This is a transcription of the two lectures given by Eli Stein at the 2011 Princeton summer school in analysis and geometry, delivered on July 11 and July 12. All errors in this post are those in my transcription and interpretation of the lectures.

## 1. The Fourier Transform

We shall restrict our attention to the Euclidean space $$\mathbb{R}^d$$ with the standard metric. $$L^p$$ shall denote the space of measurable functions $$f:\mathbb{R}^d \to \mathbb{C}$$ such that

$\|f\|_{L^p} = \left( \int_{\mathbb{R}^d} |f|^p \, dx \right)^{1/p}$

is finite. Each $$L^p$$ $$(1 \leq p\ leq \infty)$$ is a Banach space, and $$L^p$$ is isometrically isomorphic to $$L^q$$ whenever $$1 < p < \infty$$ and $$1/p+1/q = 1$$. We expect the reader to be familiar with Hölder's inequality and Minkowski's integral inequality.

We define the Fourier transform of $$f \in L^1$$ to be the integral

$\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x) e^{-2 \pi i x \cdot \xi} \, dx.$

The Fourier inversion formula for suitable $$f$$ is

$f(x) = \int_{\mathbb{R}^d} \hat{f}(\xi) e^{2 \pi i x \cdot \xi} \, d\xi.$

Plancheerel's theorem states that the Fourier transform is an isometry on $$L^2$$, viz.,

$\|\hat{f}\|_{L^2} = \|f\|_{L^2}.$

An important multiplication operation in harmonic analysis is convolution, which is defined by the integral

$(f \ast g)(x) = \int f(x-y) g(y) \, dy = \int f(y) g(x-y) \, dy.$

The Fourier transform of a convolution is the pointwise product of the Fourier transform:

$\widehat{f \ast g} = \hat{f}\hat{g}.$

## 2. Three continuous operators, part 1: the maximal function

We shall consider the following symmetries of $$\mathbb{R}^d$$:

• Translations $$x \mapsto x + h$$, $$h \in \mathbb{R}^d$$
• Dilations $$x \mapsto \delta x$$, $$\delta > 0$$
• Rotations $$x \mapsto R(x)$$, where $$R$$ is a linear map such that $$|R(x)| = |x|$$.

We will exploit these special symmetries of $$\mathbb{R}^d$$ again and again.

Definition 2.1. The maximal operator of a function $$f$$ is \begin{align*} M(f)(x) &= \sup_{r > 0} \frac{c}{r^d} \int_{|y| \leq r} |f(x-y)| \, dy \\ &= \sup \frac{1}{m(B(x))} \int_{B(x)} |f(y)| \, dy, \end{align*} where the supremum is taken over all balls $$B(x)$$ centered at $$x$$.

The maximal operator came to be when G. H. Hardy wanted to study crickets. We observe that the maximal operator is invariant under all three symmetries.

Theorem 2.2. Let $$f:\mathbb{R}^d \to \mathbb{C}$$. a) For each $$1 < p \leq \infty$$, we have $\|Mf\|_{L^p} \leq A_p \|f\|_{L^p},$ where $$A_p$$ only depends on $$p$$ and $$d$$. b) For each $$\alpha > 0$$, the Hardy–Littlewood maximal inequality $m\{x : (Mf)(x) > \alpha\} \leq \frac{A}{\alpha} \|f\|_{L^1}$ holds, where $$A$$ only depends on $$d$$.

Note that, by definition, $$Mf$$ is bigger than $$f$$. (a) shows that $$Mf$$ is, in the $$L^p$$ sense, not much bigger. This estimate is substituted by a weak-type inequality (b) in $$L^1$$ (cf. Chebychev's inequality).

We remark further that the maximal function is the "mother of all averages." if $$\Phi$$ is radial, viz.,

$\Phi(x) = \Phi_0(|x|)$

with $$\Phi_0$$ positive, decreasing, and $$\int \Phi = 1$$, then we have

$\left| \int f(x-y) \Phi(y) \, dy \right| \leq M(f)(x).$

To see this, we approximate $$\Phi$$ by step functions

$\sum c_k \frac{1}{m(B_k)}{\xi_{B_k}},$

where $$B_k$$ are balls centered at origin and $$\sum c_k = 1$$.

Proof of Theorem 2.2b. Let $$E_\alpha = \{x : M(f)(x) > \alpha\}$$, and $$K$$ a compact subset of $$E_\alpha$$. We shall exploit the inner regularity.

We pick a cover of $$K$$ by fixing a ball at each point of $$K$$, and extract a finite cover $$B_i = B_i(x_i)$$ centered at $$x_i$$ with

$K \subseteq \bigcup_{i=1}^N B_i$

and

$\frac{1}{m(B_i)} \int_{B_i} |f| > \alpha.$

We now appeal to the Vitali covering lemma, from which we can select pairwise disjoint balls $$B_{i_1}, B_{i_2}, \ldots, B_{i_k}$$ that "almost cover" $$K$$, viz.,

