Homework 9 - Math 321, Spring 2015
1. Let f be a differentiable 2π-periodic function with continuous first derivative. Is f the uniform limit of its partial Fourier sums?
Solution. Yes. Using integration by parts and periodicity of f ,
π
Z π
Z
1
1
ik π
0
−ikt
−ikt
0
[
f (k) =
f (t)e dt =
f (t)e +
f (t)e−ikt dt
2π −π
2π
2π −π
−π
= ik fˆ(k)
By the Cauchy-Schwarz inequality,
X
2 X X
2
1
2 [
[
k |f (k)|
|f (k)| ≤
k 2 k∈Z
k∈Z
k∈Z
P
The series k∈Z k12 is convergent. By Parseval’s identity and continuity of f 0 ,
X
X
2
2
0 (k)| = kf 0 k2 < ∞.
k 2 |f[
(k)| =
|f\
2
k∈Z
k∈Z
P
Hence k |fb(k)| < ∞. Since f is continuous, it follows from a result proved in class that its
partial fourier sum converges uniformly to f .
2. Determine whether the following statement is true or false: If f : R → R is 2π-periodic and
Riemann-integrable on [−π, π], then ||f − f ||2 → 0 as → 0. Here f denotes the translated
function f (x) = f (x + ).
Solution. We employ a density argument. If g is assumed to be continuous and 2π-periodic
on [−π, π], it follows from uniform continuity (write out a proof of this!) that
(1)
lim sup g (x) − g(x) = 0, hence ||g − g||2 ≤ ||g − g||∞ → 0,
→0 x∈[−π,π]
so the result is verified for continuous functions.
Now let f be a general 2π-periodic Riemann integrable function on [−π, π]. Then by
problem 4 of HW set 8, for any given κ > 0, there exists a continuous and 2π-periodic
function g on [−π, π] such that ||f − g||2 < κ2 . In particular, this means that ||f − g ||2 < κ2 .
Therefore, using the first part of the argument, namely (1), we find that
||f − f ||2 ≤ ||f − g ||2 + ||f − g||2 + ||g − g||2 ≤ κ
as → 0.
Since κ was arbitrary, this completes the proof of the theorem.
3. (a) Obtain the Fourier series of the 2π-periodic function that coincides with f (t) = (π − t)2
on [0, 2π].
(b) Does f match its Fourier series? Give reasons for your answer.
P∞ −2 π2
(c) Use your results from above to derive the identity:
= 6.
n=1 n
1
2
Solution. Using Fourier expansion
∞
(π − t)2 =
a0 X
+
ak cos(kt) + bk sin(kt)
2
k=1
where
Z
1 π
8π 2
a0 =
(π − t)2 dt =
π −π
3
Z π
1
4(π 2 k 2 − 1) sin(kπ) + πk
ak =
(π − t)2 cos(kt)dt =
π −π
πk 3
Z
1 π
4π(πk cos(kπ)) − sin(kπ)
bk =
(π − t)2 sin(kt)dt =
π −π
πk 2
The series converges uniformly to the function since it is differentiable with continuous
derivative using Question 1 (one may only have to check this at t = 0 using definition of
P
1
π2
derivative). Putting t = π in the fourier series gives ∞
k=1 k2 = 6
4. (a) Show that the Fourier series of a function f can alternatively be written in the form
Z π
∞
X
1
ikx
f (t)e−ikt dt
fb(k)e ,
where fb(k) =
2π −π
k=−∞
is referred to as the kth Fourier coefficient.
(b) Determine the relation of fb(k) with gb(k) in each of the following cases:
(i) g is a translate of f , namely g(x) = f (x + α).
(ii) g is a modulation of f , namely g(x) = f (x)e−iαx .
(c) Given two bounded, 2π-periodic functions f and g, both of which are Riemann-integrable
on [−π, π], define their convolution h = f ∗ g as follows,
Z π
1
h(x) = f ∗ g(x) =
f (x − t)g(t) dt.
