Let us today discuss some notions and properties
on convergence of functions. So, we discussed
the properties of what continuity means in
what uniform continuous functions are. Let
us extend this notion to convergence and then
connect convergence with continuity, ok.
So, before we delve into the details why this
is useful because since we are studying Fourier
series some of the properties we want acclimatize
ourselves with with notions of convergence.
So, keeping this in mind let us get started.
So, first is point wise convergence. A sequence
of functions you can interchangeably use functions
you can use signals defined 
on a set some S converges point wise to a
function f defined on S if limit as n goes
to infinity f n of x equals f of x holds for
all x belonging to this set.
So, we will go a little deeper into what this
this definition means right. So, you have
a sequence of functions that are defined on
a set S and that converges point wise to a
function f defined on S if this condition
holds.
Now, in other words what this means is f n
converges point wise to f on this set S, if
for every x belonging to this set S and for
every epsilon greater than 0 there exists
some capital N that belongs to the set of
natural numbers. This script n is basically
set of natural numbers such that for every
small n which is greater than or equal to
capital N the absolute value of f suffix n
small nx minus f of x is less than epsilon
and here N, capital N depends on both epsilon
and x.
So, f n there is a sequence of functions converges
point wise to some function on that set S
if we pick some x belonging to that set S
and then for every epsilon that is greater
than 0 there exists some capital N a natural
number such that for every small n greater
than or equal to capital N this function is
within within epsilon of the limiting function
right f n x minus f of x absolute value is
less than epsilon. So, n here depends upon
both epsilon and x. So, pick the initial point
x pick epsilon the capital N depends upon
both epsilon and x.
Now, let us see an example. Consider the graph
of a continuous function f of x equals x power
n over this interval minus 1 to plus 1 this
is semi open I mean it says open on the origin
on the negative side and it is closed from
the positive side 1 is included and minus
1 is not included. So on this set that is
over the semi open interval minus 1 to 1 f
of x equals 0 for minus 1 less than x less
than 1 and 1 when x equal to 1 right, as n
goes to infinity because these are bounded
between minus 1 and one 1 minus 1 and plus
1 not included in this range basically decays
to 0 and then when x equals 1 because 1 is
part of the set 1 power n is basically 1.
Now, if you observe this function f of x the
limiting function right I mean this should
be if nx here, f n x is x power n. Now, if
you observe the limiting function right the
limiting function 
that is f of x is discontinuous and this was
part of your homework exercise that you know
if you have a jump like this right it is a
discontinuous function I gave you a homework
exercise last time and this is basically this
continuous function. So, what it implies is
the point wise limit of a continuous function
need not be continuous right, we may start
with a continuous function like f n x which
is x power n and then if we look at the limiting
value of this function this function need
not be continuous.
So, there are many pathological cases that
you can see with point wise limits if you
look at the point wise limit of a sequence
of differentiable functions they need not
be differentiable. Similarly the point wise
limit of a sequence of integrable functions
need not be integral, right.
Another example would help us here to understand
what we are talking about with these limits.
Consider a sequence of piecewise linear functions
and these are called tent functions because
I will sketch this graph how they look.
So, this is x this is you can call a triangular
function retained function because it looks
like a tent here this is 1 upon n, this is
2 upon n and this is n, this is f and x, and
it is piecewise linear as you can observe
is an an up ramp and this is a down ramp.
So, you can describe this function as n square
x in the interval 0 less than or equal to
x less than 1 upon n, it is 2 n minus n square
x in the interval 1 upon n less than or equal
to x less than or equal to 2 upon n and it
is 0 for 2 upon n less than or equal to x
less than or equal to 1.
So, now, let us investigate into how f n of
x dives point wise over the interval 0 to
1, ok. So, let us examine, let us examine
if f n of x goes to 0 point wise on the interval
0 to 1 it is basically closed on both sides.
Now, if x belongs to this semi closed interval
then f n of x equals 0 for all x greater than
2 upon n because you can choose your n and
x depending on n such that this is a 0 function.
If x equals 0 f n of 0 equals 0 for all n
because you have n square x you plug x equals
0 then for all such n this is basically 0
therefore, this function f n of x goes to
the 0 function point wise over this interval
0 to 1.
Now, we have to investigate the following.