$\bigcup_j B_{i_j}^\ast \supseteq K,$

where \$B_{ij}^\ast$$is the "three-times dilated" balls of$$B{i_j}. It then follows that \begin{align*} m(K) &\leq \sum_{j} m(B_{i_j}^\ast) \\ &= 3^d \sum_j m(B_{i_j}) \\ &\leq \frac{3^d}{\alpha} \sum_j \int_{B_{i_j}} |f|, \end{align*} as was to be shown. $$\square$$ Definition 2.3. Suppose $$F \geq 0$$. The distribution function of $$F$$ is $\lambda(\alpha) = m\{x: F(x) > \alpha\}.$ We note that $\int_0^\infty \lambda(\alpha) \, d\alpha = \int F(x) \, dx$ and that \begin{align*} \int_0^\infty \lambda(\alpha^{1/p}) \, d\alpha &= \int(F(x))^p \, dx \\ &= p \int_0^\infty \lambda(\alpha) \alpha^{p-1} \, dx. \end{align*} Proof of Theorem 2.2a. We will strengthen $m \{ x : (Mf)(x) > \alpha \} \leq \frac{A}{\alpha} \frac{A}{\alpha} \int_{\mathbb{R}^d} |f| \, dx$ by $m\{x : (Mf)(x) > \alpha\} \leq \frac{A'}{\alpha} \int_{|f| > \alpha/2} |f| \, dx.$ In fact, $$|f(x) \leq |f_1(x)| + \alpha/2$$, where $f_1(x) = \begin{cases} f(x) & \mbox{ if } |f(x)| > \alpha/2 \\ 0 & \mbox{ otherwise}\end{cases}$ Then we have $$Mf \leq Mf_1 + \alpha/2$$ and $$\{Mf > \alpha\} \subseteq \{Mf_1 > \alpha/2\}$$, and so we have $m\{x: Mf(x) > \alpha\} \leq \frac{2A}{\alpha} \int_{|f(x)| > \alpha/2} |f| \, dx.$ Take $$F = Mf$$ and $$\lambda(\alpha) = m\{x:F(x) > \alpha\}$$. Then \begin{align*} \int(Mf)^p &\leq A_p' \int_0^\infty \alpha^{p-1} \alpha^{-1} \left( \int_{|f| > \alpha / 2} |f| \, dx \right) \, d\alpha \\ &= c \int |f| \left( \int_0^{2|f|} \alpha^{p-2} \, d\alpha \right) \, dx \\ &= c' \int |f| |f|^{p-1} \, dx \\ &= c' \int |f|^p \, dx, \end{align*} as was to be shown. $$\square$$ ## 3. Three continuous operators, part 2: the fractional integral Definition 3.1. The Riesz potential of $$f$$ is $I_\alpha(f)(x) = \frac{1}{\gamma_\alpha} \int_{\mathbb{R}^d} f(x-y)|y|^{-d+\alpha} \, dy,$ where $$0 < \alpha < d$$ and $\gamma_\alpha = \pi^{d/2} 2^{\alpha} \frac{\Gamma(\frac{\alpha}{2})}{\Gamma(\frac{d-\alpha}{2})}.$ These are also called fractional integrals. We can write the above as $I_\alpha(f) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x-y) K_\alpha(y) \, dy$ where $$K_\alpha$$ is locally integrable. Why is this operator interesting? Formally, $$\widehat{I_\alpha(f)}(\xi) = (2\pi |\xi|)^{-\alpha} \hat{f}(\xi)$$, so $$I_\alpha(f) = (-\Delta)^{-\frac{\alpha}{2}}$$, where $$\Delta$$ is the Laplacian. Here, $$\Lambda$$ is to be considered as a "basic differentiation operator of order 2", whence $$(-\Delta)^{-\alpha/2}$$ is the "basic differentiation operator of order $$\alpha$$. We identify $$\widehat{K_\alpha}(\xi) = (2\pi|\xi|)^{-\alpha}$$ in the sense of distributions. In particular, if $$\alpha = 2$$ and $$d \geq 3$$, then $$I_\alpha$$ is the fundamental solution operator for the Laplacian $$\Delta$$. When $$d = 2$$, $$I_\alpha$$ is the Newtonian potential, whose definition, we note, makes use of the inverse Laplacian $$\Delta^{-1}$$. The expression of $$f$$ in terms of the partial differentials can be given as $f(x) = \frac{1}{\omega_d} \int_{\mathbb{R}^d} \sum_{j=1}^d \frac{\partial f_j}{\partial x_j}(x-y) \frac{y_j}{|y|^d} \, dy,$ where $$\omega_d$$ is the area of the unit sphere in $$\mathbb{R}^d$$. Therefore, $|f(x)| \leq cI_1(|\nabla f)|)(x),$ where $$\nabla f$$ is the usual gradient of $$f$$. Theorem 3.2. Suppose that $$1 < p < q < \infty$$ with $$\frac{1}{q} = \frac{1}{p} - \frac{\alpha}{\delta}$$. Then $\|I_\alpha(f)\|_{L^q} \leq A_{p,q} \|f\|_{L^p},$ where $$A_{p,q}$$ depends only on $$d$$, $$p$$, and $$q$$. Observe that the fractional integral operator exhibits a "relative" dilation invariance: if $$f_\delta(x) = f(\delta x)$$, then $I_\alpha(f_\delta) = \delta^{-\alpha} (I_\alpha(f))_\delta.$ Indeed, $\int f(x-y) \frac{dy}{(y)^{d-\alpha}}$ is invariant, hence the factor $$\delta^{-\alpha}$$ comes to play. Furthermore, $\|f_\delta\|_{L^p} = \delta^{-\frac{d}{p}}\|f\|_{L^p}.$ It thsu follows that we have the relation $\frac{1}{q} = \frac{1}{p} - \frac{\alpha}{d}.$ Proof of Theorem 3.2. We claim that $I_\alpha(f)(x) \leq c(Mf(x))^\theta \|f\|_{L^p}^{1-\theta}$ with $$0 < \theta 1$$ and $$\theta = p/q$$. To see this, it sufficies to consider the special case $$d = 1$$, $$\alpha = 1/4$$, p = 2$$, and$$q = 4, which then yields the inequality

