Final Exam: Solutions
Problem 1. (a) Find all continuous functions f on the line segment
[0, 1] which satisfy
Z 1
f (x)xn dx = 0, n = 0, 1, . . .
0
(b) What will be the answer if one requires only that
Z 1
f (x)xn dx = 0, n = 2014, 2015, . . .?
0
R1
Solution. (a) By linearity, we have that 0 f (x)p(x) dx = 0 for
every polynomial p. By Weierstrass’ theorem, there exists a sequence
of polynomials pn such that supx∈[0,1] |pn (x) − f (x)| → 0 as n → ∞.
P
k
∗
We
that if p(z) = m
k=0 ak z is a polynomial then p (z) =
Pmalso∗ observe
k
k=0 ak z is also a polynomial, and that for real x we have
p(x)∗ =
m
X
a∗k xk = p∗ (x).
k=0
Therefore,
Z 1
Z 1
Z 1
∗
∗
≤
f
(x)f
(x)
dx
−
f
(x)p
(x)
dx
|f (x)||f (x)∗ −pn (x)∗ | dx
n
0
0
0
Z 1
Z 1
|f (x)| dx → 0
|f (x)||f (x) − pn (x)| dx ≤ sup |f (x) − pn (x)|
=
x∈[0,1]
0
as n → ∞. Since
Z 1
1
Z
∗
f (x)p∗n (x) dx = 0,
f (x)pn (x) dx =
0
0
0
we obtain that
Z
1
2
1
Z
f (x)f (x)∗ dx = 0.
|f (x)| dx =
0
0
Since f is continuous on [0, 1], we obtain that f (x) = 0 identically.
(b) Let g(x) = x2014 f (x). Then we have that
Z 1
Z 1
n
g(x)x dx =
f (x)xn+2014 dx = 0
0
0
1
2
for n = 0, 1, . . .. By part (a), g(x) = 0 identically. Therefore, f (x) =
1
g(x) = 0 for x ∈ (0, 1]. By continuity, we also obtain that f (0) = 0.
x2014
Therefore, f (x) = 0 identically.
Problem 2. Express the following properties of the function f : T →
C in terms of its Fourier coefficients:
(a) f (t + π) = f (t) for every t ∈ T.
(b) f (t + 2π/k) = λf (t) for some k ∈ Z and λ ∈ C and for every
t ∈ T; for which λ there exist non-zero functions f satisfying this
condition?
Solution. (a) We have
Z 2π
Z 2π
−ikt
ˆ
f (k) =
f (t)e
dt =
f (t + π)e−ikt dt
0
0
Z 3π
Z 3π
−ik(t−π)
ikπ
f (t)e
dt = e
f (t)e−ikt dt
=
π
π
Z 2π
= eikπ
f (t)e−ikt dt = eikπ fˆ(k) = (−1)k fˆ(k).
0
Therefore, fˆ(k) = 0 for all odd k.
Conversely, suppose fˆ(k) = 0 for all odd k. Suppose that f is continuous and that the Fourier series of f converges to f pointwise. Then
P
P
2nit
2ni(t+π)
b
b
= f (t) for every
= ∞
f (t + π) = ∞
n=−∞ f (2n)e
n=−∞ f (2n)e
t. If one just requires continuity of f , then a similar argument works
with Fejer sums replacing Fourier sums.
(b) We have
Z
Z 2π
−int
ˆ
f (n) =
f (t)e
dt =
f (t + 2π/k)e−in(t+2π/k) dt
−2π/k
0
Z
2π−2π/k
2π−2π/k
=
−in(t+2π/k)
λf (t)e
−2nπi/k
Z
2π−2π/k
dt = λe
−2π/k
f (t)e−int dt
−2π/k
= λe−2nπi/k
Z
2π
f (t)e−int dt = λe−2nπi/k fˆ(n).