Observe the look at the integral of the function
f n of x and look at the limit of the integral
of this function and look at the integral
of the limiting function and see if what we
can infer from these two cases. So, one thing
that you need to observe 
integral from 0 to 1 f n of x, dx equals 1
for all n this is because you are looking
at the area of the tenth function here which
is half times the base is 2 upon n times the
height is n right and this evaluates to 1.
Now, the limit as n goes to infinity of the
integral of this function 
this is basically 1. So, 1 is basically limit
of limit as n goes to infinity integral of
this function because this integral is is
1, this is let us examine if it is equal or
not equal.
Now, consider the integral of the limit function
the limiting function as we discussed earlier
is the 0 function because it goes to 0 point
wise. So, therefore, if you integrate the
0 is basically which is 0 therefore, these
two limits are not same. So, therefore, the
limit of the integral of this function is
not equal to the integral of the limiting
function. So, this is an important observation
that one might make mind while dealing with
point wise limits. So, you may wonder if point
wise limits are or anything useful at this
stage, ok.
There is a useful notion called uniform convergence,
and let us discuss uniform convergence a sequence
functions f n defined on a set S converges
uniformly to a function f if for every epsilon
greater than 0. There exists some number capital
N whose belonging to the set of ah natural
numbers such that for small n greater than
or equal to capital N 
absolute value of f n of x minus f of x is
less than epsilon holds for all x belonging
to this set S. So, you have to look at the
order here it is very important, which means
for every epsilon greater than 0 there exists
some capital N belonging to the set of natural
numbers such that for every small n greater
than or equal to capital N and for every x
belonging to this set S absolute value of
f n of x minus f of x is strictly less than
epsilon.
That is capital N depends on epsilon, but
not on the initial point, x this is a very
important difference when we think about uniform
convergence versus point wise convergence.
That means, for point wise convergence this
function is bounded within epsilon and his
bound depends; that means, for every point
x and a given tolerance epsilon you can you
can figure out some integer capital or some
natural number n which can satisfy this condition.
But if capital N depends only on the tolerance
epsilon and not on the initial point it is
very important and not on the point then it
is uniform convergence and this is a very
important important idea and we will see this
quite often very useful for us.
So, we have a note here it is very important
if f n converges to f uniformly on S then
f n converges to f point wise as well and
and it is not the other way around. And one
of the reasons for studying uniform convergence
is basically to study if continuity is inherited
from a sequence of functions you might be
given a sequence of functions that are continuous
and you want to look at a functional series
and then you want to see if that helps to
you know if there's uniform convergence. And
if there is uniform convergence then it implies
continuity and is a very important thing and
continuous functions are very useful particularly
when we deal with certain aspects in the Fourier
series. So, we will prove this result as we
go through later part of this lecture, but
let us revisit quickly an example.
Let us examine if the sequence of functions
f n given by say nx square plus 1 upon an
x plus 1 is uniformly convergent over the
interval 1 comma 3 this is basically the close
interval 1 and 3 are included as part of the
interval. So, first let us take 
the point wise limit, ok. Now, limit as n
goes to infinity of nx square plus 1 upon
nx plus 1 equals limit as n goes to infinity
I just pull the n outside which is basically
x square plus 1 upon n divided by x plus 1
upon n and this is basically x, right. So,
that is f n converges to x point wise over
the interval 1 comma 3 right, it basically
heads to this ramp function defined over 1
comma 3.
Now, let us examine uniform convergence ok.
For this let us consider the absolute value
of f n of x minus this limiting function f
of x which is f n x minus x, right. So, plug
in f n x we get nx square plus 1 upon nx plus
1 minus x which is basically absolute value
of 1 minus x. So, just multiply nx plus 1
with x and then you cancel out nx square.
So, you get 1 minus x upon nx plus 1 absolute
value and then x is positive. So, therefore,
the denominator we can say it is an x plus
1 and this is 1 plus mod x this is basically
strictly less than or equal to because I would
like to say this is mod of 1 minus x is definitely
upper bounded by 1 plus mod x right I take
the positive quantity here. So, this is true.