$I_1(f) \leq c(M(f))^{1/2} \|f\|_{L^2}^{1/2}.$

Note that the exponent 4, in the sense of conjugate exponents, is the half-way in between the exponent 2 and the exponent $$\infty$$.

Now, we split the integral in the following manner:

$I_1(f) = c \left(\int_{|y| \leq R} |y|^{-3/4} f(x-y) \, dy + \int_{|y| > R} |y|^{-3/4} f(x-y) \, dy\right).$

But

\begin{align*} \int_{|y| \leq R} |y|^{-3/4} f(x-y) \, dy & \leq \left( \int_{|y| \leq R} |y|^{-3/4} \, dy \right) Mf(x) \\ & \leq cR^{1/4} Mf(x). \end{align*}

Next, by Schwarz's inequality,

\begin{align*} \int_{|y| > R} |y|^{-3/4} f(x-y) \, dy &\leq \left( |y|^{-3/2} \, dy \right) \|f\|_{L^2} \\ &= cR^{-1/4} \|f\|_{L^2}. \end{align*}

As a result,

$I_1(f)(x) \leq c \left( R^{1/4}Mf(x) + R^{-1/4} \|f\|_{L^2} \right).$

Choose

$R = \frac{\|f\|_{L^2}^2}{(Mf(x))^2},$

so that

$I_1(f) \leq c'm ((Mf)(x))^{1/2} \|f\|_{L^2}^{1/2},$

whence the desired result follows. $$\square$$

Note that Theorem 3.2 fails if $$p = 1$$. By a duality argument, $$q = \infty$$ fails as well. If we replace $$|y|^{-d+\alpha}$$ in the definition of fractional integral by any kernel with the same distribution as $$|y|^{-d+\alpha}$$, Theorem 3.2 still holds.

## 4. Three continuous operators, part 3: the singular integral

Definition 4.1 (Mihlin, Calderón, Zygmund). The singular integral of $$f$$ is \begin{align*} T(f)(x) &= \mbox{p.v.} \int_{\mathbb{R}^d} f(x-y)K(y) \, dy \\ &= \lim_{\varepsilon > 0} \int_{|y| > \varepsilon} f(x-y) K(y) \, dy, \end{align*} such that
• $$K$$ is homogeneous of degree $$-d$$, i.e., $$K_\delta = \delta^{-d} K$$
• $$k$$ is smooth when $$x \neq 0$$
• $$\int_{|x| = 1} K(x) \, d \sigma = 0$$.

For example, consider the "forbidden kernel" $$K(x) = |x|^{-d}$$.

As another example, we define the Hilbert transform in $$d = 1$$ as

$H(f)(x) = \frac{1}{\pi} \mbox{p.v.} \int_{-\infty}^{\infty} f(x-y) \, \frac{dy}{y}.$

The principal value (p.v.) is interpreted as

$\mbox{p.v.} \int f(x-y) \, \frac{dy}{y} = \lim_{\varepsilon \to 0} f(x-y) \, \frac{dy}{y}.$

In a sense, the Hilbert transform is the "fundamental operator" of one-variable Fourier analysis. $$H$$ is uniatry on $$L^2(\mathbb{R})$$. Moreover, $$H^\ast = -H$$ and $$H^2 = -I$$. The Hilbert transform also manifests itself in complex analysis. Indeed, the Hilbert transform maps a harmonic function to its harmonic conjugate.

More examples:

$R_{ij}(f) = \frac{\partial^2}{\partial x_i \partial x_j} I_2(f)$

and the Riesz transform

$R_{j}(f) = \frac{\partial}{\partial x_j} I_1(f).$

Theorem 4.2. If $$T$$ is a singular integral operator, then $$T$$ is a bounded operator on $$L^p$$ ($$1 < p < \infty$$).

We remark that the cases $$p = 1$$ and $$p = \infty$$. Indeed, "all interesting theorems should fail for $$p = 1$$; if a theorem holds for $$p = 1$$, then it is easy or semi-trivial." (Direct quote from Stein)

The method of proof is called the Calderón–Zygmund paradigm. Deviating from the classical method (of M. Riesz) of exploting the power of complex analysis, we turn to the tools of real analysis. The proof has two parts: we show that the operator is bounded in $$L^2$$, and extend the result to the $$L^1$$ theory.

