Exam Final - Math 139
Dec 16th
Instructor
Web page
Mauro Maggioni
www.math.duke.edu/ mauro/teaching.html
You have 3 hours. You may not use books, internet, notes, calculators. The exam should
be stapled, written legibly, with your name written at the top of every page, on one side
of each page only, and must contain the rearmation of the Duke community standard.
You can invoke any Theorems that we discussed in class and are in the sections of the
book we studied. No partial credit is given for statements for which proof is not provided.
Problems are roughly ordered in increasing diculty, and value (in terms of points). (*)
preceeds parts of problems that are for extra credit: most of these are not particularly
hard, but do take extra time.
1. Let f C 1 ([a, b]) and f (x) > 0 for all x [a, b]. Prove that f is strictly monotonically
increasing, i.e. f (x) < f (y ) for all x < y , x, y [a, b].
[5 pts.]
Fix x, y [a, b]. Without loss of generality we may assume that x < y . By the Mean
Value Theorem, applied on [x, y ] to the C 1 function f , we have
f (y ) f (x) = f ( )(y x)
for some point (x, y ). Since f is strictly positive everywhere and y x > 0, the right
hand side is strictly positive, proving that f (y ) > f (x). Since x, y were arbitrary, we are
done.
2. Show that equation arctan(x) = 1 x has at least one real solution.
[5 pts.]
Let f (x) = arctan(x) (1 x). f is continuous on the whole real line, f (0) = 1 and
f (x) (1 ) =: L > 0 as x + since 1 < 0. Therefore there exists
2
2
2
M > 0 such that f (x) > 0 for x M by denition of limit as x + and the fact that
the limit is positive (in fact, there exists M such that |f (x) L| < L/2 and therefore
f (x) > L/2 > 0 for x M ). In particular f (M ) > 0, and by the Intermediate Value
Theorem applied in [0, M ] to the continuous function f , there exists (0, M ) such that
f ( ) = 0, i.e. arctan( ) = 1 .
3. Dene
(a) what a metric space (X, ) is;
(b) what does it mean for a sequence in (X, ) to converge;
(c) what does it mean for a sequence in (X, ) to be Cauchy;
(d) what does it mean for (X, ) to be complete.
State an interesting Theorem that crucially uses completeness.
1
Finally, consider the set A = +=1 {(x, n ) R2 : x R} in R2 . Show that A is not
n
complete (with respect to the Euclidean distance in R2 ).
(*) Complete A (with respect to the Euclidean distance in R2 ).
[10 pts.]
(a) A metric space (X, ) is a pair consisting of a set X and a metric , i.e. a function
: X X [0, +) satisfying the following conditions:
(x, y ) 0 for all x, y X , and (x, y ) = 0 if and only if x = y ;
(x, y ) = (y, x) for all x, y X ;
(x, z ) (x, y ) + (y, z ) for all x, y, z X .
(b) A sequence {xn }n X (i.e. a map N X written as n xn ) is convergent to a
point x X if for every > 0 there exists N = N () such that for all n N we have
(xn , x) < .
(c) A sequence {xn }n X is Cauchy if for every > 0 there exists N = N () such that
for all n, m N we have (xn , xm ) < .
(d) A metric space is complete if every Cauchy sequence is convergent to some point of
X.
We state the contraction mapping theorem: let (X, ) be a complete metric space and
T a contraction X X , i.e. (T (x), T (y )) (x, y ) for all x, y X , for some < 1
independent of x, y ). Then there exists a unique xed point x X for T , i.e. such that
T (x ) = x . Moreover, if x0 is any point in X , the sequence xn = T n (x0 ) converges to
x .
1
Consider the sequence {xn }n := {( n , 0)}+=1 , which is contained in A. If the sequence
n
converges to x A, then it also converges to x R2 since A has the metric of R2
1
restricted to A. But xn (0, 0) in R2 , since ||xn 0||R2 = n 0, therefore it converges
to (0, 0) also as a sequence in A. But (0, 0) A.
/
(*) The completion of A is A := A {(x, 0) : x R|}. Clearly every point in A has to be
in the completion of A, since (x, 0) A is the limit of at least one Cauchy sequence, for
example of the sequence {(x, 1/n)}n . A is complete: in fact, let {(xn , yn )}n be a Cauchy
sequence in A. Such a sequence is also Cauchy in R2 : let (x , y ) R2 be its limit (R2
is complete). Then xn x and yn y (in R). Since the rst coordinate of points in
A can be any real number, there is no issue with x . As for y , it can be either 0 or 1/n
for some n, since yn {1/n : n N}. To see this, notice that either there exists > 0
such that yn > for all ns, in which case y = 1/n for some n, or for all > 0 there
exists n such that yn < , in which case there is a subsequence of {yn }n converging to 0,
and since {yn }n is Cauchy, the whole sequence {yn }n converges to 0. We conclude that
(x , y ) A, so that A is complete, and it is the smallest complete space including A.