0
If f is not identically zero, then fˆ(n) =
6 0 for at least one n. This
2nπi/k
−2(n+jk)πi/k
implies that λ = e
. Since e
= e−2nπi/k for every integer
j, there is no contradiction in the identity fˆ(m) = λe−2mπi/k fˆ(m) for
m = n + jk with any j ∈ Z. On the other hand, if m 6= n + jk for all
j ∈ Z, then we must have fˆ(m) = 0.
The converse direction can be proved similarly to that of part (a).
3
Problem 3. The Hermite polynomials are defined by
2 /2
Hn (t) = (−1)n et
dn −t2 /2
(e
),
dtn
n = 0, 1, . . .
(a) Prove the recurrence relation
Hn+1 (t) = tHn (t) − Hn0 (t).
(b) Verify that Hn is a polynomial of n-th degree with the leading
coefficient 1.
(c) Prove the orthogonality relations hHn , Hm i = 0 for n 6= m and
find the norm of Hn in the space of continuous square integrable func2
tions on R with weight φ(t) = e−t /2 , i.e., in the inner product
Z
∞
hf, gi =
2 /2
f (t)g(t)e−t
dt.
−∞
Solution. (a) We have
Hn0 (t)
n+1
n
−t2 /2
t2 /2 d
−t2 /2
t2 /2 d
(e
)+e
(e
)
= (−1) te
dtn
dtn+1
dn
dn+1
2
2
2
2
= (−1)n tet /2 n (e−t /2 ) − (−1)n+1 et /2 n+1 (e−t /2 )
dt
dt
= tHn (t) − Hn+1 (t).
n
Thus, Hn+1 (t) = tHn (t) − Hn0 (t).
(b) We will prove the statement by induction. We have H0 (t) = 1,
H1 (t) = t, so that the statement is true for n = 0, 1. Suppose that Hn
is a polynomial of degree n with the leading coefficient 1. Then Hn0 is a
polynomial of degree n − 1, and tHn (t) is a polynomial of degree n + 1
with the leading coefficient 1. Therefore, Hn+1 (t) = tHn (t) − Hn0 (t) is
also a polynomial of degree n + 1 with the leading coefficient 1.
4
(c) We first observe that if k ≤ n then, applying k times integration
by parts, we obtain
Z ∞
dn
2
2
2
k
tk (−1)n et /2 n (e−t /2 )e−t /2 dt
ht , Hn (t)i =
dt
−∞
Z ∞
dn
2
= (−1)n
tk n (e−t /2 ) dt
dt
−∞
Z ∞
n−1
∞
dn−1
2
n k d
−t2 /2 n−1
ktk−1 n−1 (e−t /2 ) dt
= (−1) t n−1 (e
)
+ (−1)
dt
dt
−∞
−∞
Z ∞
dn−1
2
= (−1)n−1
ktk−1 n−1 (e−t /2 ) dt
dt
−∞
Z ∞ n−k
d
2
n−k
= . . . = (−1) k!
(e−t /2 ) dt,
n−k
−∞ dt
2
where we use the fact that the function e−t /2 tends to 0 as t → ±∞,
as well as all its derivatives (which have the form“a polynomial times
2
e−t /2 ”). If k < n, then we deduce that
∞
dn−k−1
2
= 0.
htk , Hn (t)i = (−1)n−k k! n−k−1 (e−t /2 )
dt
−∞
If k = n, then we deduce that
Z ∞
√
2
n
e−t /2 dt = n! 2π.
ht , Hn (t)i = n!
−∞
Since Hn ⊥ tk for all k < n, and Hm is a polynomial of degree m, we
obtain that Hn ⊥ Hm for m < n. For the same reason, we obtain that
√
kHn k2 = hHn , Hn i = htn , Hn i = n! 2π,
√ √
and therefore kHn k = n! 4 2π.
Problem 4. (a) Prove Abel’s summation formula
q
X
n=p
an b n =
q−1
X
An (bn − bn+1 ) + Aq bq − Ap−1 bp ,
n=p
P
where an , bn ∈ C, n = 0, 1, . . ., 0 ≤ p ≤ q, A−1 = 0, An = nj=0 aj , and
P
the sum q−1
n=p An (bn − bn+1 ) is defined to be zero if p = q.