Proof of Theorem 4.2. We first consider the $$L^2$$ theory: we claim that

$\widehat{(T f)}(\xi) = m(\xi) \hat{f}(\xi)$

where $$m = \hat{K}$$, taken in the sense of distributions. By Plancherel's theorem, it will then suffice to prove that

$|m(\xi)| \leq c$

for all $$\xi$$. IF so, then we will have

\begin{align*} \|T(f)\|_{L^2} &= \| \widehat{T(f)} \|_{L^2} \\ &= \|m(\xi) \hat{f}(\xi)\|_{L^2} \\ &\leq c \|\hat{f}\|_{L^2} \\ & \leq c\|f\|_{L^2}. \end{align*}

Let $$\eta$$ be a radial cut-off function: $$\eta$$ is $$C^\infty$$ and $$\eta(x) = 1$$ if $$0 < |x| \leq 1$$, and $$\eta(x) = 0$$ if $$|x| > 1$$. Then we shall break the integral into two parts

$m(\xi) = \mbox{p.v.} \int e^{-2 \pi i x \cdot \xi} K(x) \, dx = I + II$

where

$I = \mbox{p.v.} \int e^{-2 \pi i x \cdot \xi} \eta(x/r) K(x) \, dx$

and

$II = \mbox{p.v.} \int e^{-2 \pi i x \cdot \xi} (1 - \eta(x/r)) K(x) \, dx.$

Since $$\eta$$ is a radial cut-off function,

$I = \int (e^{-2 \pi i x \cdot \xi} - 1) \eta(x/r)K(x) \, dx,$

and so

\begin{align*} |I| &\leq c \int_{|x| \leq r} |x||\xi||K(x)| \, dx \\ &\leq c'|\xi| \int_{|x| \leq r} |x|^{-d+1} \, dx \\ &\leq c', \end{align*}

where $$r = 1/|\xi|$$.

Also,

$II = \int e^{-2 \pi i x \cdot \xi} (1 - \eta(x/r)) K(x) \, dx.$

Suppose that $$|\xi_j| \geq |\xi|/d^{1/2}$$. Then, since

$e^{-2 \pi i x \cdot \xi} = \frac{-1}{2 \pi i \xi_j} \frac{\partial}{\partial x_j} (e^{-2 \pi i x \cdot \xi}),$

it follows that

$|II| \leq \frac{c}{|\xi_j|} \int_{|x| \geq r/2} |x|^{-d-1} \, dx \frac{c}{|\xi_j|} r^{-1} \leq c'',$

thsu establishing the easy part of the proof (via the Fourier transform).

Now, we turn to the $$L^1$$ theory. We claim that

$m\{x: |Tf(x)| > \alpha\} \leq \frac{A}{\alpha} \|f\|_{L^1}$

for all $$\alpha > 0$$, reminiscent of the weak-type estimate for the maximal function.

The two main points of the proof are as follows:

a) The "atoms" (We can't actually call them atoms, since it is a term used in another part of harmonic analysis). Suppose $$f \in L^1$$ is supported in a ball $$B$$ and $$\int_B f \, dx = 0$$. Let $$B^\ast$$ be the twice-dilated ball of $$B$$. Then

$\int_{{}^cB^\ast} |Tf| \, dx \leq c \int_B |f| \, dx,$

where $${}^cB^\ast$$ is the complement of the ball $$B^\ast$$. In fact, we suppose that $$B$$ is centered at the origin, for "everything is translation-invariant". Then

\begin{align*} T(f)(x) &= \int K(x-y)f(y) \, dy \\ &= \int (K(x-y) - K(x)) f(y) \, dy, \end{align*}

so

$|T(f)(x)| \leq \int_B |K(x-y) - K(x)||f(y)| \, dy.$

Now, if $$y \in B$$ and $$x \in {}^c B^\ast$$ with $$r$$ the radius of $$B$$, then

$|K(x-y) - K(x)| \leq r \max_{x \in L} |\nabla K(x)|$

where $$L$$ is the line segment joining $$x$$ to $$x-y$$. Since $$|x| \geq 2r$$ and $$|y| \leq r$$ and $$|\nabla K| \leq A/|x|^{d+1}$$,

$\int_B |K(x-y) - K(x)| \, dx \leq cr \int_{|x| \geq 2r} |x|^{-d-1} \, dx = c'$

if $$|y| \leq r$$.

b) The Calderón–Zygmund decomposition. Given $$f \in L^1$$ and $$\alpha > 0$$, we can decompose $$f = g + b$$, where

$b = \sum_k b_k$

such that $$|g(x)| \lesssim \alpha$$ ($$|g(x)| \leq C\alpha$$ for some $$C > 0$$) and $$g(x) = f(x)$$ except when $$x \in E_\alpha$$ with $$m(E_\alpha) \lesssim \frac{1}{\alpha} \int |f| \, dx$$.

Each $$b_k$$ is supported in a ball $$B_k$$, $$\int b_k \, dx = 0$$, and

$\int |b_k| \, dx \lesssim \alpha m(B_k).$

(Therefore, each $$b_k$$ is an "atom".) Furthermore,

$\sum m(B_k) \lesssim \frac{1}{\alpha} \int |f| \, dx.$

See chapters 1 and 2 of Stein, Singular Integrals and Differentiability Properties of Functions for details.

Here is an alternative approach to the Calderón–Zygmund decomposition that does not make use of the maximal function. We slice the Euclidean space into cubes with the mesh so large that the mean value is small:

$\frac{1}{m(Q)} \int_Q |f| \, dx \leq \alpha.$

We subdivide each cube into $$2^d$$ pieces by bisection. If the mean value is larger than $$\alpha$$ on a subdivision, then we disregard that cube. We subdivide the "good cubes" again by bisection, and continue the process. In each step, we have

$\alpha < \frac{1}{m(Q)} \int_Q |f| \leq 2^d \alpha.$

We have divided the space into two sets: the union of good cubes, and the union of rejected cubes. The rejected cubes are disjoint and has the "right kind of estimate".

