So, we have seen so far what is meant by saying
that sequence of functions or a series of
function converges either point wise or uniformly
and we have seen that uniform convergence
has many desirable properties. For example,
what we have proved in the last class is that
when each f n is continuous the limit and
if f n converges to f uniformly then f is
also continuous.
We have also seen that if a sequence converges
uniformly it also converges point wise, but
the converse is false and there is one theorem
which puts some additional conditions along
with the point wise convergence. That implies
uniform convergence and that is a theorem
which we shall discuss today, before proceeding
further it is a fairly well known theorem.
It is known as Dini's theorem and, so there
are few additional things till, now whenever
we talked about convergence of a sequence
of functions either point wise or uniformly
we did not say anything about the set E on
which the functions were defined. But, now
we shall require those things we will have
to assume that E is a compact set in some
metric space, E is compact set in some metric
space. You can take it as a compact metric
space and f n and f these are functions from
E to R and f n converges to f point wise that
is will say f n x converges to f x as n tends
to infinity for every x.
So, we want to say under what additional conditions
f n converges to f uniformly of course this
one additional condition is already there
compact set. But, that is not sufficient we
will have to put a few more things first thing
is that we have to assume that f n and f all
of them are continuous that is, that is f
n and f those are continuous function. That
is f n each f n is in continuous, f is also
continuous and this f n is a monotonically
increasing sequence or monotonically decreasing
sequence. It has to be a monotonic sequence
if we assume that, then we can show that f
n converges to f uniformly.
Let us take one of the cases suppose it is
monotonically increasing and let us say that
f n x is less not equal to f n plus 1, x for
every x in E and n in n then f n converges
to f uniformly. This is the theorem that is
what are the additional requirements, E is
a compact set f n as well as f they are all
continuous functions and f n is monotonic.
Here, we have taken monotonically increasing
sequence, but a similar thing can proved if
you take a monotonically decreasing sequence.
Let us, now look at the proof, let us take
some x in E, now again you want to proof f
n converges to f uniformly means by definition
what we have to do is that given epsilon we
have to produce some n 0 such that whenever
n is bigger than or equal to n 0, mod f n
x minus f x is less than epsilon. That should
happen for every x in E that n 0 should work
for every x that is the idea, so this such
a proof has to start with let epsilon be bigger
than 0, let epsilon be bigger than 0. Let
us consider some x in E, consider some x in
E then since we know that f n x converges
to f x corresponding to this epsilon there
will exist some n 0.
So, that where ever is n is bigger than n
0 whatever happens we say will happen, but
that n 0 will depend on x because we have
only assumed the point wise convergence, so
let us call it n suffix x. So, we can say
that there exist n suffix x in fact our whole
point is to find n 0 which does not depend
that is the whole idea of the proof. So, starting
initially we do not know that, so let us take
there exist n suffix n in such that n bigger
than or equal to n suffix x this implies mod
f n x minus f x less than epsilon. Now, let
us also use the fact that this is monotonically
increasing sequence for each x f n, x is a
monotonically increasing sequence of real
numbers.
We know that a monotonically increasing sequence
converges to its supremum monotonically, so
each f n x is less not equal to f x and f
x is the supremum of all those f n x. So,
let us rewrite this and I will use for this
number n suffix x, I will take is equal to
n suffix x, so what do we know this n suffix
x we of course we know that f n x at x this
is always less not equal to f x. But, because
f n x minus f x is less than this, what this
means is that f n x lies between f x minus
epsilon to f x plus epsilon. So, this must
be f n x also must be bigger than this is
f x will obviously less than f x plus epsilon,
but this will be f n x will be bigger than
f x minus epsilon f n x is bigger than f x
minus epsilon I will just look at this part
of the inequality.
That is, that is the one which is useful,
now and what it means is that f x minus this
f n suffix x at x, remember everything is
happening at x this f depends x I am taking
that particular function f suffix n suffix
x evaluating that at x, evaluating n at x,
evaluating f at x. What follows from this
is that the difference between this is less
than epsilon, difference between this is less
than epsilon. Now, we know that each f n as
well as f is a continuous function each f
n as well as f is a continuous function. So,
what I will do is that I will take the set
of all those num points in E for which this
happens that is what is by I will, I will
call that and that will of course, depend
on this x that will of course depend on this
x.
