CHAPTER 1 Tempered distributions and the Fourier transform^{[T empered]}

Microlocal analysis is a geometric theory of distributions, or a theory of geometric distributions. Rather than study general distributions – which are like general continuous functions but worse – we consider more specific types of distributions which actually arise in the study of differential and integral equations. Distributions are usually defined by duality, starting from very “good” test functions; correspondingly a general distribution is everywhere “bad”. The conormal distributions we shall study implicitly for a long time, and eventually explicitly, are usually good, but like (other) people have a few interesting faults, i.e. singularities. These singularities are our principal target of study. Nevertheless we need the general framework of distribution theory to work in, so I will start with a brief introduction. This is designed either to remind you of what you already know or else to send you off to work it out.^{1}Proofs of some of the main theorems are outlined in the problems at the end of the chapter.

1.1. Schwartz test functions^{[S.Schwartz.T est]}

To fix matters at the beginning we shall work in the space of tempered distributions. These are defined by duality from the space of Schwartz functions, also called the space of test functions of rapid decrease. We can think of analysis as starting off from algebra, which gives us the polynomials. Thus in ℝ^{n} we have the coordinate functions, x_{1}, …, x_{n} and the constant functions and then the polynomials are obtained by taking (finite) sums and products:

(1.1) ϕ(x) = ∑_{|α|≤k} p_{α}x^{α}, p_{α} ∈ ℂ, α ∈ ℕ_{0}^{n}, α = (α_{1}, …, α_{n}),

where x^{α} = x_{1}^{α_{1}} … x_{n}^{α_{n}} = ∏_{j=1}^{n} x_{j}^{α_{j}} and ℕ_{0} = {0, 1, 2, … }.^{[1.1]}

A general function ϕ : ℝ^{n} ⟶ ℂ is differentiable at x̄ if there is a linear function

ℓx̄(x) = c + ∑_{j=1}^{n}(x_{j} − x̄_{j})d_{j} such that for every ϵ > 0 there exists δ > 0 such that

|ϕ(x) − ℓx̄(x)| ≤ ϵ|x − x̄| ∀ |x − x̄| < δ̄.^{[1.2]} (1.2)

14 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM^{[T EMP ERED]}

The coefficients d_{j} are the partial derivative of ϕ at the point x̄. Then, ϕ is said to be differentiable on ℝ^{n} if it is differentiable at each point x̄ ∈ ℝ^{n}; the partial derivatives are then also functions on ℝ^{n} and ϕ is twice differentiable if the partial derivatives are differentiable. In general it is k times differentiable if its partial derivatives are k − 1 times differentiable.

1I suggest Friedlander’s little book [3] (there is also a newer edition) as a good introduction to distributions. Volume 1 of Hörmander’s treatise [7] has all that you would need and a good deal more; it is a good general reference.

If ϕ is k times differentiable then, for each x̄ ∈ ℝ^{n}, there is a polynomial of degree k,

p_{k}(x; x̄) = ∑_{|α|≤k} a_{α}i^{|α|}(x − x̄)^{α}/α!, |α| = α_{1} + ⋯ + α_{n},

such that for each ϵ > 0 there exists δ > 0 such that

|ϕ(x) − p_{k}(x, x̄)| ≤ ϵ|x − x̄|^{k} if |x − x̄| < δ.^{[1.3]} (1.3)

Then we set

D^{α}ϕ(x̄) = a_{α}.^{[1.4]} (1.4)

If ϕ is infinitely differentiable all the D^{α}ϕ are infinitely differentiable (hence continuous!) functions.

[1.5]

Definition 1.1. The space of Schwartz test functions of rapid decrease consists of those ϕ : ℝ^{n} ⟶ ℂ such that for every α, β ∈ ℕ_{0}^{n}

sup_{x∈ℝ^{n}} |x^{β}D^{α}ϕ(x)| < ∞;^{[1.6]} (1.5)

it is denoted S(ℝ^{n}).

From (1.5) we construct norms on S(ℝ^{n}) :

∥ϕ∥_{k} = max_{|α|+|β|≤k} sup_{x∈ℝ^{n}} |x^{α}D^{β}ϕ(x)|.^{[1.7]} (1.6)

It is straightforward to check the conditions for a norm:

1. ∥ϕ∥_{k} ≥ 0, ∥ϕ∥_{k} = 0 ⟺ ϕ ≡ 0

2. ∥tϕ∥_{k} = |t|∥ϕ∥_{k}, t ∈ ℂ
3. ∥ϕ + ψ∥_{k} ≤ ∥ϕ∥_{k} + ∥ψ∥_{k} ∀ ϕ, ψ ∈ S(ℝ^{n}).

The topology on S(ℝ^{n}) is given by the metric

d(ϕ, ψ) = ∑_{k} 2^{−k} \frac{∥ϕ − ψ∥_{k}}{1 + ∥ϕ − ψ∥_{k}} .^{[1.8]} (1.7)

See Problem 1.4.

[1.9]

Proposition 1.1. With the distance function (1.7), S(ℝ^{n}) becomes a complete metric space (in fact it is a Fréchet space).

Of course one needs to check that S(ℝ^{n}) is non-trivial; however one can easily see that

exp(−|x|^{2}) ∈ S(ℝ^{n}). (1.8)

In fact there are lots of smooth functions of compact support and

(1.9) C_{c}^{∞}(ℝ^{n}) = {u ∈ S(ℝ^{n}); u = 0 in |x| > R = R(u)} ⊂ S(ℝ^{n}) is dense.^{[1.2.2000.266]}

The two elementary operations of differentiation and coordinate multiplication give continuous linear operators:

\begin{matrix}x_{j}: & S(ℝ^{n}) ⟶ S(ℝ^{n}) \\ D_{j}: & S(ℝ^{n}) ⟶ S(ℝ^{n}). \end{matrix}^{[31.1.2000.263]} (1.10)

1.2. LINEAR TRANSFORMATIONS^{[S.LINEAR.T RANSF ORMAT IONS]} 15

Other important operations we shall encounter include the exterior product,

\begin{matrix}S(ℝ^{n}) × S(ℝ^{m}) ∋ (ϕ, ψ) ↦ ϕ ⊠ ψ ∈ S(ℝ^{n+m}) \\ ϕ ⊠ ψ(x, y) = ϕ(x)ψ(y). \end{matrix}^{[1.10]} (1.11)

and pull-back or restriction. If ℝ^{k} ⊂ ℝ^{n} is identified as the subspace x_{j} = 0, j > k, then the restriction map

π_{k}^{∗} : S(ℝ^{n}) ⟶ S(ℝ^{k}), π_{k}^{∗}f(y) = f(y_{1}, …, y_{k}, 0, …, 0)^{[31.1.2000.264]} (1.12)

is continuous (and surjective).

