10.5.3 Fourier Transforms of Functions In L 2
Consider Ff and F − 1 f for f ∈ L 2 (ℝ n ). First note that the formula given for Ff and F − 1 f when
f ∈ L 1
will not work for
f ∈ L 2 (
ℝ n ) unless
f is also in
L 1 (
ℝ n ). Recall that
a + ib =
a − ib .
Theorem 10.5.13 For ϕ ∈G , || Fϕ || 2 = || F − 1 ϕ || 2 = || ϕ || 2 .
Proof: First note that for ψ ∈G ,
-- ------- −1 -- -----
F (ψ ) = F −1(ψ ),F (ψ) = F (ψ ). (10.10)
(10.10)
This follows from the definition. For example,
-- −n∕2∫ −it⋅x--
F ψ (t) = (2π) ℝn e ψ (x )dx
--------∫-------------
= (2π)−n∕2 eit⋅xψ(x)dx
ℝn
Let
ϕ,ψ ∈G . It was shown above that
∫ ∫
n(Fϕ)ψ(t)dt = n ϕ(F ψ)dx.
ℝ ℝ
Similarly,
∫ ∫
−1 −1
ℝn ϕ(F ψ)dx = ℝn(F ϕ)ψdt. (10.11)
(10.11)
Now, 10.10 - 10.11 imply
∫ ∫ -- ∫ -------- ∫ ---
|ϕ |2dx = ϕϕdx = ϕF −1(F ϕ)dx = ϕF(F ϕ)dx
ℝn ∫ℝn --- ℝn ∫ ℝn
= Fϕ(F ϕ)dx = |Fϕ |2dx.
ℝn ℝn
Similarly
Lemma 10.5.14 Let f ∈ L 2
and let ϕ k → f in L 2 where ϕ k ∈G . (Such a sequence exists
because of density of G in L 2 .) Then Ff and F − 1 f are both in L 2 and the following limits take
place in L 2 .
lim F (ϕ ) = F (f), lim F−1(ϕ ) = F−1(f).
k→ ∞ k k→ ∞ k
Proof: Let ψ ∈G be given. Then
∫
Ff (ψ) ≡ f (F ψ) ≡ f (x)F ψ(x)dx
∫ ℝn ∫
= kli→m∞ ℝn ϕk(x)Fψ (x)dx = lk→im∞ ℝn F ϕk(x)ψ (x )dx.
Also by Theorem
10.5.13 k =1 ∞ is Cauchy in
L 2 and so it converges to some
h ∈ L 2 .
Therefore, from the above,
which shows that F
∈ L 2 and
h =
F . The case of
F − 1 is entirely similar.
■
Since Ff and F − 1 f are in L 2
, this also proves the following theorem.
Theorem 10.5.15 If f ∈ L 2 (ℝ n ), Ff and F − 1 f are the unique elements of L 2
such that for all
ϕ ∈G ,
∫ ∫
ℝn F f(x)ϕ(x )dx = ℝn f(x)F ϕ(x )dx, (10.12)
(10.12)
∫ ∫
n F −1f(x)ϕ(x)dx = n f(x )F −1ϕ(x)dx. (10.13)
ℝ ℝ
(10.13)
Theorem 10.5.16 (Plancherel)
||f||2 = ||F f||2 = ||F −1f||2. (10.14)
(10.14)
Proof: Use the density of G in L 2
to obtain a sequence,
converging to
f in
L 2 . Then by
Lemma
10.5.14
||F f||2 = lim ||F ϕk||2 = lim ||ϕk||2 = ||f||2 .
k→ ∞ k→ ∞
Similarly,
The following corollary is a simple generalization of this. To prove this corollary, use the following
simple lemma which comes as a consequence of the Cauchy Schwarz inequality.
Lemma 10.5.17 Suppose f k → f in L 2
and g k → g in L 2 . Then
∫ ∫
lim f g dx = fgdx
k→ ∞ ℝn k k ℝn
Proof:
|∫ ∫ | |∫ ∫ |
|| f g dx− f gdx||≤ || f g dx− f gdx||+
| ℝn k k ℝn | | ℝn k k ℝn k |
|∫ ∫ |
|| fkgdx − fgdx||
| ℝn ℝn |
≤ ||fk||2||g− gk||2 + ||g||2||fk − f||2.
Now
2 is a Cauchy sequence and so it is bounded independent of
k . Therefore, the above expression is
smaller than
ε whenever
k is large enough.
■
Corollary 10.5.18 For f,g ∈ L 2 (ℝ n ),
∫ ∫ --- ∫ -----
fgdx = F fF gdx = F−1fF −1gdx.
