So, as I’ve already indicated, I’m planning to talk about PDEs for the next month or so, both the general theory and specific posts on the equations of mathematical physics.  There are some preliminaries I’ll have to do first such as Fourier transforms.  Today, I’ll get up to the inversion formula for Schwarz functions.

The Schwarz Class

The Schwarz class ${\mathcal{S}}$ consists of smooth functions ${f: \mathbb{R}^n \rightarrow \mathbb{R}}$ such that for all multi-indices ${\alpha=(\alpha_1, \dots, \alpha_n), \beta=(\beta_1, \dots, \beta_n)}$,

$\displaystyle x^{\alpha} D^{\beta} f := x_1^{\alpha_1} \dots x_n^{\alpha_n} \left( \frac{\partial}{\partial x_1}\right)^{\beta_1} \dots \left( \frac{\partial}{\partial x_n}\right)^{\beta_n}f$

is bounded. For instance, the function ${e^{-|x|^2}}$ is in ${\mathcal{S}}$, as is any ${C^{\infty}}$ function with compact support. Elements in ${\mathcal{S}}$ are loosely speaking, functions that decrease rapidly at ${\infty}$ with all their partial derivatives.

There is a way to make the space ${\mathcal{S}}$ into a Frechet space, by the countable family of seminorms

$\displaystyle ||f||_{a,b} := \sup |x^{a} D^{b} f(x) |.$

It is not a Banach space, but these seminorms do induce a topology on ${\mathcal{S}}$. A neighborhood basis at ${f \in S}$ can be given by

$\displaystyle \{ g: ||g-f||_{a,b} < \epsilon \ \mathrm{for} a+b \leq n \}.$

It follows that ${\mathcal{S}}$ becomes a metric space if we define

$\displaystyle d(f,g) = \sum_{a,b} 2^{-a-b} \frac{ ||f-g||_{a,b} }{ ||f-g||_{a,b} + 1}.$

The space ${\mathcal{S}}$ has many pleasant properties with respect to this topology that make it useful to work with. For instance, it is complete in the above metric. It is also a topological vector space because addition and scalar multiplication (and inversion) are continous. Moreover, differentiation and multiplication by a function in ${\mathcal{S}}$ is a continuous map.

The Fourier transform

Given ${f \in L^1(\mathbb{R}^n)}$, we define the Fourier transform ${\hat{f}}$:$\displaystyle \hat{f}(x) = \int_{\mathbb{R}^n} f(y) e^{-2 \pi i x.y} dy.$

It is easy to check that this is a continuous function on ${\mathbb{R}^n}$, bounded by ${||f||_1}$

Let’s now restrict to ${f \in \mathcal{S}}$.The key property of the Fourier transform is that it converts differentiation into multiplication by a power of ${x}$ and vice versa, which makes it especially useful in dealing with PDEs. To solve a PDE with constant coefficients one can consider the much nicer problem obtained by a Fourier transform.This is what we’ll now prove, and it shows that in particular, ${\hat{\cdot}}$ maps ${\mathcal{S}}$ into itself.In particular, if we abbreviate ${D_j = \frac{\partial}{\partial x_j}}$,$\displaystyle D_j \hat{f}(x) = -2\pi i x_j \int_{\mathbb{R}^n} f(y) e^{-2 \pi i x.y} dy$

which implies inductively that

$\displaystyle D^a \hat{f}(x) = (-2\pi i)^{|a|} \widehat{ x^a f} . \ \ (|a|=a_1+\dots+a_n)$

To go the other way, note that by integration by parts

$\displaystyle \widehat{ D_j f}(x) = \int_{\mathbb{R}^n} D_j f(y) e^{-2 \pi i x.y} dy = \int_{\mathbb{R}^n} f(y) e^{-2 \pi i x.y} dy = -2\pi i x_j \int_{\mathbb{R}^n} f(y) e^{-2 \pi i x.y} dy$

so by induction

$\displaystyle \widehat{ D^a f}(x) = (-2\pi i)^{|a|} x^a \hat{f}.$

If we combine these two we find:

$\displaystyle \boxed{ \widehat{ x^a D^b f}(x) = (-2 \pi i )^{-|a|} D^a \widehat{D^b f }=(-2\pi i)^{|b|-|a|} D^a( x^b \hat{f}).}$

There is a similar identity with the ${D}$ and the ${x}$ reversed. As a result, it follows that the Fourier transform maps ${\mathcal{S}}$ into ${\mathcal{S}}$