So, suppose I call that let us say U suffix
x, what is this U suffix x, this is defined
as let us say set of all y in E, set of all
y in E such that f at y minus f suffix n suffix
x at y. This should be less than epsilon,
is it obvious that x belongs to this x because
that is what this says if you take. So, first
obviously that x belongs to U suffix x 
is it also clear that U suffix x is open 
this y is less than epsilon that is it is
inversely match as under the continuous function
f minus f n x of the interval.
You can take minus infinity to epsilon of
course actually 0 to epsilon, so it is an
inverse image of an open set under a continuous
function. So, this is an open set, so U suffix
and that is where we are using a continuity
of f n and f suffix n x. So, what do we say
that since f n f suffix n x are continuous,
U x is open, U suffix is open and non empty
of course because x belongs to u suffix x.
What did we do given a point x in, we have
constructed an open set containing that open
or we just call open neighbourhood of x. Then
what is to be done after this because this
is a standard compactness argument of that
is we have started with e as a compact set
cover E with all such sets. So, that, so E
is contained in, E is contained in union of
U suffix x union is taken over x in E, in
other words what it means is that.
This suppose it take this family U suffix
x, x belonging to E this is an open cover
of E, this is an open cover of E, yes why
it is an open cover obviously each U x is,
U x is open set and every x in E is contained
in some U x. So, you collect all such U x,
E will be a subset of this union, so that
means is open cover of E you can make a slight
change in this definition. Suppose, E itself
is you are taking as a metric space then you
can take these, otherwise you can take suppose
E is a subset of some metric space x instead
of taking y belonging to E you can take y
in that x, you can take y in that x. But,
usually for compactness argument that does
not matter, if E is a compact set in any metric
space then it is a compact set regarded if
you regard E itself as a metric space.
So, that really does not matter what follows
after this because E is compact this should
have a finite sub cover, since E is compact
you can say its finite sub cover means there
exist some finite number of points. Let us
say x 1, x 2, x k there exist x 1, x 2, let
us say suppose it happens up to x k such that
E is contained in union of this U suffix x
suffix k, sorry U suffix let us say x suffix
j, j going from 1 to k, j going from 1 to
k. Now, what you do is you look at the corresponding
ns, for each x j you have n suffix x suffix
j and then take n 0 as the maximum of all
of them. So, let n 0 maximum of maximum of
n suffix x 1, n suffix 2 etcetera n suffix
x k then let us click, suppose we take any
n bigger than equal to n 0.
Let us say what happens, let n bigger than
or equal to n 0 then what we want to show
is that for if you if you look at n f t for
any n bigger than or equal to n 0, then we
want to say that mod f n f n t minus f t is
less than epsilon. Let n bigger than or equal
to n 0 and t or x or whatever you take n bigger
than or equal to n 0 and x in E 
if x is in E x is in one of this U x, 1 U
x, 2 U x k because this is E is contained
in the union of U x 1, U x 2, U x k etcetera.
So, every point x in E is one of these open
sets, so then x belongs to this U suffix x
suffix j for some j, for some j, j going from
1 to K 1 of does, but if x belongs to U x
j means, what it means.
That I think in order to avoid the confusion,
I will change this notation, here instead
of x let me take y, so that you do not confuse
that with this x we start actually consideration
of this x is over, here U suffix x does not
matter. So, let us take some y in E then y
belongs to U suffix x j, but what is the meaning
of saying that y belongs to u suffix x j,
look at what is the definition of u suffix
x it set of all those y in E for which f y
minus f n x y is less than now. Here, you
have n suffix x suffix j, so y belongs to,
y belongs to U x j this means f y minus f
n suffix x j at y is less than epsilon let
us rewrite this is, this is same as saying
that that is f y minus epsilon is less than
n f suffix n suffix x suffix j of y. Now,
till now we have not used this fact only once
we have used that f n is a increasing sequence,
now for each n we know that f n x is less
not equal to f n plus 1 x this is less not
equal to f n plus 2 x. So, whenever n is less
than m f n x is less than f m x, now what
is relationship between this n x j and n 0
this is less not equal to n 0 is maximum of
this two. So, I can say that this is less
not equal to f n 0 of y and we have taken
n bigger not equal to n 0 E have taken n bigger
not equal to n 0. So, f n 0 of y this is less
not equal to f n of y, can we always say that
f whatever be y f n y is always less not equal
to f y because again it is a monotonically
increasing sequence, and f y is the supremum
of all of all of those f n by and obviously
f y will be less than f y plus epsilon.
