Master’s Thesis
The following link includes a draft of my master’s thesis. It is still in progress and may contain some typos and errors.
The following link includes a draft of my master’s thesis. It is still in progress and may contain some typos and errors.
In this article, we will investigate interplays between various kinds of convergence of sequence of random variables. Specifically, we will discuss convergence a.s., convergence in probability, convergence in th mean and convergence in distribution.
We can naturally formalize the notions of sequence of random variables being Cauchy in these contexts of convergence. We say: is Cauchy in probability if
as
;
is Cauchy a.s. if
is Cauchy everywhere except a set of measure zero;
is Cauchy in
if
as
. Now we give some basic facts about the advantages of using Cauchy’s criterion.
Proposition 1
a.s. if and only if
for every
as
.
Proof: Let us consider the formulation of the set of points where ; that is
In fact, we can use the denseness of real numbers to write
So
Proposition 2
is Cauchy a.s. if and only if
as
for every
; or equivalently
as
for every
.
Proof: Using a similar argument, we obtain a chain of equivalences,
where the last equivalence is obtain from the following inequality
This completes the proof.
Remark: Similarly one can formulate another equivalent definition of Cauchy a.s., that is if for all
as
, then
is Cauchy a.s. Let
. Then there is an integer
such that
whenever
. Sending
gives
which is just the previous proposition.
1. Partial fractions
Theorem 1 (Mittag-Leffler) Let
be a sequence of complex numbers with
and let
be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with poles at the points
and the corresponding singular parts
. Moreover, the most general meromorphic function of this kind can be written in the form
where
‘s are suitably chosen polynomials and
is analytic in the whole plane.
Proof: The idea is to subtract a certain portion of analytic part of to ensure convergence without adding any poles. For each
, the singular part
is analytic in the ball
. So we can obtain Taylor expansion
of
around the origin up to the degree
. Using estimate for the remainder term, we have
where and
denotes the maximum of
in
. In particular, this estimate implies that
For each , we may pick
large enough to ensure
, which can be done by making
. In doing so, we obtain
Given , we shall show that the series converges uniformly in
. We may split the series
into two categories: the first one with
and the second one with
. Then
represents a meromorphic function with poles at
because it is a finite sum. In the latter category, since
, we have
in the disk.Thus above estimate implies
This completes the proof. Let us try to investigate several standard examples. Consider the function
Recall that and so
. We then can work out a first few terms in the Laurant expansion at the origin which is
Thus the singular part of at 0 is
. Moreover, as
has period 1 namely
for all
, we can see that the singular part at integers are
. Clearly
converges on all compact subsets which do not contain integers. It follows that we may write
where is entire. Write
we see that
. From the uniform convergence of the series, we can see that
converges to 0 as
. Using the idendity
we see that uniformly as
. In particular, this implies
is bounded in the strip
. By periodicity,
is bounded in the entire complex plane. Then Liouville’s theorem says the
reduces to a constant. In this case,
must vanish as the limit is zero. Therefore, we have the identity
Integrate on both sides, we obtain that
In order to make the right-hand side converge, we need to substract extra terms from the Taylor series of . In this case, subtracting all constant term is enough. That is
which is comparable to
. Thus we have
We may bracket terms of and
together, and obtain
Now is necessary because both sides are odd.
Next we exploit this fact to investigate the sum
We may split into odd terms and even terms
Clearly we know that and
. Thus in the limit we have
2. Product representation
An infinite product of complex numbers
is evaluated by taking limits of the partial products . WLOG, we say that
converges if and only if at most a finite number of the factors are zero and if the partial products formed by the nonvanishing factors tend to a finite nonzero limit. By writing
, we see that a necessary condition is that
. Thus WLOG, we may write
so that is a necessary condition for convergence. If no factor is zero, it is natural to compare with
We then have the following result:
Lemma 2 The infinite product
converges simultaneously with the series
whose terms represent the values of the principal branch of the logarithm.
Proof: Suppose converges. Denote the partial sum
. Then clearly
and by continuity of
we have
.
Lemma 3 The product
converges absolutely if and only if
converges.
