LECTURE 3, 18.155, 15 SEPTEMBER 2011
(1) Since I will need them later, let me start with a little disussion
of ‘bump’ functions. First, the function
(
exp( |x|21−1 ) |x| < 1
(1)
f (x) =
∈ S(Rn ).
0
|x| ≥ 1
Since this function vanishes in |x| ≥ 1 be definition, the only
issue is its smoothness. It is smoot in |x| < 1 since |x|2 is
1
smooth and exp( t−1
) is smooth in t in t < 1. Moreover, as t ↑ 1,
1
exp( t−1 ) → 0 rapidly, since the argument of the exponential is
tending to −∞. Differentiating f (x) gives a mess but it is, by
induction, of the form
∂ α f (x) =
(2)
1
Pα (x)
exp( 2
)
2|α|
− 1)
|x| − 1
|x|2
1
where Pα is a polynomial. A product such as (t−1)−2k exp( t−1
)→
0 as t ↑ 1 so all the derivatives converge to zero at |x| = 1. It
follows from this that f ∈ S(Rn ).
(2) We can turn this into a true bump function by considering a
complementary function such as
(
1
exp( 1−4|x|
|x| > 1/2
2)
(3)
g(x) =
∈ C ∞ (Rn )
0
|x| ≤ 1/2
by the same argument. Since f ≥ 0, g ≥ 0 and f > 0 in |x| < 1,
g > 0 in |x| > 1/2, f + g > 0 so 1/(f + g) ∈ C ∞ (Rn ). It follows
that
(4)
χ(x) =
f (x)
1
∈ S(Rn ), χ(x) = 0 in |x| ≥ 1 and χ(x) = 1 in |x| < .
f (x) + g(x)
2
(3) Now we use this to prove a ‘division property’.
Lemma 1. If φ ∈ S(Rn ) then there exist φj ∈ S(Rn ) such that
(5)
2
φ(x) = φ(0) exp(−|x| ) +
n
X
j=1
1
xj φj (x).
2
LECTURE 3, 18.155, 15 SEPTEMBER 2011
If you want a little exercise, show (you can use the proof
here) that one can actually find continuous linear maps Tj :
S(Rn ) −→ S(Rn ) such that (5) holds with φj = Tj φ.
(4) To prove this lemma we can start by using Taylor’s theorem. In
fact let me do this constructively so just consider the obvious
identity which is just the Fundamental Theorem of Calculus
again:
Z 1
d
(6)
φ(x) = φ(0) +
φ(sx) ds.
ds
0
(7)
Expanding out the derivative, this becomes
Z 1
X
φ(x) = φ(0) +
xj φ̃j (x), φ̃j (x) =
∂j φ(sx)ds.
0
j
(8)
The problem here is that φ̃j ∈ C ∞ (Rn ) from properties of the
integral, but there is no reason at all why they should decay at
infinity, so they are not in S(Rn ). So, instead, we just cut them
off and write
X φ(x) = φ(0)χ(x) +
xj χ(x)φ̃j (x) + φ0 (x)
j
(9)
(10)
where χ is the cut-off constructed above. It follows that φ0 = 0
in |x| < 21 and φ0 ∈ S(Rn ) – the latter because all the other
terms are in S(Rn ) and the former because (7) holds in |x| < 12 .
So, now consider the function
(
|x|−2 ψ 0 (x) |x| > 41
ψ(x) =
∈ S(Rn ).
0
|x| ≤ 41
It is smooth since the first definition makes the function zero in
|x| < 21 . So, now we can replace (8) by
X φ(x) = φ(0)χ(x) +
xj χ(x)φ̃j (x) + xj ψ .
j
This is what we want, except that in (3) χ is replaced by
exp(−|x|2 ). However, this is not a problem since
(11)
µ = χ(x) − exp(−|x|2 ) ∈ S(Rn ), µ(0) = 0.
Then applying (10) to µ gives some more terms and leads us to
(5).
LECTURE 3, 18.155, 15 SEPTEMBER 2011
3
(5) Now, we will normalize the Fourier transform as
Z R
Z R
Z
−ix·ξ
e−ix·ξ φ(x)dx1 . . . dxn
...
e
φ(x)dx = lim
(12) φ̂(ξ) =
R→∞
Rn
−R
−R
thinking of it as an ‘improper iterated Riemann integral’.
(6) Before the limit, everything is smooth since we can differentiate
under Riemann integrals (when the integrand is smooth). We
find for instance that
(13)
Z R
Z R
Z R
Z R
−ix·ξ
α
e−ix·ξ (−ix)α φ(x)dx1 . . . dxn ,
...
e
φ(x)dx1 . . . dxn =
∂ξ
...