So, now, over this interval 1 comma 3 1 plus
mod x upon nx plus 1 can be upper bounded
to 4 upon n plus 1. I mean if I want this
to be maximized the numerator should be more
and the denominator should be less. So, the
denominator has to be less I choose x to be
the lesser value because this is a ramp I
choose x equals 1. So, I get n plus 1 and
the numerator has to be maximum so therefore,
I choose you know mod x x to be 3 here, so
therefore, I get 4 in the numerator and the
denominator is n plus 1 right. This is for
all x belonging to 1 comma 3 you have gotten
rid of the x then I compute this this quantity.
So, now, what it means is if epsilon is greater
than 0 some positive quantity chosen. There
exists some number capital N which is a natural
number such that for all small n greater than
or equal to capital N 4 upon n plus 1 is strictly
less than epsilon. Which means for every small
n greater than or equal to capital N absolute
value of f n of x minus f of x is strictly
less than epsilon this is over all points
belonging to the interval 1 comma 3 this proves
uniform convergence.
Now, why do we care upon these these ideas
in in an analysis, right?
So, let us see the connection in some applications.
So, we know that the Fourier series for a
2 pi periodic function is given by a dc term
plus the even the cosine and the sin harmonics.
Now, in functional series form the above can
be written as summa k equals 0 to infinity
summation S k of x.
Now, this if you write it in a limiting form
is limit capital N going to infinity summation
S k of x k equals 0 to capital N that is I
take a partial sum from 0 to capital N, I
sum and then I take the limit and you see
you ask questions if the limit exists right
the question is here if limit exists. And
for different values of x, I can have different
limits right I mean it is also obvious 
that different values of x can give different
limits and this is exactly what you are sort
of dealing with in the notions of our conversions,
and does it converge to some function, and
is it point wise conversions or uniform conversions
and this is there we are sort of alluding
towards right. We will discuss all these things
in detail when we delve into Fourier series,
but I am sort of setting up a basic background
into ah basically the notions of convergence
as part of this background.
Now, before we go further let us recall the
basic definitions of supremum and infimum.
You might have done this in your basic math,
but we will just recall this definition because
we will use this definition subsequently.
Let S be a set which is contained in R 
the supremum of S denoted by sup S is the
smallest number a belonging to the set of
real numbers such that x is less than or equal
to small a for all x belonging to the set
S pick an x belonging to this set S which
is contained in R and x is less than or equal
to a.
Now, supremum of S in in the set form is basically
minimum over all a belonging to the set of
reals such that x is less than or equal to
a for every x belonging to the set S right.
So, similar to the supremum we have something
called the infimum as well I think you can
see the connection.
Now, similarly infimum of S is the largest
number b belonging to the set of reals such
that I am just flipping it x is greater than
or equal to b for all x belonging to the set
S. So, infimum of S is basically maximum over
all b belonging to R set of reals such that
x is greater than or equal to b for every
x belonging to the set S.
So, pick an x belonging to the set S right
and I have an x, I have a b such that b is
greater than or equal to x and b belongs to
are not necessarily belonging to the set S
it could belong to the set S or it may it
need not belong to the set S, but b belongs
to R.
Now, with this let us get slightly a little
more into depth here using these notations
that we already described. So, let us define
for a real valued function 
on a non-empty set 
denoted by S the supremum on these set S given
by 
norm of f on the set S is basically supremum
x belonging to the set S absolute value of
f of x ok.
Now, if f is a bounded function on S then
supremum over x belonging to S absolute value
of x is basically supremum if you write this
in this form absolute value of f of x such
that x belongs to S. And this this if f is
a bounded function on the set S then this
supremum exists and you know we have to observed
that mod f of x is less than or equal to the
supremum of this function on the set S for
every x belonging to S and means values that
mod f will take is very close to the supremum
that is what it it means, ok.
So, with this we can slightly think about
revisit, revisit revisiting the uniform convergence
and let us check if uniform convergence implies
point wise convergence.
let us check this I think I made a note that
uniform convergence implies point wise convergence
let us see this. So, from the definition that
we have for uniform convergence; that means,
the absolute value of f of x, if f n of x
minus f of x is within epsilon and that does
not it and you can choose some number capital
N that does not depend on x, but on epsilon
n depends only on epsilon it does not depend
on the initial point right. So, absolute value
of f n x minus f of x is less than or equal
to the supremum for all x belonging to this
set S absolute value of f n of x minus f of
x which is basically in our notation f n minus
f on this set S. So, that f n heads to f uniformly
on S as n goes to infinity.
