Schwartz Functions

Video. Schwartz Functions

Schwartz functions are a category of “extremely nice” functions which behave well with respect to most, if not nearly all, basic operations.

Definition

We say that a function \(f\) on \({\mathbb R}^n\) belongs to the Schwartz Class \(\mathcal S\) when it is infinitely differentiable and every mixed partial derivative \(\partial^\alpha f\) satisfies an inequality of the form

\[ |\partial^\alpha f(x)| \leq \frac{C_{\alpha,N}}{(1 + ||x||)^N}\]

for all \(x \in {\mathbb R}^n\) and all natural numbers \(N\), where \(C\) is a constant that depends only on \(\alpha\), \(f\), and \(N\) (but not on \(x\)). This criterion is sometimes described informally as saying that \(f\) and all of its partial derivatives are “rapidly decaying.”

Example

The Gaussian function \(e^{-x^2}\) is a Schwartz function on the real line.

Proof

Meta (Key Ideas)

It helps to prove the following fact: For each nonnegative integer \(k\), there is a polynomial \(p_k(x)\) such that

\[ \frac{d^k}{dx^k} e^{-x^2} = p_k(x) e^{-x^2}. \]

This is fairly easy to prove by induction, because the product rule dictates that \(p_{k+1}(x) = p_k'(x) - 2x p_k(x)\), so all that is really necessary to observe is that when \(p_k\) is a polynomial, so is \(p_{k+1}\).

It also helps to prove that \(|x^\ell| \leq (1 + |x|)^N\) for all \(x \in {\mathbb R}\) whenever \(N \geq \ell\). Then by the triangle inequality, it follows that every polynomial \(p\) of degree at most \(k\) will satisfy \(|p(x)| \leq C (1+ |x|)^N\) for some constant \(C\), any fixed \(N \geq k\), and all \(x \in {\mathbb R}\).

After these observations, it suffices to show that for each \(N\), there is a constant \(C_N\) such that

\[ |e^{-x^2}| \leq C_N (1 + |x|)^{-N} \text{ for all } x \in {\mathbb R}. \]

One approach is to combine two separate observations:

\(|e^{-x^2}| \leq 1\) for all \(x \in {\mathbb R}\).
\(|e^{-x^2}| = 1/(e^{x^2}) \geq n!/x^{2n}\) for any nonnegative integer \(n\) (because the terms in the Taylor series of \(e^{x^2}\) are all nonnegative). Therefore \(x^{2n} e^{-x^2} \leq n!\) for all \(x \in R\). This means that
\[{}|x|^n e^{-x^2}{}\]
\[{}= (x^{2n}e^{-x^2})^{1/2} (e^{-x^2})^{1/2}{}\]
\[{}\leq \sqrt{n!}{}\]
for each nonnegative integer \(n\). Now try summing these inequalities with coefficients:
\[ (1 + |x|)^N e^{-x^2} \leq \sum_{k=0}^N \binom{N}{k} \sqrt{k!}\]

The constant on the right-hand side is not important; what matters is merely that it is some constant not depending on \(x\). To conclude, divide by \((1+|x|)^N\) on both sides.

A useful Lemma is the following:

Lemma

A function \(f\) belongs to \(\mathcal S\) if and only if \(x^\beta \partial^\alpha f(x)\) is a bounded function for any multiindices \(\alpha\) and \(\beta\).

Proof

Meta (Main Ideas)

If \(f\) is in the Schwartz class, it suffices to show that

\[ \frac{x^\beta}{(1 + ||x||)^N}\]

is a bounded function of \(x\) if \(N\) is chosen sufficiently large depending on \(\beta\). That can be done by fixing \(N = |\beta|\) and observing that each coordinate \(x_i\) satisfies \(|x_i| \leq (1 + ||x||)\).

