So, in the yesterday’s class we proved that
if you consider a sequence of functions and
if each function f n is integrable and if
f n converges to f uniformly then f is also
integrable. The sequence of numbers integral
f n that will converge to integral of f, let
us also see the interpretation of this results
in language of this metric spaces.
We have already seen that, for example if
you take this space C a b that is space of
all continuous function with this matrix d
suffix infinity. We have seen that our earlier
result about the uniform convergence and continuity
that implies that this is a complete metric
space and we have also observed quite early
that convergence, uniform convergence is basically
convergence this metric.
Now, let us see how our yesterday’s result
of integration can be interpreted in this,
now we also know that this C a b is a vectors
space. So, one can define a map phi from this
C a b to R, let us talk of a Riemann integrals
only for the time being as follows at define
phi of f as integral a to b f x d x. 
Then you already know that this is a linear
functional, in fact have approved that if
following the various theorems of the integration
that we have proved phi of f 1 plus f 2 is
phi f 1 plus phi f 2 phi of alpha times f
will be alpha times.
So, phi is a linear function phi that is of
course that is something that is already know
phi is a linear functional and what did we
proved yesterday that if f n convergence to
f n uniformly, f n convergence to f n uniformly
then integral of f n converges to integral
of f. In other words, if f n converges to
f in this d infinity that is, that is d infinity
f n f tends to 0 this implies phi of f n tends
to phi of f, remember phi f and phi f n are
numbers phi f and phi f n. So, that means
if f n converges to f in this metric then
phi f n converges to phi f, now what is the
meaning of this it means that phi from this
C a b to R is a continuous function.
That is one of the ways of describing continuities
is that whenever x n converges to x f of x,
n converges to f of x that is equivalent of
saying that f is continuous. So, that is what
we have said, here whenever f n converges
to f in this metric phi f n converges to phi
f that is same as saying that phi is a continuous
function. Since it is already a linear functional
it means that the phi is a continuous linear
functional, then there is one more thing.
Here, while this while describing the results
about the integrals we have take b this upper
limit as b, but that is not very much essential
one can let this upper limit vary also.
For example, I can consider integral a to
x f n t d t, suppose I can say that integral
a to x f n t d t and then this will converge
to integral a to x f t d t, if f n converges
to f uniformly. Whatever you take x in a b
then integral a to x f n t d t will converge
to integral a to x f t d t, so it means that
this upper limit need not big it can be any
number. Now, whatever you say about the upper
limit, one can say the same thing about the
lower limit also instead of a, you can take
some other number.
So, consider integral from y to x f n t d
t where y and x are any numbers lying in a
b that will converge to integral y to x f
t d t, so there is nothing very particular
about this numbers a and b. In other words,
you should take any sub interval of the a
b that integral of f n over that sub interval
will converge to integral of f over that same
sub interval that is about the integration.
Now, let us go to the discussion of uniform
converges and differentiation 
and differentiation one can ask an analogous
question. Here, suppose we have a sequence
f n, f n from a b to R and suppose f n converges
to f uniformly, f n converges to f uniformly
then suppose each f n is differentiable. Suppose
each f n is differentiable then what we might
except that if each f n is differentiable
then we can expect, we might expect that f
also is differentiable and not only that we
should also expect that f n prime converges
to f prime. We have already seen that this
is false as far as point wise converges is
concern, but which is also true is that this
is the question is suppose each f n is differentiable.
The question is this is f differentiable is,
f differentiable and thus f n prime converges
to f prime in whatever sense point wise uniformly
whatever and answer to both the questions
are no. That even if f n converges to f uniformly
it does it is not necessary that f must differentiable
and even when f is differentiable this may
or may not happen. It is fairly easy to construct
examples in fact subsequently we will prove
a very powerful theorem, perhaps you may have
already heard of that theorem. It is called
Weierstrass theorem, Weierstrass theorem says
that given any continuous function you can
find a sequence of polynomials which converges
to that continuous function uniformly.
So, you can obviously find the function which
is continuous, but not differentiable and
such a function also you can a sequence of
polynomials. Now, every polynomials is differentiable,
but the limit will not be differentiable also
it can happen that f n converges uniformly.
But, f n prime need not be convergent all
f n prime need not be convergent all that
is that is fairly easy to show suppose we
take this example f n, f n at x is let us
say sine n x by root n, sine n x by root n.
Then is it, is it obvious that this tends
to 0 uniformly whatever interval you take,
this tends to 0, uniformly 
you can take m suffix n will be less not equal
to 1 by root n.
