# Welcome.

## March 4, 2011

### Harmonic Analysis Lecture Notes 16

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:54 pm

Method of Stationary Phase

Proposition (non-stationary phase) Let ${a \in C^\infty_c}$ and ${\phi \in C^\infty}$ (and real-valued) with ${\phi \geq 1}$ on ${supp(a)}$, then

$\displaystyle \bigg| \int e^{\pm i \lambda \phi(x)} a(x) \; dx \bigg| \lesssim_N \lambda^{-N}$

for every integer ${N \geq 0}$ and ${\lambda > 0}$.

Proof: Let us take the ${e^{+ i \lambda \phi}}$ case,

$\displaystyle I(\lambda) := \int e^{i\lambda \phi(x)} a(x) \; dx.$

In the spirit of integration by parts,

$\displaystyle \begin{array}{rcl} \displaystyle I(\lambda) &=& \displaystyle \int \left[ \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} e^{i\lambda \phi(x)}\right] a(x) \; dx \\ \\ &=& \displaystyle \int e^{i \lambda \phi(x)} \frac{d}{dx} \left( \frac{i}{\lambda \phi'(x)} a(x) \right) \; dx. \end{array}$

Thus, by the quotient rule,

$\displaystyle |I(\lambda)| \leq \frac{1}{\lambda} \left\{ \left\| \frac{a'}{\phi'}\right\|_{L^1} + \left\| \frac{a\phi''}{(\phi')^2}\right\|_{L^1} \right\}$

More generally

$\displaystyle e^{i\lambda \phi} = \left( \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} \right)^N e^{i\lambda \phi}$

and so

$\displaystyle I(\lambda) = \int e^{i\lambda \phi} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right)^N a(x) \; dx.$

Thus

$\displaystyle | I(\lambda) | \lesssim \lambda^{-N} \sum_{k=0}^N \;\;\; \sum_{\tiny \begin{matrix} \alpha_1 + \cdots + \alpha_k + \beta = N \\ \alpha_j \geq 1 , \beta \geq 0 \end{matrix} } \left\| \frac{(\partial^\beta a)(\partial^{\alpha_1} \phi') \cdots (\partial^{\alpha_k} \phi')}{(\phi')^{N+k}} \right\|_{L^1}$

$\Box$

The ideas we use work for sums as well. Traditionally, dealing with sums are much harder.

Theorem (van der Corput lemma) Suppose ${\phi \in C^\infty}$ and ${\phi^{(k)}(x) \geq 1}$ throughout the interval ${[a,b]}$ for some ${k \geq 1}$. If ${k =1}$, then we additionally assume that ${\phi'}$ is monotone. Then

$\displaystyle \bigg| \int_a^b e^{\pm i \lambda \phi(x) } \; dx \bigg| \lesssim_k \lambda^{-\frac{1}{k}}$

independent of ${a,b,}$ and ${\phi}$.

Remark

1. If ${k = 1}$ then the phase is non-stationary; however, we cannot do better because of the end-point contributions from integration by parts (alternatively, the “amplitude function” is ${\chi_{[a,b]}}$, which is not smooth). If ${\phi(x) = x}$, we can compute exactly and see that we cannot do better.
2. To get a bound which is independent of the length of ${[a,b]}$, we need to assume ${\phi'}$ is monotone (or at least some additional hypothesis). Recall

$\displaystyle \text{Im}\; \int_a^b e^{i \phi(x)} \; dx = \int_a^b \sin(\phi(x)) \; dx.$

We define ${\phi}$ heuristically, slow down when ${\sin(x)}$ is positive and speed up when ${\sin(x)}$ is negative. Doing this appropriately, we may make the right side arbitrarily large.

