Alright, so let us start looking at Pointwise
convergence, Convergence of sequences of functions.
So, let us take a sequence fn, a sequence
of functions. Each fn is defined on X to R,
so we say that fn converges pointwise 
to a function f X to R, if fn x converges
to f of x for every x belonging X. So, at
every point you look at the value of the function
fn that gives the sequence of real numbers
and that converges to f of x.
So, that is pointwise so that is same as saying
in terms of definition of convergence of sequences,
that is for every epsilon bigger than 0 there
exist some stage n which will depend on epsilon
as well as on x, such that fn x minus f of
x is less than epsilon for every n bigger
than this number n. So, this stage may depend
upon epsilon of course, and it may depend
on the point. So, at different points the
stage may be a different. So, we had looked
at some examples last time.
So we looked at one example was I gave you
the graph of the function, 0 to 1 so here
is 1 by n and you look at the graph to be
say, you can start at 0 itself does not matter
where you start. So you can go up to, so this
is a point 1, if you like you can go to minus
1 does not matter. So, this is the graph of
the function. So, if that is your fn, then
this fn converges to f and what is that f,
f of x is equal to 1 if x is bigger than or
equal to so if you are including 1 over n
then bigger than or equal to 0 and is equal
to 0 if x is less than 0.
So, that meant that so fn converges to fx
for every x. Each fn is continuous but f is
not continuous. We look at some other examples
also, so second example we looked that was
I think fn x equal to Sin of nx divided by
n, n bigger than or equal to 1. Then fn x
converges to f of x which is equal to 0 identically
0 for every x. So, converges pointwise so
the function because Sin is always bounded.
And third but this example, this was an example
particularly it is saying that, if I look
at fn dash of x exist for every n. But, fn
dash of x is not convergent. I think I should
need to change that for the, if we want not
convergent so let us put it square root of
n here. Then the derivative will be square
root of n into Cos nx which will be not be
convergent.
I am just repeating the examples that we have
done last time. So, that essentially says
that fn differentiable need not imply f differentiable.
I think so let me give you one more example
of, let us look at I gave one example last
time let me give another one here. Let us
look at rationales between 0 and 1, let what
shall I write, let A be the set of x belonging
to 0, 1 x rational. So, look at all the rational
points between 0 and 1, A is a countable set,
A is a countable set. So, let us write A to
be equal to sum r1, r2, rn and so on. So,
give an enumeration of the rationales, it
is a countable set so call it r1, r2, rn we
are not saying increasing or decreasing or
anything, just enumeration of rationales.
So define fn to be equal to the indicator
function of the set r1, r2, rn. So, what does
that mean? What is the meaning of indicator
function? If you recall that meant that fn
of x is equal to 1 if x is equal to r1, r2,
r up to rn and is 0s otherwise. Of course
x is between so we are looking at this in
the interval as a function 0, 1 to R. So,
look at this is a subset of 0, 1 so given
a subset A which is a countable set rationales
in 0, 1 define fn to be the indicator function
of the rationales r1 up to rn in that enumeration
and then each fn is integrable. Because each
fn is a function which has only discontinuity
at the points r1, r2, rn everywhere else is
a constant function. 0 only at this point
itís the value is one so it has only jump
discontinuities at this finite number of points.
So, it is a finite number of discontinuities
so f is integrable.
Where does fn x converge to, what is f of
x it is nothing but the it will be 1 if x
is any one of the rationales r1, r2, rn otherwise
it will be 0. So, it is the indicator function
of, so let me write it is 0, it is 1 if x
is equal to rn and 0 otherwise. So, it is
the indicator function of the set A. A is
the set of all rationales in 0 and 1. So,
what is the indicator function of rational
mean? It is 1 at all rationales in 0, 1 and
0 at all irrational points. It is discontinuous
everywhere and if you look at the upper sums
or lower sums with-respect-to any partition,
the upper sum will be equal to 1, the lower
sum will be equal to 0.
So, this function is not f is, each fn is
integrable f is not integrable. So, this very
simple way of looking at convergence of sequences
namely at every point, look at the sequence
fn x and look at the limit if it exist that
function. So, then you say fn converges pointwise.
So what these examples illustrate that pointwise
convergence does not preserve continuity,
does not preserve differentiability and does
not preserve integrability. So, is a reasonably
bad way of looking at convergence of sequences.
