Hints on Fourier Analysis∗
General Properties of the Fourier Modes
Exercise 1.
(1.1)
i. Notice that eiϑ = cos(ϑ ) + i sin(ϑ ), thus, for n , 0,
fˆ (n )einx + fˆ (−n )e−inx = fˆ (n ) + fˆ (−n ) cos(nx ) + i fˆ (n ) − fˆ (−n ) sin(nx ).
The results follows if we couple the n-th and −n-th modes of the Fourier series of f .
ii. Let n ∈ Z. Then
2π fˆ (n ) =
(1.2)
π
Z
f (x )e−inx dx
−π
Z
x =−y
π
(1.4)
=
Z
=
(1.5)
= 2π fˆ (−n ).
(1.3)
f (−y)einy dy
−π
π
f (y)einy dy
π
iii. Proceed similarly to ii.
iv. Let n ∈ Z be odd. Then
2π fˆ (n ) =
(1.6)
Z
π
f (x )e−inx dx
−π
(1.8)
Z 0
e−inπ
f (y + π )e−iny dy
−2π
Z π
−inπ
=e
f (y)e−iπ dy
(1.9)
= e−inπ 2π fˆ (n ).
x =y+π
=
(1.7)
−π
Since n is odd, e−inπ = −1, and the equality above implies that fˆ (n ) = 0.
v. Assume first that f ∈ R. Then, for n ∈ Z,
Z
2π fˆ (n ) =
(1.10)
f (x )einx dx
Z−ππ
=
(1.11)
π
f (x )einx dx
−π
= 2π fˆ (−n ).
(1.12)
If fˆ (n ) = fˆ (−n ), for every n ∈ Z, then
SN f (x ) = SN f (x ),
(1.13)
for every N ≥ 0, and by [1][Theorem 2.1] f (x ) = f (x ), for every x ∈ [−π, π ], which implies that
f ∈ R.
Exercise 2. Let n ∈ Z. Then
(2.1)
(2.2)
(2.3)
Z π
−inx
ˆ
ˆ
(fk (x ) − f (x )) e
2π |fk (n ) − f (n )| = dx Z −ππ
≤
|fk (x ) − f (x )| |e−inx | dx
−π
Z π
=
|fk (x ) − f (x )| dx.
−π
Since the last bound is independent of n and converges to 0 we obtain the appropriate result.
∗ The document contains material I collected myself while I was running the support classes on Fourier Analysis during
the academic year 2015-2016. Though it might cover several aspects of the module, it should not be considered a
representative of the module’s exam.
1
Decay of the Fourier Modes
Exercise 3. Since the function f is square integrable we have that (see [1, Section 3])
X
(3.1)
|fˆ (n )|2 = kf k2L 2 (T) ,
n ∈Z
which implies that the sequence (fˆ (n ))n ∈Z converges to 0 as |n | → ∞.
Exercise 4. We give a proof for k = 1. For k ≥ 2, the assertion follows in a similar way.
Let n ∈ Z. Then
2π fˆ (n ) =
(4.1)
Z
π
f (x )e−inx dx
−π
!0
e−inx
dx
−in
−π
π
Z π
e−inx 1
= f (x )
+
f 0 (x )e−inx dx
(−in ) −π in −π
=
(4.2)
(4.3)
Z
π
f (x )
1
= 2π
(4.4)
in
fˆ0 (n ),
where the last equality holds because of periodicity. Given that f 0 is bounded, we obtain that
1
|fˆ (n )| ≤ sup f 0 (x ) : x ∈ [−π, π ]
,
|n |
(4.5)
which completes the proof.
To check if fˆ (n ) ∼ o (1/|n |) we should investigate the convergence of |fˆ (n )n | as |n | → ∞. But since
the function f 0 is continuous, it is also square integrable and we can apply Exercise 3.
Exercise 5. Since f is monotone we can prove (exercise!) that there exists a sequence of step
functions {φk }k ∈N such that φk → f uniformly and
Z
1
(5.1)
π
|f (x ) − φk (x )| dx ≤
2π
−π
1
k
,
for all n ∈ N. We can also prove (another exercise!) that there exists M > 0 such that |φˆk (n )| ≤
M/|n |, for all k ∈ N, n ∈ Z. Then, for n ∈ Z,
(5.2)
(5.3)
π
Z
2π |n fˆ (n )| = n
Z−ππ
≤ |n |
−π
(5.4)
≤ 2π
|n |
k
f (x )e−inx dx Z
|f (x ) − φk (x )| dx + |n | π
−π
φk (x )e−inx dx + 2πM,
where the last quantity converges to M as k → ∞ and proves the assertion.
Exercise 6.
