So, let us just recall we were looking at
Pointwise and Uniform Convergence of sequences
of functions. So, let me just recall that
we said that fn is a sequence of functions
defined on n. We say, fn converges to f pointwise
if fn of x, also f is also a function from
x to R, converges to f of x for every x belonging
to the domain.
And that means, we should not forget what
that means; means for every epsilon bigger
than 0, there is a stage n naught which may
depend upon, in general it will depend upon
epsilon and the point x, such that 
mod of fn x minus f of x is less than epsilon
for every n bigger than or equal to that natural
number n naught, which may depend upon n and
x. So, that stage may depend upon epsilon
of course and also on x.
So, then we define what is called fn converges
to f uniformly. If, same thing if for every
epsilon bigger than 0 there is a stage n naught
which now will not depend on the point, but
it will depend only upon epsilon. There is
a stage, there is a natural number n naught,
such that fn x minus f of x is less than epsilon
for every n bigger than n naught and for every,
so probably we should mention that, for every
x this happens.
So, we gave a lot of examples of sequences
which converge pointwise, which converge uniformly
and we had started looking at properties of
uniform convergence. We said that, pointwise
convergence need not preserve various properties,
namely if each fn is continuous f may not
be continuous, each fn is differentiable then
f may not be differentiable, and if each fn
is integrable, f may not be integrable.
So, to analyze these properties for uniform
convergence functions let us recall, if f
is a function defined on X to R and f is bounded.
We define what is called the infinity norm,
that is supremum x belonging to X of mod f
x. So for, and we observed that this is a
norm and gives a metric or so I think we called
it B X, R on the set of bounded functions
with domain X and taking values in R. So what
is that metric, that metric is d infinity
f, g is equal to norm of f minus g.
So, the relation of uniform convergence with
this norm is, let us put that as a theorem.
We proved it last time, that 
if fn, if fn X to R are bounded and f X to
R is also bounded then fn converges to f uniformly,
if and only if norm of fn minus f goes to
0. So, we had proved this theorem.
Let me also point out that here, so let us
observe a few things, some observations. One,
if fn converges to f uniformly and each fn
is bounded then f is also bounded. So, if
a sequence of functions converges uniformly
to a some function f, and each fn is bounded,
then the function f is also bounded. And hence,
you can apply earlier criteria because there
we assumed f is also bounded.
So, one way, if f is uniformly convergent,
each all fn’s are bounded then f is also
bounded. And hence, you can write and hence
norm of fn minus f converges to 0. That is
a consequence of the earlier theorem.
So, let us prove this. So, fn converges to
f uniformly. So that means, what does that
mean, that means for every epsilon bigger
than 0, there is a stage n naught which depends
only on epsilon, such that norm of, absolute
value of fn x minus f of x such that for every
x belonging to X is less than epsilon for
every n bigger than n naught. That is the
definition of uniform convergence.
So, let us specialize it. So, in particular,
it is not necessary but anyway, let us take
epsilon equal to 1, then for every x belonging
to X we have fn x minus f of x will be less
than 1 for every n bigger than that stage
n naught 1.
That means fn is close to f by distance 1
and fn’s are bounded anyway, so I can just
apply triangle inequality. Hence, for n greater
than n naught epsilon fixed, so fix one, any
one of the numbers, mod of fx I want to show
it is bounded, it is bounded by some scalar
for every x, then for every x belonging to
X. I know f of x is close to fn x, and fn’s
are bounded anyway. So, I can use the triangle
inequality.
Then for every x belonging to X this is true,
this is less than or equal to 1 plus supremum
over x belonging to X of fn x, that exists.
So, you can call this number as M if you like,
so implies mod fx is less than or equal to
1 plus M, for every x. So hence, f is bounded.
So, if a bounded 
sequence of real valued functions converges
uniformly then the limit also is a bounded
function. So, that is 1.
Let us look at, second property that we looked
at last time, so let us do it again anyway.
That if, fn converges to f uniformly, f continuous,
each fn is continuous at x equal to c, then
f is also, then f is also continuous at x
is equal to c. So, continuity is also preserved
under uniform convergence. So, let us prove
that.