1.2. Linear transformations^{[S.Linear.transformations]}

A linear transformation acts on ℝ^{n} as a matrix^{2}

L : ℝ^{n} ⟶ ℝ^{n}, (Lx)_{j} = ∑_{k=1}^{n} L_{jk}x_{k}.^{[1.2.2000.267]} (1.13)

The Lie group of invertible linear transformations, GL(n, ℝ) is fixed by several equivalent conditions

L ∈ GL(n, ℝ) ⟺ det(L) ≠ 0
⟺ ∃ L^{−1} s.t. (L^{−1})Lx = x ∀ x ∈ ℝ^{n}
⟺ ∃ c > 0 s.t. c|x| ≤ |Lx| ≤ c^{−1}|x| ∀ x ∈ ℝ^{n}.

[1.2.2000.268] (1.14)

Pull-back of functions is defined by

L^{∗}ϕ(x) = ϕ(Lx) = (ϕ ∘ L)(x).

The chain rule for differentiation shows that if ϕ is diffferentiable then^{3}

(1.15) D_{j}L^{∗}ϕ(x) = D_{j}ϕ(Lx) = ∑_{k=1}^{n} L_{kj}(D_{k}ϕ)(Lx) = L^{∗}((L_{∗}D_{j})ϕ)(x),

L_{∗}D_{j} = ∑_{k=1}^{n} L_{kj}D_{k}.^{[1.2.2000.269]}

From this it follows that

(1.16) L^{∗} : S(ℝ^{n}) ⟶ S(ℝ^{n}) is an isomorphism for L ∈ GL(n, ℝ).^{[1.2.2000.270]}

To characterize the action of L ∈ GL(n, ℝ) on S′(ℝ^{n}) consider, as usual, the distribution associated to L^{∗}ϕ :

(1.17) T_{L^{∗}ϕ}(ψ) = ∫_{ℝ^{n}} ϕ(Lx)ψ(x)dx

= ∫_{ℝ^{n}} ϕ(y)ψ(L^{−1}y)| det L|^{−1}dy = T_{ϕ}(| det L|^{−1}(L^{−1})^{∗}ψ).^{[1.2.2000.271]}

elements e_{j} ∈ ℝ^{n}, Le_{j} = ∑_{k=1}^{n} L_{kj}e_{k}.

2This is the standard action, but it is potentially confusing since it means that for the basis

3So D_{j} transforms as a basis of ℝ^{n} as it should, despite the factors of i.

Since the operator | det L|^{−1}(L^{−1})^{∗} is an ismorphism of S(ℝ^{n}) it follows that if we take the definition by duality

(1.18) L^{∗}u(ψ) = u(| det L|^{−1}(L^{−1})^{∗}ψ), u ∈ S′(ℝ^{n}), ψ ∈ S(ℝ^{n}), L ∈ GL(n, ℝ)

⟹ L^{∗} : S′(ℝ^{n}) ⟶ S′(ℝ^{n})^{[1.2.2000.272]}

is an isomorphism which extends (1.16) and satisfies

D_{j}L^{∗}u = L^{∗}((L_{∗}D_{j})u), L^{∗}(x_{j}u) = (L^{∗}x_{j})(L^{∗}u), u ∈ S′(ℝ^{n}), L ∈ GL(n, ℝ),^{[1.2.2000.273]} (1.19)

as in (1.15).

1.3. Tempered distributions^{[S.T empered.distributions]}

As well as exterior multiplication (1.11) there is the even more obvious multiplication operation

\begin{matrix}S(ℝ^{n}) × S(ℝ^{n}) ⟶ S(ℝ^{n}) \\ (ϕ, ψ) ↦ ϕ(x)ψ(x) \end{matrix}^{[1.11]} (1.20)

which turns S(ℝ^{n}) into a commutative algebra without identity. There is also integration

∫ : S(ℝ^{n}) ⟶ ℂ.^{[1.12]} (1.21)

Combining these gives a pairing, a bilinear map

S(ℝ^{n}) × S(ℝ^{n}) ∋ (ϕ, ψ) ⟼ ∫_{ℝ^{n}} ϕ(x)ψ(x)dx. (1.22)

If we fix ϕ ∈ S(ℝ^{n}) this defines a continuous linear map:

T_{ϕ} : S(ℝ^{n}) ∋ ψ ⟼ ∫ ϕ(x)ψ(x)dx.^{[1.13]} (1.23)

Continuity becomes the condition:

∃ k, C_{k} s.t. |T_{ϕ}(ψ)| ≤ C_{k}∥ψ∥_{k} ∀ ψ ∈ S(ℝ^{n}). (1.24)

We generalize this by denoting by S′(ℝ^{n}) the dual space, i.e. the space of all continuous linear functionals

u ∈ S′(ℝ^{n}) ⟺ u : S(ℝ^{n}) ⟶ ℂ

∃ k, C_{k} such that |u(ψ)| ≤ C_{k}∥ψ∥_{k} ∀ ψ ∈ S(ℝ^{n}). [1.14]

Lemma 1.1. The map

S(ℝ^{n}) ∋ ϕ ⟼ T_{ϕ} ∈ S′(ℝ^{n})^{[1.15]} (1.25)

is an injection.

Proof. For any ϕ ∈ S(ℝ^{n}), T_{ϕ}(ϕ) = ∫ |ϕ(x)|^{2}dx, so T_{ϕ} = 0 implies ϕ ≡ 0.

