Fall 2014 MATH3060 Mathematical Analysis III Solution to Midterm Examination 1. Let x

Fall 2014
MATH3060 Mathematical Analysis III
Solution to Midterm Examination
1. Let x1 and x2 be two distinct points in the metric space (X, d). Find a bounded,
continuous function ϕ in X satisfying ϕ(x1 ) = 0 and ϕ(x2 ) = 1.
Solution. There are many such functions. The simplest one is
f (x) =
d(x, x1 )
.
d(x, x2 ) + d(x1 , x2 )
Some tried g(x) = d(x, x1 )/d(x2 , x1 ), but that is no good since it may not be
bounded. The modified one
n d(x, x ) o
1
h(x) = min
,1
d(x2 , x1 )
is good.
2. Consider C[−1, 1] with the metric induced by the sup-norm. Is the set
n
o
f ∈ C[−1, 1] : f is differentiable at 0
closed in C[−1, 1]? Justify your answer.
Solution.
No, it is not a closed set. A simple example is to consider the
function g(x) which is equal to x for x ∈ [0, ∞) and vanishes on (−∞, 0). It is
not differentiable at 0 but, as we shift it, gn (x) = g(x − 1/n) is differentiable at
0 and gn converges uniformly to g on [−1, 1] as n → ∞. The sequence of smooth
functions (x2 + 1/n)1/2 , which converges uniformly to |x| on [−1, 1], is another
example.
3. Let A be a non-empty set in (X, d) and define
g(x) = inf{d(x, y) : y ∈ A}.
Show that
|g(x) − g(y)| ≤ d(x, y),
Solution.
x, y ∈ X.
See Solution to Assignment.
4. Let {ϕj }∞
j=1 be an orthonormal set in R[−π, π]. Show that it does not admit any
convergent subsequence in L2 -distance.
√
Solution. Let {fn } be such a sequence. We have kfn − fm k2 = 2 for all distinct
n, m. On the other hand, if {fnk } is convergent, it is Cauchy and in particular
there is some n0 such that kfnk − fnj k2 < 1, for all nk , nj ≥ n0 . Contradiction
holds.
∞
to
5. Give a definition of the fractional derivative L ≡ (d/dx)1/3 so that it maps C2π
itself and
df
∞
L3 f =
, ∀f ∈ C2π
,
dx
holds.
Solution. As fb0 (n) = infˆ(n) = eπi/2 fˆ(n), it suggests us to define
X
Lf (x) =
eπi/6 n1/3 cn einx ,
n
where cn is the n-th Fourier coefficient of f . First of all, we need to show this is
a well-defined smooth function. Indeed, as f is smooth and 2π-periodic, |cn | =
◦(n−k ) for all k. It follows that |eπi/6 n1/3 cn | = ◦(n−k ) for all k. By M-test the
RHS in the definition of Lf defines a smooth function. Now, we can easily check
P
that L3 f = f 0 . You may also define Lf = n (−i)n1/3 cn eix , since the cube root of
i is not unique. (I prefer the root in the first quadrant.)
(a) (10 marks) Find the Fourier series of h(x) = |x| on [−π, π].
Solution. By direction computation,
∞
π
4 X cos(2k + 1)
h(x) ∼ −
.
2 π k=0 (2k + 1)2
(b) (5 marks) Prove
∞
1
π2 X
=
.
8
(2k + 1)2
k=0
Quote a theorem to justify your derivation.
Solution. A theorem states that the Fourier series of a Lipschitz continuous
function converges to the function uniformly. (You should state this result.
The key words Lipschitz continuous or piecewise C 1 should come up.) Now
h is Lipschitz continuous. By taking x = 0 in (a), we get
∞
4X
1
π
,
0= −
2 π k=0 (2k + 1)2
and the identity follows.
(c) (5 marks) Find the sum
∞
X
k=0
1
.
(2k + 1)4
Quote a theorem to justify your derivation.
Solution. As h is continuous, it is integrable on [−π, π]. (Should point out
integrable first in order to apply Parseval. This identity does not come for free.
