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Math 131B-1: Analysis
Instructor: Alp´ar R. M´esz´aros
Final exam, June 14th, 2017
Name (use a pen):
Student ID (use a pen):
Signature (use a pen):
Rules:
- Duration of the exam: 180 minutes.
- By writing your name and signature on this exam paper, you attest that you are the person indicated and will adhere to the UCLA Student Conduct Code.
- No calculators, computers, cell phones (all the cell phones should be turned off during the exam), notes, books or other outside material are permitted on this exam. If you want to use a scratch paper, you should ask for it from one of the proctors. Do not use your own scratch paper!
- Please justify all your answers with mathematical precision and write rigorous and clear proofs. You may lose points in the lack of justification of your answers.
- Theorems from the lectures and homework assignments may be used in order to justify your solution. In this case state precisely the theorem you are using.
- The problems are not necessarily ordered w.r.t. difficulty.
- This exam has 4 problems and is worth 33 points. Adding up the indicated points you can observe that there are 44 points, which means that there are 11 “bonus” points. This permits to obtain the highest score 33, even if you do not answer some of the questions. On the other hand nobody can be bored during the exam. All scores higher than 33 will be considered as 33 in the gradebook.
- I wish you success!
Problem Score
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Total
Exercise 1 (8 points).
Let us consider the trigonometric functions sin, cos : R → R. In this exercise you might use all the properties of these functions, which were derived during the lectures or showed in homework exercises. We define tan : R \ {(k + 1/2)π : k ∈ Z} → R as
tan(x) :=
sin(x) cos(x)
Notice that this function is well-defined, since cos is vanishing only at points of the form (k + 1/2)π, k ∈ Z.
(1) Show that tan is π-periodic and differentiable on its domain of definition. Give a formula for its derivative as well!
(2) Show the identity cos^2 (x) = 1/(1 + tan^2 (x)) for all x ∈ (−π/ 2 , π/2).
(3) Show that tan is strictly increasing on (−π/ 2 , π/2). Compute
lim x→−π/ 2
tan(x) and lim x→π/ 2
tan(x).
(4) Argue why is arctan differentiable. Derive the formula for its derivative. Hint: you might use tan(arctan(x)) = x for all x ∈ R and (2).
(5) Show that tan is a bijective map from (−π/ 2 , π/2) onto R, hence tan−^1 : R → (−π/ 2 , π/2) exists. We denote tan−^1 as arctan.
(6) Using eventually the geometric series, write the power series expansion for arctan′^ (the derivative of arctan) with center 0 for all x such that |x| < 1.
(7) Using the appropriate theorem for power series, deduce a power series expansion for arctan on (− 1 , 1). Argue why is arctan is real analytic on (− 1 , 1). Hint: you might use a homework exercise to justify the last question.
(8) Using Abel’s theorem and (7), show that arctan is continuous at x = 1, (although we know that the function is differentiable by (5), show the continuity of it at x = 1, as it is asked, using Abel’s theorem). Hint: you have to discuss the convergence of an alternating series.
(9) Using the fact that tan(π/4) = 1 and (8), write π/4 as the sum of a convergent series.
Solution (1) From the definition of Trigonometric functions, we have
cos(z + π) =
ei(z+π)^ + e−i(z+π) 2
−eiz^ − e−iz 2
= − cos(z) (1)
sin(z + π) = ei(z+π)^ − e−i(z+π) 2 i
−eiz^ + e−iz 2 i
= − sin(z) (2)
Using the properties above, we conclude that
tan(z + π) =
cos(z + π) sin(z + π)
− cos(z) − sin(z)
= tan(z) (3)
Note that tan(z) is well defined on the domain. Moreover, it is differentiable since sin and cos are differentiable on the domain. Use Theorem 4.7.2 in the textbook and differentiation formula for fractions to conclude that
(tan(z))′^ = (sin(z))′^ cos(z) − sin(z)(cos(z))′ cos^2 (z)
sin^2 (z) + cos^2 (z) cos^2 (z)
cos^2 (z)
Exercise 2 (Fourier series – 9 points).