If we wish to consider a topology on S′(ℝ^{n}) it will normally be the weak topology, that is the weakest topology with respect to which all the linear maps

S′(ℝ^{n}) ∋ u ⟼ u(ϕ) ∈ ℂ, ϕ ∈ S(ℝ^{n}) (1.26)

1.4. TWO BIG THEOREMS^{[S.T W O.BIG.T HEOREMS]} 17

are continuous. This just means that it is given by the seminorms

S(ℝ^{n}) ∋ u ⟼ |u(ϕ)| ∈ ℝ^{[1.16]} (1.27)

where ϕ ∈ S(ℝ^{n}) is fixed but arbitrary. The sets

{u ∈ S′(ℝ^{n}); |u(ϕ_{j})| < ϵ_{j}, ϕ_{j} ∈ Φ}^{[1.101]} (1.28)

form a basis of the neighbourhoods of 0 as Φ ⊂ S(ℝ^{n}) runs over finite sets and the ϵ_{j} are positive numbers.

[1.17]

Proposition 1.2. The continuous injection S(ℝ^{n}) ↪ S′(ℝ^{n}), given by (1.25), has dense range in the weak topology.

See Problem 1.8 for the outline of a proof.

The maps x_{i}, D_{j} extend by continuity (and hence uniquely) to operators

x_{j}, D_{j} : S′(ℝ^{n}) ⟶ S′(ℝ^{n}).^{[1.18]} (1.29)

This is easily seen by defining them by duality. Thus if ϕ ∈ S(ℝ^{n}) set D_{j}T_{ϕ} = T_{D_{j}ϕ}, then

T_{D_{j}ϕ}(ψ) = ∫ D_{j}ϕψ = − ∫ ϕD_{j}ψ, (1.30)

the integration by parts formula. The definitions

D_{j}u(ψ) = u(−D_{j}ψ), x_{j}u(ψ) = u(x_{j}ψ), u ∈ S′(ℝ^{n}), ψ ∈ S(ℝ^{n}) (1.31)

satisfy all requirements, in that they give continuous maps (1.29) which extend the standard definitions on S(ℝ^{n}).

1.4. Two big theorems^{[S.T wo.big.theorems]}

The association, by (1.25), of a distribution to a function can be extended considerably. For example if u : ℝ^{n} ⟶ ℂ is a bounded and continuous function then

T_{u}(ψ) = ∫ u(x)ψ(x)dx (1.32)

still defines a distribution which vanishes if and only if u vanishes identically. Using the operations (1.29) we conclude that for any α, β ∈ ℕ_{0}^{n}

(1.33) x^{β}D_{x}^{α}u ∈ S′(ℝ^{n}) if u : ℝ^{n} ⟶ ℂ is bounded and continuous.

Conversely we have the Schwartz representation Theorem:

[1.19]

Theorem 1.1. For any u ∈ S′(ℝ^{n}) there is a finite collection u_{αβ} : ℝ^{n} ⟶ ℂ of bounded continuous functions, |α| + |β| ≤ k, such that

u = ∑_{|α|+|β|≤k} x^{β}D_{x}^{α}u_{αβ}. (1.34)

Thus tempered distributions are just products of polynomials and derivatives of bounded continuous functions. This is important because it says that distributions are “not too bad”.

The second important result (long considered very difficult to prove, but there is a relatively straightforward proof using the Fourier transform) is the Schwartz

18 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM^{[T EMP ERED]}

kernel theorem. To show this we need to use the exterior product (1.11). If K ∈ S′(ℝ^{n+m}) this allows us to define a linear map O_{K} : S(ℝ^{m}) ⟶ S′(ℝ^{n}) (1.35) by O_{K}(ψ)(ϕ) = ∫ K · ϕ ⊠ ψ dxdy. (1.36)

[1.20]

Theorem 1.2. There is a 1-1 correspondence between continuous linear operators A : S(ℝ^{m}) ⟶ S′(ℝ^{n}) (1.37) and S′(ℝ^{n+m}) given by A = O_{K}.

Brief outlines of the proofs of these two results can be found in Problems 1.15 and 1.16.

1.5. Examples^{[S.Examples]}

Amongst tempered distributions we think of S(ℝ^{n}) as being the ‘trivial’ examples, since they are the test functions. One can say that the study of the singularities of tempered distributions amounts to the study of the quotient S′(ℝ^{n})/S(ℝ^{n}) (1.38) which could, reasonably, be called the space of tempered microfunctions.

The sort of distributions we are interested in are those like the Dirac delta “function” δ(x) ∈ S′(ℝ^{n}), δ(ϕ) = ϕ(0). (1.39)

The definition here shows that δ is just the Schwartz kernel of the operator S(ℝ^{n}) ∋ ϕ ⟼ ϕ(0) ∈ ℂ = S(ℝ^{0}). (1.40) This is precisely one reason it is interesting. More generally we can consider the maps S(ℝ^{n}) ∋ ϕ ⟼ D^{α}ϕ(0), α ∈ ℕ_{0}^{n}. (1.41) These have Schwartz kernels (−D)^{α}δ since (−D)^{α}δ(ϕ) = δ(D^{α}ϕ) = D^{α}ϕ(0).^{[1.21]} (1.42) If we write the relationship A = O_{K} ⟷ K as (Aψ)(ϕ) = ∫ K(x, y)ϕ(x)ψ(y)dxdy (1.43) then (1.42) becomes D^{α}ϕ(0) = ∫ (−D)^{α}δ(x)ϕ(x)dx. (1.44) More generally, if K(x, y) is the kernel of an operator A then the kernel of A · D^{α} is (−D)_{y}^{α}K(x, y) whereas the kernel of D^{α} ∘ A is D_{x}^{α}K(x, y).

1.6. TWO LITTLE LEMMAS^{[S.T W O.LIT T LE.LEMMAS]} 19

1.6. Two little lemmas^{[S.T wo.little.lemmas]}

Above, some of the basic properties of tempered distributions have been outlined. The main “raison d’être” for S′(ℝ^{n}) is the Fourier transform which we proceed to discuss. We shall use the Fourier transform as an almost indispensable tool in the treatment of pseudodifferential operators. The description of differential operators, via their Schwartz kernels, using the Fourier transform is an essential motivation for the extension to pseudodifferential operators.

Partly as simple exercises in the theory of distributions, and more significantly as preparation for the proof of the inversion formula for the Fourier transform we consider two lemmas.