So, we would like to define a notion of convergence
which is reasonably well behaved so that is
called uniform convergence. So, let fn X to
R be let fn X to R be a sequence of functions,
I should say n bigger than or equal to 1.
We say fn the sequence fn converges uniformly
to a function f on X to R, if 
we want something stronger than pointwise
convergence. So, it says if for every epsilon
bigger than 0, there exist some stage n which
depends only on epsilon such that for every
x belonging to x, fn x minus f of x is less
than epsilon for every n bigger than n epsilon.
So, what we are saying is convergence of fn
x at any point brings it closer to f of x
and that closeness is not affected by the
point x. So, same stage works for all the
points, so for every x 
so for every x this is true. Let us look at
some examples, I click that we have one example
ready here namely, that fn x which was defined
as Sin nx divided by square root of n. So,
x belonging to R.
So, we want claim that fn converges to f which
is identically 0 uniformly 
so that only we have note that, if I take
fn x minus f of x. So, what is that? F is
identically 0, so it is absolute value of
Sin nx divided by square root of n which is
less than or equal to 1 over square root of
n. So, irrespective where is a point x, the
distance between fn x and f of x is always
less than 1 over square root n. So, that implies
given we can choose say stage n epsilon such
that fn x minus f of x is less than or equal
to or is less than epsilon for every n bigger
than n epsilon.
Because we can choose, so given epsilon n
is such 1 over square root n is less than
epsilon that is all we have to do. So, choose
n larger than so that 1 over square root of
n is less than epsilon. So, then this will
imply that this is true so fn converges to
f uniformly. There are many examples of functions
which converge uniformly. Let us look at,
will give more examples later on.
Let us just note one thing so note, what is
meaning of saying fn does not converge, fn,
the sequence fn does not converge uniformly
to f, all functions are x to R. So, fn is
X to R and f is from X to R saying that fn
does not converge uniformly to f means what?
So, once again I am trying to bring your attention
to the point that, if you want to understand
something is true, you should also understand
when something is not true equally important
to understand both ways.
So, let us write once again fn converges to
f uniformly meant was equivalent to saying
for every epsilon bigger than 0, for every
x belonging to X, there exist a stage n epsilon
such that fn x minus f of x is less than epsilon
for every n bigger than n naught or than n
epsilon. So, what is fn not converging to
f uniformly? So that is equivalent to saying
so for every, we should change it to, so there
exist epsilon bigger than 0 so that for every
x the statement is not true. And what is statement
that is not true? There is a stage after which
something is small, that should not happen
that means after every stage I am able to
find a point where this thing goes back.
So, that means there exist a sequence of point
nk and there is a sequence nk of a natural
numbers and a sequence xk of points in X such
that, mod of fn k places where the things
are going bad, so that is same as nk xk minus
f of xk is bigger than or equal to epsilon,
is it okay? For every x the same stage works
so it does not work for x the same stage does
not work. That means there are points xk where
this thing does not work. For every k there
is a point where this does not work. That
means there is a stage nk say that f of nk
minus f of xk is not less than epsilon. It
is bigger than or equal to epsilon.
SirÖ (()) (inaudible)
For every x it does not happen so there is
a sequence xk for which things go bad. And
what goes bad? Goes bad this is going bad
right that means for every n, I am able to
find some stage after which the things go
bad. So, that stage is for every k nk so f
of nk xk minus f of xk is not less than epsilon
so, two things are going bad. Something was
happening for every epsilon, for every epsilon
that is taken care of for every x and for
every here. When things go bad that means
there should be points where things are going
bad, so that a point xk. And for xk what is
going bad? fn of xk minus f of xk is not less
than epsilon for what n.
At least, every given any stage I can find
something after which. So, that is same as
a finding a subsequence xn k and k. So, this
is what is it means saying that, the sequence
does not converge uniformly. So, this gives
us a way of testing many something not converging
uniformly. For example, I will take ordinary
sequences of numbers saying that a sequence
a n converges to a that means what, a n comes
closer to a after some stage. If a n, is not
converging to a that means what?
That means there is a epsilon such that whatever
stage you give me, there is a something stage
after which, so there is a subsequence s a
and k. So, sequence not converging to a means
there is at least one subsequence it is not
converge. But, here this was happening for
every point also so there is a sequence of
points where things are also going bad.