2π −π
Note that the nth partial Fourier sum sn f and the nth Cesàro sum σn f are both given
in terms of convolutions of f with appropriate kernels. Find b
h in terms of fb and gb.
(d) Use this and the Fourier coefficients of Kn to derive an explicit formula for the nth
Cesàro sum σn f of a Fourier series.
ix
−ix
ix
−ix
Solution. (a) Substitute sin x = e −e
and cos x = e +e
into the the partial fourier sum
2i
2
(also in ak , bk )
X
a0 X
+
ak cos(kx) +
bk sin(kx)
2
k≥1
k≥1
To see that it is
X
|k|≤n
ck e
ikx
1
, cn =
2π
Z
π
−π
f (t)e−ikt dt
3
(b) If g(x) = f (x + α) then
Z π
Z π
1
−ikx
ikα 1
ĝ(k) =
f (x + α)e
dx = e
f (x + α)e−ik(x+α) dx
2π −π
2π −π
Since f is 2π-periodic, this becomes
Z π
iαk 1
e
f (y)e−iyk dy = eiαk fˆ(k)
2π −π
If g(x) = f (x)e−iαx them
Z π
Z π
1
1
−iαx −ikx
ĝ(k) =
f (x)e
e
dx =
f (x)e−i(α+k)x dx = fˆ(k + α)
2π −π
2π −π
(c) Since f, g are 2π periodic,
Z π
Z π
1
1
[
f (y)g(x − y)dy e−inx dx
f ∗ g(n) =
2π −π 2π
−π
Z π
Z π
1
1
−iny
−in(x−y)
=
f (y)e
g(x − y)e
dx dy
2π −π
2π −π
Z π
Z π
1
1
−iny
−inx
f (y)e
g(x)e
dx dy
=
2π −π
2π −π
= fˆ(n)ĝ(n)
(d) We know from definition that
X
|j| ijt
Kn (t) =
1−
e
n
|j|≤n
Hence
(
cn (j) = 1 −
K
0
|j|
n
By the previous part of the exercise
(
1−
σ
cn = f\
∗ Kn (j) =
0
if |j| ≤ n
if |j| > n.
|j| b
f (j)
n
if |j| ≤ n
if |j| > n.
By uniqueness of Fourier series (in this case they are trigonometric polynomials)
X
|j| b
σn (t) =
1−
f (j)eijt
n
|j|≤n
5. Let f ∈ C 2π , the class of continuous 2π-periodic functions on [−π, π]. We have seen that
the partial Fourier sums sn (f ) are excellent approximations of f in L2 norm, but so far only
have indirect evidence that they may not be very good approximations in the sup norm. This
problem attempts to make this intuition precise.
4
(a) Show that for every n ≥ 1, there exists fn ∈ C 2π such that ||fn ||∞ = 1 and supj |sj fn (0)| >
n.
(b) Use the functions fn in part (a) to find a single f ∈ C 2π whose Fourier series diverges
at 0.
(c) Now modifiy your construction in part (b) to create a continuous 2π-periodic function
whose Fourier series diverges at a dense set of points.
Solution. (a) Let gj = sgn(Dj ), where Dj is the Dirichlet kernel. We have seen in class that
Z
1 π
4
(2)
sj gj (0) =
|Dj (t)| dt ≥ 2 log j.
π −π
π
Let Gj denote the set of discontinuities of the function gj , along with the endpoints
±π. By the nature of the function Dj , the set Gj consists of finitely many points, say
Gj = {x1 = −π < x2 < · · · < xNj = π}. Picking j > 0 to be a constant small enough so
that
4
1
2
1
Nj j (j + ) < 2 log j,
j < min{|xk − x` | : xk 6= x` ; xk , x` ∈ Gj } and
2
π
2
π
let us define a continuous function hj as follows: hj is linear on the interval Ik = (xk −
j , xk + j ) for each k = 1, · · · , Nj , and hj = gj otherwise. Then ||hj ||∞ = 1 for each j.