(b) Let a(p) ≥ a(p + 1) ≥ . . . ≥ a(q) ≥ 0. Prove that for every
t ∈ T : t 6= 0,
q
X
a(p)
int a(n)e ≤
.
| sin(t/2)|
n=p
5
(c) Let a(n) ≥ a(n + 1) for all n ∈ N, and let limn→∞ a(n) = 0.
Prove that the series
∞
X
a(n)eint
n=1
converges for each t ∈ T : t 6= 0, and that the convergence is uniform
on any closed subset of (0, 2π).
Solution. (a)
q
X
q
q
q
X
X
X
An−1 bn
An bn −
(An − An−1 )bn =
an b n =
=
n=p
n=p
n=p
q
X
An bn −
q−1
X
An bn+1 =
An (bn − bn+1 ) + Aq bq − Ap−1 bp .
n=p
n=p−1
n=p
q−1
X
n=p
(b) Denoting an = eint , bn = a(n) and applying the formula from
part (a), we obtain
q
X
int
a(n)e
=
q−1
X
n=p
An (a(n) − a(n + 1)) + Aq a(q) − Ap−1 a(p),
n=p
where
An =
n
X
eijt =
j=0
ei(n+1)t − 1
ei(n+1)t/2 ei(n+1)t/2 − e−i(n+1)t/2
=
·
eit − 1
eit/2
eit/2 − e−it/2
(n+1)t
2
.
sin 2t
int/2 sin
=e
Since |An | ≤
1
,
| sin 2t |
we obtain that for any t 6= 0
q
q−1
X
X
int a(n)e ≤
|An |(a(n) − a(n + 1)) + |Aq |a(q) + |Ap−1 |a(p)
n=p
n=p
≤
1
2a(p)
.
t (a(p) − a(q) + a(q) + a(p) =
| sin 2 |
| sin 2t |
(I guess, there is a typo in the book where this problem is taken from
— the “2” in the numerator is missing there.)
(c) Since a(n) → 0 as n → ∞, we obtain that
q
X
2a(p)
int a(n)e ≤
→0
| sin 2t |
n=p
6
P
int
as p → ∞. Therefore, for any t 6= 0, the series ∞
conn=0 a(n)e
verges. Moreover, this convergence is uniform on any closed subset of
(0, 2π), since on such subset | sin 2t | is not zero and therefore, attains
its minimum.
Problem 5. Let f, g ∈ `1 (Z) have “compact support”, i.e., there
are integers A, B, C, D for which f [n] = 0 except for A ≤ n ≤ B and
g[n] = 0 except for C ≤ n ≤ D. Recall that the convolution of f and
g is defined as
∞
X
(f ∗ g)[n] =
∞
X
f [n − k]g[k] =
k=−∞
f [k]g[n − k].
k=−∞
(a) (a fast convolution algorithm) Let N ≥ B − A + D − C + 1. Then
τ−A f and τ−C g have their support in the set {0, 1, . . . , N − 1} (here
(τ−A f )[t] = f [t + A] and (τ−C g)[t] = g[t + C]). Let F = FN (τ−A f )
and G = FN (τ−C g) be the DFTs of τ−A f and τ−C g on ZN , respectively
(here FN denotes the DFT operator). Prove that if A+C ≤ n ≤ B+D,
then
(f ∗ g)[n] = (τA+C FN−1 (F G))[n],
and that (f ∗ g)[n] = 0 elsewhere.
(b) Prove that a direct implementation of the definition of convolution requires (B − A + 1)(D − C + 1) multiplications.
(c) Using the FFT algorithm (Section 3.9 in Benedetto’s book), estimate the number of multiplications necessary for the computation of
f ∗ g by the method of part (a) in the case where N = 2k , k ∈ N, and
compare this estimated number with the result of part (b).