So to prove this, what we have to do, so we
want to make for, we want to analyze the distance
between f of x and f of c. We want to say
that this distance can be made small whenever
x is close to c, for continuity. And what
we know is, fn is converging uniformly to
f. So close to f, there is a fn, at every
point and fn is also continuous.
So both these facts, so let us note this I
can add, so let me write less than or equal
to, f of x minus fn of x for any n, plus,
now fn x minus fn at c, that will be small
because of continuity of fn, so plus fn of
c, and the last is f of c.
Why I am doing this, I want the left hand
side to be small. But close to f there is
a fn, because fn is converging uniformly,
so I can estimate change f to fn. And fn’s
are given to be continuous, so I can use continuity
for the second term of fn. And finally, once
again, close to fn there is a f, fn is close.
So, all these 3 terms can be made small. So
how do I write it, so note for every n, for
every x this is true.
So, by uniform continuity, by uniform convergence,
by uniform convergence. So, this is how you
think and now how you write for uniform convergence,
given epsilon bigger than 0 choose the stage
n naught such that mod fn x minus f of x is
less than epsilon for every n bigger than
n naught and for every x.
So, that is by uniform convergence. So now,
let us fix some n bigger than n naught, bigger
than this n naught fix. By continuity of f
at, by continuity of f at the point x is equal
to c, given epsilon as before, that is already
fix, there is a delta such that mod of x minus
c less than delta implies f of x minus f of
c, fn so fn of c is less than epsilon.
So, there is continuity of fn, we are given
fn is continuous. So, we have fixed n bigger
than n naught, already epsilon is fixed, so
find the delta size that this happens. Then
for every x with x minus c less than delta
f of x minus f of c will be less than or equal
to, so go back, our starting point. Let us
call that as star, when we said that f is
close to fn, fn is continuous so that triangle
inequality, the first inequality. So, that
is what motivated us, so in the star use these
things, is less than, at least strictly less
than 3 epsilon.
And if you wanted nice thing, then you could
have gone back and modified everything, this
by epsilon by 3, this by epsilon by 3 and
this by epsilon by 3 and that could have been
epsilon, because epsilon is arbitrary. So,
you can always make it as small as you want.
So, that will prove.
So that proves, basically keep in mind what
we want to show, we want to make this thing
small, and we are given that fn’s are converging
to f uniformly and each fn is continuous.
So, bring in fn’s. So that is, continuity
is preserved under uniform convergence.
So, let us look at next, something which is
also preserved. So, third, fourth, I don't
know what is it, property number 3. Let us
look at integrability. Let, f belong to R
a, b, fn’s belong to R a, b and greater
than equal to 1, or Riemann integrable functions
on the interval a, b and fn converges to f
uniformly. Then f is also Riemann integrable,
so then f belongs to R a, b, and integral
fn d mu, dx converges to integral f dx, a
to b.
By the way, I think, let me just make a point
here that this theorem when we said uniform
limit of continuous functions is continuous,
another way of writing that. So let me put
a note; fn converges to f uniformly fn continuous
implies, see we look at limit fn of x, n going
to infinity, that is f of x, and as x goes
to c that is continuity.
So, limit x going to c is same as, you can
write it as, you can take the limit inside.
So, it is limit x going to, n going to infinity,
let me write n going to, n going to infinity
limit x going to c of fn of x. So, limit fn
will be f of c and f is continuous.
So this, conclusion of this you can write
it as like this. It is like interchanging
2 orders of taking limiting operations, limit
x going to c, limit n going to infinity is
same as limit n going to infinity and limit
x going to c, interchange of limits operations
are possible whenever the convergence is uniform.
So, that is another way of writing this theorem,
so which is useful way of observing. There
are 2 limit operations, so you can interchange
whenever there is uniform convergence, that
is what it says
So, that is something, something coming here
also, so I can write this as, so it says a
to b limit n going to infinity fn dx. What
is the right hand side, f is the limit, and
the right hand side f is the limit, so here,
this f is the limit. So, I can write as limit,
that is same as, this converges so that is
limit n going to infinity, integral a to b
fn x dx.
Again, there is interchange of limit and integral
here now. Integral of the limit is the limit
of the integrals. So again, so it essentially
says under uniform convergence integration
is a continuous operation kind of, integration
is continuity operation. Whenever there is
interchange, something implies continuity
of something.