4. Let f C ([0, 1]). Prove or disprove:
(a) if
1
0
f (t)dt = 0, then f = 0.
(b) if f (x) 0 for every x [0, 1] and
(c) if for every x [0, 1],
x
0
1
0
f (t)dt = 0, then f = 0.
f (t)dt = 0, then f = 0.
[10 pts.]
(a) False. For example f (x) = sin(2x) is not identically 0 and integrates to 0 on the
interval [0, 1].
(b) True. By contradiction, assume that there exists a point s.t. f ( ) > 0. Assume
> 0, otherwise we slightly modify the proof to allow for asymmetric intervals around .
By continuity, there exists an interval ( , + ) on which f is larger than f ( )/2 > 0.
But then
1
0
+
f (x)dx
f (x)dx f ( )/2 2 > 0 ,
contradicting our assumptions (here we used the fact that g (x) h(x) for all x [0, 1]
implies g h).
(c) True. By contradiction, suppose that f = 0, i.e. there exists a point [0, 1] :
f ( ) = 0. Without loss of generality we may assume that f ( ) > 0. Proceeding as in (b),
assuming > 0 (otherwise we slightly modify the proof to allow for asymmetric intervals
around ), f (x) > f ( )/2 on some interval ( , + ). But then
+
f (t)dt > f ( )/2 2
and at the same time
+
+
f (t)dt =
0
f (t)dt
by our assumptions. This is a contradiction.
0
f (t)dt = 0 0 = 0
5. Consider a bounded sequence {xn }n in R. Let x be the set of limit points of {xn }n .
(a) Write the denition of x .
(b) What can you say about the cardinality of x ? Can it be 0 (i.e. x = )? Finite?
Countably innite? Uncountably innite?
(c) Is x bounded?
Let {yn }n be a second bounded sequence in R, and y the corresponding set of limit
points. Let a x and b y . Is a b necessarily a limit point of {xn yn }n ?
(*) Prove that x is closed.
[10 pts.]
(a) x = {y R : there exists a subsequence of {xn }n that converges to y }.
(b) x cannot be empty: since {xn }n is bounded, it is contained in an interval [M, M ]
for some M > 0, and by Bolzano-Weierstrass {xn }n has a convergent subsequence, whose
limit is by denition a point in x .
x may be nite: for example if {xn }n converges to a point x R, then x = x . By
picking the union of k sequences converging to k dierent points x , . . . , x we obtain a
1
k
set of limit points of cardinality exactly k .
x may be countable: for example consider the sequence
{1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, 1, 1/2, 1/3, 1/4, 1/5, 1/6, . . . } .
Then x = A := {1/n : n N+ } {0}. Indeed, for each n N we can clearly nd a
subsequence which is constant and equal to 1/n and therefore 1/n x , so N x . Also
there is clearly at least one subsequence converging to 0, so 0 x . On the other hand
no other point can be in x : if y R \ A is in x , then no point in sequence has distance
to y less than the (positive!) distance of y to the nearest point in A (which is either the
point 1/n where n is the nearest integer to 1/y , or 0), so that y cannot be the limit of
any subsequence.
x may be uncountable: for example consider the set A = Q [0, 1]: since A is countable
(it is contained in the countable set Q) we may list the elements of A in a sequence {xn }.
Then we claim x = [0, 1]: in fact for every point x [0, 1] there exist innitely many
elements of A in the interval (x 1/k, x + 1/k ) [0, 1] (this is a consequence of the
Archimedean property, which we discussed in classed and proved in the homework): for
each k we may select one such point xnk , making sure that nk+1 > nk (this is possible
since there are innitely many points of A in (x 1/k, x + 1/k ) [0, 1], so that they
cannot have been exhausted by xn1 , . . . , xnk ). Then this is a subsequence of {xn } that
converges to x .
(c) x is bounded. In fact, let M > 0 be such that |xn | M for all n (such an M exists
because {xn } is bounded). x Let x and {xnk }k a subsequence of {xn } converging to
x . Since |xnk | M , necessarily, by taking limits, |x | M . Since x was arbitrary in
x , x [M, M ].