Solution. (a) We first observe that
(f ∗ g)[n] =
∞
X
k=−∞
f [k]g[n − k] =
B
X
f [k]g[n − k].
k=A
More over, if n < A + C then for A ≤ k ≤ B we have n − k < C and
g[n − k] = 0, thus (f ∗ g)[n] = 0. Likewise, if n > B + D then for
A ≤ k ≤ B we have n − k > D and g[n − k] = 0, thus (f ∗ g)[n] = 0.
7
Next, if A + B ≤ n ≤ C + D then
(τA+C FN−1 (F G)[n]) = FN−1 (F G)[n − A − C]
N −1
1 X
=
F [j]G[j]ei(n−A−C)j/N
N j=0
=
N −1 N −1
N
−1
X
1 XX
f [k + A]e−ijk/N
g[` + C]e−ij`/N ei(n−A−C)j/N
N j=0 k=0
`=0
N −1 B
D
X
1 XX
−ij(k−A)/N
=
f [k]e
g[`]e−ij(`−C)/N ei(n−A−C)j/N
N j=0 k=A
`=C
N −1
1 X ij(n−k−l)/N
e
.
=
f [k]
g[`]
N j=0
k=A
`=C
B
X
D
X
P −1 ij(n−k−l)/N
equals 1 if n = k + `, and 0 otherwise, we
Since N1 N
j=0 e
obtain that
B
X
−1
(τA+C FN (F G)[n]) =
f [k]g[n − k],
k=A
which coincides with the expression for (f ∗ g)[n].
(b) Since there are B − A + 1 nonzero values of f and D − C + 1
nonzero values of g, the implementation of the direct formula for f ∗ g
requires (B − A + 1)(D − C + 1) multiplications in terms f [k]g[n − k].
(c) Suppose that N = 2k , B −A+1 = D−C +1 = N/2 = 2k−1 . Then
the FFT algorithm for the computation of F (or G) requires N4 log2 N2
multiplications. The computation of the product F G requires N/2
more multiplications, and the computation of the inverse DFT requires
additional N4 log2 N2 multiplications. Altogether, the computation of
the convolution as in part (a) requires
3N
N
N
3N
N
log2
+
=
log2 N −
4
2
2
4
4
multiplications. Comparing with the direct method which requires
N2
4
multiplications, this (indirect) convolution algorithm is much faster for
large N .
(B − A + 1)(D − C + 1) =
Problem 6. Let {fk }k=1,2,... be a sequence of continuous functions
which converges to a continuous function f uniformly on T.
8
(a) Show that the sequence of Fourier coefficients fbk (n) converges to
fb(n) as k → ∞ uniformly in n ∈ Z.
(b) Find an example of a sequence {fk }k=1,2,... of continuous functions
which converges to a continuous function f pointwise (not uniformly)
on T, but the sequence {fbk (n)}k=1,2,... does not converge to fb(n) for at
least one n ∈ Z.
Solution. (a) Suppose that fk (t) → f (t) uniformly on T as k → ∞,
i.e., supt∈T |fk (t) − f (t)| → 0 as k → ∞. Then
Z
−int
dt
|fbk (n) − fb(n)| = (fk (t) − f (t))e
T
Z
≤ |fk (t) − f (t)| dt ≤ 2π sup |fk (t) − f (t)| → 0
t∈T
T
as k → ∞ uniformly in n ∈ Z.
(b) Let f (t) = 0 identically on T, and

2k 2 t,

fk (t) = −2k 2 (t − k1 ),

0,
define
1
t ∈ [0, 2k
],
1 1
t ∈ ( 2k , k ],
t ∈ ( k1 , 2π].
Then fk is a continuous function on T for each k = 1, 2, . . .; fk (t) → 0
for every t ∈ T (since the sequence eventually stabilizes at 0); however
Z
1
fbk (0) = fk (t) dt = ,
2
T
i.e.,
1
1
lim fbk (0) = lim = 6= 0 = fb(0).
k→∞ 2
k→∞
2