Proof: Recall that . Then if either the product of the series converges, we see that
. Then we employ the following inequality
to derive the result. Next we prove Weierstrass products formula.
Theorem 4 There exists an entire function with prescribed zeros
provided in the infinitely many zeros case that
. Every entire function with these zeros and no other zeros can be written in the form
where the product is taken over all
,
is an entire function.
Proof: First we shall show that the product converges. This is direct because
By taking , we see that for
, the series
as we may take
sufficiently large enough to make
.
Corollary 5 Every function which is meromorphic in the whole plane is the quotient of two entire function.
Proof: Suppose is meromorphic in the whole plane. Let
be an entire function with poles of
for zeros. Then
is an entire function. Then
as desired.
The product representation becomes more interesting if it is possible to pick a uniform
. The previous proof shows that the product
converges to an entire function if the series converges for all
. That is to say
. Assume that
is the smallest integer for which this series converges; the previous product expression is called the canonical product associated with the sequence
and
is the genus of the canonical product. If the exponent
in the product representation of an entire function
with
reduces to a polynomial, then is of finite genus. The genus of
is equal to
. For instance, an entire function of genus zero is of the form
with . The canonical representation of an entire function of genus 1 is either of the form
with ; or of the form
with .
Next we discuss an example: . It has zeros at
. Since
adn
, we must take
so that
In order to determine we form logarithmic derivative on both sides and find
This shows that and
reduces to a constant. Since
as
, we must have
identically. Hence we can write
3. Hadamard product
Recall that an entire function is said to of order is
is the smallest number such that
for all as soon as
is sufficiently large. Here
denotes the maximum modulus of
on
. The genus and the order are closely related as seen by the following theorem.
Theorem 6 The genus
and the order
of an entire function satisfies
Proof: Suppose that is an entire function of order
. First we claim that
for any given
. This will address the issue of convergence of the canonical product. Given
. Recall Jensen’s formula
and the fact
Apply these formulae with and we will get
This implies as
. Moreover, note we also have the relation
This implies
Pick so that
will show that
.
In this post, we will discuss the product representation of entire functions. In order to do so, we first discuss the convergence of infinite products.
1. Infinite products
Given a sequence of complex numbers , we say the product
converges if the limit
exists.
Theorem 1 If
, then
converges. Moreover, the product converges to 0 if and only if one of its factors is 0.
Proof: WLOG, we may assume that for all
. Thus we may use power series to define logarithm
so that
. Then we obtain the following partial products
For each , we make the estimate
Using continuity of , we obtain the first result. To prove the second assertion, we note that if
for all
, then clearly the products converges to
which is nonzero.
Similarly we have convergence theorem for functions.
Theorem 2 Suppose
is a sequence of holomorphic functions in an open set
. If for all
we have
then
converges to a holomorphic function. Moreover, if
does not vanish in
for any
, then
2. Weisterass product
Theorem 3 Given any sequence
of complex numbers with
as
, there exists an entire function
that vanishes at all
and nowhere else. Any other entire function is of the form
where
is entire.
In general the function may not converge, so we need to insert extra factors to make it converge without adding any zeros. For each integer
, we define
and
. Then we prove the following estimate:
Lemma 4 If
, then
for some constant
.
Proof: Since , we obtain the taylor expansion
. Thus we have
Clearly . In particular,
, thus
This provides a desired bound. Proof: Define
. Fix
, we will show that
converges uniformly on
. Decompose
and we only need to consider the latter factor. Clearly
for all
and all
and by previous lemma we see that the product converges uniformly in
. Moreover, they never vanish in
which implies the limit does not vanish also. Hence the function
provides the desired function.
3. Hadamard product
In order to refine the result, we need some elementary results. First tool is Jensen’s formula.
Theorem 5 Let
be an open set contains the closure of a disk
and suppose
is holomorphic in
,
, and never vanishes on the circle
. If
denote the zeros of
inside the disc, then
Proof: First note that if the theorem holds true for and
then it holds true for
. Consider
which is holomorphic function in
that vanishes nowhere. So we only need to show that the theorem is true for
and
. Since
has no zeros we need to show
. This is clearly true by mean-value property of the harmonic function
. It remains to show
This boils down to show for any
. Making change of variables
shows this is equivalent to
. Consider the function
which is holomorphic and vanishes nowhere in the unit disk. Thus applying mean-value property to
we have
Next we claim that if
and
does not vanish on the circle
, then
This is immediate if we prove
Lemma 6 If
are the zeros of
inside
then
Proof: We start from the right-hand side.