No. For example let xn = (1)n and yn = (1)n+1 . Then xn yn = (1)2n+1 = 1 for
all ns, so that the set of limit points of {xn yn } is the set {1}. On the other hand
1 x and 1 y , but 1 1 is not a limit point of {xn yn }.
Pick a point x which is the limit of a sequence {x }n in x . We want to show that
n
x x , i.e. that the exists a subsequence of our original sequence {xn } that converges
to x . We construct it as follows: for n = 1, let xn1 be an element of {xn } such that
|xn1 x | < 1: such an element exists (and in fact there are innitely many) because
1
x x and therefore it is the limit of a subsequence of {xn }n . By induction, having
1
constructed xn1 , . . . , xnk , we let xnk+1 be an element of o{xn }n>nk such that |xnk+1
x +1 | < 1/(k + 1): such element certainly exists since x +1 is a limit point of {xn }.
k
k
Then {xnk }k is a subsequence of {xn } (since {nk }k is constructed to be increasing) and
converges to x since |xnk x | |xnk x | + |x x | 1/k + |x x | 0 as k +.
k
k
k
6. Suppose f C (R) and |f (x)| Cx5 for some C > 0 and all x R. Find the domain
D of the function
n
F (x) = lim
f (t)dt .
n +
x
Then, not necessarily in the following order, show that: F uniformly continuous on
D , F is continuously dierentiable and F is uniformly bounded on D (i.e. M > 0 :
|F (x)| < M for all x D ). [Hint: be careful in explaining the manipulations you do
with all the limits involved.]. Find F .
+
(*) Show that F (x) = x f (t)dt.
[10 pts.]
Since f is continuous, it is Riemann integrable on any bounded interval. |f | is also
bounded by, say, M on [1, 1], since f is continuous, so |f | is upper bounded by the
function g which is M on [1, 1] and equal to Cx5 for x [1, 1]. We know that g
/
is Riemann integrable on R, and since |f | is upper bounded by g , |f | is also Riemann
integrable on R and therefore f is. In particular f is Riemann integrable on [x, +) for
any x R, so that the domain F is R. This also proves (*).
We start by showing that F is dierentiable everywhere and its derivative is bounded.
n
Consider the function Fn = x f (t)dt. Then Fn converges uniformly to F :
+
|Fn (x) F (x)| =
n
+
f (t)dt
n
Cx5 cn4
for some constant c, and the righthand side goes to 0 uniformly in x (since it does
not depend on x). Similarly, Fn converges uniformly to f (x), since Fn (x) = f (x)
(independently of n). We can therefore invoke the Theorem about uniform convergence
of a sequence of functions and their derivatives to conclude that F is dierentiable, and
F = f (x). But f is bounded on R by the reasoning above (f is dominated by g which
is bounded by max{M, C/M 5 } on R). Since F is bounded uniformly on R, we conclude
that F is uniformly continuous, since by the mean value theorem we have
|F (x) F (y )| = |F ( )| |x y | ||F || |x y | .
7. A set K in a metric space is compact if for every sequence {xn } K there exists a
convergent subsequence whose limit is also in K .
(a) Give examples of compact and non-compact sets in R (with the usual distance),
and describe a large class of compact subsets of R. Is [0, +) compact?
(b) Show the the union of two compact sets is a compact set. (*) Generalize this
statement to the union of nitely many compact sets. (*) Does the statement hold
for the union of innitely many compact sets?
(c) Consider C ([0, 1]) as a metric space with the distance induced by || || , i.e.
d(f, g ) = ||f g || = sup |f (x) g (x)|
,
x[0,1]
for f, g C ([0, 1]) .
Prove that the unit ball {f C ([0, 1]) : ||f || 1} is bounded (i.e. M > 0
such that ||f || M for all f B ), closed (i.e. the limit point of any convergent
sequence in B is contained in B ), but not compact. [Hint: nd a sequence in B
which has no convergent subsequence.]
[15 pts.]
(a) By Bolzano-Weierstrass, any bounded closed interval [a, b] is compact. (0, 1] is not
compact, since {1/n} tends to 0 but 0 (0, 1]. [0, +) is not compact since {n}n
/
[0, +) does not have any convergent subsequence, since such subsequence would be
Cauchy, but any two elements would be at distance at least 1 from each other.