Let us consider the function f : R → R defined as f (x) = x for all x ∈ [0, 1), and then we extend it periodically to the whole real line. For a given metric space (X, d), we denote by C(R/Z; X) the space of continuous functions defined on R with values in X that are Z-periodic.
(1) Is f an element of C(R/Z; R)? Justify your answer!
(2) Find the Fourier coefficients of f and the Fourier series associated to f. Hint: you might use either the definition with the complex valued characters, or the equivalent definition with sin and cos.
(3) Is it possible to study the uniform convergence of the series found in (2), by Weierstrass’ M-test? Justify your answer!
(4) Study the pointwise convergence of the series found in (2) at x = 0 and x = 1. What do you observe? Is the Fourier series of f converging uniformly to f on R? Justify your answers!
(5) Show the pointwise convergence of the series at x = 1/2 and x = 1/4 to f (1/2) and f (1/4) respectively. Hint: for the latter one, you might use Exercise 1(9).
(6) Show that the Fourier series found in (2) is converging in the L^2 sense to f , meaning that the sequence of partial sums is converging w.r.t. the dL^2 metric to f. Recall that for two complex valued Z-periodic square integrable functions g, h , we define the dL^2 metric as
dL 2 (g, h) :=
0
|g(x) − h(x)|^2 dx
Hint: compute the integral by hand. You might use the fact that
n≥ 1
1 n^2 =^ π
Solution (1) No. limx→ 1 − f (x) = 1 6 = 0 = f (0) = f (1) (2) Note that e^2 πin^ = 1 When n 6 = 0,
fˆ (n) =
0
f (x)e−^2 πinxdx
0
xe−^2 πinxdx
= x ·
− 2 πin
e−^2 πinx
∣^1
0
− 2 πin
e−^2 πinxdx
− 2 πin
e−^2 πin^ − 0) −
(2πin)^2
e−^2 πinx
∣^1
0
= −
2 πin
For n = 0, fˆ (0) =
0 xdx^ =^
1
- Thus, the Fourier series associated to^ f^ is ∑^ ∞
n=−∞
f^ ˆ (n)e^2 πinx^ =^1 2
n 6 =
2 πin e^2 πinx^ =
∑^ ∞
n=
πn sin(2πnx) (15)
(3) No. | −
2 πin
e^2 πinx| =
2 πn
which is not summable. Therefore, we cannot apply the Weierstrass’ M-test to study the uniform con- vergence of the series.
(4) At x = 0 and x = 1, sin(2πnx) = 0. Thus, the series converges pointwisely to 12. If the Fourier series of f were converging uniformly to f on R, the function f should be continuous. On the other hand, in part (1), we observed that f is not continuous. Thus, the Fourier series of f does not converge uniformly to f on R. (5) At x = 12 , sin(2πnx) = 0 for all n ≥ 1. Thus, the series converges to 12.
At x = 14 , sin(2πnx) =
(−1)k^ if n = 2k + 1 0 if n = 2k
∑^ ∞
n=
πn
sin(2πn
∑^ ∞
k=
(−1)k (2k + 1)π
π
π 4
(6) Use (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc to conclude
(x −
∑^ k
n=
πn
sin(2πnx))^2 = x^2 +
∑^ k
n=
πn
sin(2πnx))^2 − x + 2
∑^ k
n=
x πn
sin(2πnx) −
∑^ k
n=
πn
sin(2πnx)
(18) It is easy to check that ∫ (^1)
0
sin(2πnx) sin(2πmx)dx =
1 2 if^ n^ =^ m 0 if n 6 = m
0
x sin(2πnx)dx = −
2 πn
(n 6 = 0) (20) ∫ (^1)
0
sin(2πnx)dx = 0 (n 6 = 0) (21)
A combination of the equalities yields
∫ (^1)
0
|x −
∑^ k
n=
πn
sin(2πnx)|^2 dx
0
x^2 − x +
dx +
∑^ k
n=
πn
)^2 + 2
∑^ k
n=
πn
2 πn
2 π^2
∑^ k
n=
n^2
→
2 π^2
π^2 6
= 0 as k → ∞
Since the set E is dense, for each i ∈ { 1 ,... , m}, there exists xi ∈ E ∩ B(yi; δε), moreover, for any y ∈ X, there exists i ∈ { 1 ,... , m} such that y ∈ B(yi; δε). Now fix any y ∈ X and take the corresponding yi and xi. We have
dY (gn(y), gm(y)) ≤ dY (gn(y), gn(xi)) + (gn(xi), gm(xi)) + dY (gm(xi), gm(y)) ≤ ε + ε + ε = 3ε,
where the first and the third ε is coming from (23) and the second inequality is due to (22), provided n, m ≥ max{Nε,xi : i ∈ { 1 ,... , m}}. Thus, taking the supremum w.r.t. y ∈ X the sequence (gn)n is Cauchy w.r.t. the uniform convergence of continuous functions. Last, the space of continuous functions is complete, provided the image space is complete. (Y, dY ) is compact, so in particular complete, hence the sequence (gn)n converges uniformly to some continuous function.