## 5. The Role of Curvature

Definition 5.1. A (smooth) curve in $$\mathbb{R}^d$$ is a $$C^\infty$$-function $$\gamma: [-1, 1] \to \mathbb{R}^d$$ with $$\gamma(0) = 0$$ and $$|\gamma'(t)| > 0$$ for all $$t \in [-1, 1]$$.

We consider operators on a curve $$\gamma$$: for example, the Hilbert transform

$H_\gamma(f)(x) = \mbox{p.v.} \int_{-1}^1 f(x-\gamma(t)) \, \frac{dt}{t}$

and the maximal function

$M_\gamma(f)(x) = \sup_{0 < h \leq 1} \frac{1}{2h} \int_{|t| \leq h} |f(x - \gamma(t))| \, dt.$

Is there an $$L^p(\mathbb{R}^d)$$ theory for $$H_\gamma$$ and $$M_\gamma$$?

1. Yes, if $$\gamma$$ is a straight-line segment;
2. No, in general, but:
3. Yes, if $$\gamma$$ has some "curvature" near $$t = 0$$.

Why is there a good theory for straight lines? We can always rotate the line to be axis-parallel, whence the case is reduced to the theory on $$\mathbb{R}$$. Note, however, that "local flatness" is decidedly not enough: we can construct a curve which is "very flat" at every given point which nevertheless does not admit a good theory for any $$p$$.

What is a good model of non-flat curve with an $$L^p$$-theory? The more precise version of hypothesis 3 can be stated as follows: for some $$N \in \mathbb{N}$$, the vectors

$\gamma'(0), \gamma''(0), \ldots, \gamma^{(n)}(0)$

span $$\mathbb{R}^d$$.

Here is a typical example. $$\gamma(t) = (t, t^2)$$, the parabola in $$\mathbb{R}^2$$. Note that $$\gamma'(0) = (1,0)$$ and $$\gamma''(0) = (0,2)$$, whence the hypothesis is satisfied.

Theorem 5.2. Under the curvature hypothesis 3 above, both $$H_\gamma$$ and $$M_\gamma$$ are bounded operators for $$1 < p < \infty$$.

The methods of proving boundedness results for the standard maximal function, fractional integral, and singular integrals fail. In particular, no results are known for (p = 1) for $$H_\gamma$$ and $$M_\gamma$$. We do not know, for example, if the weak-type estimates hold.

The key tools here are oscillatory integrals. Here is the one-dimensional example:

$I(\lambda) = \int_a^b e^{i \lambda \Phi(x)} \, dx.$

Here is a variant:

$J(\lambda) = \int_a^b e^{i \lambda \Phi(x)} \psi(x) \, dx.$

$$\Phi$$ (real-valued) is the phase. $$\psi$$ is the amplitude.

We would like to know how the integrals behave as $$\lambda$$ tends to infinity. In particular, we would like to know how the kind of estimates we can make that are independent of the integral.

Here is an outline of the idea of the proof of Theorem 5.2 in the case when $$\gamma$$ is the parabola in $$\mathbb{R}^2$$ and $$p = 2$$: 1) oscillatory integrals, 2) boundedness of $$H_\gamma$$, and 3) boundedness of $$M_\gamma$$.

Theorem 5.3. Let $I(\lambda) = \int_a^b e^{i \lambda \Phi(x)} \, dx,$ where $$\Phi$$ is of class $$C^2$$. If $$|\Phi'(x)| \geq 1$$ and $$\Phi'$$ monotonic, then $$|I(\lambda)| \leq c |\lambda|^{-1}$$. Alternatively, if $$|\Phi''(x)| \geq 1$$, then $$|I(\lambda)| \leq c|\lambda|^{-1/2}$$ as $$|\lambda| \to \infty$$. Here the bound $$c$$ is independent of $$a$$ of $$b$$.

If $$\Phi(x) = x$$, then

$\int_a^b e^{i \lambda x} \, dx = \frac{1}{i \lambda},$

whence the decay is clearly of order $$|\lambda|^{-1}$$. Examine also $$\Phi(x) = x^2$$.

Proof of Theorem 5.3. We first write

$e^{i \lambda \Phi(x)} = \frac{1}{i \lambda \Phi'(x)} \frac{d}{dx} \left( e^{i \lambda \Phi(x)} \right),$

so that

$I(\lambda) = \frac{1}{i \lambda} \int_a^b \frac{1}{\Phi'(x)} \frac{d}{dx} \left( e^{i \lambda \Phi(x)} \right) \, dx,$

whence we can integrate by parts. We now take the absolute value of each term. The $$1/ \Phi'(x)$$ looks bad, but fear not: $$\Phi'(x)$$ is monotonic, so we can remove the absolute value. The estimate now follows from simple computation.