So, what did we prove just look at this f
y minus epsilon that is we taken n bigger
than or equal to n 0 and y is in E, then f
y minus epsilon less than f n y and then less
than f y plus epsilon. So, suppose you combine
these three inequalities 
is less than epsilon and this is true for
every y and for the same n 0, remember we
did not change n 0, this n 0 does not depend
on y, once this n 0 is chosen as maximum of
this that is fixed. So, if you take any n
bigger than or equal to 0 and any y in E then
this inequality holds f y minus f n, y less
than epsilon and that is same as saying that
f n converges to f uniformly.
This is what we wanted to show f n converges
to f uniformly, so let me again remind you
that what are the additional things that we
required E is a compact set then f n and n.
They are all continuous functions and f n
f n is monotonically increasing sequence and
each of these things we have used in the proof.
For example, compactness we have used their
f n is monotonically increasing that we have
used, here and to show to say that this is
an open set each of these are continuous functions
all those things are used here. Now, if you
have read the introduction of Seaman's book
there is one remark that he has made, here
that is whenever you learn a theorem each
theorem has certain number of hypothesis.
Some conclusions that there will be a proof
will have several steps and suppose you understand
how n plus 1 step follows from n step. Suppose
you understand this for each n that does not
really mean that you have understood the proof
completely. So, what is required is to understand
what is the basic idea of the proof and what
is the test for this test for this is that
you ask yourself whether each hypothesis in
the theorem is essential, whether it is required.
Now, how does one say that for example saying
that this hypothesis used in this proof that
is not a good enough answers it will only
mean that this proof requires that. There
may be some other proof without which does
not use and still may be possible to prove
the theorem without using the hypothesis.
So, there is only one wants to settle such
questions and what is the way you just drop
that particular hypothesis written all other
hypothesis and whether the conclusion still
follows. If your answer no for that question
there is only one way to settle this, you
have to have an example, you have to have
an example where all the hypothesis are satisfied.
Except the particular one which you are testing,
which you are testing and the conclusion is
false, and the conclusion is false. So, in
order to under, suppose we apply that thing,
here what are the additional things we have,
we have I should t is compact of whether that
is essential. We have to see, we have to do
this by talking some non compact set whether
f n and f continuous that is essential. Again,
we have to see by dropping that, so certain
examples we have already seen, let me just
remind you.
So, suppose I take E as 0 to 1, E as 0 to
1 this is in fact this is I think the starting
example we have taken f n x as x to the power,
n we have taken f n x as x to the power n
and what was f x f x was 0 for 0 less not
equal to x less than 1, for x equal to 1.
Now, look at the theorem, here what were the
hypothesis E is compact that is, now each
f n is continuous that is fine f is not continuous,
is f n monotonically increasing, is it not.
But, it is monotonically decreasing that does
not matter we have seen that whether it is
increasing or decreasing, that does not matter
is the convergence uniform we have seen, that
it is not we have seen by several basis it
is not uniform.
So, what does it mean that f n and f are continuous
that f is continuous that hypothesis, you
cannot draw that hypothesis you cannot draw.
Let us take let us take one more example of
similar type, suppose I want to check whether
compactness can be dropped, so you take E
as instead of this suppose I take E as open
interval 0 to 1 then that is not compact.
Now, suppose I take f n x of x as 1 by 1 plus
n x, f n x as 1 by 1 plus n x those this converse
point wise to what suppose you fix x and let
n vary, let n go to infinity what will happen
this will go to 0.
So, this goes to 0 as n tends to infinity
point wise, so if you take f as a constant
function 0 that is f x is 0 for x in 0 to
1.Then f n tends to f point wise 
f n tents to f point wise, is it monotonic
suppose you fix x, how are f n plus 1 f n
x related, it is it is decreasing, it is a
decreasing sequence. So, everything else is
satisfied except E is compact, is this convergence
uniform, how do you settle that, I have told
you one easiest way of settling the uniform
convergence you look at M n as supremum of
mod f n x minus f x in x n.
So, look at in fact that is the one which
will work in most of the examples, so let
M n be equal to supremum of mod f n x minus
f x where x is in E. So, what does this mean
in our situation everything is positive, so
f x is 0, so mod f n x and f n x is it is
just 1 by 1 plus n x, 1 by 1 plus n x for
E is 0. So, 0 less than x less than one supremum
of 1 by 1 plus n x and what is and what do
we know about uniform convergence f n will
converge to f uniformly if and only if M n
goes to 0 M n goes to 0. Suppose you take
n to be, suppose you take x equal to 1 by
n can you do that 1 by n lies between 0 to
1 for every n, then what will happen to this
it should be 1 by 2.