Hence |gj − hj | ≤ 2 everywhere, from which we obtain the estimate
Z
1 π
|sj hj (0) − sj gj (0)| ≤
|Dj (y)||gj (y) − hj (y)| dy
π −π
X
1
≤ ||Dj ||∞
length of Ik
π
k
(3)
1
1
2
(Nj )(2j )(2)(j + ) < 2 log j,
π
2
π
1
where we have used the fact that ||Dj ||∞ ≤ j + 2 at the second step. Combining (2) and
(3) we obtain that |sj hj (0)| ≥ |sj gj (0)| − |sj (hj − gj )(0)| ≥ π22 log j. Picking the index
j = jn so that π22 log j > n and setting fn := hjn gives the desired result.
≤
(b) Without loss of generality, after convolving with an appropriate Fejér kernel if necessary
(verify this!), we may assume that each fn in part (a) is a trigonometric polynomial, say
of degree dn , which increases with n. Let us pick a fast-increasing sequence of integers
Mn and a fast-decaying sequence of positive numbers n < 2−n obeying the following
rules: for every n ≥ 1,
Mn > dn−1 ,
n (2Mn−1 + 1) < 2−n−1 .
n Mn > 2n ,
P
Set f = n n fMn . The condition ||fn ||∞ ≤ 1 ensures that the series converges uniformly,
hence f is a continuous function. Further, for every n ≥ 1,
X
sMn f (0) ≥ n sMn fMn (0) −
k sMn fMk (0)
k6=n
≥ n Mn −
X
k |fMk (0)| − 2
k<n
≥ 2n −
X
k<n
2−k −
X
k>n
X
k>n
2−k
k (2Mn + 1)
5
≥ 2n − 1.
The second step above uses the fact that sMn fMk = fMk if k < n, since dk < Mn . We
have also used the fact that ||fn ||∞ = 1, from which it follows that ||sk fn ||∞ ≤ 2(2k + 1),
since the absolute value of every Fourier coefficient is at most 2. Thus we conclude that
the sequence {sj f (0) : j ≥ 1} diverges.
(c) Let {qn } be an enumeration of the rationals in [−π, π]. Using part (b), first find functions
ϕn such that the sequence {sj ϕn (qn ) : j ≥ 1} diverges. Now use an argument similar to
(b) to construct ϕ meeting the specifications of part (c).
6. This exercise is designed to study a curious property of a certain class of Fourier series,
known as Gibbs phenomenon. Discovered by Wilbraham (1849) and studied by Gibbs (1899),
this phenomenon refers to the manner in which the Fourier series of a piecewise continuously
differentiable periodic function behaves at a jump discontinuity. The nth partial Fourier
sums oscillate near the jump point, which is understandable, but the strange thing is that
the oscillation might result in increasing the maximum of the partial sum above that of the
function itself. Even more strange is the fact that the overshoot does not die out as you take
larger and larger sums (i.e. the frequency increases), but approach a finite nonzero limit!
Here is an example where you can see Gibbs phenomenon in action.
Let f (x) = sgn(x), which takes on the value 1, -1 or 0 according as x is positive, negative
or zero.
(a) Show that
∞
4 X sin(2n − 1)x
f (x) =
π n=1 (2n − 1)
for every x ∈ [−π, π].
(b) Denote by sn the nth partial sum of the above series. Show that
Z
2 x sin 2nt
sn (x) =
dt.
π 0 sin t
(c) Examine the local maxima and minima of sn , and deduce that the largest value of sn is
π
attained at 2n
.
π
(d) Interpret sn ( 2n
) as a Riemann sum and prove that
π 2 Z π sin t
lim sn
=
dt.
n→∞
2n
π 0
t
The value of this limit is about 1.179. Thus, although f has a jump equal to 2 at the
origin, the graphs of the approximating curves sn tend to approximate a vertical segment
of length 2.358 in the vicinity of the origin!