Conversely, if the boundedness property holds, then any polynomial times \(f\) must be bounded. So then observe that

\[ |f(x)| \leq A \]

and

\[ |(x_1^2 + \cdots + x_n^2)^N f(x)| \leq B_N \]

for all \(x \in {\mathbb R}^n\). Taking the square root of each inequality and multiplying gives that

\[ ||x||^N |f(x)| \leq \sqrt{A B_N} \text{ for all } x \in {\mathbb R}^n. \]

As in the example above, this means

\[ (1 + ||x||)^N |f(x)| \leq \sum_{k=0}^N \binom{N}{k} \sqrt{A B_k}. \]

1. Pointwise Operations

Theorem

Suppose that \(f\) and \(g\) are two Schwartz functions on \({\mathbb R}^n\), then for any constant \(c\), \(cf + g\) is a Schwartz function. Moreover \(fg\) is also a Schwartz function. If \(p\) is any polynomial, then \(f(x)p(x)\) is also a Schwartz function. If \(h \in {\mathbb R}^n\), then \(e^{2 \pi i x \cdot h} f(x)\) is also a Schwartz function.

Proof (Hint)

It helps to prove the following fact: If \(\alpha\) is any multiindex and \(f\) and \(g\) belong to \(C^{|\alpha|}\), then

\[ \partial^{\alpha} (f g) = \sum_{\alpha_1+\alpha_2 = \alpha } C_{\alpha_1,\alpha_2} \partial^{\alpha_1} f(x) \partial^{\alpha_2} g(x) \]

for some constants \(C_{\alpha_1,\alpha_2}\) where \(\alpha_1\) and \(\alpha_2\) are multiindices with \(|\alpha_1|, |\alpha_2| \leq |\alpha|\). This is not difficult to prove by induction on \(|\alpha|\).

2. Transformations of the Domain

Theorem

Suppose that \(f\) is a Schwartz function on \({\mathbb R}^n\). Let \(h \in {\mathbb R}^n\) and let \(M\) be any invertible \(n \times n\) real matrix. Then \(\tau_h f(x) := f(x-h)\) and \(\delta_M f(x) := f(M^{-1} x)\) are both Schwartz functions

3. Calculus Operations

Theorem (Schwartz Functions and Differentiation)

If \(f\) is a Schwartz function, then \(\partial^\alpha f\) is a Schwartz function for any multiindex \(\alpha\).
If \(e_j\) is any standard unit vector in \({\mathbb R}^n\), then
\[ \frac{f(x+he_j) - f(x)}{h} \rightarrow \partial_j f(x) \]

uniformly in \(x\) as \(h \rightarrow 0\).

Proof (Hints)

In the second property, use the integral identity

\[{}\frac{f(x+he_j) - f(x)}{h}{}\]

\[{}= \int_0^1 \partial_j f (x + t h e_j) dt{}\]

and the identity

\[ = \partial_j f(x) = \int_0^1 \partial_j f(x) dt. \]

Take the difference of the identities and use the Mean Value Theorem to conclude that

\[{}\left| \frac{f(x+he_j) - f(x)}{h} - \partial_j f(x) \right|{}\]

\[{}\leq \frac{|h|}{2} \sup_{y \in {\mathbb R}^n} |\partial_j^2 f(y)|.{}\]

Theorem (Schwartz Functions and Integration)

For any Schwartz function, the limit

\[ \lim_{R \rightarrow \infty} \int_{||x|| \leq R} f(x) dx \]

exists. For convenience, the limit is denoted \(\int f(x) dx\). It is also the case that the limit

\[ \lim_{R \rightarrow \infty} \int_{||x|| \leq R} |f(x)| dx \]

exists, and this is similarly denoted by \(\int |f(x)| dx\). Then

If \(f,g \in \mathcal S\) and \(c\) is a constant, then
\[ \int (cf + g) \, dx = c \int f \, dx + \int g \, dx \]

and

\[ \left| \int f(x) dx \right| \leq \int |f(x)| dx. \]

For any \(h \in {\mathbb R}^n\):
\[ \int f(x-h) dx = \int f(x) dx. \]
For any Schwartz function \(f\),
\[ \lim_{h \rightarrow 0} \int |f(x-h) - f(x)| dx = 0. \]
For any invertible matrix \(M\),
\[ \int f(M^{-1} x) dx = |\det M| \int f(x) dx. \]
For any Schwartz functions \(f\) and any multiindex \(\alpha\), if \(g\) is of class \(C^{|\alpha|}\) and all derivatives of \(g\) have at most moderate growth (i.e., are bounded in magnitude by \(C (1 + ||x||)^m\) for some fixed \(m\)), then
\[{}\int (\partial^\alpha f(x)) g(x) dx{}\]
\[{}= (-1)^{|\alpha|} \int f(x) (\partial^\alpha g(x)) dx. {}\]

