The Dirichlet Test
Theorem (The Dirichlet Test) Let X be a metric space. If the functions fn : X C,
gn : X IR, n IN obey
n
Fn (x) =
fm (x) is bounded uniformly in n and x
m=1
gn+1 (x) gn (x) for all x X and n IN
g n ( x)
nIN
converges uniformly to zero
Math 321 Assignment 11 Solutions
Q1: Let f : R R be a measurable function such that f 0. Show that for any number A > 0
we have
1
m(cfw_x : f (x) A)
f dm .
A R
Solution: Put E = cfw_x : f (x) A then
f dm
f dm Am(E) .
R
E
The rst inequality is because f
Midterm February 26th 2014
Mathematics 321
Page 1 of 5
This midterm has 4 questions on 5 pages, for a total of 100 points.
Duration: 50 minutes
Read all the questions carefully before starting to work.
Continue on the back of the previous page if you ru
Mathematics 321
Practice Exam April 22nd 2014
Page 1 of 7
This exam has 6 questions on 7 pages, for a total of 100 points.
Duration: 2.5 hours
Read all the questions carefully before starting to work.
Continue on the back of the previous page if you run
Homework #6  Practice midterm due on February 24th 2014
Math 321
Page 1 of 5
This midterm has 4 questions on 5 pages, for a total of 100 points.
Duration: 50 minutes
Read all the questions carefully before starting to work.
Continue on the back of the
Midterm Solutions  Math 321
1. Give complete denitions of the following terms:
(a) an equicontinuous family of functions in C[0, 1].
Solution. A family of functions F C[0, 1] is said to be equicontinuous if for every
> 0, there exists > 0 such that
(1)

Homework 6 Solutions  Math 321, Spring 2015
1. Given a nonconstant nondecreasing function : [a, b] R, let R [a, b] denote the collection
of all bounded functions on [a, b] which are RiemannStieltjes integrable with respect to .
Is R [a, b] a vector spa
Homework 5 Solutions Math 321, Spring 2015
1. The classical Weierstrass approximation theorem says that the class of polynomials is dense
in C[a, b]. The StoneWeierstrass theorem, on the other hand, provides a necessary and
sucient condition for a subal
Homework 2 Solutions  Math 321,Spring 2015
(1) For each a [0, 1] consider fa B[0, 1] i.e. fa : [0, 1] [0, 1] dened by
fa (t) =
1 if t = a
0 if t = a
There are uncountably many such fa as [0, 1] is uncountable. Also if a = b then
fa fb = 1 (if it is clear
Homework 1 Solutions  Math 321,Spring 2015
1a Pointwise convergence: for a x x [0, ) If x = 0, fn (x) = 0 for all n so it converges
to 0. If x = 0, we have
nx
= lim
n 1 + n2 x2
n
lim
x
n
1
n2
+
x2
=
0
=0
0 + x2
The convergence is not uniform on [0, ): If
Homework 3 Solutions  Math 321,Spring 2015
(1) 1a. We want to nd a sequence cfw_pn of polynomial such that pn f uniformly
on any bounded subset of R. By The First Weierstrass Approximation Theorem,for
each positive integer N , we can nd a sequence of po
Midterm Review  Math 321, Spring 2015
1. Give an example of an equicontinuous family of nonconstant functions that is not totally
bounded.
Sketch of solution. The function class LipK [0, 1] \ cfw_constant functions for any xed K provides an example. Thi
Alternate Proof of Integrability
Theorem. Let : [a, b] IR be monotone and f : [a, b] IR be continuous. Then f R() on [a, b]. That
b
is, the integral a f d exists.
Proof: We use the Cauchy criterion, which says
Let , f : [a, b] IR. Then f is integrable wit
Products of Riemann Integrable Functions
For these notes, let < a < b < and : [a, b] IR be nondecreasing. We shall prove
Theorem 1 Let f : [a, b] [M, M ] be Riemann integrable with respect to on [a, b] and let : [M, M ]
IR be continuous. Then f (which is
Reduction to the Riemann Integral
Theorem. Let a < b. Let f : [a, b] IR be bounded and : [a, b] IR
have a continuous derivative on [a, b]. Then
f R on [a, b] f R() on [a, b]
and, if either integral exists,
b
b
f (x) d(x)
f (x) (x) dx =
a
a
Corollary. In t
RiemannStieltjes Integrals with a Step Function
Theorem. Let : [a, b] IR be a step function with discontinuities at s1 < < sn ,
where a s1 and sn b. Let f : [a, b] IR be continuous at each sj , 1 j n. Then
f R() on [a, b] and
n
b
f (sj ) (sj +) (sj )
f d
Interchanging the Order of Summation
Corollary (Interchanging the Order of Summation)
If
then
ajk <
j =1 k=1
j =1
k=1
ajk =
j =1 k=1
Remark. The hypothesis
ajk
k=1 j =1
ajk < really means that
for each j IN,
ajk = Mj <
and
Mj <
j =1
k=1
The two double
A Nowhere Dierentiable Continuous Function
These notes contain a standard(1) example of a function f : IR IR that is continuous everywhere but
dierentiable nowhere. Dene the function : IR IR by the requirements that (x) = x for x [1, 1]
and that (x + 2)
The Change of Variables x = g (y )
Theorem. Let a < b and c < d. Let g : [c, d] [a, b] be continuous, strictly monotonic
and obey g (c) = a and g (d) = b. Let f, : [a, b] IR. Set
h(y ) = f g (y )
(y ) = g (y )
If f R() on [a, b], then h R( ) on [c, d] an
Dinis Theorem
Theorem (Dinis Theorem) Let K be a compact metric space. Let f : K IR be a
continuous function and fn : K IR, n IN, be a sequence of continuous functions. If
cfw_fn nIN converges pointwise to f and if
fn (x) fn+1 (x)
for all x K and all n IN
Equicontinuous Functions
Theorem (Arzel`Ascoli(1) ) Let K be a compact metric space, with metric dK (p, p ), and let C (K ) denote
a
the space of real (or complex) valued continuous functions on K . If fn nIN is a sequence in C (K ) obeying
(H1) for each
Functions of Bounded Variation
Our main theorem concerning the existence of RiemannStietjes integrals assures us that the integral
f (x) d(x) exists when f is continuous and is monotonic. Our linearity theorem then guarantees that
b
the integral a f (x) d
Homework 4 Solutions  Math 321, Spring 2015
1. Weierstrasss second theorem states that any continuous 2periodic function f on R is
uniformly approximable by trigonometric polynomials. The aim of this exercise is to prove
this statement.
(a) Deduce Weier