(6.1)
i. Let n ∈ N. Substituting x = y + (π/n ) we get
2π fˆ (n ) =
Z
π
f (x )e−inx dx
−π
(6.2)
(6.3)
(6.4)
=
Z
π − nπ
f y+
−π − nπ
= e−iπ
Z
=−
Z
π
π
n
f y+
−π
π
f y+
−π
2
π
n
e−in (y+ n ) dy
π
n
π
e−iny dy
e−iny dy
and writing fˆ (n ) = fˆ (n ) + fˆ (n ) /2 we obtain that
fˆ (n ) =
(6.5)
1
π
Z
4π
π
f (x ) − f x +
n
−π
e−inx dx.
ii. By (i),
|fˆ (n )| ≤
(6.6)
(6.7)
≤
(6.8)
=
Z
1
4π
π
−π
C
π
f (x ) − f x + π dx
n !α
2 |n |
C̃
,
|n |α
where we also use the α-Hölder continuity of f .
iii. We only prove α-Hölder continuity since the other assertion is trivial for a function f with
such a Fourier series. Let h > 0. Then
∞
k
X
−
kα
i
2 x
i
2k (x +h ) e
−e
|f (x ) − f (x + h )| = 2
k =0
∞
X
k 2−kα 1 − ei2 h ≤
(6.9)
(6.10)
k =0
≤
(6.11)
kX
0 −1
2−kα |2k h | +2
|k =0 {z
∞
X
2−αk ,
k =k0
}
I1
| {z }
I2
where k0 ∈ N is the smallest non-negative
integer
such that 2k0 |h | > 1 and in the last
k i
2
h
≤ |2k h |. Notice that
inequality above we exploit the fact that 1 − e
I1 = |h |
(6.12)
2(1−α )k0 − 1
21−α − 1
21−α
(1−α )(k0 −1)
≤ |h |2
(6.13)
!
21−α − 1
(6.14)
≤ |h ||h |−(1−α)
(6.15)
= |h |α
2
1−α
21−α − 1
21−α
21−α − 1
,
where in the second inequality we use that 2(1−α )(k0 −1) ≤ |h |−(1−α ) . We also have that
I2 = 2−k0 α
(6.16)
1
1 − 2−α
1
≤ |h |α
.
1 − 2−α
(6.17)
Combining the bounds on I1 and I2 we obtain α-Hölder continuity for f .
Exercise 7. We can show that (exercise!) there exists a subsequence {εkn }n ∈N of {εn }n ∈N such that
X
(7.1)
|εkn | < ∞.
n ∈N
If we let f (x ) =
(7.2)
P
n ∈N
εkn eikn x (notice that the series converges absolutely) we have that
fˆ (kn ) = εkn .
3
Convergence of the Fourier Series
Exercise 8. By Exercise 4 we see that SN f converges absolutely whereas continuity implies
uniform convergence of SN f to f (see [1][Corollary 2.3]).
Exercise 9. Since f 0 is continuous, Parseval’s identity implies that
X
(9.1)
|fˆ0 (n )|2 < ∞.
n ∈Z
In the proof of Exercise 4 we see that |n fˆ (n )| = |fˆ0 (n )|. Using the Cauchy-Schwartz inequality we
obtain that
2 



X 1  X


X
 |fˆ (n )| ≤ 
  |n fˆ (n )|2  ,
n2
(9.2)
n ,0
n ,0
n ,0
where the last series converges by Parseval’s identity. Hence, uniform convergence of SN f to f is
obtained as in Exercise 8.
Exercise 10. Let ε > 0. There exists δ > 0 such that
(10.1)
|f (x − y) − f (x + )| ≤
(10.2)
|f (x − y − f (x − )| ≤
Using that
(10.3)
Rπ
−π
ε
2
ε
2
, for all y ∈ (0, δ ).
, for all y ∈ (−δ, 0).
Fn (y) dy = 2π and that FN is even we have that
2π f ∗ FN (x ) −
Z
0
f (x + ) − f (x − )
!
2
=
FN (y)(f (x − y) − f (x + )) dy +
π
Z
−π
FN (y)(f (x − y) − f (x + )) dy.
0
Thus
(10.4)
Z −δ
Z 0
ε
f (x + ) − f (x − ) |FN (y)| dy +
|FN (y)| dy
2π f ∗ FN (x ) −
≤ 2A
2
2 −δ
−π
Z δ
Z π
ε
+
|FN (y)| dy + 2A
|FN (y)| dy,
2
0
δ
where A = supx ∈[−π,π ] |f (x )|. The result follows since FN is a good kernel which implies that
Z
|FN (y)| dy → 0.