−R
−R
−R
−R
α
\α φ.
∂ = (−ix)
Now, xα φ ∈ S(Rn ) so we can estimate as before and get
|∂ξα φ̂(ξ)| ≤ Ckφ|2n+2+|α| .
(14)
(15)
Similarly, integration by parts is possible with boundary terms
which are rapidly decreasing, so disappear in the limit as R →
∞ and it follows that
Z
β
e−ix·ξ (−i)|β| ∂ β φ(x)dx.
ξ φ̂(ξ) =
Rn
(7) Combining these two identies, we see that ξ β ∂ α φ̂ is the Fourier
transform of (−i)|α|+|β| ∂ β (xα φ(x)) so is bounded and continuous. It follows that φ̂ ∈ S(Rn ) and that
kφ̂kN ≤ CN kφk2n+2+N
(16)
although the order is by no means optimal. This shows that
(17)
F : S(Rn ) 3 φ 7−→ φ̂ ∈ S(Rn )
is a continuous linear map.
(8) To examine its invertibility we consider
the element of S 0 (Rn )
R
which corresponds to composing with F. That is, we look at
Z
n
(18)
S(R ) 3 φ −→ φ̂(ξ)dξ.
(19)
We can insert (5) into this, to see that
Z
Z
\ 2 )dξ
φ̂(ξ)dξ = Cφ(0), C = exp(−|x|
(20)
since the other terms are
XZ
XZ
\
xj φj (x) =
i∂j φbj = 0.
j
j
4
LECTURE 3, 18.155, 15 SEPTEMBER 2011
(9)
Lemma 2. The constant in (19) is C = (2π)n .
(21)
Proof. Let us work out the Fourier transform
Z
e−ix·ξ exp(−|x|2 )dx.
(22)
In fact the integral is the product of the corresponding 1-dimensional
integrands, so it suffices to work out
Z
exp(−ixξ − x2 )dx.
F (ξ) =
R
(23)
Certainly this (improper Riemann) integral is rapidly convergent. For each ξ ∈ R we can complete the square and see that
Z
2
F (ξ) = exp(−ξ /4)G(ξ), G(ξ) =
exp(−(x + iξ/2)2 )dx.
R
In fact, G(ξ) is a constant. This can be seen using contour
deformation and Cauchy’s theorem, or by real variable methods.
The rapid convergence and smoothness of the integrand mean
that G(ξ) is differentiable and we can compute its derivative
(24)
Z
Z
dG(ξ)
i d
2
= (−i(x+iξ)) exp(−(x+iξ/2) )dx = (
exp(−(x+iξ/2)2 )dx = 0.
dξ
2
dx
R
R
(25)
Then the standard integral is
Z
√
exp(−x2 )dx = π.
G(0) = G(ξ) =
R
Thus we have computed the Fourier transform of the Gaussian:(26)
(27)
n
F(exp(−|x|2 ) = π 2 exp(−|ξ|2 /4) in S(Rn ).
Now the constant in (19) is the integral of this Fourier transform. Applying the same formula after a change of variables we
see that
Z
√
exp(−|ξ|2 /4) = 2 π
R
which proves the Lemma.
(10) Thus we have finally shown that
Z
(28)
φ̂ = (2π)n φ(0).
Rn
LECTURE 3, 18.155, 15 SEPTEMBER 2011
5
(29)
From this and a little more manipulation we see that
Z
n
n
−n
G : S(R ) −→ S(R ), (Gφ)(ξ) = (2π)
eix·ξ φ(x)
(30)
is a two-sided continuous inverse to F.
Indeed, if φ ∈ S(Rn ) then ψ(x) = φ(x + y), y ∈ Rn is also an
element of S(Rn ) and it has Fourier transform
Z
ψ̂(ξ) = e−ix·ξ φ(x + y) = eiy·ξ φ̂(ξ).
(31)
Applying (28) we deduce that
Z
Z
ψ̂ = eiy·ξ φ̂(ξ)dξ = (2π)n ψ(0) = (2π)n φ(y).
This is the statement GF = Id . Note that the mapping property
in (29) follows from the corresponding statement for F since
Gφ(ξ) = (2π)−n φ̂(−ξ).
Similarly the identity FG = Id follows by changing signs,
or you can just observe that F must be a bijection, since its
injectivity has been established and its surjectivity follows from
that of G.