So, I cannot say that M n than or equal to
1 by 2, always I do, I may not it is possible
to convert the exact value of M n also, but
that is not really essential. So, can we say
that M n is bigger than or equal to 1 by 2
for all n, so what does it say if M n is bigger
than or equal to 1 by 2 for all n, obviously
it cannot go to 0? That means f n does not
converge to f uniformly, f n does not converge
to f uniformly, so that means the hypothesis
of compactness also cannot be dropped, so
what remains?
Now, f n is monotonically increasing or decreasing
whatever it is I think I will leave that for
you to as an exercise, check on your own whether
this can be dropped check on your own whether
that means what you have to construct an example.
Where f n satisfies all other hypothesis except
this and the convergence is not uniform if
this is essential, if this is not essential
then by with remaining hypothesis you should
be able to proof uniform convergence, either
you have to settle it. Now, let us look at
one more property uniform convergence and
till, now we discussed about the continuity,
now let us next is differentiability and integrability.
We will first look at integrability because
that is little easier to discuss and to talk
of integrability we cannot take any arbitrary
set, now we have to take an interval.
So, let us say that a b is an interval and
suppose you consider f n from a b to R converges
uniformly to some function f from a b to R
of course convergence uniformly on a b, convergence
uniformly on a b then what we want to say
is that each f n is integrable f is also integral
and the integral of f n converges to integral
of f. This can be proved for Remap’s integrals
and this can also be proved for Remap’s
theory integrals with, without any extra works.
So, let us do it for the Remap’s theory
integrals, so let us take a monotonically
increasing function, let alpha from a b to
R be monotonically increasing. Then if f n
belongs to let us use this notation again
R a b alpha that is, that means f n is Remap’s
theory integrable with respect to this function
alpha for each n.
Then f is also Remap’s theory integrable
and limit as n tends to infinity of integral
a to b f t alpha, sorry integral a to b f
n d alpha, this is same as integral a to b
f t alpha. See we have seen an example where
this is false under point wise convergence,
we have seen example of a sequence f n converging
to f. But, integral f n does not converge
to integral of f, in fact for Remap’s integrals
also this is not true if the convergence is
not uniform, so this is a, this is a theorem
on integration and uniform convergence this
is what we were discussing, integration and
uniform convergence.
So, there are two things to be proved, here
first is we have to prove that f is Remap’s
theory integrable and then we have to also
prove that other thing we shall select to
first to prove f is Remap’s theory integrable.
We have known that one of the standard techniques
is that given epsilon you should produce some
partition such that for that partition U p
f l alpha minus L p f alpha is less than epsilon
that is what we shall try to do.
.
So, let epsilon be bigger than 0 then we know
that each f n is integrable, so for each f
n such a partition exist for each f n such
a partition exist. So the idea is that we
will take one of the portions and if n is
large enough difference between f n and f
is small, so we should naturally expect that
difference between upper some of f n and upper
some of f is small.
Similarly, difference between upper lower
some of f n and lower some of f is also small
and that is the idea that we are going to
and since in doing this kind of things we
will have to add and subtract certain terms
2 3 times. We shall use the usual technique
of the, so call epsilon by three proofs that
is let us say suppose we that is we if we
have to add whatever quantity that we want
to show as a small quantity we express that
as a sum of the three quantities and show
each of that as less than epsilon by 3 that
is the idea.
So, first step is this since, here we have
taken this alpha I will chose you can say
something else, suppose I call it eta bigger
than 0 such that eta into alpha b minus alpha
a is less than epsilon by 3 or which is same
as saying that 0 less than eta less than epsilon
by 3 into alpha b minus alpha a. This is this
is something we can always do then since f
n tends to f uniformly what we can say is
that we can always find some n 0 such that
whenever n is bigger than or equal to n 0
difference between f n x and f x is less than
this eta.