Proof: (by induction in ${k}$). For ${k = 1}$ we argue as before

$\displaystyle \begin{array}{rcl} I(\lambda) &=& \displaystyle \int_a^b e^{i\lambda\phi(x)} \; dx \\ \\ &=& \displaystyle \int_a^{b} \frac{1}{i \lambda \phi'(x)} \left( \frac{d}{dx} e^{i\lambda \phi(x)} \right) \; dx \\ \\ &=& \displaystyle \left[ \frac{e^{i\lambda \phi(x)}}{i \lambda \phi'(x) } \right]^b_a + \int_a^b e^{i\lambda \phi(x)} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right) \; dx. \end{array}$

Thus

$\displaystyle \begin{array}{rcl} I(\lambda) &\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \int_a^b \underbrace{\frac{d}{dx} \frac{1}{\phi'(x)}}_{\tiny \begin{matrix}\text{sign definite as} \\ \phi' \text{ is monotone} \end{matrix}} \; dx \bigg| \\ \\ &\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)}\bigg| \leq \frac{3}{\lambda}. \end{array}$

Now, let us do the induction step (assume the result is true for ${k}$, now show for ${k+1}$). By hypothesis, ${\phi^{(k+1)} \geq 1}$. Let us look at ${\phi^{(k)}}$. This vanishes at most at one point ${c \in [a,b]}$; if it does not vanish on ${[a,b]}$, then choose ${c}$ to be the end point where ${|\phi^{(k)}|}$ is least. Consider the partition of ${[a,b]}$ into interval ${(c- \delta, c+ \delta) \cap [a,b]}$ and its complement, which we will write as ${J}$. On ${J}$, we have ${|\phi^{(k)}| \geq \delta}$ and ${\phi^{(k)}}$ is monotone. Using the inductive hypothesis,

$\displaystyle |I(\lambda)| \lesssim_k \delta + (\lambda \delta)^{-\frac{1}{k}}.$

Choosing ${\delta \approx \lambda^{-\frac{1}{k+1}},}$ we get ${| I(\lambda)| \lesssim_k \lambda^{-\frac{1}{k+1}}}$, as desired. $\Box$

Remark A computation using ${\phi(x) = x^k}$ shows that these bounds are the best that we may expect.

Corollary Let ${\phi}$ obey the hypothesis from the van der Corput lemma and let ${\psi}$ be of bounded variation (or at least ${C^1}$). Consider

$\displaystyle I(\lambda) = \int_a^b e^{\pm i \lambda \phi(x)} \psi(x) \; dx$

then

$\displaystyle | I(\lambda) | \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)$

Proof:

$\displaystyle \begin{array}{rcl} I(\lambda) &=& \displaystyle \int_a^b \psi(x) \frac{d}{dx} \int_a^x e^{i\lambda \phi(y)} \; dy \; dx \\ \\ &=& \displaystyle \Big[ \psi(x) \int_a^x e^{i \lambda \phi(y)} \; dy \Big]_a^b - \int_a^b \psi'(x) \int_a^x e^{i \lambda \phi(y)} \; dy \; dx. \end{array}$

Using the van der Corput lemma twice,

$\displaystyle | I(\lambda) | \lesssim_k |\psi(b)| \lambda^{-\frac{1}{k}} + \|\psi'\|_{L^1} \sup_{a \leq x \leq b} \Big| \int_a^x e^{i\lambda\phi(y)} \; dy \Big| \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)$

$\Box$

An application of this lemma

Definition A sequence ${\{x_k\}_{k=1}^\infty \subseteq {\mathbb R}^z / {\mathbb Z}^n}$ is equi-distributed if

$\displaystyle \frac{1}{K} \sum_{k=1}^K f(x_k) \stackrel{K\rightarrow \infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y) \; dy$

for all continuous functions ${f}$. Analogously, a curve ${x : [0,\infty) \rightarrow {\mathbb R}^n / {\mathbb Z}^n}$ is equi-distributed if

$\displaystyle \frac{1}{T} \int_0^T f(x(t)) \; dt \stackrel{T\rightarrow\infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y)$

for all continuous functions ${f}$.

Remark

1. The notion is unchanged if we replace ${f \in C({\mathbb R}^n/{\mathbb Z}^n)}$ by ${\chi_E}$ or ${\chi_{\mathcal{O}}}$ for arbitrary closed/open sets (Urysohn’s lemma).
2. This replacement does not work for ${F_\sigma}$‘s or ${G_\delta}$‘s (heuristically because they mimic the sequences too well).
3. That ${e^{2\pi i t^2}}$ is not uniformly continuous is implied by the fact that the curve ${(t^2, (t+h)^2)}$ is equi-distributed in ${{\mathbb R}^2 / {\mathbb Z}^2}$ when ${h \neq 0}$, because

Lemma If ${x(t)}$ is equi-distributed then it is dense.