So, here it is saying that the limit of the
integral is the integral of the limit. So,
the 2 operations are interchanged. So, let
us prove this, so proof by black. So, we are
given fn converges to f uniformly. So, this
is given and each fn belong to R a, b. We
want to show that f belongs to R a, b, then
only you can write it is integral. And integral
fn d mu converges to integral f d mu. So that
is what is to be shown.
Now if you want to show that f is Riemann
integrable, then you have to first show that
it is a bounded function, f should be a bounded
function.
But that follows from the fact that just now
we have shown. So note, so let us note each
fn is Riemann integrable, so that implies
each fn is bounded. We have shown that every
Riemann integrable function is also bounded.
Implies that f is bounded because fn converges
to f uniformly. So, all are bounded functions.
The only thing that we do not know now is,
whether f is Riemann integrable or not, f
is a bounded function.
So, let us try to show, we show first f is
Riemann integrable. We have got it is bounded,
it is okay. Boundedness helps us look at upper
and lower sums for a function. That way of
defining integration by way of upper and lower
sums is possible, only when f is given to
be bounded. In the Riemann definition f is
not assumed to be bounded. But anyway, we
have got now boundedness.
For this what we to show, for this we want
to show f is Riemann integrable. That means,
for this, given epsilon bigger than 0, to
find a partition P such that, when is a function
integrable, when given any epsilon, upper
and lower can be brought close to each other.
Such that upper sum minus the lower sum is
less than epsilon. So, this is what we want
to show. So given epsilon, if we can find
a partition P such that upper minus the lower
is, difference is small then we are done.
And keep in mind, how is the upper sum defined,
upper sum is, over a partition is by looking
at the maximum value of the function in any
subinterval multiplied by the length of the
interval, sum it up.
And what is given to us, given to us is f
is, each fn is Riemann integrable and fn’s
are converging to f uniformly. So, close to
f there is a fn, and fn is Riemann integrable.
So, the idea would be the upper sums of f,
we should try to approximate it by upper sums
of fn. That is the route we should follow.
So, let us do that.
So, by uniform convergence, since fn converges
to f uniformly, given epsilon greater than
0, there is a stage n naught such that all
are bounded. So let us write, mod fn minus
f infinity is less than epsilon for every
n bigger than n naught. Because all fn’s
f are bounded, so I can now write the supremum
less than, for every x and so supremum.
So that is same as saying, hence for every
x, for every n bigger than n naught, let us
write what is hidden here is that, fn x minus
f of x is less than epsilon. That is same
as saying that supremum and hence this also
is less than.
Sometimes expanding things help. What we want
to do is, we want to relate f with fn, the
maximum value of f with fn, the minimum value
of f with fn. So this gives me, this is same
as saying, if I look at fx it is less than,
expand this inequality absolute value, so
this is fn x minus epsilon is less than fn
x plus epsilon. This is same as saying as
this. The distance of fn from f is either
this side epsilon or that side epsilon. That
is same as saying this. Now this will give
us the required thing.
So, let us choose the partition. So given,
choose, so we have used uniform convergence,
now we use the integrability. Choose a partition
P, so let us call it as a equal to x, 0 less
than x less than equal to b of a, b such that,
so let us, we are not fix, so let us, n bigger
than n naught. So, let us take n naught itself,
such that U P fn naught minus L P fn naught
is less than epsilon.
What I have used, I have used the fact that
fn naught is Riemann integrable, because,
is Riemann integrable. So, there must be,
given epsilon there must be a partition such
that things are close. Now let us, so this
is, now that equation let us keep that in
mind, so this was the equation when fn is
close to, so star.
So, let us use star n equal to n naught, so
what we have, f n naught x minus epsilon is
less than f of x is less than fn naught of
x, I am just specialized because this is true
for all n bigger than n naught. So, let us
fix n naught so that we do not have any.
See this is what we are saying, fn naught
x is close to f of x by distance epsilon.
So, I can take the minimum values, so implies,
so let us look at the minimum value mi, let
me write fn naught what is this mi of f n
naught, that is the minimum value of the function
f n naught in the interval xi minus 1. So
if, let me write, it is equal to minimum of
fn naught x, x belonging to xi minus 1 and
xi then what does upper one give me? What
will this equation give me?