In fact m suffix n will be and at m n tends
to 0, so that is easy to show, but what about
f n prime x. That will be n, n cos n x divide
by root n, so that will be root n cos n x
root n cos n x and you can say that this becomes
unbounded even if you take x is equal to 0,,
this will this sequence is simply root n.
So, this is not a converges sequence at all
and, so as far the derivatives are concern
then things are somewhat bad that is even
with uniform convergence you cannot assure
various things. So, for the derivatives what
we have to do is that we have assumed something
more namely that not only that f n comma just
to f uniformly we have also assumed that f
n prime converges uniformly.
Once you assume that f n prime converges uniformly,
that is fairly strong assumption then you
do not need, even this we can in fact simply
assume that f n converges to f point wise
or even at one particular point also that
will be sufficient. So, let us just see what
is the what does the theorem say.
The theorem is the following, suppose f n
are functions from a b to R 
each f n is differentiable f n is differentiable
in a b, as far as the convergence is concerned
we will say that two things for some x naught
f n x naught converges to f x naught. Let
us say that is f n x naught converges to f
x naught, for some x naught in a b and f n
prime converges to g uniformly f n from we
have to take this 3 function. Also f n, f
and g all these are functions from a b to
R f and d, so what are the assumptions each
f n is differentiable if at some point f n
at x naught comma just to x naught then f
n prime converges to g uniformly.
Where f n prime converges to g uniformly then
what is the conclusion, conclusion is that
if that is the case f is f is differentiable
and its derivative is same as g. Then these
are all hypothesis then f is differentiable,
f is differentiable. In fact even before that
I would say that, since that is not we have
assume f n then f n converges to f uniformly,
f n converges to f uniformly f is differentiable
and f prime is equal to g.
Actually if we make one more stronger assumption,
suppose we also assume that f n prime is also
continuous for each n then the proof becomes
very easy. Then all we can do is that we can
use this previous theorem we can, since f
n x naught is not converges to f x naught
for any x you integrate from x naught to x.
So, consider x integral of x naught to x f
n prime t g d t that goes to g t d t and then
show that because of the earlier theorem integral
of x naught because we have seen that upper
limit and the lower limit can be anything.
So, I can take integral of x naught to x,
instead of f n I will take f n prime by this
theorem it will mean that integral of f n
prime t d t will converges to integral g t
d t. Then it is easy to show that that is
same as an integral of f n prime between x
naught to x is nothing but f n of x minus
f n of x naught and we have already known
that f n of x naught comma is just to f n
of x naught. So, using that you can clearly
easy to show that f is differentiable and
its derivative is same as g, but we if do
not assume that f n is f prime is continuous
then the proof is slightly involved.
In the sense we need some more work to do
and that is what we shall, we shall do again
the idea is that let us, let us take the things
one by one, first thing is we want f n converges
to f uniformly, f n converges to f uniformly.
Then we have seen that one of the ways to
show that sequence of function converges uniformly
is that you show that it satisfies Cauchy
criteria you, sorry it satisfies Cauchy criteria
then what do we know about f n. We only know
that f n, at x naught if converges at other
points we do not know to show that it satisfies
Cauchy criteria what will I do to show is
that.
That is if you look at f n x minus f m x 
we have to show that this quantity can be
made arbitrary small for by choosing n and
m sufficiently large. That small number should
be independent of x, in other words what I
should show is that let us say that mod f
n x converges f n x is less not equal to some
alpha m for every x in a b. Then that alpha
m goes to 0, but what we know is that f n
x naught converges to f x naught, so if I
take f n x naught minus f m x naught that
is going to be small. So, our idea is basically
we add and subtract and, similarly what we
know is that f n prime converges to g uniformly,
so this f n prime satisfies the Cauchy criteria.
So, instead of f n f n x minus f m x, if I
have taken f n prime x minus f m prime x that
that is easy to make that particular f n is
arbitrary small, but basically they use all
these facts. So, what is the idea will take
this minus this f n at x naught minus f m
at x naught suppose we are able to show that
this can be made arbitrary small an independent
of x. Then is it clear that then it will be
in this is also small because I am only adding
the small quantity, here I am adding a small,
but what can we say about this.
These is nothing but if you look at the function
suppose if we look at the function f n minus
f m if you look at the function f n minus
f m. It is a difference between the values
of this function at x and x naught it is the
difference between and, moreover this is a
differentiable function. So, one can use mean
value theorem, one can use mean value theorem
and say that this will be nothing but x minus
x naught into the derivative of this at some
point lying between x and x naught. So, we
can say that it is same as by mean value theorem
we can say that by Lagrange’s mean value
theorem this is same as x minus x naught into
f n minus f m.