Proof: Choose ${f}$ to be a bump in the open set that ${x(t)}$ misses. $\Box$

Theorem (Weyl’s criterion) The curve ${x(t)}$ is equi-distributed in ${{\mathbb R}^n/{\mathbb Z}^n}$ if and only if

$\displaystyle \frac{1}{T}\int_0^T e^{2\pi i \vec m \cdot \vec x(t)} \; dt \rightarrow 0$

for all ${\vec m \in {\mathbb Z}^n\setminus\{0\}}$.

Proof: The forward implication is a fortiori clear. For the reverse implication, trigonometric polynomials are dense. $\Box$

Theorem Let ${p_1(t), \ldots, p_n(t)}$ be rationally independent “of constants”: that is, if ${\vec m \cdot \vec p(t) := \sum m_j p_j(t) \equiv const}$ with ${\vec m \in {\mathbb Z}^n}$, then ${\vec m \equiv 0}$. Then the curve ${\vec p (t)}$ is equi-distributed in ${{\mathbb R}^n / {\mathbb Z}^n}$.

Remark Conversely, if ${\vec m \cdot \vec p \equiv c}$ where ${\vec m \in {\mathbb Z}^n \setminus\{ 0\}}$, then the curve is no equi-distributed: it lives in a co-dimension one sub-manifold.

Proof: By Weyl’s criterion, we need to show that ${\frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \stackrel{T \rightarrow \infty}{\longrightarrow} 0}$ for all ${\vec m \in {\mathbb Z}^n \setminus\{0\}}$. For fixed ${\vec m}$, we know that ${\vec m \cdot \vec p(t)}$ is a non-constant polynomial of degree ${k \geq 1}$, so ${\vec m \cdot \vec p(t) = a_k t^k + a_{k-1}t^{k-1} + \cdots + a_0}$ with ${a_k \neq 0}$. Changing variables ${t = uT}$ gives

$\displaystyle \frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt = \int_0^1 e^{2\pi iT^k \phi_T(u)} \; du$

where

$\displaystyle \phi_T(u) = T^{-k} \vec m \cdot \vec p(Tu) = a_ku^k + a_{k-1} T^{-1} u^{k-1} + \cdots + a_0T^{-k}.$

Now, note that ${\Big| \frac{d^k \phi_T}{du^k} \Big| = |a_k| > 0}$ uniformly in ${T}$. By Van der Corput’s lemma,

$\displaystyle \Big| \frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \Big| \lesssim |a_k T^k|^{-1/k}$

uniformly in ${T}$. This goes to zero as ${T \rightarrow \infty}$. $\Box$

Proposition (Multi-dimensional non-stationary phase) Fix ${a \in C_c^\infty({\mathbb R}^d)}$ and ${\phi \in C^\infty({\mathbb R}^d)}$ with ${\nabla \phi \neq 0}$ on ${\mathrm{supp}(a)}$. Then

$\displaystyle \Big| \int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx \Big| \lesssim_{N,a,\phi} \lambda^{-N}$

for any integer ${N \geq 0}$.

Proof: By introducing a partition of unity, we can assume that ${| \frac{\partial \phi}{\partial x_j} | \geq \frac{1}{2d} \min_{\mathrm{supp}(a)} |\nabla \phi| > 0}$ on the support of ${a}$ for some fixed ${1 \leq j \leq d}$. Now writing, with ${x'=(x_1, \ldots, \hat{x_j}, \ldots, x_d)}$,

$\displaystyle \int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx = \int_{\text{compact set}} \left( \int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j \right)\; dx'.$

Now, we apply the one dimensional result to ${\int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j }$ to get ${O(\lambda^{-N})}$ with a constant that depends continuously on ${x'}$. $\Box$

An alternative proof is to use ${\frac{\nabla \phi}{i \lambda |\nabla \phi|^2} \cdot \nabla e^{i \lambda \phi} = e^{i\lambda \phi}}$ and integrate by parts sufficiently many times. For example

$\displaystyle \int e^{i \lambda \phi(x)} a(x) \; dx = \int e^{i\lambda \phi} \nabla \cdot \left( \frac{ia\nabla\phi}{\lambda |\nabla \phi|^2}\right) \; dx.$