First recall that if u ∈ S′(ℝ^{n}) then we have defined D_{j}u ∈ S′(ℝ^{n}) by

D_{j}u(ϕ) = u(−D_{j}ϕ) ∀ ϕ ∈ S(ℝ^{n}). (1.45)

In this sense it is a “weak derivative”. Let us consider the simple question of the form of the solutions to

D_{j}u = 0, u ∈ S′(ℝ^{n}).^{[2.1]} (1.46)

Let I_{j} be the integration operator:

\begin{matrix}I_{j} : S(ℝ^{n}) ⟶ S(ℝ^{n−1}) \\ I_{j}(ϕ)(y_{1}, …, y_{n−1}) = ∫ ϕ(y_{1}, … y_{j−1}, x, y_{j}, … y_{n−1})dx. \end{matrix}^{[2.2]} (1.47)

Then if π_{j} : ℝ^{n} ⟶ ℝ^{n−1} is the map π_{j}(x) = (x_{1}, …, x_{j−1}, x_{j+1} …, x_{n}), we define, for v ∈ S′(ℝ^{n−1}),

π_{j}^{∗}v(ϕ) = v(I_{j}ϕ) ∀ ϕ ∈ S(ℝ^{n}). (1.48)

It is clear from (1.47) that I_{j} : S(ℝ^{n}) ⟶ S(ℝ^{n−1}) is continuous and hence π_{j}^{∗}v ∈ S′(ℝ^{n}) is well-defined for each v ∈ S′(ℝ^{n−1}).

[2.3]

Lemma 1.2. The equation (1.46) holds if and only if u = π_{j}^{∗}v for some v ∈ S′(ℝ^{n−1}).

Proof. If ϕ ∈ S(ℝ^{n}) and ϕ = D_{j}ψ with ψ ∈ S(ℝ^{n}) then I_{j}ϕ = I_{j}(D_{j}ψ) = 0. Thus if u = π_{j}^{∗}v then

u(−D_{j}ϕ) = π_{j}^{∗}v(−D_{j}ϕ) = v(I_{j}(−D_{j}ϕ)) = 0. (1.49)

Thus u = π_{j}^{∗}v does always satisfy (1.46).

Conversely suppose (1.46) holds. Choose ρ ∈ S(ℝ) with the property

∫ ρ(x)dx = 1. (1.50)

Then each ϕ ∈ S(ℝ^{n}) can be decomposed as

ϕ(x) = ρ(x_{j})I_{j}ϕ(x_{1}, …, x_{j−1}, x_{j+1}, … x_{n}) + D_{j}ψ, ψ ∈ S(ℝ^{n}).^{[2.4]} (1.51)

Indeed this is just the statement

ζ ∈ S(ℝ^{n}), I_{j}ζ = 0 ⟹ ψ(x) ∈ S(ℝ^{n}) where

\begin{aligned}ψ(x) = ∫_{−∞}^{x_{j}} & ζ(x_{1}, …, x_{j−1}, t, x_{j+1}, …, x_{n})dt \\ & = ∫_{∞}^{x_{j}}ζ(x_{1}, …, x_{j−1}, t, x_{j+1}, …, x_{n})dt. \end{aligned}

Using (1.51) and (1.46) we have

u(ϕ) = u (ρ(x_{j})I_{j}ϕ(x_{1}, …, x_{j−1}, x_{j+1}, … x_{n})) . (1.52)

Thus if

v(ψ) = u (ρ(x_{j})ψ(x_{1}, …, x_{j−1}, x_{j+1}, … x_{n})) ∀ ψ ∈ S(ℝ^{n−1}) (1.53)

then v ∈ S′(ℝ^{n−1}) and u = π_{j}^{∗}v. This proves the lemma.

Of course the notation u = π_{j}^{∗}v is much too heavy-handed. We just write u(x) = v(x_{1}, …, x_{j−1}, x_{j+1}, …, x_{n}) and regard ‘v as a distribution in one additional variable’.

The second, related, lemma is just a special case of a general result of Schwartz concerning the support of a distribution. The particular result is:

[2.5]

Lemma 1.3. Suppose u ∈ S′(ℝ^{n}) and x_{j}u = 0, j = 1, … n then u = cδ(x) for some constant c.

Proof. Again we use the definition of multiplication and a dual result for test functions. Namely, choose ρ ∈ S(ℝ^{n}) with ρ(x) = 1 in |x| < \frac{1}{2}, ρ(x) = 0 in |x| ≥ 3/4. Then any ϕ ∈ S(ℝ^{n}) can be written

ϕ = ϕ(0) · ρ(x) + ∑_{j=1}^{n} x_{j}ψ_{j}(x), ψ_{j} ∈ S(ℝ^{n}).^{[2.6]} (1.54)

This in turn can be proved using Taylor’s formula as I proceed to show. Thus

ϕ(x) = ϕ(0) + ∑_{j=1}^{n} x_{j}ζ_{j}(x) in |x| ≤ 1, with ζ_{j} ∈ C^{∞}. (1.55)

Then,

ρ(x)ϕ(x) = ϕ(0)ρ(x) + ∑_{j=1}^{n} x_{j}ρζ_{j}(x) (1.56)

and ρζ_{j} ∈ S(ℝ^{n}). Thus it suffices to check (1.54) for (1 − ρ)ϕ, which vanishes identically near 0. Then ζ = |x|^{−2}(1 − ρ)ϕ ∈ S(ℝ^{n}) and so

(1 − ρ)ϕ = |x|^{2}ζ = ∑_{j=1}^{n} x_{j}(x_{j}ζ) (1.57)

1.7. FOURIER TRANSFORM^{[S.F OURIER.T RANSF ORM]} 21

finally gives (1.54) with ψ_{j}(x) = ρ(x)ζ_{j}(x) + x_{j}ζ(x). Having proved the existence of such a decomposition we see that if x_{j}u = 0 for all j then

u(ϕ) = u(ϕ(0)ρ(x)) + ∑_{j=1}^{n} u(x_{j}ψ_{j}) = cϕ(0), c = u(ρ(x)), (1.58)

i.e. u = cδ(x).

1.7. Fourier transform^{[S.F ourier.transform]}

Our normalization of the Fourier transform will be

Fϕ(ξ) = ∫ e^{−iξ·x}ϕ(x)dx.^{[2.7]} (1.59)

As you all know the inverse Fourier transform is given by

Gψ(x) = (2π)^{−n} ∫ e^{ix·ξ}ψ(ξ) dξ.^{[2.8]} (1.60)

Since it is so important here I will give a proof of this invertibility. First however, let us note some of the basic properties.