Part 2
(1) Let us consider a sequence (rn)n∈N of continuously differentiable functions with values in [− 5 , 5], i.e. there is a T > 0 given and rn : [0, T ] → [− 5 , 5] are continuously differentiable functions for all n ∈ N. We suppose that there exists M > 0 such that ∫ (^) T
0
|r′ n(t)|^2 dt < M, ∀n ∈ N.
Show that this sequence of functions has a subsequence that is converging uniformly to some contin- uous r : [0, T ] → [− 5 , 5]. Hint: show that the sequence is equicontinuous by using the fundamental theorem of calculus and a Cauchy-Schwarz inequality, then conclude by Part 1.
Solution Take s, t ∈ [0, T ], s < t, then we have rn(t) − rn(s) =
∫ (^) t s r
′ n(τ^ ) dτ,^ from where using the Cauchy- Schwarz inequality, we get
|rn(t) − rn(s)| =
∫ (^) t
s
r′ n(τ ) dτ
∫ (^) t
s
|r n′(τ )| dτ
(∫ (^) t
s
dτ
) 1 / 2 (∫ (^) t
s
|r′ n(τ )|^2 dτ
≤ |t − s|^1 /^2
T
0
|r n′(τ )|^2 dτ
M |t − s|^1 /^2.
The last inequality implies that the sequence is equicontinuous. Indeed, for any ε > 0, by choosing δε := ε^2 /M , we have that |rn(t) − rn(s)| ≤ ε, ∀n ∈ N,
whenever |t − s| ≤ δε. Since both [0, T ] and [− 5 , 5] equipped with the standard metric are compact and separable spaces, we can conclude by Part 1. Part 3
(1) Let X be a nonempty set and let dX = ddisc. Let (Y, dY ) be an arbitrary metric space. Show that any function f : X → Y is uniformly continuous.
(2) Let X be a finite set i.e. X = {x 1 ,... , xn} for some n ∈ N and dX any metric on X. Let (Y, dY ) be a compact metric space. Show that any sequence of functions (fn)n∈N, fn : X → Y is equicontinuous. Can you say that any sequence of functions like this has a uniformly convergent subsequence? Justify your answer! Draw an analogy between these sequences of functions and bounded sequences in (Rn, d`^2 ). Solution (1) Take an arbitrary function f : X → Y , take any ε > 0 and set δ := 1/2. By this choice one has that dY (f (x), f (y)) ≤ ε,
for all x, y ∈ X such that dX (x, y) ≤ δ. Indeed, the only pairs that satisfy dX (x, y) ≤ δ are such that x = y, for which f (x) = f (y), hence dY (f (x), f (y)) = 0. This shows that any function is uniformly continuous. (2) For any function f : X → Y and for any ε > 0, set δ := min{dX (x, y) : x, y ∈ X}/2, which is a well-defined positive number since X is a finite set. Now, similarly as in (1), one has that dX (x, y) ≤ δ implies x = y, thus dY (f (x), f (y)) = 0 ≤ ε. Since the choice of δ is independent of f and the points x, y ∈ X (and actually of ε as well), we have that any sequence in this setting is equicontinuous. Since any finite metric space is compact and separable, we are in the framework of Part 1, so we can conclude. This last scenario is analogous to the Bolzano-Weierstrass theorem in (Rn, d` 2 ). Indeed, any function that maps X to Y , it basically creates n−tuples in Y (because we have to describe its values at n different point only). So the whole space of function between X and Y behaves exactly as Y × Y ×... Y =: Y n. Since Y is compact, Y n^ is also compact, so any sequence in Y n^ has a convergent subsequence. This means that any function sequence between X and Y has a pointwise convergent subsequence. But all these functions are uniformly continuous and any sequence is equicontinuous, hence the convergence is uniform as well.