Proposition (Morse Lemma) Let ${\phi \in C^\infty({\mathbb R}^d)}$ have a non-degenerate critical point at ${0 \in {\mathbb R}^d}$; that is,

$\displaystyle \phi(0) = 0 \hbox{\hskip 28pt but \hskip 28pt} \det \left( \frac{\partial^2 \phi}{\partial x_i \partial x_j}\right) \neq 0.$

Then there is a local diffeomorphism ${x \mapsto y}$ in a neighborhood of ${x =0}$ with ${0 \mapsto }$,

$\displaystyle \frac{\partial y_j}{\partial x_i}(0) = \partial_{ij}$

and

$\displaystyle \phi(x) = \phi(0) + \frac{1}{2} \sum_{i,j} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0) y_i(x) y_j(x)$

Proof: By Taylor’s formula (with integral remainder)

$\displaystyle \begin{array}{rcl} \phi(x) &=& \displaystyle \phi(0) + x \cdot \nabla \phi(0) + \int_0^1 (1-t) \left( \frac{d^2}{dt^2} \phi(t,x)\right) \; dt \\ \\ &=& \phi(0) + \int_0^1 (1-t) \sum_{i,j} x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt \end{array}$

Note that, for ${|x|}$ sufficiently small, we can write

$\displaystyle \int_0^1 (1-t) x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt = \frac{1}{2}\big(T(x)^t \phi''(0) T(x) \big)_{i,j} = \frac{1}{2} \sum_{k,l} T_{kj}(x) \frac{\partial^2 \phi}{\partial x_k \partial x_l}(0) T_{li}(x)$

with ${T(x)}$ a smooth, non-singular matrix with ${T(0) = \mathrm{Id}}$. (Eg. first make a linear change of variables to reduce ${\phi''(0)}$ to ${\mathrm{diag}(\pm 1, \cdots, \pm 1)}$, then explicitly do the completing the square argument).

Now, set ${y_i(x) = \sum_k T_{ik}(x) x_k}$. Note that ${y(0) = 0}$ and ${\frac{\partial y_i}{\partial x_j}(0) = \sum_{k} T_{ik}(0) \delta_{kj} = \delta{ij}}$ $\Box$

Theorem Let ${a \in C_c^\infty}$, and ${\phi \in C^\infty}$ with a non-degenerate critical point at ${x =0}$ but otherwise, ${\nabla \phi}$ does not vanish on ${\mathrm{supp}(a)}$. Then

$\displaystyle \int_{{\mathbb R}^d} e^{i\lambda \phi(x)} a(x) \; dx = e^{i\lambda \phi(0)} a(0) \lambda^{-\frac{d}{2}} \det \left( \frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)\right)^{-\frac{1}{2}} + O_{a,\phi}(\lambda^{-\frac{d+2}{2}})$

as ${\lambda \rightarrow \infty}$.

Note that the ${\frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)}$ (wait… I forgot what I wanted to say here)

Remark One can use a partition of unity (and spatial translations) to generalize to multiple stationary phase points.

Proof: By the Morse lemma, we can find some small neighborhood of ${x = 0}$ where we can change variables ${x = \Phi(y)}$ to simplify ${\phi}$. Now use a smooth partition of unity to write ${a(x) = a_1(x) + a_2(x)}$ where ${a_1}$ is supported in the neighborhood of ${x = 0}$ where the change of variables is defined, while ${a_2(x}$ vanishes in a neighborhood of ${x=0}$. (${a_1}$, ${a_2 \in C_c^\infty}$ with ${\mathrm{supp}(a_j) \subseteq \mathrm{supp}(a))}$. By the non-stationary phase position (and ${\nabla \phi \neq 0}$ on ${\text{supp}(a_2)}$),

$\displaystyle \int e^{i \lambda \phi(x)} a_2(x) \; dx = O(\lambda^{-N})$

for ${N}$ as large as we wish. This leaves

$\displaystyle \int e^{i\lambda \phi(x)} a_1(x) \;dx = \int e^{i \lambda ( \phi(0) + \frac{1}{2} y^t \phi''(0) y)} a_1 \circ \phi(y) \det(\Phi' x) |.$

Lets call ${a_1 \circ \Phi(y) \det(\Phi'(y)) = \tilde a(y)}$; it is ${C_c^\infty}$ and ${\tilde a(0) = a_1 \Phi \circ \phi(0) = a(0)}$.