4. Functions of finite order
Let be an entire function. We call
has order of growth
if
In particular, the order of is
.
Theorem 7 If
is an entire function that has order of growth
, then
for sufficiently large
. Moreover if
are zeros of
, then
converges for any
.
Proof: We shall use Jensen’s formula. Note that is an increasing function of
. Now choose
, then clearly we have
By Jensen’s formula we have
This proves the first part. Then we directly estimate
5. Hadamard products
Theorem 8 Suppose
is entire and has order of growth
. Let
be the integer such that
. If
are zeros of
, then
where
is a polynomial of degree
,
is the order of zero at
.
Proof: The convergence of the canonical products is now easily solved. Using Lemma 2.2, we have . Thus the convergence of
implies the convergence of the product. In order to show that
is a polynomial of degree
, we need more efforts.
Lemma 9 We have the following estimates
Proof of Lemma. If
, we directly have
If
, then
Lemma 10 For any
, we have
except possibly for
.
Proof of Lemma. Again we use usual decomposition
We first estimate the second product. Note that for all
. By previous lemma, we have
Note that for some constant
as
. Thus we have
because . To estimate the first prouct we use previous lemma to obtain
Using a similar trick we have . Thus
It remains to estimate . But using the property that
, we have
Now we consider
This completes the proof of the lemma.
Lemma 11 There exists a sequence of complex numbers
with
such that
Proof of Lemma. First we note that converges so we may take
. For each positive integer
we may find
such that the the disk
does not intersect the forbidden areas in previous lemma. Suppose it is not the case, then there is a particular
such that all balls of radius
with
will intersect the forbidden area. In particular, by rotating the those balls on the real line so that the diameter will lie in the positive real axis, we see that
. This implies
which is a contradiction.
Now we prove the Hadamard’s theorem. Let . Previously we’ve shown that
converges to a holomorphic which has zeros as
does and does not vanish everywhere else. Thus we have
and in particular
This implies on
. Finally we show that this is enough to guarantee that
is a polynomial of degree less than
.
Lemma 12 Suppose
is entire and
satisfies
for a sequence of positive real numbers
that tends to infinity. Then
is a polynomial of degree
.
Proof of Lemma. First we write . Let
denote the circle centerer at the origin with radius
. By Cauchy’s integral formula, we have
whenver
. This implies
and
which further implies
Thus
Recall that whenever
. Thus for
, we have
Letting shows that
if
. Hence
is a polynomial of degree
.
Definition: If there exists a 1-1 mapping of onto
, we say that
and
can be put in 1-1 correspondence , or that
and
have the same cardinal number , or briefly, that
and
are equivalent, and we write
. This relation clearly has the following three properties:
Definition: For any positive integer , let
be the set whose elements are the integers,
; let
be the set of all positive integers. For any set
, we say:
Remark: A finite set cannot be equivalent to one of its proper subsets. However, this is possible for infinite sets. In fact, we could restate the definition (b): is infinite if
for some
.
Theorem 1 Every infinite subset of a countable set
is countable.
Proof: Suppose and
is infinite. Arrange the elements of
which are all distinct in a sequence
. Then we construct a sequence
as follows:
Let be the smallest number such that
. Note
exists because
is well-ordered. After choosing
, we let
be the smallest number greater than
such that
.
Then consider the function where
. We then check
is indeed a bijection. It is one-to-one because
is a distinct sequence; then for any
, we have
.
must be surjective for otherwise, there exists
is not in the image of
contradicting the fact that
.
\hfill
Corollary 2 Every subset of an at most countable set is at most countable.
Proof: Let be an at most countable set and
. If
is finite, then
is must also be finite. So let us assume
is countable. Then
cannot be uncountable by theorem 1.
\hfill
Corollary 3 An at most countable set that contains a countable subset is countable.