(b) Let K = K1 K2 with K1 , K2 compact. Let {xn }n be a sequence in K . Then
{xn }n has innitely many elements at least one of the two sets K1 and K2 . Without
loss of generality, let K1 contain innitely many elements of {xn }n , that we may list as a
subsequence {xnk }k . This subsequence is completely contained in K1 and therefore has
a convergent subsequence to an element of K1 , which is also an element of K . Such a
subsequence is a subsequence of the original sequence {xn }n , thereby proving that K is
compact.
(*)Clearly this may be extended to any nite union of compact sets, since we only used
the fact that at least one of them needs to contain countably many elements of the
original sequence.
(*) No. For example let, for every n N, Kn = {n} (the singleton set containing the
integer n). Clearly Kn is compact, +=1 = N, but N is not compact (the sequence
n
{n}n = {1, 2, 3, . . . } has no convergent subsequence).
(c) The unit ball - lets call it B - is clearly bounded by denition, as we may choose
M = 1. It is also closed: if {fn } is a sequence in B that converges to f in the metric d,
i.e. uniformly (by a Theorem proved in class), then f is in B , since
||f || lim ||f fn || + ||fn || 1 .
On the other hand B is not compact: consider the sequence of functions {fn }n with
fn (x) = xn . This sequence converges pointwise to the discontinuous function f which
is 0 for x [0, 1) and 1 at x = 1. Let fnk be any subsequence: it will also converges
pointwise to f , but certainly it cannot converge uniformly to f since fnk is continuous
and f is not. We have therefore exhibited a sequence in B which has no convergent
subsequence, so B cannot be compact.
8. Let C 1 ([0, 1]) be the set of continuously dierentiable functions on [0, 1]:
C 1 ([0, 1]) := {f : [0, 1] R : f is continuously dierentiable } .
For f C 1 ([0, 1]) dene
||f ||C 1 := |f (0)| + ||f || = |f (0)| + sup |f (x)| .
x[0,1]
(a) Prove that ||f ||C 1 is a norm on C 1 ([0, 1]).
(b) (*) Prove that C 1 ([0, 1]) is complete with respect to (the distance associated with)
this norm.
(c) Does
+
sin(nx)en
n=0
converge in this norm?
[15 pts.]
(a) Let ||h|| = supx[0,1] |h(x)| in what follows. Clearly ||f ||C 1 0, and if ||f ||C 1 = 0
the |f (0)| = 0 and ||f || = 0 so that f (0) = 0 and f is constant, and therefore f is
identically 0. Also, ||f ||C 1 = ||(|f (0)| + ||f || ) = || ||f ||C 1 for any R. Finally
||f + g ||C 1 = |f (0) + g (0)| + ||f + g || |f (0)| + ||f || + |g (0)| + ||g || = ||f ||C 1 + ||g ||C 1
(where we used the fact that || || satises the triangle inequality), proving the triangle
inequality.
(b) (*) Let {fn } be a Cauchy sequence in || ||C 1 . Then
|fn (0) fm (0)| ||fn fm ||C 1
implies that the numerical sequence {fn (0)} is Cauchy and therefore convergence to some
number that we call f (0); moreover
||fn fm || ||fn fm ||C 1
implies that the sequence of continuous functions {fn } is Cauchy and therefore uniformly
convergent to some continuous function g . By applying the Theorem on uniform convergence of derivates, which also needs convergence at one point (0 in this case), we deduce
that {fn } converges uniformly to a function f and that f exists and is equal to g ; in
particular f C 1 ([0, 1]). In fact, fn converges to f in || ||C 1 , since
||fn f ||C 1 |fn (0) f (0)| + ||fn f || 0 .
This proves completeness, since we showed that any Cauchy sequence converges to a
point in C 1 ([0, 1]).
n
k
Let fn (x) =
k =0 sin(kx)e : by Weierstrass M -test, fn converges uniformly, since
k
k
k
| sin(kx)e | e = (1/e) is the k -th term of a geometric sequence with ratio 1/e < 1,
therefore convergent. Analogously, fn also converges uniformly, by applying again Weierk
k/2 k/2
strass M test: |k cos(kx)e | ke
e
Cek/2 since kek/2 0 and therefore it
is bounded, and the upper bound is the k -th term in geometric series of ratio 1/ e < 1.
Let f be the limit of fn : then f is C 1 and moreover f = lim fn (by the same Theorem
used above). Then
||f fn ||C 1 = |f (0) fn (0)| + ||f fn || 0
since both terms go to 0 (the rst by pointwise convergence at 0, which is implied the
uniform convergence of fn to f , and the second since fn converges to f ). This proves
that the series converges in || ||C 1 .