Remembering that ${\int fg = \int \hat f \check g}$, we see that

$\displaystyle \mathrm{above} = e^{i\lambda \phi(0)} \det \left( \lambda \frac{\phi''}{2\pi i}\right)^{-\frac{1}{2}} \iint e^{-2\pi^2 i\lambda^{-1} \xi^t (\phi'')^{-1} \xi} \check{\tilde{a}}(\xi) \; d\xi$

which has the correct asymptotics. $\Box$

Application of the theorem

For the following discussion, we shall denote the Fourier transform (and its inverse) by

$\displaystyle \begin{array}{rcl} \hat f(x) &=& \displaystyle (2\pi)^{-d/2} \int e^{-ix \cdot \xi} f(x) \; dx \\ \\ f(x) &=& (2\pi)^{-d/2} \int e^{+ix \cdot \xi} f(x) \; dx \end{array}$

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

Example (Schrödinger’s equation) Consider the long-time behavior of the solutions to

$\displaystyle i \partial_t \psi = -\frac{1}{2} \Delta \psi$

for ${\psi: {\mathbb R}_t \times {\mathbb R}_x^d \rightarrow {\mathbb C}}$ with ${\psi(0,x) = \psi_0(x)}$.

As the equation is space-translation invariant (it is also time-translation invariant, but we will not use that here), we take the spatial Fourier transform to find

$\displaystyle i \partial_t \hat \psi (t,\xi) = \frac{1}{2} |\xi|^2 \hat \psi (t,\xi)$

with ${\hat \psi(0,\xi) = \hat \psi_0(\xi)}$, which we take (for now) to be Schwartz ${\mathcal{S}({\mathbb R}^d)}$. Therefore ${\hat \psi(t,\xi) = e^{-t|\xi|^2/2}\hat \psi_0(\xi)}$, i.e. ${\psi(t,x) = e^{-it|\nabla|^2/2} \psi_0 = e^{it\Delta/2}\psi_0}$. Taking inverse Fourier transforms, we get

$\displaystyle \psi(t,x) = (2\pi)^{-d/2} \int e^{i x \cdot \xi - \frac{1}{2} i t|\xi|^2} \hat \psi_0(\xi) \; d\xi$

and so we may apply the stationary phase theorem. Looking for stationary phase points,

$\displaystyle 0 = \nabla \phi = x - t\xi \iff \xi = \frac{x}{t}.$

Also,

$\displaystyle \mathrm{Hessian}(\phi) = \partial_{\xi_i} \partial_{x_j} \left[ -\tfrac{1}{2} |\xi|^2 + \frac{x}{t} \cdot \xi \right] = -\delta_{ij}$

So the theorem says

$\displaystyle \psi(t,x) = [e^{it\Delta/2}\psi_0](x) = (2\pi)^{-d/2} \cdot e^{\frac{i}{2} |x|^2/t} \cdot \hat \psi_0\left(\tfrac{x}{t}\right) \cdot e^{-i\frac{d\pi}{4}} \cdot \left(\tfrac{1}{2\pi}\right)^{-d/2} t^{-d/2} + O(t^{-\frac{d+2}{2}})$

with the big-${O}$ notation for ${\frac{x}{t}}$ constant and ${t \rightarrow +\infty}$ (i.e., this describes the decay if we travel along a world-line). Simplifying the above, we get

$\displaystyle \psi(t,x) = e^{-i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{smaller"}$

with the interpretation that ${\xi \equiv \mathrm{momentum} \equiv \mathrm{velocity}}$ and so ${\mathrm{mass} = 1}$. Note that for ${t \rightarrow -\infty}$, the asymptotics pick up a “discrete” change

$\displaystyle \psi(t,x) = e^{+i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{smaller"}.$

Proposition We have, uniformly in ${t}$ and ${\psi_0}$, the dispersive estimate,

$\displaystyle \|e^{it\Delta/2}\psi_0 \|_{L_x^\infty} \lesssim |t|^{-d/2} \|\psi_0\|_{L_x^1}$

and the “Fraunhofer formula”

$\displaystyle \left\|e^{it\Delta/2} \psi_0(x) - e^{-i\frac{d\pi}{4} + i \frac{|x|^2}{2t}} t^{-d/2} \hat \psi_0\left(\tfrac{x}{t}\right) \right\|_{L_x^2} \stackrel{t\rightarrow \infty}{\longrightarrow} 0$

for every ${\psi_0 \in L^2({\mathbb R}^d)}$.

