10.5.2 Fourier Transforms of Functions In L^{1}
First suppose f ∈ L^{1}
. As mentioned, you can think of it as being in
G^{∗} and so one can take its Fourier
transform as described above. However, since it is in
L^{1}, there is a natural way to define its Fourier
transform. Do the two give the same thing?
Theorem 10.5.10 Let f ∈ L^{1}
. Then Ff =
∫
_{ℝn}gϕdt where
( )n∕2∫
g(t) = 1 e−it⋅xf (x )dx
2π ℝn


and F^{−1}f
=
∫
_{ℝn}gϕdt where g =
^{n∕2} ∫
_{ℝn}e^{it⋅x}fdx. In short,
∫
F f(t) ≡ (2π)−n∕2 e−it⋅xf (x)dx,
ℝn


∫
F− 1f (t) ≡ (2π)−n∕2 n eit⋅xf(x)dx.
ℝ


Proof: From the definition and Fubini’s theorem,
∫ ∫ ( ) ∫
1 n∕2 −it⋅x
Ff (ϕ) ≡ ℝn f (t)F ϕ(t)dt = ℝn f (t) 2π ℝn e ϕ(x)dxdt
∫ ( ( )n∕2∫ )
= 1 f (t)e− it⋅xdt ϕ(x)dx.
ℝn 2π ℝn
Since
ϕ ∈G is arbitrary, it follows from Theorem
10.5.8 that
Ff is given by the claimed formula. The
case of
F^{−1} is identical.
■
Here are interesting properties of these Fourier transforms of functions in L^{1}.
Theorem 10.5.11 If f ∈ L^{1}
and f_{k} −f_{1} → 0
, then Ff_{k} and F^{−1}f_{k} converge uniformly to Ff
and F^{−1}f respectively. If f ∈ L^{1}, then F^{−1}f and Ff are both continuous and bounded. Also,
lim F−1f(x) = lim Ff(x) = 0. (10.7)
x→ ∞ x→ ∞
 (10.7) 
Furthermore, for f ∈ L^{1}
both Ff and F^{−1}f are uniformly continuous.
Proof: The first claim follows from the following inequality.
∫
F fk(t) − Ff (t) ≤ (2π)−n∕2 e−it⋅xfk(x)− e−it⋅xf(x)dx
∫ℝn
= (2π)−n∕2 f (x)− f (x)dx
ℝn k
= (2π)−n∕2f − fk .
1
which a similar argument holding for
F^{−1}.
Now consider the second claim of the theorem.
∫
′ −n∕2 −it⋅x −it′⋅x 
Ff (t)− Ff (t ) ≤ (2π) ℝn e − e f(x)dx


The integrand is bounded by 2
, a function in
L^{1} and converges to 0 as
t^{′}→ t and
so the dominated convergence theorem implies
Ff is continuous. To see
Ff is uniformly
bounded,
∫
Ff (t) ≤ (2π)− n∕2 n f (x)dx < ∞.
ℝ


A similar argument gives the same conclusions for F^{−1}.
It remains to verify 10.7 and the claim that Ff and F^{−1}f are uniformly continuous.
 
 −n∕2∫ − it⋅x 
Ff (t) ≤ (2π) ℝn e f(x)dx


Now let ε > 0 be given and let g ∈ C_{c}^{∞}
such that
^{−n∕2}_{1} < ε∕2. Then
∫
F f (t) ≤ (2π)−n∕2 f(x)− g(x)dx
ℝn
 −n∕2 ∫ −it⋅x 
+(2π) ℝn e g(x )dx
 ∫ 
≤ ε∕2 + (2π)−n∕2 e−it⋅xg(x)dx.
ℝn
Now integrating by parts, it follows that for
_{∞}≡ max
> 0
 ∫ n   
F f (t) ≤ ε∕2+ (2π)−n∕2 1 ∑ ∂g(x)dx (10.8)
t∞ ℝnj=1 ∂xj  
 (10.8) 
and this last expression converges to zero as
_{∞}→∞. The reason for this is that if
t_{j}≠0, integration by
parts with respect to
x_{j} gives
∫ ∫
(2π)−n∕2 e−it⋅xg(x)dx = (2π)−n∕21 e−it⋅x∂g(x)dx.
ℝn − itj ℝn ∂xj


Therefore, choose the j for which
_{∞} =
and the result of
10.8 holds. Therefore, from
10.8, if
_{∞}
is large enough,
< ε. Similarly, lim
_{t
→∞}F^{−1} = 0. Consider the claim about uniform
continuity. Let
ε > 0 be given. Then there exists
R such that if
_{∞} > R, then
< . Since
Ff is
continuous, it is uniformly continuous on the compact set
^{n}. Therefore, there exists
δ_{1} such
that if
_{
∞} < δ_{1} for
t^{′},t ∈^{n}, then
F f (t)− F f (t′) < ε∕2. (10.9)
 (10.9) 
Now let 0 < δ < min
and suppose
_{
∞} < δ. If both
t,t^{′} are contained in
^{n}, then
10.9
holds. If
t ∈^{n} and
t^{′}^{n}, then both are contained in
^{n} and so this
verifies
10.9 in this case. The other case is that neither point is in
^{n} and in this case,
Ff (t)− F f (t′) ≤ Ff (t)+ F f (t′)
ε ε
< 2 + 2 = ε.■
There is a very interesting relation between the Fourier transform and convolutions. Recall
∫
f ∗ g(x) ≡ f (x− y)g (y )dy
ℝn


and part of the problem is in showing that this even makes sense. This is dealt with in the following
theorem.
Theorem 10.5.12 Let f,g ∈ L^{1}(ℝ^{n}). Then f ∗ g ∈ L^{1} and F(f ∗ g) =
^{n∕2}FfFg.
Proof: Consider
∫ ∫
n n f (x − y)g(y)dydx.
ℝ ℝ


The function,
→ is Lebesgue measurable and so by Fubini’s theorem,
∫ ∫ ∫ ∫
ℝn ℝn f (x − y)g(y)dydx = ℝn ℝn f (x − y )g(y)dxdy = f1 g1 < ∞.


It follows that for a.e. x, ∫
_{ℝn}
dy < ∞ and for each of these values of
x, it follows that
∫
_{ℝn}fgdy exists and equals a function of
x which is in
L^{1},f ∗ g. Now
∫
F(f ∗g)(t) ≡ (2π)−n∕2 e−it⋅xf ∗ g(x)dx
∫ ∫ ℝn
−n∕2 −it⋅x
= (2π) ℝn e ℝn f (x− y)g (y )dydx
−n∕2 ∫ −it⋅y ∫ −it⋅(x−y)
= (2π) n e g(y) n e f (x− y)dxdy
n∕2 ℝ ℝ
= (2π) F f (t)F g(t). ■
There are many other considerations involving Fourier transforms of functions in L^{1}
. Some others
are in the exercises.