Sadly most of you did not. I did not subtract any points this time though.)
A theorem states that Parseval’s identity holds in this case. Therefore, we
have
∞
X
2
2
khk2 = 2πa0 + π
a2k ,
k=1
which yields
∞
X
k=0
1
π4
,
=
(2k + 1)4
96
after some manipulations.
7. (a) (10 marks) Let f ∈ R2π , that is, it is 2π-periodic and integrable on [−π, π],
and Sn f the n-th partial sum of its Fourier series. Derive the formula
Z
1 π sin n + 12 z
f (x + z) dz .
(Sn f )(x) =
π −π 2 sin 12 z
Solution. (a). See Notes.
(b) (10 marks) Let g ∈ R2π which is equal to f on (−ρ, ρ) for some small ρ > 0.
Show that limn→∞ Sn g(0) = g(0) if limn→∞ Sn f (0) = f (0).
Solution. Let I = [−δ, δ] and Dn be the Dirichlet kernel. We write
Z
Z
Sn g(0) = Dn (z)g(z)dz +
Dn (z)g(z)dz
I
[−π,π]\I
and similarly
Z
Sf (0) =
Z
Dn (z)f (z)dz +
I
Dn (z)f (z)dz.
[−π,π]\I
Using f = g on I, we have
Z
Dn (z)(g(z) − f (z))dz|,
|Sn g(0) − Sn f (0)| = |
[−π,π]\I
whose RHS tends to 0 as n → ∞ as a result of Riemann-Lebesgue lemma.
Note: This is Riemann’s localization principle. It tells us that although the
Fourier series of a function depends on its values on [−π, π], its convergence
to the function at a particular point is only determined by its values in a
small neighborhood of this point.
8. (20 marks) For f ∈ R2π and a ∈ R, define fa (x) = f (x + a) ∈ R2π . Show that
Z π
|fa (x) − f (x)| dx → 0, as a → 0.
−π
Solution. I will be sketchy. Let I = [−π, π] and J = [−2π, 2π]. First approximate f by a step function on J (see the proof of Riemann-Lebesgue lemma in
notes) and then approximate the step function by a continuous function g on J
so that their L1 -distance is very small. (There are finitely many discontinuity for
a step function, so approximation is possible.) Therefore, for ε > 0, there is a
continuous function g such that
Z
ε
|f (x) − g(x)|dx <
.
8π
J
On the other hand, g is continuous on J and hence uniformly continuous, we
can find a small ρ > 0 less than π such that |g(y) − g(x)| < ε/4 for x, y ∈
[−2π, 2π], |x − y| < ρ. Therefore, for a, |a| < ρ,
Z
Z
|fa (x) − f (x)|dx =
|f (x + a) − f (x)|dx
I
Z
ZI
|f (x + a) − g(x + a)|dx + |g(x + a) − g(x)|dx
≤
I
IZ
+ |g(x) − f (x)|dx
I
< 2π
ε
ε
ε
+ + 2π
< ε.
8π 4
8π
Note: We need to choose a slightly bigger J because x + a may not stay in I for
x ∈ I. The proof is based on the same idea by which Riemann-Lebesgue lemma
was proved in notes. You may use the cut-off function if you like.
9. (20 marks) Show that every open set in (−∞, ∞) can be written as a disjoint
union of open intervals.
Solution. Let G be open. Since G is open, for every x ∈ G, there exists some
open interval containing x that belongs to G. Denote all these open intervals by
S
{Ix }. Let Jx = Ix . Then Jx is an open interval containing x that belongs to G.
It is the largest one in the sense that when I is an open interval containing x that
belongs to G, I ⊂ Jx . It follows that for x, y ∈ G, either Jx = Jy or Jx ∩ Jy = φ.
If one introduces a relation on G × G by x ∼ y if y ∈ Jx . It is readily checked that
S
∼ is an equivalence relation. So G = [x] Jx is a disjoint union.
Saying more, the union is a countable one. For, if there are uncountably many
inequivalence classes [x], picking one rational number in each Jx will produce
uncountably many rational numbers, which is absurd. Therefore, the union is
countable.