Remark The motivation for the dispersive estimate comes from the combination of our stationary phase computation and the a priori estimate ${\| \hat \psi_0\|_{L^\infty} \leq (2\pi)^{-d/2} \|\psi_0\|_{L^1}}$.

Proof: As a Fourier multiplier, ${e^{it\Delta/2}}$ is also just a convolution operator. To compute the convolution kernel, take a ${\psi_0 \in \mathcal{S}({\mathbb R}^d)}$ and find

$\displaystyle \begin{array}{rcl} \psi(t,x) &=& \displaystyle (2\pi)^{-d/2} \int e^{i \xi \cdot x - i t|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ &=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int e^{i \xi \cdot x - (\epsilon+i t)|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ &=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int \int e^{i \xi \cdot (x-y) - (\epsilon+i t)|\xi|^2/2} \psi_0(y) \; dy \; d\xi \\ \\ &\vdots& \scriptsize\hbox{using Fubini and Gaussian integrals} \\ &=& e^{-i\pi d/4}(2\pi t)^{-d/2} \int e^{+ i|x-y|^2/(2t)} \psi_0(y) \; dy \end{array}$

In particular, the above computation gives the dispersive estimate

$\displaystyle \|e^{it\Delta/2} \psi_0\|_{L_x^\infty} \leq (2\pi t)^{-d/2} \| \psi_0\|_{L_x^1}.$

Furthermore, we get

$\displaystyle [e^{it\Delta} \psi_0] (x) = (2\pi t)^{-d/2} e^{-i\pi d/2 + i|x|^2/(2t)} \int e^{i x\cdot y/t} \underbrace{e^{i y^2/(2t)}}_{\scriptsize \begin{matrix} \text{the only factor making}\\ \text{Fraunhofer non-exact} \end{matrix}} \psi_0(y) \; dy.$

Thus,

$\displaystyle \begin{array}{rcl} \text{LHS (Fraunhofer)} &=& \displaystyle \Big\| (2\pi t)^{-d/2} \int (1 - e^{iy^2/(2t)}) e^{i x\cdot y/t} \psi_0(y) \; dy \Big\|_{L_x^2} \\ \\ \tiny \begin{bmatrix} \text{Plancherel and changing}\\ \text{variables to } x' = \frac{x}{t} \end{bmatrix} &=& \| (1-e^{i|y|^2/(2t)}) \psi_0(y)\|_{L_x^2}, \end{array}$

which, by the dominated convergence theorem with dominating function ${4|\psi_0(y)|^2}$, goes to ${0}$. $\Box$

As an aside, we discuss the result in the setting of optics. If we shine a laser through a screen with an aperture, let us denote ${\psi_0(y_1,y_2) = (\text{amplitude }) e^{i \text{phase}}}$ with ${y_1, y_2}$ being the horizontal and vertical axes, respectively. At the far screen we have

$\displaystyle \psi(x_1, x_2) = \int_{{\mathbb R}^2} e^{i \sqrt{L^2 + |x-y|^2}} \; \psi_0(y) \; dy$

where ${\sqrt{L^2 + |x-y|^2}}$ is the distance from the source point to the observation point. Because ${\sqrt{L^2 + |x-y|^2} \approx \sqrt{L^2 + x^2} - \frac{x \cdot y}{\sqrt{L^2 + x^2}} + (\text{smaller})}$ and that ${\sqrt{L^2 + x^2} \approx L + \frac{x^2}{2L} + \cdots}$, the dominant term is essentially the Fourier transform.

Also

$\displaystyle \begin{array}{rcl} 0 &=& (\partial_t^2 - \Delta_{x,y,z}) ( e^{i\omega (t-z)} \psi(x,y,z)) \\ &=& \big( 2i\omega \psi_z + \underbrace{\psi_{zz}}_{\text{negligible''}} - \Delta_{x,y} \psi \big) e^{i\omega(t-z)} \end{array}$

cf Paraxial approximations.