The next basic property of the Fourier transform, which is actually valid for any two ${L^1}$ functions, is that it turns convolution into multiplication. Recall that if ${f,g \in L^1(\mathbb{R}^n)}$, then the convolution ${f \ast g}$ is defined by$\displaystyle (f \ast g)(x)= \int_{\mathbb{R}^n} f(y) g(x-y) dy = \int_{\mathbb{R}^n} f(x+y) g(-y).$

By the second equality, this can be thought of as a weighted average of translations of ${f}$ with respect to ${g(-y)}$. It can be checked that ${||f \ast g||_1 \leq ||f||_1 ||g||_1}$

Now we can prove the identity$\displaystyle \boxed{\widehat{f \ast g} = \hat{f} \hat{g}.}$

Indeed:

$\displaystyle \widehat{f \ast g}(x) = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} e^{-2\pi i x.u} f( y)g(u-y) dy du ,$

which we change variables into

$\displaystyle \iint e^{-2\pi i x.y} f( y) e^{-2\pi i x.(u-y)} g(u-y) dy du = \iint e^{-2\pi i x.y} f( y) e^{-2\pi i x.t} g(t) dy dt,$

which now makes the identity clear.

Fourier inversion

The Fourier transform can be thought of as a resolution of a function into continuous wave frequences, kind of like Fourier series are a resolution of a function on the circle into discrete frequencies. In the same way, a function can—in a certain sense—be recovered from its Fourier transform.The more general theory here is that of Fourier analysis on locally compact abelian groups, which I’m not going to discuss (yet).Given ${f \in L^1(\mathbb{R}^n)}$, we define the inverse Fourier transform by combining suitable frequences at levels determined by ${f}$:$\displaystyle \tilde{f}(x) = \int_{\mathbb{R}^n} e^{2 \pi i x.y} f(y) dy = \hat{f}(-x).$

We will now establish the Fourier inversion formula for ${f \in \mathcal{S}}$:

Theorem 1 If ${f \in \mathcal{S}}$, then ${\tilde{\hat{f}} = \hat{\tilde{f}} = f}$.

We will prove the first identity; the second is similar. The idea behind the proof is to check it for a certain convenient function whose Fourier transform is easily calculated. Then, using an approximation to the identity argument, we can represent any function using it and its dilations and translations.The function in question is ${\phi(x) = e^{- \pi |x|^2}}$. I claim:$\displaystyle \hat{\phi} = \tilde{\phi} = \phi.$

By evenness it suffices to prove the first statement. We need to evaluate the integral

$\displaystyle \int_{\mathbb{R}^n} e^{- \pi |x|^2 -2\pi i x.y} dy = e^{-\pi|y|^2} \int_{\mathbb{R}^n} e^{- \pi |x|^2 -2\pi i x.y+ \pi |y|^2 } dy$

The integral in question is actually

$\displaystyle \int_{\mathbb{R}^n} e^{- |x+iy|^2 } dy = \int_{\mathbb{R}^n} e^{- |iy|^2 } dy = 1$

by a shift in contours (justifiable because these are rapidly decreasing functions).

Now that we’ve proved the claim, we can do the inversion formula. Recall that ${\phi_t(x) := t^{-n} \phi(t^{-1} x)}$. As ${t \rightarrow 0}$, this becomes kind-of-like the Dirac delta function—a point mass at ${0}$. (This convergence is in the sense of distributions, to be covered shortly.) The Fourier transform of this is ${x \rightarrow \phi(tx) }$ which approaches the function identically 1. This is part of a general principle that a function supported near the origin has a spread out Fourier transform, and vice versa.So$\displaystyle \tilde{\hat{f}}(x) = \iint e^{2\pi i x.y} f(u) e^{-2\pi i u.y} dy du$

which becomes by the formula for ${\hat{\phi}}$, and an elementary change of variables,

$\displaystyle \lim_{t \rightarrow 0} \iint e^{2\pi i (x-u) .y} \phi(ty) f(u) du dy= \lim_t \int \phi_t(x-u) f(u) du .$

But the last integral is just

$\displaystyle (\phi_t \ast f)(x),$

and from basic facts on approximations to the identity, this tends to ${f(x)}$ as ${t \rightarrow 0}$. In detail,

$\displaystyle (\phi_t \ast f)(x)- f(x) = \int \phi_t(-u) ( f(u+x) - f(x)) du = \int_{B_r(0)} + \int_{B_r(0)^C} ;$

the integral in ${B_r(0)}$ is bounded by something tending to zero depending on the continuity of ${f}$, and the other part goes to zero because ${\phi_t(x) \rightarrow 0}$ uniformly for ${x}$ in a compact set not including the origin.