This is f n naught x, so it will be bigger
than the minimum value. So, mi of f n naught
minus epsilon will be less than f of x will
be less than mi fn naught plus epsilon. This
equation, this true for all x. So, let us
look x in the subinterval xi minus 1 to xi.
So, this will be bigger than the minimum value,
it is less than f n naught x, for every x.
So, take the minimum over that. So, this equation
gives me this.
Now from the minimum, value how do you go
to the lower sums, by multiplying by the length
and adding up. So, let us do that, so implies
sigma mi of fn naught multiplied by the length,
so xi minus xi minus 1, i equal to 1 to n.
When I multiply epsilon by those lengths,
those lengths will add up, summation, so it
will be just epsilon times b minus a.
This, this part is this is less than f of
x multiplied by the length is less than the
corresponding thing, so that is mi fn naught
xi minus xi minus 1 plus epsilon times b minus
a. What happened to the sum, here is the sum.
So, i equal to 1 to n.
The previous equation I have multiplied throughout
by xi into xi minus 1 and summed up. So, the
first term is mi, so this first term is, let
me just underline it so this thing, if you
multiply by xi minus 1 to xi minus 1, the
length of the interval and sum it up, when
you multiply by epsilon, that should be giving
you epsilon times P minus a. In between, I
should put the sum also, I forgot to put the
sum here, 1 to n, there is sum everywhere.
So, what does this give me now? Why f of x?
No. Actually, I can put here also the minimum
of f of x. So, let me do that. Anyway this,
I should have, I can do that. So, from here
when I multiply mi less than, I can put here
also the minimum of f.
See from this equation, let me just clarify,
this is true for every x. Now in the first
part here, take the minimum over fn naught.
This is for every x, so minimum of the function
fn naught in the interval xi minus 1 to xi,
so that will be less than or equal to f of
x for every x anyway. And there you can take
the minimum of f also.
So, you can get here minimum, so you will
get this part. Take first the minimum over
x, you get this quantity. This is less than
f of x. Take the minimum of fx over that interval.
So, that gives you the minimum function of
fi, mi, so what is that notation I am writing,
we are not writing that, let me just write,
what is that quantity, so that quantity is
mi of f.
So, I am saying I can just write that, is
f of x, this quantity is less than f of x,
so it will be less than equal to minimum also.
And that minimum will also be less than or
equal to minimum of that. So, I can take minimum
everywhere in this inequality, that is what
I am saying.
One at a time, but you should do, one at a
time to justify that. First take the minimum
over this part, this is for every x, so minimum
only in the left hand side, so you will get
this quantity. Less than equal to f of x for
every x, so take the minimum in that interval,
so that is mi. This is less than equal to
fn naught of x, so take the minimum over that,
so as is mi. So, I should have, so this f
of x I can replace it by mi of f.
Basically, the idea is fn is close to f by
margin of epsilon this side or that side.
So, I can take the minimum over the interval,
this is happening for every x, so I can take
the minimums.
So that means, so what is the meaning of this,
this means, this is the lower sum, so lower
sum of small mi, so lower sum of fn naught
with respect to P minus epsilon times b minus
a, this first part is less than or equal to
the lower sum of f with respect to P, f with
respect to P and the last term this one is
less than or, why less than or equal to, is
less than actually, it is strictly less than.
It does not matter actually, is less than
L fn naught, it is the minimum over f n naught,
so fn naught of P plus epsilon times b minus
a.
Basically, what I am saying is this equation,
if I take the minimum over all x in the interval
xi minus 1 to xi, so equation holds for minimum
also. Multiply by the length of the interval
and add up, that gives you the lower sums
minus epsilon times this.
So similarly, the upper ones. So, that means
what, the lower sum of f and the lower sum
of this is less than, that means mod of L,
we are writing fn first, fn naught P minus
the lower sum f, P is less than epsilon. In
the absolute value I can write it this way.
This minus b minus a, epsilon times b minus
a. I am writing that in equality back in terms
of absolute value, nothing more than that.
So similarly, you will have upper sum fn naught
P minus the lower sum of f, P will be less
than epsilon times b minus a. Instead of taking
minimum, you can take the supremum and the
same thing. So hence, upper sum P, f minus
the lower sum, I want to estimate this quantity,
for f.