Derivative of this at some point zi let s
u say which is same as saying that f n prime
zi minus f m prime zi for some zi in zi between
x and x naught let me see because I do not
know which is bigger and which is smaller
between x and x naught. Now, the idea is this
we know that this quantity f n prime zi minus
f m prime zi that can be made arbitrary small.
That will be arbitrary small independent of
this zi whatever it is zi because we know
that f n prime converges uniformly, f n prime
converges uniformly. So, this satisfies Cauchy
criteria so let us start from this that is
he starting point of the proof.
So, we can say that, let me, let me write
it here, so let us say that let epsilon be
bigger than 0 since f n prime converges uniformly,
since f n prime converges uniformly. You can
say there exist n 0 in n such that m and n
bigger not equal to n 0, m and n bigger not
equal to n 0 this implies mod f n prime. Let
me call it y f n prime y minus f m prime y
this can be made less not equal to any arbitrary
number for every y in m I will take that number
as because here we had to add and subtract
this quantity.
So, it will make that less than epsilon by
2, but here the multiplication is there also
by x minus x naught, so this will remain bounded
by b minus a. So, we will say that this is,
we shall we shall choose n 0 such that whenever
m and n are bigger not equal to n 0, this
is less than epsilon by 2 times b minus a.
Now, consider some m and n which are bigger
than not equal to this n 0, consider m and
n bigger not equal to this n 0.
Let me take it, here consider m and n bigger
not equal to n 0 then for those m and n we
have done all these calculations of course
this is true for any m and n. But, now will
be apply for those m and n bigger not equal
to n 0, i will take f 2 value everywhere.
So, this mod x minus x naught x and x naught
both are points at the interval a b, so mod
x minus x naught is less not equal to b minus
a mod x minus x naught is less not equal to.
So, this less not equal to b minus a and this
is less than epsilon by 2 into b minus a,
so which is the whole things is less than
in fact this is strictly less.
So, I can take it strictly less than epsilon
by 2 alright what next we can say that since
f n converges to f x naught I can again do
the same I can find, I find some. Let us say
n 0 prime such that, here itself let me call
it b n 0 prime, so let us this is say n 0
double prime such that this I think this primes
are unnecessary complicating. Let us call
it n 1, let us call it n 1, so consider this
m and n bigger not equal to n 1 what I would
say is that there will exist n 2 such that
whenever m and n are bigger than n 2, f n
x naught minus f n x naught is less than epsilon
by 2.
So, what is the argument since f n x naught
since the argument is f n x naught is a convergence
sequence hence it is a Cauchy sequence. Since
f n x naught is a Cauchy sequence, Cauchy
sequence 
there exist into n such that m and n bigger
not equal to n 2 or this implies mod of f
n x naught minus f m x naught is less than
epsilon by 2. What is to be done after, this
is clear you take n 0 as maximum of n 1 and
n 2, let n 0 as maximum of n 1 and n 2 then
for m and n which is bigger not equal to this
n 0 both of this inequalities then satisfies.
Let us continue here, so what we say is for
m and n bigger not equal to n 0 consider a
term and x is a b consider mod f n x minus
f n x and all that we do is add and subtract
f n x naught minus f n x naught add and subtract.
So, this is less not equal to mod f n x minus
f m x minus f n at x naught minus f m at x
naught and plus mod f n at x naught minus
f m at x naught. Each of this is less than
epsilon by 2, this is less than epsilon by
2 because of these considerations, here remember
the first part is less than epsilon by 2 because
of this second assumption f n prime converges
to g uniformly.
Second is less than epsilon by 2 because of
the first assumption that f n x naught is
converges to f of x naught, so this is less
than epsilon by 2 plus epsilon by 2 which
is epsilon. So, what we will show that given
any epsilon there exist n 0 such that wherever
m and n is bigger not equal to n 0 mod f n
x minus f m x is less than epsilon for every
x in n which is same as saying that f n is
uniformly Cauchy. So, this implies that f
n is uniformly Cauchy or f n satisfies Cauchy
criteria, Cauchy criteria 
and hence converges uniformly, and hence converges
uniformly 
to let us say some f.
So, converges uniformly to sub that is whatever
it function it converges that function we
shall call that function we shall call f.
Now, what remains is to show that this f is
differentiable and its derivative is same
as that function g, what is the function g.