So, this implies absolute value of f n of
x minus f of x this heads to 0 for each x
belonging to S, this is very important for
every x pick in x for each x this f n x minus
f of x absolute value heads to 0 right this
implies 
f n heads to f as n goes to infinity point
wise on on S. So, uniform convergence implies
point wise convergence, but point wise may
not imply uniform convergence.
Now, there is a link between uniform convergence
and continuity and let us establish this result
via a theorem.
So, I will state the theorem and I will also
prove this result. Suppose f n x is a sequence
of continuous functions 
on an interval S suppose f n of x converges
uniformly to f of x on S. Then the limit function
f of x is also continuous this is a very important
very important theorem I have a sequence of
continuous functions defined on an interval
S and I know that f n of x converges uniformly
to f of x on that interval then the limit
function is also continuous a very powerful
statement. So, if it is not continuous then
what are the implications right, I mean it
the convergence of a point of discontinuity
would have to be invoked and all these subtle
notions have to be really revisited. So, therefore,
this uniform convergence is a very powerful,
ah powerful idea.
Now, let us try to establish the proof of
this result, ok. So, we need to establish
that f of x heads to f of a when x heads to
a. So, when x heads to a I want establish
that f of x goes to f of a for every x and
a belonging to this interval ok.
Now, let us start with 
let us start with absolute value of f of x
minus f of a. Now, for any n greater than
or equal to 0 that is n equals 0 1 2 dot dot
dot absolute value of f of x minus f of a
can be written as absolute value of f of x
minus f n of x plus I add and subtract f n
of x and f n of a very conveniently. And I
think if you are seeing the trick here basically
you can see that you can invoke triangle inequality
for this for this term that I have written
here right. So, what I have done is f of x
minus a I add and subtract f n x and f n a.
Now, invoking triangle inequality I have this
is less than or equal to absolute value of
f of x minus f n of x plus absolute value
of f n of x and f n of a plus absolute value
of f n of a minus f of a. But now, I can bound
f of x minus f n of x absolute value this
is basically the superior norm of f minus
f n same thing holds for this term as well.
So, I can conveniently state f of x minus
f of a is less than or equal to 2 times the
norm of f minus f n soup norm absolute value
of f n of a minus f of a because mod f of
x minus f n of x is less than or equal to
norm if minus f n is an absolute value of
f of a minus f n of a 
is less than or equal to f minus f n superior
ok supremum of f minus f n. Now, we are seeing
the trick here.
Now, we choose a positive number epsilon which
is greater than 0 which is arbitrarily small
such that f minus f n heads to 0 as n goes
to infinity. Therefore, there exists some
natural number n greater than 0 for which
this norm f n minus f, I mean I am interchangeably
using f n minus f and f minus f n it is basically
the same two superior supermom is less than
some epsilon by 3. So, let us assume that
this is some epsilon by by 3 for every small
n greater than or equal to capital N ok. Then
choose and fix that small n say n equal to
capital N and then which ensures that this
is less than epsilon upon 3.
Now, here is a important part if suffix capital
N of x is continuous because we said all of
these are continuous functions and there exist
some small n which is equal to capital N in
this that fits this bill and gives you f suffix
capital N of x is continuous. Now, this is
continuous, so for any choice of epsilon which
is greater than 0 there is an interval 
center around the point a, so that absolute
value of f suffix capital N of x minus f suffix
capital N of a is less than epsilon upon 3
whenever x belongs to that interval. This
interval is not necessarily yes. There exists
some interval because it because we want a
no continuity here right there exists some
interval that is centered around a such that
this is holding true that is absolute value
of f n of x minus f n of a is within epsilon
upon 3.
So, formally since f suffix capital N of x
heads to f suffix capital N of a for x tending
to a for every epsilon which is greater than
0 there is a corresponding delta which is
greater than 0. So, that absolute value of
f n of x minus f n of a is strictly less than
epsilon upon 3 whenever mod x minus a is less
than delta, right.