Difference between f n x and f x is less than
this eta, so since f n tends to f uniformly
on a b, there exist n 0 in n such that mod
of f n x minus f x is less than eta, mod of
f n x minus f x is less than eta. Remember
that this is for every x in a b that is where
the uniform convergence comes into picture,
this is for every x in a b this is for every
x in 
a b. Now, consider one such n you can even
take n equal to n 0 or n 0 plus 1 or anything,
so consider n bigger than or equal to consider
some n bigger than or equal to n zero. Now,
this f n is integrable, this f n is integrable,
so there will exist some partition p such
that difference between upper and lower sum
for that partition is less than, whatever
you want is less than let us say epsilon by
3.
So, there exist a partition p in script p
such that U p f n alpha minus L p f n alpha
is less than epsilon by 3, but of course this
is not what we are interested, what we want
to show is that the difference between U p
f alpha and L p f alpha. That is small and
as I told you the idea is that we shall use
this that f mod f n x minus f x is less than
eta, idea is the following. We want to show
this, we want to show this is small U p l
alpha minus L p f alpha, we want to show this
is small, so what we will do is that we will
add this terms U p f n alpha minus L p f n
alpha that is we shall write this as U p f
alpha minus U p f n alpha.
That is adding and subtracting U p f n alpha
than plus u p f n alpha minus L p f n alpha
and finally plus L p f n alpha minus L p f
alpha and we will show that each of this less
than epsilon by 3, each of this less than
epsilon by 3. Out of which we already know
about the middle term, here we know that the
U p f n alpha minus L p f n alpha is less
than epsilon by 3, only thing remains to be
shown at these two are also less than epsilon
by 3 that is where we shall use this, now
to do that let us look at this partition p.
So, suppose p is let us say as usual a is
equal to x naught less than x 1 less than
x n is equal to b and what is U p f alpha,
for U p to consider U p f n alpha or U p f
alpha you have to look at the corresponding
supremum in the sub interval x i minus 1 to
x i. So, suppose let us say that M i is equal
to supremum of let us say f n x for x in x
i minus 1 to x i and let us say M i star is
equal to supremum of f x for x in x i minus
1 to x i.
Look at this what this says is mod f n x minus
f x is less than eta for every x in a b, let
us rewrite it what this means is that f n
x lies between f x minus eta f x plus eta
that is, I will just rewrite this inequality
what does this mean. That f x minus eta is
less than f n x is less than f x plus eta
this is true for remember that is important
that is true for x in a b because that is
what follows from that is what follows from
uniform convergence for every x in a b. Hence,
for every x in this x i minus 1 to x i also
that means each f n x, here will be less than
f x plus eta for every, so can I say from
here that M i must be less than M i star plus
eta M i.
So, I will say that, so what I will get is
M i is less than because remember, here you
are taking supremum of f x. Here, we taking
supreme of f n x and f n x is less than f
x plus eta, so we can say that M i less than
M i star plus eta this is for every i, for
every i then, so if you look at U p f alpha
or let me write U p f n alpha that is nothing
but sigma pi going from 1 to n. If it is f
n it is M i, M i into delta x i, not delta
x i delta alpha i, M i into delta alpha i
and U p f alpha that is nothing but sigma
i going from one to n M i star into delta
alpha i.
So, if each of this M i is less than M i star
plus eta I can say that this is less than
sigma i going from 1 to n M i star plus eta
into delta alpha i. But, what is this suppose
I split this it is sigma M i delta alpha i
that is nothing but U p f alpha, that is nothing
but U p f alpha and plus eta times sigma delta
alpha i going from 1 to n. This is something
we have calculated several times sigma delta
alpha i is nothing but alpha b minus alpha
a, alpha b minus alpha a and, so this is we
have chosen eta in such a way that eta times
alpha b minus alpha a is less than epsilon
by 3. So, that is the, so what is the that
result, that if you look at the difference
between U p f alpha and U p f n alpha that
is in another words what I want to say is
this U p f alpha minus U p f n alpha.
This difference is less than eta into alpha
b minus alpha a, it is the other way less
than eta into alpha b minus alpha a and this
is less than epsilon by 3, this is less than
epsilon by 3. Now, in a similar way if you
look in a similar way, if you look at the
lower sums instead of taking supremum we are
taking infimum instead of taking lower sums.
So, a similar relationship will be true, there
a similar relationship will also be true there
and with in a similar way we can show that
the difference between the lower sum of f
n and lower sum of f that is also be less
than epsilon by 3.