Proof: Let be at most countable and
is countable. If
is countable then we are done. So we need to prove that
cannot be finite. Suppose
is finite then there exists a bijection
. Let
. Then the composition
must be injective. However, by pigeonhole’s principle we know this cannot happen, thus a contradiction.
\hfill
Theorem 4 Countable unions of countable sets is countable.
Corollary 5 At most countable unions of at most countable sets is at most countable.
Let be an arbitrary family of sets. We define the Cartesian product
to be all the functions
such that
. We can think of
as an evaluation of
at the
-th entry. For example if
,
is simply the first entry of the Cartesian product.
Theorem 6 Finitely many Cartesian products of countable sets is countable.
Theorem 7 Countable Cartesian products of countable sets is uncountable.
Proof: Let where
is countable for each
. Let
be an enumeration. For each
we pick two different element
. Consider the following function:
Note obviously. However, note
for all
because
differs from every
at least at
. Thus
in not countable.
\hfill
Corollary 8 Let
be the set of all sequences whose elements are the digis 0 and 1. Then
is uncountable.
Proof: Remember the set is indeed identified with the Cartisian product
. Although we cannot directly apply theorem 7 because
is finite, we still can use the same Cantor trick. Let
be an enumeration of
. We define the following function
Again obviously, but
differs from
for every
. Thus
is uncountable.
Theorem 9 There exists no surjection from a set
to
.
Proof: We prove by contradiction. Let be a surjection. By definition, for any
there exists
such that
. Consider the following set
So has an pre-image in
. Let
such that
. It is either
or
. If it is the first case, then
and
by definition, but his contradicts the fact that
. If it is the second case, we have
, so
and
, contradiction again. Hence
cannot be a surjection and the proof is complete.
\hfill
Finally the principle states that every convergent sequence nearly converges uniformly. Let’s figure out what the word “nearly” means in this context and this brings us to the Egorov’s theorem.
Theorem 1 (Egorov)Suppose
is a sequence of measurable functions defined on a measurable set
with
, and assume
almost everywhere on
. Given
, we can find a closed set
such that
and
uniformly on
.
Proof: We may assume for every
since we can always find a subset
which differs with
by a measure zero set such that the convergence holds everywhere on
, then we just rename
by
. Define
Note for fixed we have the relation
and
as
. Then by the continuity of measures we know
which implies we can find an integer
such that
. By construction, we have
whenever
and
. Then we choose
so that
and let
.
First we notice the relation . Next for any
, we can choose
such that
when
. Now we indeed have completed our job as for any
we can find an integer
such that for all
we \left|f_{j}(x)-f(x)\right|<\delta<\frac{1}{n} for all
.
Finally, as a technical requirement, we can find a closed subset such that
which leads us to the desired property
.
The second principle asserts that every function is nearly continuous. Now we investigate the meaning of “nearly” in terms of mathematics.
Theorem 2 (Lusin) Suppose f is measurable and finite valued on E with E of finite measure. Then for every
there exists a closed set
with
and such that
is continuous.
Proof: Since by assumption, is measurable and finite valued on E, we can find a sequence of step functions,
such that
almost everywhere. Note that each
is a step function which means it is continuous almost everywhere as it contains only at most countably many discontinuities which have Lebesgue measure 0. So without loss of generality, say for each
we can find sets
with
so that
is continuous outside
. Now to proceed, we need something extra which is obtained from the third littlewood’s principle, i.e., Egrovff’s theorem. It enables us to find a set
with
such that
uniformly on
. Then for the integer N such that
, we consider
For every , the step function
is continuous on
; therefore, the uniform limit
is also continous on
. To complete the proof, notice that
is also measurable so we can approximate from below by a closed set
such that
which gives us the desired
.
3. Measures and Outer measures
Let be a measurable space.
Definition: A measure on is a function
such that:
(1)
(2) For all families of disjoint measurable sets,
Remarks:
Proposition 6 : Let
, we have the following:
(1) If, then
.
(2) Ifand
, then
(3)
(4) Iffor all
and
for all
, then
.
(5) Iffor all
and
for all
, then
.
(6) If, then
.
To be updated.