Both F and G give continuous linear maps

F, G : S(ℝ^{n}) ⟶ S(ℝ^{n}).^{[2.9]} (1.61)

To see this observe first that the integrals in (1.59) and (1.60) are absolutely convergent:

|Fϕ(ξ)| ≤ ∫ |ϕ(x)|dx ≤ ∫ (1 + |x|^{2})^{−n} dx × sup_{x∈ℝ^{n}}(1 + |x|^{2})^{n}|ϕ(x)|, (1.62)

where we use the definition of S(ℝ^{n}). In fact this shows that sup |Fϕ| < ∞ if ϕ ∈ S(ℝ^{n}). Formal differentiation under the integral sign gives an absolutely convergent integral:

D_{j}Fϕ(ξ) = ∫ D_{ξ_{j}}e^{−ixξ}ϕ(x)dx = ∫ e^{−ix·ξ}(−x_{j}ϕ)dx^{[2.10]}

since sup_{x}(1 + |x|^{2})^{n}|x_{j}ϕ| < ∞. Then it follows that D_{j}Fϕ is also bounded, i.e. Fϕ

is differentiable, and (1.7) holds. This argument can be extended to show that Fϕ is C^{∞},

D^{α}Fϕ(ξ) = F((−x)^{α}ϕ).^{[2.11]} (1.63)

Similarly, starting from (1.59), we can use integration by parts to show that

ξ_{j}Fϕ(ξ) = ∫ e^{−ixξ}ξ_{j}ϕ(x)dx = ∫ e^{−ix·ξ}(D_{j}ϕ)(x)dx

i.e. ξ_{j}Fϕ = F(D_{j}ϕ). Combining this with (1.63) gives

ξ^{α}D_{ξ}^{β}Fϕ = F(D^{α} · [(−x)^{β}ϕ]).^{[2.12]} (1.64)

Since D_{x}^{α}((−x)^{β}ϕ) ∈ S(ℝ^{n}) we conclude

sup |ξ^{α}D_{ζ}^{β}Fϕ| < ∞ ⟹ Fϕ ∈ S(ℝ^{n}). (1.65)

This map is continuous since

\begin{matrix}sup |ξ^{α}D_{ξ}^{β}Fϕ| ≤ C · sup_{x} |(1 + |x|^{2})^{n}D_{x}^{α}[(−x)^{β}ϕ] \\ ⟹ ∥Fϕ∥_{k} ≤ C_{k}∥ϕ∥_{k+2n}, ∀ k. \end{matrix}

The identity (1.64), written in the form

\begin{matrix}F(D_{j}ϕ) = ξ_{j}Fϕ \\ F(x_{j}ϕ) = −D_{ξ_{j}}Fϕ \end{matrix}^{[2.13]} (1.66)

is already the key to the proof of invertibility:

[2.14]

Theorem 1.3. The Fourier transform gives an isomorphism F : S(ℝ^{n}) ⟷ S(ℝ^{n}) with inverse G.

Proof. We shall use the idea of the Schwartz kernel theorem. It is important not to use this theorem itself, since the Fourier transform is a key tool in the (simplest) proof of the kernel theorem. Thus we consider the composite map

G ∘ F : S(ℝ^{n}) ⟶ S(ℝ^{n}) (1.67)

and write down its kernel. Namely

\begin{matrix}K(ϕ) = (2π)^{−n} ∫∫∫ e^{iy·ξ−ix·ξ}ϕ(y, x)dxdξdy \\ ∀ ϕ ∈ S(ℝ_{y}^{n} × ℝ_{x}^{n}) ⟹ K ∈ S′(ℝ^{2n}). \end{matrix}^{[2.15]} (1.68)

The integrals in (1.68) are iterated, i.e. should be performed in the order indicated. Notice that if ψ, ζ ∈ S(ℝ^{n}) then indeed

(1.69) (G · F(ψ))(ζ) = ∫ ζ(y)(2π)^{−n} (∫ e^{iy·ξ} ∫ e^{−ix·ξ}ψ(x)dx dξ) dy dξ dy

= K(ζ ⊠ ψ)^{[2.16]}

so K is the Schwartz kernel of G · F.

The two identities (1.66) translate (with essentially the same proofs) to the conditions on K :

\begin{cases}(D_{x_{j}} + D_{y_{j}})K(x, y) = 0 \\ (x_{j} − y_{j})K(x, y) = 0 \end{cases} j = 1, …, n.^{[2.17]} (1.70)

Next we use the freedom to make linear changes of variables, setting

\begin{matrix}K_{L}(x, z) = K(x, x − z), K_{L} ∈ S′(ℝ^{2n}) \\ i.e. K_{L}(ϕ) = K(ψ), ψ(x, y) = ϕ(x, x − y) \end{matrix}^{[2.18]} (1.71)

where the notation will be explained later. Then (1.70) becomes

D_{x_{j}}K_{L}(x, z) = 0 and z_{j}K_{L}(x, z) = 0 for j = 1, … n^{[2.19]} (1.72)

This puts us in a position to apply the two little lemmas. The first says K_{L}(x, z) = f(z) for some f ∈ S′(ℝ^{n}) and then the second says f(z) = cδ(z). Thus

K(x, y) = cδ(x − y) ⟹ G · F = c Id . (1.73)

It remains only to show that c = 1. That c ≠ 0 is obvious (since F(δ) = 1). The easiest way to compute the constant is to use the integral identity

∫_{−∞}^{∞} e^{−x^{2}}dx = π^{\frac{1}{2}} (1.74)

to show that^{4}

\begin{matrix}F(e^{−|x|^{2}}) = π^{\frac{n}{2}} e^{−|ξ|^{2}/4} \\ ⟹ G(e^{−|ξ|^{2}/4}) = π^{− \frac{n}{2}} e^{−|x|^{2}} \\ ⟹ G · F = Id . \end{matrix} (1.75)

Now (2π)^{n}G is actually the adjoint of F :

∫ ϕ(ζ)\overline{Fψ}(ζ)dζ = (2π)^{n} ∫ (Gϕ) · \overline{ψ}dx ∀ ϕ, ψ ∈ S(ℝ^{n}).^{[1.103]} (1.76)

It follows that we can extend F to a map on tempered distributions

\begin{matrix}F : S′(ℝ^{n}) ⟶ S′(ℝ^{n}) \\ Fu(\overline{ϕ}) = u((2π)^{n}\overline{Gϕ}) ∀ ϕ ∈ S(ℝ^{n}) \end{matrix} (1.77)

Then we conclude

[2.20]

Corollary 1.1. The Fourier transform extends by continuity to an isomorphism

F : S′(ℝ^{n}) ⟷ S′(ℝ^{n}) (1.78)

with inverse G, satisfying the identities (1.66).