ℝn ℝn ℝn
Proof: First note the above formula is obvious if f,g ∈G . To see this, note
---------------------
∫ --- ∫ ---1---∫ −ix⋅t
ℝn Ff Fgdx = ℝn Ff (x)(2π)n∕2 ℝn e g(t)dtdx
∫ ∫ ∫
= ---1--- eix⋅tFf (x)dxg-(t)dt = (F−1 ∘F) f (t)g(t)dt
ℝn (2π)n∕2 ℝn ℝn
∫ ----
= n f (t)g(t)dt.
ℝ
The formula with
F − 1 is exactly similar.
Now to verify the corollary, let ϕ k → f in L 2
and let
ψ k → g in
L 2 . Then by Lemma
10.5.14
∫ ∫ ∫ ∫
--- ---- --- -
ℝn F fFgdx = kli→m∞ ℝn F ϕkF ψkdx = kli→m∞ ℝn ϕkψkdx = ℝn fgdx
A similar argument holds for F − 1 . ■
How does one compute Ff and F − 1 f ?
Theorem 10.5.19 For f ∈ L 2 (ℝ n ), let f r = f X E r where E r is a bounded measurable set with E r ↑ ℝ n .
Then the following limits hold in L 2
.
−1 −1
Ff = rli→m∞ F fr,F f = rl→im∞ F fr.
Proof: || f − f r || 2 → 0 and so || Ff − Ff r || 2 → 0 and || F − 1 f − F − 1 f r || 2 → 0 by Plancherel’s Theorem.
■
What are Ff r and F − 1 f r ? Let ϕ ∈G
∫ ∫
n Ffrϕdx = n frFϕdx
ℝ ℝ n∫ ∫
= (2π )− 2 fr(x )e−ix⋅yϕ (y)dydx
∫ ℝn ∫ℝn
= [(2π)− n2 fr(x)e− ix⋅ydx]ϕ(y)dy.
ℝn ℝn
Since this holds for all
ϕ ∈G , a dense subset of
L 2 (
ℝ n ), it follows that
n ∫
Ffr(y) = (2π)−2 fr(x)e−ix⋅ydx.
ℝn
Similarly
∫
F −1f (y) = (2π)− n2 f (x )eix⋅ydx.
r ℝn r
This shows that to take the Fourier transform of a function in L 2
, it suffices to take the limit as
r →∞ in
L 2 of (2
π )
− ∫
ℝ n f r (
x )
e − i x ⋅ y dx . A similar procedure works for the inverse Fourier
transform.
Note this reduces to the earlier definition in case f ∈ L 1
. Now consider the convolution of a
function in
L 2 with one in
L 1 .
Theorem 10.5.20 Let h ∈ L 2
and let f ∈ L 1 . Then h ∗ f ∈ L 2 ,
− 1 n∕2 −1 − 1
F (h∗f ) = (2π) F hF f,
n∕2
F (h∗f ) = (2π) F hFf,
and
||h∗ f||2 ≤ ||h||2 ||f||1. (10.15)
(10.15)
Proof: An application of Minkowski’s inequality yields
( )
∫ (∫ )2 1∕2
n n |h(x − y)||f (y)|dy dx ≤ ||f||1||h ||2. (10.16)
ℝ ℝ
(10.16)
Hence ∫
dy < ∞ a.e.
x and
is in L 2
. Let
E r ↑ ℝ n , m < ∞ . Thus,
hr ≡ XE h ∈ L2(ℝn)∩ L1 (ℝn ),
r
and letting ϕ ∈G ,
∫
≡ (hr ∗ f)(Fϕ)dx
∫ ∫ ∫
= (2π)−n∕2 hr (x − y)f (y )e− ix⋅tϕ (t)dtdydx
∫ ∫ ( ∫ )
= (2π)−n∕2 hr(x− y) e− i(x−y)⋅tdx f (y)e−iy⋅tdyϕ(t)dt
∫
n∕2
= (2π) Fhr (t)F f (t)ϕ(t)dt.
Since
ϕ is arbitrary and
G is dense in
L 2 ,
F (hr ∗ f) = (2π)n∕2F hrFf.
Now by Minkowski’s Inequality, h r ∗ f → h ∗ f in L 2
and also it is clear that
h r → h in
L 2 ; so, by
Plancherel’s theorem, you may take the limit in the above and conclude
F (h∗ f) = (2π)n∕2F hF f.
The assertion for F − 1 is similar and 10.15 follows from 10.16 . ■