So, thus absolute value of f of x minus f
of a is less than or equal to 2 times the
superior of f minus f n plus absolute value
of f n of x minus f n of a this is less than
or equal to 2 times this quantity is epsilon
upon 3 and this is also epsilon upon 3 therefore,
this is equal to epsilon for all mod x minus
a less than delta.
So, therefore, this implies f of x heads to
f of a as x tends to a which means f of x
is continuous. So, this is a very very important
relationship because if you consider a sequence
of functions and these functions are continuous
and then there's uniform convergence then
it implies that the limit function is continuous.
So, this also gives us an idea for a quick
test towards uniform convergence.
So, if the sequence of functions f n are continuous
then f n and and f n converges point wise
to f of x however, if the limit function f
of x is not continuous implies f n x does
not converge 
uniformly to f of x this is a very important
quick test. So, I considered a sequence of
continuous functions and I assume f n of x
small n of x converges point wise to f of
x; that means, each each individual function
converges point wise. However, if the limit
function is not limiting function is not continuous
then the sequence does not converge uniformly
to f of x and this is a very useful result
when we have to touch upon subtle aspects
in the convergence of the Fourier series.
I will conclude with a theorem, another theorem.
Suppose f n of x is a sequence of continuous
functions which converges uniformly to a continuous
function f of x on a bounded interval this
is a small twist here it is a bounded interval.
Let us suppose it is the closed interval a
to b we have limit as n goes to infinity integral
a to b, f n of x dx is integral from a to
b limit n goes to infinity f n of x dx which
is basically the integral from a to b of the
limiting function f of x.
Recall in one of the earlier examples when
I took the integral of a continuous function
and took the limit of that versus taking the
limit of the function and then integrating
it there was a case when they were not the
same right we looked into that example right
and that was this tenth function. So, now,
and and and we saw the issues there, right.
So, now, we will see under what conditions
that we can take the limit of the integral
as the same as the integral of the limit right
and that happens when there is uniform convergence
over a bounded interval. So, we will we will
prove this result carefully.
So, consider the norm I mean absolute value
of 
this quantity which is basically 
the integral of the limit minus the integral
of the sequence. So, this is basically absolute
value of f n of x minus f of x dx. I just
wrote the integral sign before and I just
took f n of x minus f of x.
Now, this quantity here is certainly less
than or equal to integral a to b absolute
value of f n of x minus f of x dx absolute,
but I think I have to be careful here, I have
to put the dx outside. So, basically I take
this deviation I integrate it and then I take
the mod this is definitely less than or equal
to if I did absolute first and then I integrate
it over the interval a to b that is what it
it means right. And this because there is
uniform convergence I can write this as the
integral from a to b. The supremum exists
and I can pull the supremum outside and this
is basically the integral over this interval
right because this is just one quantity which
is maximum over all x belonging to that set,
right. So, or x belonging to R and this is
a subset of R.
Now, I can I can get the super supremum outside
this is just the integral and this is bounded.
So, therefore, it is f n minus f times b minus
a and this hits 0 as n goes to infinity and
this proves this result right. I mean this
is this norm basically heads to 0 as n goes
to infinity because of uniform convergence
and therefore, if you take the limit of the
integral of this take the step is a function
in the sequence integrate it look at the limit
that is the same as the integral over the
limiting function and this is a very important
result and this holds for uniform conversions.
So, if you look at a sequence of functions
which uniformly converge to a limiting function
over a bounded interval then limit of the
integral of the sequence is basically integral
of the limit function ok, and that is basically
equivalent is equal to the integral of the
limiting function over that interval is a
very very important result.
So, with this we have sort of given a basic
background into the notions of convergence
and the implication of uniform convergence
to a limit function which is continuous. And
these subtleties we will we will see when
we discuss the Fourier series summation of
the for; I mean I mean if I look at the Fourier
series does the limit converge. And if the
limit converges to what does this limit converge
to and if there is a discontinuity point then
what is the limit and some of these subtle
subtle aspects have to be dealt with when
you deal with Fourier series and for this
some of the notions of the convergence of
the functional series is very very important
ok.
So, a lot of other details will be dealt in
functional analysis or if you graduate course
in an analysis all these would be covered.
So, I am giving you basically a background
into this, into these aspects because you
will require this in signal processing.
So, we will end this lecture here.