Although I have not discussed Lebesgue integrability I assume familiarity with the basic Hilbert space

L^{2}(ℝ^{n}) =

{u : ℝ^{n} ⟶ ℂ; f is measurable and ∫_{ℝ^{n}} |f(x)|dx < ∞} / ∼,

f ∼ g ⟺ f = g almost everywhere.

This also injects by the same integration map (1.102) with S(ℝ^{n}) as a dense subset

S(ℝ^{n}) ↪ L^{2}(ℝ^{p}) ↪ S(ℝ^{n}).

[1.102]

Proposition 1.3. The Fourier transform extends by continuity from the dense subspace S(ℝ^{n}) ⊂ L^{2}(ℝ^{n}), to an isomorphism

F : L^{2}(ℝ^{n}) ⟷ L^{2}(ℝ^{n})

satisfying ∥Fu∥_{L^{2}} = (2π)^{\frac{1}{2} n}∥u∥_{L^{2}}.

4See Problem 1.9.

Proof. Given the density of S(ℝ^{n}) in L^{2}(ℝ^{n}), this is also a consequence of (1.76), since setting ϕ = Fu, for u ∈ S(ℝ^{n}), gives Parseval’s formula

∫ Fu(ζ)\overline{Fv(ζ)} = (2π)^{n} ∫ u(x)\overline{v(x)}dx.

Setting v = u gives norm equality (which is Plancherel’s formula).

An outline of the proof of the density statement is given in the problems below.

1.8. Differential operators^{[S.Differential.operators]}

The simplest examples of the Fourier transform of distributions are immediate consequences of the definition and (1.66). Thus

F(δ) = 1 (1.79)

as already noted and hence, from (1.66),

F(D^{α}δ(x)) = ξ^{α} ∀ α ∈ ℕ_{0}^{n}. (1.80)

Now, recall that the space of distributions with support the point 0 is just:

(1.81) {u ∈ S′(ℝ^{n}); sup(u) ⊂ {0}} = {u = ∑_{finite} c_{α}D^{α}δ}.^{[2.21]}

Thus we conclude that the Fourier transform gives an isomorphism

(1.82) F : {u ∈ S′(ℝ^{n}); supp(u) ⊂ {0}} ⟷ ℂ[ξ] = {polynomials in ξ}.^{[2.22]}

Another way of looking at this same isomorphism is to consider partial differential operators with constant coefficients:

\begin{matrix}P (D) : S(ℝ^{n}) ⟶ S(ℝ^{n}) \\ P (D) = ∑ c_{α}D^{α}. \end{matrix}^{[2.23]} (1.83)

The identity becomes

F(P (D)ϕ)(ξ) = P (ξ)F(ϕ)(ξ) ∀ ϕ ∈ S(ℝ^{n})^{[2.24]} (1.84)

and indeed the same formula holds for all ϕ ∈ S′(ℝ^{n}). Using the simpler notation û(ξ) = Fu(ξ) this can be written

{ \atop P (D)u(ξ)} = P (ξ)û(ξ), P (ξ) = ∑ c_{α}ξ^{α}.^{[2.25]} (1.85)

The polynomial P is called the (full) characteristic polynomial of P (D); of course it determines P (D) uniquely.

It is important for us to extend this formula to differential operators with variable coefficients. Using (1.59) and the inverse Fourier transform we get

P (D)u(x) = (2π)^{−n} ∫∫ e^{i(x−y)·ξ}P (ξ)u(y)dydξ^{[2.26]} (1.86)

where again this is an iterated integral. In particular the inversion formula is just the case P (ξ) = 1. Consider the space

C_{∞}^{∞}(ℝ^{n}) = {u : ℝ^{n} ⟶ ℂ; sup_{x} |D^{α}u(x)| < ∞ ∀ α} (1.87)

the space of C^{∞} function with all derivatives bounded on ℝ^{n}. Of course

S(ℝ^{n}) ⊂ C_{∞}^{∞}(ℝ^{n}) (1.88)

1.9. RADIAL COMPACTIFICATION^{[SECT.RADIAL.COMP ACT IF ICAT ION]} 25

but C_{∞}^{∞}(ℝ^{n}) is much bigger, in particular 1 ∈ C_{∞}^{∞}(ℝ^{n}). Now by Leibniz’ formula

D^{α}(uv) = ∑_{β≤α} \binom{α}{β}D^{β}u · D^{α−β}v (1.89)

it follows that S(ℝ^{n}) is a module over C_{∞}^{∞}(ℝ^{n}). That is,

u ∈ C_{∞}^{∞}(ℝ^{n}), ϕ ∈ S(ℝ^{n}) ⟹ uϕ ∈ S(ℝ^{n}). (1.90)

From this it follows that if

P (x, D) = ∑_{|α|≤m} p_{α}(x)D^{α}, p_{α} ∈ C_{∞}^{∞}(ℝ^{n}) (1.91)

then P (x, D) : S(ℝ^{n}) ⟶ S(ℝ^{n}). The formula (1.86) extends to

P (x, D)ϕ = (2π)^{−n} ∫ e^{i(x−y)·ξ}P (x, ξ)ϕ(y)dydξ^{[2.27]} (1.92)

where again this is an iterated integral. Here

P (x, ξ) = ∑_{|α|≤m} p_{α}(x)ξ^{α} (1.93)

is the (full) characteristic polynomial of P.