In the second alternative, we shall write

$I(\lambda) = I + II,$

such that

\begin{align*} I &= \int_{c-\lambda}^{c + \lambda} e^{i \lambda \Phi(x)} \, dx \\ II &= \mbox{the complementary part of I}, \end{align*}

where $$c$$ is the unique point where $$\Phi'(c) = 0$$. By doing so, we have singled out the "bad point." We first make the trivial estimate

$|I| \leq 2 \delta,$

which will turn out to be good enough. (We shall make $$\delta$$ very small.) Moreover, we have

$|II| \leq \frac{2}{|\lambda|^\delta}.$

We still have not picked $$\delta$$ yet. The right choice is

$\delta = |\lambda|^{-1/2}$

which yields the desired estimate. $$\square$$

Corollary 5.4. Let $J(\lambda) = \int_a^b e^{i \lambda \Phi(x)} \psi(x) \, dx.$ If $$|\Phi'(x)| \geq 1$$ and $$\Phi'$$ monotonic, then $|J(\lambda) \leq \frac{c}{|\lambda|} \left( |\psi(b)| + \int_a^b |\phi'(x)| \, dx \right).$ If, instead, $$|\Phi'(x)| \geq 1$$, then $|J(\lambda)| \leq \frac{c}{|\lambda|^{1/2}} \left( |\psi(b)| + \int_a^b |\phi'(x)| \, dx \right).$

Proof. We write

\begin{align*} I^u(\lambda) &= \int_a^u e^{i \lambda \Phi(x)} \, dx \\ J(\lambda) &= \int_a^b \frac{d}{du} (I^u(\lambda)) \psi(u) \, du \end{align*}

and integrate by parts. $$\square$$

We now examine the Hilbert transform on the parabola $$\gamma = (t, t^2) \in \mathbb{R}^2$$:

$H_\gamma(f) = \mbox{p.v.} \int_{-\infty}^\infty f(x-t, y-t^2) \, \frac{dt}{t}.$

This is a convolution-type integral, and so we are interested in Fourier transform of the above integral: indeed, we have

$\widehat{H_\gamma(f)} = m(\xi, \eta) \hat{f}(\xi, \eta)$

where

$m(\xi, \eta) = \mbox{p.v.} \int_{-\infty}^\infty e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t}.$

Theorem 5.5. The above $$m(\xi, \eta)$$ satisfies constant bound $|m(\xi, \eta)| \leq A$ for all $$(\xi, \eta)$$.
Corollary 5.6. We can write $H_\gamma(f) = \mbox{p.v.} \int_{-\infty}^\infty f(x-t, y-t^2) \, \frac{dt}{t}$ and $$f \mapsto H_\gamma(f)$$ is bounded on $$L^2(\mathbb{R}^2)$$.

Proof. We consider new "dilations"

$\delta \circ (\xi, \eta) = (\delta \xi, \delta^2 \eta)$

for each $$\delta > 0$$. We also consider a new "norm"

$\rho(\xi, \eta) = |\xi| + |\eta|^{1/2}.$

These are the natural notions of dilation and norm for the parabola. Indeed,

$\rho(\delta \circ (\xi, \eta)) = \delta \rho (\xi, \eta).$

We decompose the integral into an infinite sum of diadic parts:

$\mbox{p.v.} \int_{-\infty}^\infty e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t} = \sum_{-\infty < k < \infty} \int_{2^k}^{2^{k+1}} \int_{-\infty}^\infty e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t}.$

Note that

$\int_{2^k}^{2^{k+1}} e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t} = \int_1^2 e^{-2 \pi i (2^k \xi t + 2^{2k} \eta t^2)} \, \frac{dt}{t}$

by introducing a proper dilation: here, each integral is taken over half-open interval $$[a, b)$$. We then have $$m_0(2^k\xi, 2^{2k} \eta)$$, where

$m_0(\xi, \eta) = \int_1^2 e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t}.$

We shall write

$m_0(\xi, \eta) = \int_1^2 e^{i \lambda \Phi(t)} \psi(t) \, dt.$

Claim 1) $$|m_0(\xi, \eta)| \leq A$$ for all $$(\xi, \eta)$$.

Claim 2) $$m_0(\xi , \eta)$$ is a smooth function with $$m_0(0, 0) = 0$$; therefore, $$|m_0(\xi, \eta)| \leq c \rho(\xi, \eta)$$.

Claim 3) $$|m_0(\xi, \eta)| \leq c (\rho(\xi, \eta))^{-1/2}$$.

Claim 1 and Claim 2 are easy. To show Claim 3, we write

\begin{align*} m_0(\xi, \eta) &= \int_1^2 e^{i \lambda \Phi(t)} \psi(t) \, dt \\ &= \int_1^2 e^{-2 \pi i (\xi t + \eta t^2)} \, \frac{dt}{t}. \end{align*}

We now make an estimate using the first and the second derivative tests. We first consider the case when $$\xi$$ is very large, say $$|\xi| \geq 10 |\eta|$$. Then

$\Phi(t) = -2 \pi \left( t + \frac{\eta}{\xi} \cdot t^2 \right),$

$$\lambda = \xi$$ and $$\psi(t) = \frac{1}{t}$$. Therefore, $$|\Phi'(t)| \geq 1$$ and

$|m_0(\xi, \eta)| \leq \frac{c}{|\xi|} \leq \frac{c}{\rho(\xi, \eta)^{1/2}}$

if $$|\xi| \geq 10 |\eta|$$, and so $$\rho(\xi, \eta) \geq 1$$.