(10.5)
|y|>δ
Kernel and Integral Computations
Exercise 11. Let n > 0. Then, if we let λ = eix ,
(11.1)
Dn (x ) =
X
λm =
(11.3)
(11.4)
λm +
m =0
m ≤n
(11.2)
n
X
=
=
=
−1
X
λm
m =−n
1 − λn +1
1−λ
+
λ−n − 1
1−λ
λ−n −1/2 − λn +1/2
λ−1/2
sin n + 12 x
sin(x/2)
4
.
λ1/2
Thus, for N ≥ 0, we have that
NFN (x ) =
(11.5)
=
(11.6)
=
(11.7)
(11.8)
=
(11.9)
=
N
−1
X
1
sin(x/2)
1
sin n +
x
2
n =0
N
−1
X
1
2 sin(x/2)2
1
n+
2 sin(x/2) sin
N
−1
X
2 sin(x/2)2
x
2
n =0
1
(cos(nx ) − cos((n + 1)x ))
n =1
1
(1 − cos(Nx ))
2 sin(x/2)2
sin(Nx/2)2
,
sin(x/2)2
where we use the trigonometric identities
2 sin(z ) sin(y) = cos(z − y) − cos(z + y),
(11.10)
1 − cos(z )
sin(z/2)2 =
(11.11)
2
.
Exercise 12. Recall that the n-th Dirichlet kernel is given by
Dn (x ) =
(12.1)
and that
Rπ
−π
sin n + 12 x
sin (x/2)
Dn (x ) dx = 2π. Then
(12.2)
2π =
Z
π
sin n + 12 x
x/2
−π
dx +
π
Z
g(x ) sin
n+
−π
1
2
x dx,
1
where g(x ) = sin(1x/2) − x/2
. Notice that there exists (why?) a continuous extension g̃ of g on [−π, π ]
and by the Lebesgue–Riemmann Lemma
Z
π
1
g̃(x ) sin
(12.3)
n+
x dx → 0,
2
−π
which implies that
Z
π
lim
(12.4)
n →∞
sin n + 21 x
x/2
−π
dx = 2π.
We also notice that
Z
π
(12.5)
−π
sin n + 21 x
x/2
dx = 2
Z
π
sin n + 21 x
x/2
0
=2
(12.6)
Z (n + 1 )π
2
2 sin(x )
x
0
thus
(12.7)
lim
n →∞
Z (n + 1 )π
2
2 sin(x )
x
0
dx =
which implies that
Z
(12.8)
0
∞
sin(x )
x
5
dx =
π
2
.
π
2
,
dx
dx,
Heisenberg’s Uncertainty Principle
Exercise 13. Using integration by parts we have that
∞
Z
1=
(13.1)
|psi (x )|2 dx
Z−∞
∞
d |ψ(x )|2 dx
dx
Z−∞
∞
=
x2Re ψ0 (x )ψ(x ) dx,
=
(13.2)
(13.3)
x
−∞
thus we get the bound
∞
Z
|xψ(x )ψ0 (x )| dx
1≤2
(13.4)
−∞
Z
2
≤2
(13.5)
! 21 Z
∞
2
∞
0
x |ψ(x )| dx
2
! 12
|ψ (x )| dx
−∞
,
−∞
where we also use the Cauchy–Schwartz inequality. Using Plancherel identity and the fact that
ψ0 (x ) 7→ 2π iξ ψ̂(ξ ) we obtain that
kψ0 kL 2 = k2π i(·)ψ̂kL 2 .
(13.6)
Combining all the above we get
∞
Z
1
(13.7)
16π 2
∞
Z
x 2 |ψ(x )|2 dx
≤
−∞
ξ 2 |ψ̂(ξ )|2 dξ.
−∞
If equality holds, then we should have equality when we use the Cauchy–Schwartz inequality (see
(13.5)), which implies that
ψ0 (x ) = Cxψ(x ),
(13.8)
for some C ∈ R. If we solve the above differential equation we obtain solutions of the form
2
ψ(x ) = Ae−Bx , for A, B satisfying the appropriate relations.
Exercise 14.
i. Assume first that hf, f i = 1. By Exercise 14,
Z
1
(14.1)
16π 2
(14.2)
≤2
Z
=
∞
2
2
! 12 Z
∞
x |f (x )| dx
! 21
−∞
−∞
∞
x |f (x )| dx +
2
ξ |fˆ (ξ )|2 dxi
2
Z
∞
ξ 2 |fˆ (ξ )|2 dξ,
2
−∞
−∞
where we use 2ab ≤ (a + b ). Using integration by parts we have that
2
Z
∞
(14.3)
−∞
2
∞ Z ∞ Z ∞
2
df
df
(x ) dx +
(x )f (x ) +
x 2 |f (x )|2 dx.