1.9. Radial compactification^{[Sect.radial.compactification]}

For later purposes, and general propaganda, consider the quadratic radial compactification of ℝ^{n}. The smooth map

QRC : ℝ^{n} ∋ x ⟼ \frac{x}{(1 + |x|^{2})^{\frac{1}{2}}} ∈ ℝ^{n}^{[1.104]} (1.94)

is 1-1 and maps onto the interior of the unit ball, 𝔹^{n} = {|x| ≤ 1}. Consider the subspace

Ċ^{∞}(𝔹^{n}) = {u ∈ S(ℝ^{n}); supp(u) ⊂ 𝔹^{n}}.^{[1.105]} (1.95)

This is just the set of smooth functions on ℝ^{n} which vanish outside the unit ball. Then the composite (‘pull-back’) map

QRC^{∗} : Ċ^{∞}(𝔹^{n}) ∋ u ⟼ u ∘ QRC ∈ S(ℝ^{n})^{[1.106]} (1.96)

is a topological isomorphism. A proof is indicated in the problems below.

The dual space of Ċ^{∞}(𝔹^{n}) is generally called the space of ‘extendible distributions’ on 𝔹^{n} – because they are all given by restricting elements of S′(ℝ^{n}) to Ċ^{∞}(𝔹^{n}). Thus QRC also identifies the tempered distributions on ℝ^{n} with the extendible distributions on 𝔹^{n}. We shall see below that various spaces of functions on ℝ^{n} take interesting forms when pulled back to 𝔹^{n}. I often find it useful to ‘bring infinity in’ in this way.

Why is this the ‘quadratic’ radial compactification, and not just the radial compactification? There is a good reason which is discussed in the problems below. The actual radial compactification is a closely related map which identifies Euclidean space, ℝ^{n}, with the interior of the upper half of the n-sphere in ℝ^{n+1} :

(1.97) RC : ℝ^{n} ∋ x ⟼ (\frac{1}{(1 + |x|^{2})^{\frac{1}{2}}} , \frac{x}{(1 + |x|^{2})^{\frac{1}{2}}} )

∈ 𝕊^{n,1} = {X = (X_{0}, X′) ∈ ℝ^{n+1}; X_{0} ≥ 0, X_{0}^{2} + |X′|^{2} = 1}^{[1.2.2000.275]} Since the half-sphere is diffeomorphic to the ball (as compact manifolds with boundary) these two maps can be compared – they are not the same. However it is true that RC also identifies S(ℝ^{n}) with Ċ^{∞}(𝕊^{n,1}).

1.10. Problems^{[S.P roblems.1]}

[P 1.1]

Problem 1.1. Suppose ϕ : ℝ^{n} ⟶ ℂ is a function such that for each point x̄ ∈ ℝ^{n} and each k ∈ ℕ_{0} there exists a constant ϵ_{k} > 0 and a polynomial p_{k}(x; x̄) (in x) for which

|ϕ(x) − p_{k}(x; x̄)| ≤ \frac{1}{ϵ_{k}} |x − x̄|^{k+1} ∀ |x − x̄| ≤ ϵ_{k}. (1.98)

Does it follow that ϕ is infinitely differentiable – either prove this or give a counterexample.

[P 1.2]

Problem 1.2. Show that the function u(x) = exp(x) cos[e^{x}] ‘is’ a tempered distribution. Part of the question is making a precise statement as to what this means!

[P 1.3]

Problem 1.3. Write out a careful (but not necessarily long) proof of the ‘easy’ direction of the Schwartz kernel theorem, that any K ∈ S′(ℝ^{n+m}) defines a continuous linear operator

O_{K} : S(ℝ^{m}) ⟶ S′(ℝ^{n}) (1.99)

[with respect to the weak topology on S′(ℝ^{n}) and the metric topology on S(ℝ^{m})] by

O_{K}ϕ(ψ) = K(ψ ⊠ ϕ). (1.100)

[Hint: Work out what the continuity estimate on the kernel, K, means when it is paired with an exterior product ψ ⊠ ϕ.]

[31.1.2000.260]

Problem 1.4. Show that d in (1.6) is a metric on S(ℝ^{n}). [Hint: If ∥ · ∥ is a norm on a vector space show that

\frac{∥u + v∥}{1 + ∥u + v∥} ≤ \frac{∥u∥}{1 + ∥u∥} + \frac{∥v∥}{1 + ∥v∥}.]

[31.1.2000.261]

Problem 1.5. Show that a sequence ϕ_{n} in S(ℝ^{n}) is Cauchy, resp. converges to ϕ, with respect to the metric d in Problem 1.4 if and only if ϕ_{n} is Cauchy, resp. converges to ϕ, with respect to each of the norms ∥ · ∥_{k}.

[31.1.2000.262]

Problem 1.6. Show that a linear map F : S(ℝ^{n}) ⟶ S(ℝ^{p}) is continuous with respect to the metric topology given in Problem 1.4 if and only if for each k there exists N(k) ∈ ℕ a constant C_{k} such that

∥F ϕ∥_{k} ≤ C_{k}∥ϕ∥_{N(k)} ∀ ϕ ∈ S(ℝ^{n}).

Give similar equivalent conditions for continuity of a linear map f : S(ℝ^{n}) ⟶ ℂ and for a bilinear map S(ℝ^{n}) × S(ℝ^{p}) ⟶ ℂ.

1.10. PROBLEMS^{[S.P ROBLEMS.1]} 27

[31.1.2000.265]

Problem 1.7. Check the continuity of (1.12).

[P 1.4]

Problem 1.8. Prove Proposition 1.2. [Hint: It is only necessary to show that if u ∈ S′(ℝ^{n}) is fixed then for any of the open sets in (1.1), B, (with all the ϵ_{j} > 0) there is an element ϕ ∈ S(ℝ^{n}) such that u − T_{ϕ} ∈ B. First show that if ϕ′_{1}, … ϕ′_{p} is a basis for Φ then the set

B′ = {v ∈ S′(ℝ^{n}); |⟨v, ϕ′_{j}⟩| < δ_{j} (1.101)

is contained in B if the δ_{j} > 0 are chosen small enough. Taking the basis to be orthonormal, show that u − ψ ∈ B′ can be arranged for some ψ ∈ Φ.]

[1.2.2000.278]

Problem 1.9. Compute the Fourier transform of exp(−|x|^{2}) ∈ S(ℝ^{n}). [Hint: The Fourier integral is a product of 1-dimensional integrals so it suffices to assume x ∈ ℝ. Then

∫ e^{−iξx}e^{−x^{2}}dx = e^{−ξ^{2}/4} ∫ e^{−(x+ \frac{i}{2} ξ)^{2}}dx.