Solution. (a) Verify that the series on the right is the Fourier series of f , so it remains to
verify that the series converges pointwise to f (x) for every x ∈ [−π, π]. Clearly, equality
holds for x = 0. For any x 6= 0,
Z
sin(n + 21 )y
1 π
dy
f ∗ Dn (x) − f (x) =
f (x − y) − f (x)
π −π
sin y2
6
2
=−
π
Z
I
sin(n + 21 )y
dy
sin y2
where I is the union of two intervals not containing 0. Integrating by parts with u =
1/ sin(y/2) and dv = sin((n + 12 )y) leads to the integral being bounded from above by
C/n, where C is a constant depending on n. Thus f ∗ Dn (x) → f (x) as n → ∞, which
is the desired conclusion.
(b) We estimate sn as follows,
4
sn (x) =
π
Z
0
x
n
X
cos(2k − 1)t dt
k=1
Z
4 x sin 2nt
dt,
=
π 0 sin t
since sin t cos(2k − 1)t = sin 2kt − sin(2k − 2)t.
(c) Since s0n (x) =
1). Further,
4 sin 2nx
,
π sin x
the critical points of sn in [−π, π] are x = ± kπ
for k = 1, · · · , (2n−
2n
π 00 kπ
cos kπ
2n(−1)k
,
sn (± ) = 2n
=
4
2n
sin ± kπ
sin ±πk
n
2n
from which we conclude that the local maxima are attained at x =
The other critical points are local minima.
π
, − 2π
, 3π , − 4π
2n
2n 2n
2n
···.
Since the function sn (x) is odd, the largest value of sn is the maximum of |sn ( (2k−1)π
)|
2n
for k = 1, · · · , n. Suppose further that 2k + 1 ≤ n; then,
Z (2k+1)π
2n
π
π 4
sin 2nt
sn (2k + 1)
− sn (2k − 1)
=
dt
2n
2n
π (2k−1)π
sin
t
2n
Z (2k+1)π
4
sin s
ds
=
s
π (2k−1)π sin 2n
Z
Z (2k+1)π
i
sin s
4 h 2kπ
sin s
=
ds
+
ds
s
s
π (2k−1)π sin 2n
sin 2n
2kπ
Z
Z 2kπ
i
4 h 2kπ
sin u
sin s
=
ds
−
du
s
π+u
π (2k−1)π sin 2n
(2k−1)π sin
2n
"
#
Z 2kπ
4
1
1
−
ds
=
sin s
s
π (2k−1)π
sin 2n
sin π+s
2n
≤ 0,
where the last step follows from the fact that the first factor of the integrand is nonpositive, while the second one is non-negative on the domain of integration. This in turn
is a consequence of the assumption that both ns , π+s
∈ [0, π2 ], and the non-decreasing
n
nature of the function sin x in the range [0, π2 ]. On the other hand, if 2k − 1 ≥ n, then
setting k = n − r, we obtain that 2r + 1 ≤ n. The same set of calculations as above now
yields that
"
#
Z
π
π 4 2kπ
1
1
−
ds
sn (2k + 1)
− sn (2k − 1)
=
sin s
s
2n
2n
π (2k−1)π
sin 2n
sin π+s
2n
7
Z
"
2kπ
≤
sin s
(2k−1)π
Z
(2r+1)π
≤
sin t
2rπ
Z
sin
"
sin
"
(2r+1)π
sin t
2rπ
1
1
sin
t
2n
s
2n
−
1
2nπ−t
2n
−
#
1
sin
−
π+s
2n
#
1
sin
t−π
2n
π+2nπ−t
2n
dt
#
1
sin
ds
dt
≤ 0,
since the first factor of the integrand is non-negative and the second factor non-positive
in the domain of integration. Thus we have shown that among the values {sn ( π(2k−1)
):
2n
π
k = 1, · · · , n}, the largest one is sn ( 2n ).
(d) Use the power series expansion of the sine function to show that
1
1
sup −
s → 0 as n → ∞.
s 2n sin( )
s∈[0,π]
2n
This implies that
π
Z
Z
π 2 Z 2n
sin 2nt
2 π sin s ds
2 π sin s
sn
=
dt =
→
ds.
s
2n
π 0
sin t
π 0 sin 2n
2n
π 0
s