So, suppose you combine all these you get
finally that the difference between U p f
alpha minus L p f alpha is less than epsilon
that will show that each i, each f n is integrable
f is also remains in this integrable again
recall see what is the idea. Here, we are
writing this U p f alpha minus L p f alpha
as the difference as the sum of these three
quantities and U p if for large n U p f n
alpha minus L p f n alpha. That can be, that
can be made less than epsilon by 3 because
in, because f n is integrable because f n
is integrable.
When n is large the difference between mod
f n x minus f x can be made arbitrarily small
and that is why the difference between the
corresponding lower sums. The corresponding
upper sums can be made arbitrarily small that
is the idea that is the idea, to what remains
we have to show this that limit we already
know that this integral exist. Now, we have
to show that limit of integral a to b f n
t alpha limit of that is same as this, now
to do that let us, let us recall something
that we have proved earlier.
Do you remember we have proved this that if
you look at any integral from a to b f d alpha
we have prove that if f is integrable mod
f is also integrable and absolute value of
this is less not equal to integral a to b
mod f d alpha this what we have proved. Then
what I will say is that again consider a let
us say this n 0 eta at everything is as already
chosen, here keep this same things and again
choose some n bigger not equal to n 0. Consider
n bigger not equal to n 0, consider n bigger
not equal to n 0 and look at the difference
between these two integrals integral a to
b f d alpha minus integral a to b f n d alpha
I can write this as follows.
Integral of this, as integral of a to b minus
f n d alpha 
that is from the properties of integral, now
I use this, so what is means is that this
is less not equal to integral a to b mod f
minus f n d alpha. Now, again look at what
we have said here for every x in if n is bigger
not equal to n 0 for every x in a b mod f
n x minus f x is less than eta mod f n x minus
f x is less than eta. So, what follows from
that this function if you look at this function
mod f minus f n x it is value is less than
eta in every x in a b. So, what can we say
about the integral it will, it will integral
will less than eta into alpha b minus alpha
a, so this will be less than eta into alpha
b minus alpha a.
We have already seen that this must be less
than epsilon by 3, than epsilon by 3 and of
course epsilon by 3 is always less than epsilon,
so what did we show that given any epsilon
bigger than 0 there exist n 0. So, whenever
n is bigger not equal to n 0 the difference
between these two numbers is less than epsilon
that is same as showing that this, that is
same as showing this. So, again recall what
we have shown that if f n converges to f uniformly
and if each f n is integrable then f is also
integrable and integral of f n converges to
integral of f.
Now, let us write an implication of this or
the series because that is something, that
is something more useful in practice and that
is what you require very often and what is
that. Suppose sigma f n, n going from 1 to
infinity is uniformly convergent let us say
this is converges uniformly, converges uniformly.
Since we use f already we will use something
else, since converges uniformly let us say
to some function g 
then what does this mean that if you take
the sequence of partial sums s n then s n
converges to g uniformly. Now, let us assume
that sub, let us assume the same suppose each
f n is integrable that it will mean that g
is integrable, g is integrable at then integral
of s n will converge to integral of g but,
what is integral of s n.
Let us say what is s n, s n is f 1 plus f
2 plus f n, so integral of s n integral a
to b s n d alpha that will be same as integral
a to b f 1 d alpha etcetera plus integral
a to b f n t alpha integral a to b f n t.
So, in other words what I can say is that
if sigma f n converges to g uniformly than
g is if g also belongs to R a b alpha that
means g is also integrable and integral a
to b g d alpha is same as sigma n going from
one to infinity integral a to b f integral.
That is integral of the sum is same as sum
of the integrals of course this is something
we already know the finite sums, the question
is about the infinite sums.
In case of infinite sums it is in general
false it holds only when the convergence is
uniform that is important it holds when the
if the series converges uniformly then the
series form by taking the integrals converges
to the integral of the sum. In other words
this is roughly express by saying that you
can integrate this series term by term that
you can interchange the operations of integration
and summation. Suppose you write this in the
full form what is the meaning of this, this
g is nothing but integral summation of suppose
I write in the full form.
This will mean integral a to b sigma f n d
alpha, this is same as this sigma this is,
here sigma n going from one to infinity sigma
n going from 1 to infinity integral a to b
this should be f n, f n d alpha. In other
words what this means is that the operations
of integration and summation can be interchanged,
can be interchanged and this can be done provided
the convergence is uniform that is what this
theorem says. So, we will stop with this,
in the next class we shall discuss the differentiation
and uniform convergence relationship between
differentiation and uniform convergence.