If $$|\xi| \leq 10 |\eta|$$, then we write

$\Phi(t) = - 2 \pi \left( t \frac{\xi}{\eta} + t^2 \right),$

so that $$\lambda = \eta$$. Then $$|\Phi'(t)| \geq 1$$ and

$m_0(\xi, \eta) \leq c \eta^{-1/2} \leq c \rho(\xi, \eta)^{-1/2}.$

Now,

\begin{align*} m(\xi, \eta) &= \sum_k m_0(2^k \xi, 2^{2k} \eta) \\ &= \sum_{2^k \rho(\xi, \eta) \leq 1} m_0(2^k \xi, 2^{2k} \eta) + \sum_{2^k \rho(\xi, \eta) > 1} m_0(2^k\xi, 2^{2k} \eta). \end{align*}

The first sum is estimated as follows:

\begin{align*} c \sum \rho(2^k \xi, 2^{2k} \eta) &= c \left( \sum_{2^k \rho(\xi, \eta) \leq } 2^k \right) \\ &\leq c'. \end{align*}

The second sum is estimated as follows:

\begin{align*} \sum_{2^k \rho(\xi, \eta) > 1} \rho(2^k \xi, 2^{2k} \eta)^{-1/2} &= \left( \sum_{2^k\rho(\xi, \eta) > 1} 2^{-k/2} \right) (\rho(\xi, \eta))^{-1/2} \\ &\leq c'. \end{align*}

$\square$

We mention some other results involving oscillatory integrals. Suppose

$P(t) = a_0 + a_1t + \cdots + a_kt^k$

is a polynomial with real coefficients. Then Corollary 5.6 implies that

$\left| \mbox{p.v.} \int_{-\infty}^\infty e^{iP(t)} \, \frac{dt}{t} \right| \leq M,$

where $$M$$ is independent of $$a_0, a_1, \ldots, a_k$$.

Corollary 5.7. We suppose that $$Q(t) = (Q_1(t), \ldots, Q_d(t))$$ is a polynomial function from $$\mathbb{R}$$ to $$\mathbb{R}^d$$. Then the "Hilbert transform" $H_Q(f)(x) = \mbox{p.v.} \int_{-\infty}^\infty f(x-Q(t)) \, \frac{dt}{t}$ is bounded on $$L^2(\mathbb{R}^d)$$.

We now turn to the boundedness of $$M_\gamma$$, where $$\gamma$$ is still fixed to be the parabola. To this end, we consider non-isotropic dilations in $$\mathbb{R}^d$$:

$x \mapsto \delta \circ x = (\delta^{a_1} x_1, \delta^{a_2} x_2, \ldots, \delta^{a_d} x_d).$

Here $$a_1,\ldots,a_d$$ are fixed strictly-positive exponents. The quantity $$a_1 + a_2 + \cdots + a_d = Q$$ is the homogeneous dimension on $$\mathbb{R}^d$$. With this definition, we have:

$d(\delta \circ x) = \delta^Q \, dx.$

In general, many of the basic results, such as the above theorems, have valid extensions where the isotropic dilations

$x \mapsto \delta x = (\delta x_1, \delta x_2, \ldots, \delta x_d)$

are replaced by non-isotropic dilations. Of course, the extensions will have to be formulated properly.

Consider the maximal function

$\widetilde{M}(f)(x) = \sup_{\delta > 0} \frac{1}{m(B_\delta)} \int_{B_\delta} |f(x-y)| \, dy$

where

$B_\delta = \delta \circ B = \{x = \delta \circ y, y \in B = \mbox{ unit ball }\}$

and

$m(B_\delta) = \delta^Q m(B) = c\delta^Q.$

While this is quite a different maximal operator, the boundedness proof essentially follows by simply imitating the proof for the isotropic dilation case. Indeed:

Proposition 5.8. a) $$\|\widetilde{M}(f)\|_{L^p} \leq A_p\|f\|_{L^p}$$ ($$1 < p \leq \infty$$).
b) $$m\{x : \widetilde{M}(f) > \alpha\} \leq \frac{A}{\alpha} \|f\|_{L^1}$$ for all $$\alpha > 0$$.

This follows from the non-isotropic Vitali lemma: there is a $$c > 0$$ such that if $$B_{\delta_1}(x_1)$$ and $$B_{\delta_2}(x_2)$$ intersect and $$\delta_1 \geq \delta_2$$, then

$B_{c\delta_1}(x_1) \supseteq B_{\delta_2}(x_2).$

Theorem 5.9. Let $M_\gamma(f)(x) = \sup_{0 < r} \frac{1}{2r} \int_{|t| \leq r} |f(x_1 - t, x_2 - t^2)| \, dt.$ Then $\|M_\gamma(f)\|_{L^2} \leq A \|f\|_{L^2}.$

Proof. It suffices to assume that $$f \geq 0$$, and establish

$\left\|\sup_{k \in \mathbb{Z}} |A_k(f)| \right\|_{L^2} \leq C \|f\|_{L^2},$

where

$A_k f = 2^{-(k+1)} \int_{|t| \leq 2^k} f(x_1 - t, x_2 - t^2) \, dt.$

These are very "singular averages."