So that is why, so now add and subtract. So,
less than equal or to absolute value triangle
inequality, upper sum P, f, but that is close
to upper sum of Pn, P, fn naught, I should
interchange, because I am writing P first
and then f, so be consistent so that is, so
we wanted to estimate f...
Sir the second equation is
This one?
Yes sir, it is U fn mod comma P minus U
fn naught, no, that is U, sure, sure, thank
you. That is upper sum. I said similarly,
the lower and similarly the upper
maximum instead of minimum
Where are in that same thing here. Now this
quantity is less than or equal to f of x,
so maximum of that must be less than or equal
to f of x, less than or equal to maximum of
f x. I am saying same inequality gives you
both, lower as well as the upper.
So, now I want to estimate this quantity.
For, so add and subtract, so this is less
than or equal to upper sum f, P is close to
upper sum of fn naught P plus, so I should
add upper sum of fn naught P minus lower sum
of fn naught P plus, the last term now left
would be lower sum of fn naught P minus lower
sum of L P, add and subtract. This technique
by now we are very familiar.
So, less than or equal to upper f and fn so
this is this equation now. Call it 2, call
it 3. Using 2 and 3, the first one is less
than epsilon times b minus a. The last also
is less than same. So, 2 times this, upper
minus the lower is less than epsilon. So,
that we have already seen, would where was
that, upper minus the lower, here.
So, you can call that as 1 if you like, where
fn was, fn naught was integrable. So, upper
minus lower is small, that is what we started.
And 2 and 3 are giving corresponding things
between fn and f, so, plus, so 1,2 and 3,
using 1,2 and 3 I get this, no problem? So,
if you like, this is epsilon times 2 of b
minus a plus 1. Because epsilon I can always
modify, go back and change wherever required,
so implies f belongs to R ab.
So, the basic idea is because of uniform convergence
fn is close to f for all x, that is important
thing. So, you can take the minimum over that
subinterval, as well take the maximum over
subinterval and then multiply by the lengths
and add up to get the corresponding.
to see that if fn converges uniformly and
f does not, f is Riemann integral then fn
must be Riemann integral.
fn’s are given to be Riemann integrable.
Here fn’s are given to be Riemann integrable,
we prove that f is Riemann integrable.
But why in general, given that fn is bounded
why it should be Riemann integrable?
I am not saying that if it is bounded, it
must be Riemann integrable. I am saying that…
What are you saying? So, here is the theorem.
So what modification you are suggesting?
What I am saying that, fn converges uniformly
f are bounded, f are Riemann integrable. Small
f’s are Riemann integrable.
Small fs meaning what? fns or what?
fns, only fns.
If the limit is Riemann integral why you should
say all corresponding sequence is Riemann
integrable? Why should it say that? All fn’s
can come to a singleton, all fn’s may not
be Riemann integrable, but they converge to
a constant function. Then the limit is Riemann
integrable. Why should the function be Riemann
integrable, each fn? That is expecting too
much, giving nothing and you are expecting
too much.
So basic property we are saying is if fns
are Riemann integrable, and fn’s converge
to f uniformly, then f also becomes Riemann
integrable and integrals converge. We have
not proved integrals converge yet. We have
only proved f is Riemann integrable.
But proving integrability is, so also look
at integral of fn d mu minus integral of f
d mu. You want to show this converges, so
you want to show that this quantity goes to
0, we want to show that this quantity goes
to 0.
But we know something about, this is less,
absolute value of, this is less than or equal
to integral of f minus fn. Using the property
at absolute value of the integral, Riemann
integral is less than equal to integral of
the absolute value and which is less than
or equal to, now each function fn x is less
than equal to norm of, into the length of
the interval b minus a.
The right hand side, because the function
f minus fn is bounded by the constant, norm
of f minus fn. So, I can take it out, less
than and this goes to 0, as n goes to infinity.
Because of uniform convergence norm goes to
0. So, that is not much of a problem.
Once you know that limit is integrable, convergence
of integrals is not a big issue. But the important
thing is that you need uniform convergence,
fn’s converging uniformly to f to say that
you can interchange limit and the integral
sign.