Part 2 Let us consider a metric space (X, d) and a nonempty subset E ⊆ X. Let us consider a family F of subsets of X with the following properties
- if A ∈ F, then int(A) 6 = ∅;
- for each B ⊂ X nonempty and open there exists A ∈ F such that A ⊆ B.
Two players P 1 and P 2 play the following game: they alternatively choose sets F 1 ⊃ F 2 ⊃ F 3 ⊃... from the family F. P 1 wins if and only if
E ∩
n
Fn
otherwise P 2 wins. Notice that this game might not end in finitely many moves.
(1) Let X = {x 1 , x 2 , x 3 } be a set with 3 distinct elements, let d = ddisc and let E = {x 1 }. Write down the family F associated to this configuration. (2) Show that in the case of (1), P 1 has a winning strategy, i.e. there is a set of moves for P 1 such that no matter what P 2 does, P 1 wins using this set of moves. (3) Show that for an arbitrary metric space (X, d) with an isolated point x 0 ∈ X, and for a subset E of X containing x 0 , P 1 has a winning strategy. In how many moves does the game end in this case? (4) Show that for an arbitrary metric space (X, d) and E ⊂ X nonempty and open, P 1 has a winning strategy. Give an example of a metric space and an open subset of it E s.t. P 2 has a winning strategy. (5) Let (X, d) be an arbitrary metric space and let U ⊆ X be a nonempty open set and let x 0 ∈ U. Show that U \ {x 0 } is an open set. (6) Let (X, d) be an arbitrary metric space that has no isolated points. Let E be a nonempty countable subset of X. Show that in this case P 2 has a winning strategy. Describe this strategy. Hint: notice that the objective of P 1 is to choose sets which contain as many elements from E as possible, while the objective of P 2 if to choose sets which contain as few elements as possible from E. (7) In the situation as in (6), what changes if E is a finite set? Can P 2 win in less moves?
Solution. (1) A possible choice for the family F is to consider all possible subsets of X, i.e. F := {{x 1 }, {x 2 }, {x 3 }, {x 1 , x 2 }, {x 1 , x 3 }, {x 2 , x 3 }, {x 1 , x 2 , x 3 }}.
Another choice that satisfies the properties is for instance F := {{x 1 }, {x 2 }, {x 3 }, {x 1 , x 3 }}. Other correct choices are also possible (2) P 1 choses {x 1 }, then P 2 cannot choose anything, moreover E ∩ {x 1 } = {x 1 }, so P 1 wins. (3) Since x 0 is an isolated point of X, {x 0 } is open. By the second property, it will be an element of F. The winning strategy for P 1 , as in (2), is to choose {x 0 } at the first step. Then the game ends in one move. (4) Let us consider (Q, d) as a metric space where d is a standard metric on R restricted to Q. Let E = Q, then P 2 has a winning strategy. Indeed, whatever P 1 choses, P 2 removes at least one element in the next step, so that the consider set is in F. Since the rationals are countable, by this procedure eventually after infinitely many steps P 2 wins the game. Actually this strategy is the same as in (6). (5) Notice that X \ (U \ {x 0 }) = (X \ U ) ∪ {x 0 }, moreover this last set is closed since it is the union of two closed sets. Thus the complement of U \ {x 0 } is closed, so this set it open. (6) Let us denote E = {x 1 , x 2 ,.. .}. Suppose that P 1 chooses the set F 1 , which has U 1 as its nonempty interior. Then U 1 \ {x 1 } is an open set, and P 2 chooses a set from F that is contained in U 1 \ {x 1 } (this is possible by the second point). After, whatever P 1 does, P 2 behaves as described before, and after each move of P 2 , at least one element of E is excluded from the chain of chosen sets. Thus, this is the winning strategy of P 2. (7) Yes. P 2 can win in a finite number of steps.
