6.Padrão_English_2019.0(3).pdf
Documento PDF (154.3KB)

Exibir conteúdo Baixar

                    Federal University of Alagoas
Graduate Program in Mathematics

Ph.D. Program Entrance Exam
Date: December 3rd, 2018

Time: 15h30 - 18h30

Candidate:
Q1- Suppose that f : Rn → Rm is continuous at a ∈ Rn . Prove that if f (a) does not belong to the
closed ball B[b; r] ⊂ Rm , then there exists δ > 0 such that f (x) does not belong to B[b; r] for all
x ∈ Rn satisfying |x − a| < δ.
Solution: Since f (a) 6∈ B[b; r], we have |f (a) − b| > r, that is, |f (a) − b| − r > 0. Since f is
continuous at a, there exists δ > 0 such that |f (x) − f (a)| < |f (a) − b| − r for all x ∈ Rn satisfying
|x − a| < δ. Then, it follows from the Triangle Inequality that
|f (x) − b| = |f (x) − f (a) + f (a) − b|
≥

|f (a) − b| − |f (x) − f (a)|

>

r

n

for all x ∈ R satisfying |x − a| < δ.
Q2- Suppose that f : Rn → R is a differentiable function such that f (x/2) = f (x)/2 for each x ∈ Rn .
Prove that f is a linear functional.
Solution: Since f (x/2) = f (x)/2 for each x ∈ Rn , we have f (0) = f (0/2) = f (0)/2, that is,
f (0) = 0. By the other hand, it follows directly from the hypothesis that
 x  f (x)
f k = k
2
2
for all k ∈ N. So, since f is differentiable, we have

f 2xk
f (tx)
0
f (0) · x = lim
= lim
= lim f (x) = f (x).
1
t→0
k→∞
k→∞
t
2k
This proves that f = f 0 (0), which is a linear functional.
Q3- Consider the function f : R2 → R given by
( 3 2
f (x, y) =

x y
x4 +y 4 , if

(x, y) 6= (0, 0)
0, if (x, y) = (0, 0),

2
a) Show that there exists ∂f
∂v (0, 0), for all v ∈ R .
b) Is the equality h∇f (0, 0), vi = ∂f
∂v (0, 0) holds?
c) What can be said about the differentiability of f at (0, 0)?

Solution:
a) Set v = (a, b). By a straightforward computation we have that
 f (ta, tb) − f (0, 0) 
∂f
a3 b2
(0, 0) = lim
= 4
.
t→0
∂v
t
a + b4
3 2

a b
So, there exists ∂f
∂v (0, 0) and it is equal to a4 +b4 .
b) False. Indeed, ∇f (0, 0) = (0, 0), however ∂f
∂v (0, 0) is not always zero.
1

c) No, because of the previous item.
Q4- Let f : Rm → Rn be a function satisfying
|f (x) − f (y)| ≤ |x − y|2
for all x, y ∈ Rm . Show that f is a constant function.
Solution: Notice that f is differentiable and that df (x) = 0 for all x ∈ Rm . Indeed,
f (x + h) = f (x) + O(h) + r(h),
where O is the identically zero map and r satisfies
kf (x + h) − f (y) − O(h)k
kr(h)k
khk2
= lim
≤ lim sup
= 0.
h→0 khk
h→0
khk
khk
h→0
lim

m
Thus, limh→0 kr(h)k
is connected,
khk = 0 and so f is differentiable and df (x) = 0. Using that R
we conclude that f is constant.

Q5- Let U ⊂ R2 be an open set in R2 . Let a(x, y) and b(x, y) be positive functions defined on U ∪ ∂U ,
where ∂U denotes the boundary of U . Suppose that the quadratic form given by the matrix


a 0
A=
0 b
is positive definite for all (x, y). Given a function of class C 2 , v, defined on U ∪ ∂U we define the
operator L by
∂2v
∂2v
Lv = a 2 + b 2 ,
∂x
∂y
which with this positivity condition is called of elliptic operator. A function f is said to be strictly
subharmonic relative to L if Lv > 0. Show that a strictly subharmonic function cannot attain its
maximum value at any point of U .
2

2

∂ v
∂ v
Solution: In an interior point of maximum we would have ∂x
Since a
2 ≤ 0 e ∂y 2 ≤ 0.
and b are nonnegative, then Lv at this point of maximum is nonpositive, which contradicts the
hypothesis of Lv > 0.