Proof (Hints)

To prove convergence of the integral, show that the limit is Cauchy. Begin by defining:

\[ I(R) := \int_{||x|| \leq R} f(x) dx. \]

If \(R_2 > R_1\), then

\[{}|I(R_2) - I(R_1)|{}\]

\[{}= \left| \int_{R_1 < ||x|| \leq R_2} f(x) dx \right|{}\]

\[{}\leq \int_{R_1 < ||x|| \leq R_2} |f(x)| dx.{}\]

Now prove an inequality of the form

\[{}\int_{R_1 < ||x|| \leq 2R_2} |f(x)| dx \leq \frac{C_{n+1}}{||x||^{n+1}} dx{}\]

\[{}\leq \frac{C_{n+1}}{R_1^{n+1}} \int_{[-2R_1,2R_1]^n} 1 dx{}\]

\[{}= \frac{4^n C_{n+1}}{R_1}{}\]

Since the annulus \(R_1 < ||x|| < R_2\) is always contained in a union of annuli \(R_1 < ||x|| \leq 2R_1\), \(2 R_1 < ||x|| \leq 4 R_1\), and so on, it follows that

\[{}|I(R_2) - I(R_1)|{}\]

\[{}\leq \sum_{k=0}^\infty \int_{2^k R_1 < ||x|| \leq 2^{k+1} R_1} |f(x) dx{}\]

\[{}\leq \frac{4^n C_{n+1}}{R_1} \sum_{k=0}^\infty \frac{1}{2^k}{}\]

which tends to zero as \(R_1 \rightarrow \infty\).

For translation, observe that

\[{}\int \chi_{||x|| \leq R} f(x-h) dx{}\]

\[{}= \int_{||x+h|| \leq R} f(x) dx, {}\]

\[{}\int \chi_{||x|| \leq R} f(x-h) dx{}\]

\[{}- \int_{||x|| \leq R} f(x){}\]

\[{}= \int (\chi_{||x+h|| \leq R} - \chi_{||x||\leq R}) f(x).{}\]

This helps because the difference of the two characteristic functions will be zero when \(||x|| \leq R- ||h||\) (because both equal \(1\)) and also zero when \(||x|| > R + ||h||\) (because both equal zero). Thus the integrand is only nonzero on the annulus \(R- ||h|| < ||x|| \leq R + ||h||\). As above, the integral of a Schwartz function on an annulus (or even simply outside a large ball) tends to zero as the inner diameter tends to infinity.

The limit property involving translations works on a similar principle: Pick a radius \(R\) large enough that the integral of \(f\) is small outside the ball of that radius. Then on the ball \(||x|| \leq R\) \(f\) is uniformly continuous, so \(f(x-h) - f(x)\) can be made uniformly small if \(h\) is sufficiently small.

For the matrix transformation, it's a consequence of change of variables combined with the fact that there must be constants \(c_1\) and \(c_2\) such that \(c_1 ||x|| \leq ||M x || \leq c_2 ||x||\) for each \(x\).

The integration by parts inequality can be proved by induction fairly directly once it's proved for a single partial derivative in any one of the coordinate directions. By Fubini,

\[{}\int_{||x|| \leq R} \partial_j (f g){}\]

\[{}= \int_{||x'|| \leq R} fg\Big|^{\sqrt{R^2-||x'||^2}}_{x_1=-\sqrt{R^2-||x'||^2}} dx' {}\]

where \(x := (x_1,x')\) and \(x' \in {\mathbb R}^{n-1}\). The boundary terms all involve evaluating the product \(fg\) at points on the sphere \(||x||=R\), and the maximum value on this sphere tends to zero so rapidly as \(R \rightarrow \infty\) that it easily counteracts the growing volume of the ball in \({\mathbb R}^{n-1}\) which must also be accounted for.