{N}f(x)=\lim_{r\rightarrow 0+} \int_{|y|>r} K_{ij}(y)f(x-y)\myd{y} -\frac{1}{n}\delta_{ij}f(x) \end{equation*}

for f\in C_{c}^\infty(\mathbb{R}^n).
Now define

    \[   \mathcal{K}_{ij}f(x)=\lim_{r\rightarrow 0+}\int_{|y|>r} K_{ij}(y)f(x-y)\myd{y}.  \]

Note that the kernel K_{ij} satisfies

    \[  |K_{ij}(y)|\leq \frac{C}{|y|^n}\quad \text{and}\quad  \int_{\mathbb{S}^{n-1}} K_{ij}\myd{\sigma}=0. \]

We call such kernel as Calderón-Zygmund singular kernel. Then  it follows from the well-known theory on singular integral operator that for 1<p<\infty, there exists a constant C=C(n,p)>0 such that

    \[   \norm{\mathcal{K}_{ij}f}{\Leb{p}(\mathbb{R}^n)}\leq C \norm{f}{\Leb{p}(\mathbb{R}^n)}\]

for all f\in C_c^\infty(\mathbb{R}^n). Hence by (2), we conclude that

    \[  \norm{D_{ij}\mathcal{N}f}{\Leb{p}(\mathbb{R}^n)}\leq C \norm{f}{\Leb{p}(\mathbb{R}^n)}. \]

We call such estimate as Calderón-Zygmund type estimate. This estimate plays a crucial roles when we study the Sobolev space theory for partial differential equations. Due to our purpose of this note, we will not go to any futher. We end this section by introducing several textbooks. When one has some interests on general theory on singular integrals, see Stein (1970). Modern theory of this theory can be found in Stein (1993). An application for Calderón-Zygmund estimate can be found in Gilbarg-Trudinger (1998) and Krylov (2008)