This is the beginning of a, I have already
mentioned I think that the Riemann integral
is not very well behaved with respect to limiting
operations, namely pointwise convergence need
not imply integrability, limit of integrals
is equal to limit of integrals.
So, that started the research for looking
for a integral which has better properties
compared to this. So, that is the another
point for the beginning of what is called
Lebesgue integration, whether this is much
better behaved. You do not need uniform convergence,
you need a slightly milder conditions to be
true.
So this is, now the question comes, see what
we are doing today is very important in mathematics
to relate ideas. Continuity is an idea about
how fn x, f x converges to something, f of
x, as x goes to c. And if you take limiting
operations, we know algebra of limits. If
f plus g are continuous then f, f and g are
continuous then f plus g is continuous, product
is continuous.
So, it says if you go on taking limits, so,
see this is a very general thing one consider.
You are looking at the class of functions,
real valued functions, say R to R. What are
the properties you would like to analyze,
you would like to analyze, you can add functions,
so you like to know under addition what is
preserved.
I can multiply, under multiplication what
is preserved, I can scalar multiply, given
a function f I can multiply by c times f,
what properties are preserved under that?
Because, on R these are the structures available,
addition, scalar multiplication, product,
you can take limits.
Under all these what are the properties which
are preserved? So, what we have shown is when
we did continuity, differentiability and so
on, if f is continuous, g is continuous, f
plus g is continuous. Differentiability we
saw that, f plus g is differentiable if f
and g are differentiable, product was differentiable.
If fn’s are differentiable, can you say
the limit is differentiable, that we saw it
is not necessarily true under pointwise. So
that is why, there is a need to put some stronger
conditions which allow us to pass over limit
and that operation, interchange those 2 operations,
uniform convergence is one.
Unfortunately, differentiability does not
even, is not preserved under uniform convergence.
One has to put slightly more stronger conditions.
So, we will not prove that theorem. Let me
just show you that theorem, so that you understand.
So, this is integrability fn converges to
f uniformly, then f is integrable. And, so
here is the, it says suppose not only fn is
converging uniformly, fn dash each function
is differentiable and the derivatives also
converge uniformly, you need something more.
Also suppose that, it converges at some point
then fn converges uniformly to a differentiable
function and derivative.
So, it is slightly more convoluted kind of
a thing. You have to put conditions on the
derivative itself to ensure that the limit
function is differentiable. So, we will not
go into that because it is not a very useful
result, in many situations. It just, simply
does not say fn’s are differentiable, converging
uniformly then limit is differentiable. So,
we will not do that.
So, let me just summarize what we have done
till now. We had looked at sequences of functions,
and tried to analyze what are the limits of,
under what is the meaning of pointwise convergence
at every point it converges. And we observe
that, at pointwise convergence is not very
good behaved, they are not very well behaved.
It does not have any one of the properties
that we can think of, continuity, differentiability
and so on. So for that, one looks at what
is called uniform convergence and it preserves
continuity, it preserves integrability, it
preserves boundedness but in some way it preserves
differentiability, not exactly straightforward
way.
So, next time what we want to do is, I will
stop here today; there is a reason for that.
But what we want to do is, start with sequences
and see what is other use of sequences. For
example, the algebraic operation of addition
of numbers, given 2 numbers you can add them,
a plus b.
Given 3 numbers a1, a2, a3 you can add a1
plus a2, any finite number of real numbers
you can add them, because there is operation
of addition, and inductively given any n,
you can add them. Can you add infinite number
of numbers? That is a question.
How does one find a way of adding, so what
does it mean, given a sequence a1, a2, a3
and so on. How do I say something like a1
plus a2 plus a3 plus dot, dot, dot? It makes
sense or not. In what way I can give a sense
to that operation, infinite addition of numbers.
And that, very natural way of doing that,
that is how do you add? a1 you add a2, so
a1 plus a2, you have got another one, add
a3. So up to an.
So, you can go on doing it but what eventually
you want is when you go on adding a1 plus,
plus an, whether they become stabilized somewhere.
So, that is limit of a1 plus a2 plus an, you
will like to consider. So, that gives you
a notion of what is called series of numbers.
So, we will do it next time. So, we will stop
here today.