Let $$\varphi$$ be a smooth bump function, viz., a $$C^\infty$$ function on $$\mathbb{R}^2$$ such that $$\varphi \geq 0$$, $$\varphi$$ supported in $$|x| \leq 1$$, and

$\int \varphi(x) \, dx = 1.$

Setting

$\varphi_k(x) = 2^{-3k} \varphi(2^{-k}x_1, 2^{-2k} x_2),$

we see that $$\int \varphi_k(x) \, dx = 1$$ and $$\varphi_k$$ is supportd on $$B_{2^k}$$ with $$m(B_{2^k}) = 2^{2k} m(B)$$.

We write

$M_k(f) = f \ast \varphi_k = \int_{\mathbb{R}^2} f(x-y) \varphi_k(y) \, dy,$

so that

\begin{align*} M_k(f)(x) & \leq c \frac{1}{m(B_{2^k})} \int_{B_{2^k}} f(x-y) \, dy \\ & \leq c \widetilde{M}(f). \end{align*}

It then follows that

$\left\| \sup_k |M_k(f)| \right\|_{L^2} \leq c\|f\|_{L^2},$

whence it suffices to compare $$M_k$$ with $$A_k$$.

This comparison is done by a square function $$\mathcal{S}$$ defined by

$(\mathcal{S}(f)(x))^2 = \sum_{k \in \mathbb{Z}} (A_k(f)(x) - M_kf(x))^2.$

Observe that

$|A_k(f)(x) - M_k(f)(x)| \leq \mathcal{S}(f)(x),$

and so

$\sup_k A_k(f)(x) \leq \mathcal{S}(f)(x) + c \widetilde{M}(f)(x).$

It thus suffices to prove

Lemma 3.10. $\|\mathcal{S}(f)\|_{L^2} \leq c\|f\|_{L^2}.$

The proof is the lemma is as follows. Let $$\hat{\varphi}(\xi)$$ be the Fourier transform of $$\varphi$$, which is

$\int_{\mathbb{R}^2} \int e^{-2 \pi i (\xi x_1 + \eta x_2)}\varphi(x_1,x_2) \, dx.$

We note that

1. $$\hat{\varphi}$$ is smooth.
2. $$\hat{\varphi}(0) = 1$$.
3. $$\hat{\varphi}$$ is rapidly decreasing at infinity, so, in particular, $$|\hat{\varphi}(\xi, \eta)| \leq c (\rho(\xi, \eta))^{-1/2}$$.
4. $$\hat{\varphi}(\xi, \eta) = \hat{\varphi}(2^k \xi, 2^{2k} \eta)$$.

Therefore,

$(\widehat{A_k - M_k})(f)(\xi, \eta) = m_k(\xi, \eta) - \hat{\varphi}_k(\xi, \eta),$

where

\begin{align*} m_k(\xi, \eta) &= 2^{-k-1} \int_{|t| \leq 2^k} e^{-2 \pi i (\xi t + \eta t^2)} \, dt \\ &= m_0(2^k \xi, 2^{2k} \eta). \end{align*}

We note that

1. $$m_0$$ is smooth.
2. $$m_0(0, 0) = 1$$.
3. $$|m_0(\xi, \eta) \leq c \rho(\xi, \eta)^{-t/2}$$.

We thus conclude that

$(\widehat{A_k - M_k})(f) = \Delta_k \cdot \hat{f},$

where

\begin{align*} \Delta_k(\xi, \eta) &= m_k(\xi, \eta) - \hat{\varphi}_k (\xi, \eta) \\ &= \Delta_0 (2^k\xi, 2^{2k} \eta). \end{align*}

Also, if $$\rho \leq 1$$,

\begin{align*} |\Delta_0(\xi, \eta)| &\leq c(|\xi| + |\eta|) \\ &\leq c \rho(\xi, \eta), \end{align*}

and, if $$\rho \geq 1$$,

$|\Delta_0(\xi, \eta)| \leq c (\rho(\xi, \eta)^{-1/2}).$

But, by Plancherel's theorem,

$\|(A_k - M_k)(f)\|_{L^2}^2 = \int_{\mathbb{R}^2} |\Delta_k(\xi, \eta)|^2 |\hat{f}(\xi, \eta)|^2 \, d\xi d\eta,$

so

\begin{align*} \|\mathcal{S}(f)\|_{L^2}^2 &= \sum_k \|(A_k - M_k)(f)\|_{L^2}^2 \\ &= \int \left( \sum_k |\Delta_k (\xi, \eta)|^2 \right) |\hat{f}|^2 \, d\xi d\eta \\ &\leq c^2 \int |\hat{f}(\xi, \eta)|^2 \, d\xi d\eta \\ &= c^2 \int |f(x)|^2 \, dx, \end{align*}

for $$\sum_k |\Delta_k (\xi)|^2 \leq c^2$$. This proves the lemma.

The proof now follows from the lemma. $$\square$$

We also note that

$\sum_{2^k \rho \leq 1} 2^{2k} \rho(\xi, \eta)^2 \leq c_1$

and

$\sum_{2^k \rho \geq 1} 2^{-k} \rho(\xi, \eta)^{-1} \leq c_1.$

Further readings in Stein, Harmonic Analysis:

• Oscillatory integrals: Chapter 8, sections 1-3
• Maximal functions and singular integrals on curved varieties: Chapter 9, sections 1.2 and 2.