−∞
dx
d
x
−∞
−∞
|
{z
}
Lf (x )f (x ) dx = −
=0
Combining the above together with the fact that kf 0 kL 2 = k2π i(·)fˆ kL 2 (see the solution of
Exercise 13 for an explanation of that identity) we obtain that
(14.4)
hLf, f i ≥ 1.
For a general f˜ the conclusion arises using the above inequality for f = f˜ /kf˜ kL 2 .
ii. Part (a.) follows easily by using an integration by parts formula. For (b.),
(14.5)
hAf, Af i = hf, A∗ Af i
(14.6)
= hA∗ Af, f i
(14.7)
= hA∗ Af, f i,
6
where we use (a.) and the fact that hAf, Af i ∈ R. For (c.),
(14.8)
!
df
df
d2 f
(
x
)
−
f
(
x
)
+
x
(
x
)
+x
(x ) + x 2 f (x )
d2 x
dx
dx
A∗ Af (x ) = −
= Lf (x ) − f (x ).
(14.9)
Combining (a.) and (b.) we get an alternative proof of i.
iii. Notice that
Π(t ) := hAt∗ At , f i
(14.10)
= kf 0 k2L 2 + t kf k2L 2 + t 2 kTf k2L 2 ,
(14.11)
where Tf (x ) = xf (x ). Since Π(t ) ≥ 0, for every t ∈ R, we should have
kf k4L 2 ≥ 4kf 0 kL 2 kTf kL 2 ,
(14.12)
which completes the proof.
Fourier Transform and Fourier Series
Exercise 15. i. Let g : [−L/2, L/2] → R such that g(x ) = f (x ) for every x ∈ [−L/2, L/2]. Then g
is a continuous function on [−L/2, L/2] such that g(−L/2) = g(L/2). Then
ĝ(n ) =
(15.1)
=
(15.2)
=
(15.3)
Z
1
L/2
g(x )e−2π inx/L dx
L
−L/2
Z
1
L/2
f (x )e−2π inx/L dx
L
1
−L/2
fˆ (n/L ).
L
Since the function f is of moderate decrease we have that
1 fˆ (n/L ) . 1
(15.4)
!
1
,
L 1 + |n |2 /L 2
L
which implies the absolute convergence of the Fourier series of g. Thus
f (x ) =
(15.5)
1X
L
fˆ (n/L )e2π inx/L ,
n ∈Z
uniformly in x.
ii. Notice that, for every δ > 0,
(15.6)
δ
X
F (nδ ) = δ
n ∈Z
X
X
F (nδ ) + lim δ
N →∞
|n |<1/δ
F (nδ ),
1/δ ≤|n |<N/δ
where
(15.7)
lim+ δ
δ →0
X
F (nδ ) =
Z
F (ξ ) dξ,
|ξ |<1
|n |<1/δ
since F is Riemann integrable and the sum is actually a Riemann sum. Moreover
(15.8)
δ
X
F (nδ ) .
1/δ ≤|n |<N/δ
1
Z
δ
.1−
(15.9)
7
N/δ
1/δ
1
N
,
1
x2
dx
which implies uniformly boundedness with respect to N, δ. Hence
(15.10)
X
lim+ lim δ
δ →0 N →∞
X
F (nδ ) = lim lim+ δ
N →∞ δ →0
1/δ ≤|n |<N/δ
F (nδ )
1/δ ≤|n |<N/δ
Z
= lim
F (ξ ) dξ
N →∞ 1<|ξ |<N
Z
=
F (ξ ) dξ.
(15.11)
(15.12)
|ξ |>1
Combining the above we obtain that
(15.13)
lim δ
δ →0+
X
F (nδ ) =
Z
F (ξ ) dξ +
|ξ |<1
n ∈Z
Z
F (ξ ) dξ.
|ξ |>1
iii. By i, for every δ < 1/2M, we have that
f (x ) = δ
(15.14)
X
fˆ (nδ )e2π ixnδ .
n ∈Z
Using ii we see that
Z
∞
(15.15)
−∞
fˆ e2π ixξ dξ = lim+ δ
δ →0
X
fˆ (nδ )e2π ixnδ
n ∈Z
(15.16)
= lim+ f (x )
(15.17)
= f (x ),
δ →0
which completes the proof.
References
[1] E. Stein, R. Shakarchi. Fourier Analysis. Princeton University Press, 2003.
[2] A. Giannopoulos. Ανάλυση Fourier και Ολοκλήρωµα Lebesgue (Greek). Lecture Notes,
University of Athens, Department of Mathematics, 2012.
Pavlos Tsatsoulis
University of Warwick
Coventry, UK
p.tsatsoulis@warwick.ac.uk
8