Interpret the integral as a contour integral and shift to the new contour where x + \frac{i}{2}ξ is real.]

[P 1.5]

Problem 1.10. Show that (1.23) makes sense for ϕ ∈ L^{2}(ℝ^{n}) (the space of (equivalence classes of) Lebesgue square-integrable functions and that the resulting map L^{2}(ℝ^{n}) ⟶ S′(ℝ^{n}) is an injection.

[P 1.6]

Problem 1.11. Suppose u ∈ L^{2}(ℝ^{n}) and that
D_{1}D_{2} ⋯ D_{n}u ∈ (1 + |x|)^{−n−1}L^{2}(ℝ^{n}),

where the derivatives are defined using Problem 1.10. Using repeated integration, show that u is necessarily a bounded continuous function. Conclude further that for u ∈ S′(ℝ^{n})

D^{α}u ∈ (1 + |x|)^{−n−1}L^{2}(ℝ^{n}) ∀ |α| ≤ k + n

⟹ D^{α}u is bounded and continuous for |α| ≤ k.

[1.2.2000.274] (1.102)

[This is a weak form of the Sobolev embedding theorem.]

[P 2.1]

Problem 1.12. The support of a (tempered) distribution can be defined in terms of the support of a test function. For ϕ ∈ S(ℝ^{n}) the support, supp(ϕ), is the closure of the set of points at which it takes a non-zero value. For u ∈ S′(ℝ^{n}) we define

(1.103) supp(u) = O^{∁}, O = ⋃ {O′ ⊂ ℝ^{n} open; supp(ϕ) ⊂ O′ ⟹ u(ϕ) = 0} .

Show that the definitions for S(ℝ^{n}) and S′(ℝ^{n}) are consistent with the inclusion S(ℝ^{n}) ⊂ S′(ℝ^{n}). Prove that supp(δ) = {0}.

28 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM^{[T EMP ERED]}

[P 2.2]

Problem 1.13. For simplicity in ℝ, i.e. with n = 1, prove Schwartz theorem concerning distributions with support the origin. Show that with respect to the norm ∥ · ∥_{k} the space {ϕ ∈ S(ℝ); ϕ(x) = 0 in |x| < ϵ, ϵ = ϵ(ϕ) > 0} (1.104) is dense in {ϕ ∈ S(ℝ); ϕ(x) = x^{k+1}ψ(x), ψ ∈ S(ℝ)} . (1.105) Use this to show that (1.106) u ∈ S′(ℝ), supp(u) ⊂ {0} ⟹ u = ∑_{ℓ, finite} c_{ℓ}D_{x}^{ℓ}δ(x).

[P 2.3]

Problem 1.14. Show that if P is a differential operator with coefficients in C_{∞}^{∞}(ℝ^{n}) then P is local in the sense that (1.107) supp(P u) ⊂ supp(u) ∀ u ∈ S′(ℝ^{n}). The converse of this, for an operator P : S(ℝ^{n}) ⟶ S(ℝ^{n}) where (for simplicity) we assume supp(P u) ⊂ K ⊂ ℝ^{n} (1.108) for a fixed compact set K, is Peetre’s theorem. How would you try to prove this? (No full proof required.)

[P 2.4]

Problem 1.15. (Schwartz representation theorem) Show that, for any p ∈ ℝ the map R_{p}: S(ℝ^{n}) ∋ ϕ ⟼ (1 + |x|^{2})^{−p/2}F^{−1}[(1 + |ξ|^{2})^{−p/2}Fϕ] ∈ S(ℝ^{n})^{[2.28]} (1.109) is an isomorphism and, using Problem 1.11 or otherwise, p ≥ n + 1 + k ⟹ ∥R_{p}ϕ∥_{k} ≤ C_{k}∥ϕ∥_{L^{2}}, ∀ ϕ ∈ S(ℝ^{n}). (1.110) Let R_{p}^{t}: S′(ℝ^{n}) ⟶ S′(ℝ^{n}) be the dual map (defined by T_{p}^{t}u(ϕ) = u(R_{p}ϕ)). Show that R_{p}^{t} is an isomorphism and that if u ∈ S′(ℝ^{n}) satisfies |u(ϕ)| ≤ C∥ϕ∥_{k}, ∀ ϕ ∈ S(ℝ^{n}) (1.111) then R_{p}^{t}u ∈ L^{2}(ℝ^{n}), if p ≥ n + 1 + k, in the sense that it is in the image of the map in Problem 1.10. Using Problem 1.11 show that R_{n+1}(R_{n+1+k}^{t}u) is bounded and continuous and hence that

u = ∑_{|α|+|β|≤2n+2+k} x^{β}D^{α}u_{α,β}^{[2.29]} (1.112)

for some bounded continuous functions u_{α,β}.

[P 2.5]

Problem 1.16. (Schwartz kernel theorem.) Show that any continuous linear operator

T : S(ℝ_{y}^{m}) ⟶ S′(ℝ_{x}^{n})

extends to a continuous linear operator T : (1 + |y|^{2})^{−k/2}H^{k}(ℝ_{y}^{m}) ⟶ (1 + |x|^{2})^{−q/2}H^{q}(ℝ_{x}^{n})

1.10. PROBLEMS^{[S.P ROBLEMS.1]} 29

for some k and q. Deduce that the operator

T̃ = (1 + |D_{x}|^{2})^{(−n−1−q)/2}(1 + |x|^{2})^{q/2} ∘ T ∘ (1 + |y|^{2})^{k/2}(1 + |D|^{2})^{−k/2} :

L^{2}(ℝ^{m}) ⟶ C_{∞}(ℝ^{n})

is continuous with values in the bounded continuous functions on ℝ^{n}. Deduce that T̃ has Schwartz kernel in C_{∞}(ℝ^{n}; L^{2}(ℝ^{m})) ⊂ S′(ℝ^{n+m}) and hence that T itself has a tempered Schwartz kernel.

[1.2.2000.301]

Problem 1.17. Radial compactification and symbols.

[P olyDouble]

Problem 1.18. Series of problems discussing double polyhomogeneous symbols.

30 1. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORM^{[T EMP ERED]}