It is the function to which this f n prime
converges uniformly that is what is remaining
to be shown. Now, to show that let us proceed
as follows, so let us, let us take some x
in a b, let us take some x in a b and we shall
procedure a function for let us for t in a
b, for t in a b and t naught equal to x for
t in a b and t naught equal to x define several
function.
First function I will define is phi of t,
phi of t that we will define as follows, if
t minus f x divided by t minus f t minus,
f x divided by t minus x and I will define
phi n of t minus f n x divided by t minus
x. You can understand why these functions
are defined because we are interested in differentiability,
so to show that we want to show that f is
differentiable we want to show that f is differentiable.
That is same as showing that limit of phi
t as t tends to x exist limit of phi t as
t tends to x exist what about phi n t, in
case of phi n t we know that limit of limit
of phi n t as t tends to x exist.
That is same as saying that f n is differentiable
that is something we have assume, so this
limit is exist and that is something that
we know already. So, let me just say that
know the idea is as follows suppose we show
that phi n converges to phi uniformly in this
interval a b except at t naught equal to x.
Then of course that then this x is a limit
point of this a b minus x, remember our theory
about the limit let me, let me first say what
we trying to do.
We will claim this you can, I will say that
phi n t converges to phi t uniformly not on
the whole interval a b, but in the interval
a b minus x uniformly on the interval a b
minus this x singleton x. 
Let me remind that out theorem about the limit
we have said that if f n converges to f uniformly
on e and x is a limit point then limit of
f n as t tends to x goes to limit of f x as
t tends to x. So, suppose this is proved,
suppose this claim is true then one can say
that limit of phi t as t tends to x and see
limit of limit of phi n t as t tends to x
is nothing but f n prime at x is that clear
that is limit of phi n t as t tends to x is
nothing but, f n prime x.
Let me remind you that theorem once again,
in case you have forgotten.
The theorem was the following f n even before
proving the theorem we have taken the functions
f n from e to e to r and t x was a limit point
of e and we have said that limit of f n t
as t tends to x. We have taken these numbers
at some l n, we have taken these numbers at
some l n and what was the assumption that
f n converges to f uniformly. If f n converges
to f uniformly then we have say was that limit
of this l n as n tends to infinity exist also
that is that limit is same limit of this l
n as n tends to infinity exist.
Limit of this f t as t tends to x exist and
that limit is same as this limit this was
our, in fact our first theorem of the uniform
convergence. This is same as limit of f t
as t tends to x 
we are going to use that theorem for the functions
phi n and phi.
Suppose we show that phi n t converges to
phi t uniformly and limit of phi n t as t
tends to x axis that is same as f n prime
x, so if this claim is prove if this. Let
us if this claim is proved or if this claim
is true if this claim is true then what must
happen is that then limit of phi t as t tends
to x this limit should exist. This limit should
exist and that should be same as limit of
this f n prime x as n tends to infinity limit
of f n prime x as n tends to infinity remember
this f n prime x is something like this l
n f n prime x is takes the place of this l
n.
That is limit of f and t as t tends to limit
of phi n t as t tends to x is f n prime x,
but, we have already assumed that f n prime
converges to g. So, this right hand side is
nothing but g prime x, right hand side is
nothing but g prime x, sorry g of x I am sorry
you are right it is nothing but g of x. So,
what it says it that this limit exists and
that limit is same as g of x which is same
as saying that f is differentiable. Saying
that this limit exist is same as saying that
f is differentiable and that limit is same
as g of x is saying that f prime x is equal
to g g x that is that is f prime x is equal
to g of x.
So, what remains, now what remains is to prove
this claim we assume this claim is proved
then whatever we wanted would be proved first
this claim is true, once the claim is true
the proof or theorem is over. Now, what does
what is involve, here we want to show that
phi n converges, see remember phi n t converges
to phi t point wise there is no problem about
that, about that. There is no problem because
we have already shown that f n comma just
to f uniformly, so f n t will converge to
f t f n x converges to f x.
So, only thing reality to show is that phi
n uniform convergence and we have the same
thing we have done in the earlier that is
we shall show that phi n is a satisfies Cauchy
criteria phi n, satisfies Cauchy criteria.
So, if we consider let us say phi n x minus
phi m x phi n x minus phi m x then that is
same as f n x minus f n t then minus f m x
minus f m t this divided by t minus x. We
shall use the same thing or same technique
which we have used earlier, namely mean value
theorem because if look at this f n x minus
f n t and f m x minus f m t these are nothing.
But, the values of the function f n minus
f m at x and t if you look at again the function
f n minus f m you have f n x minus f m x and
then minus f n t minus f m t. So, what we
can say is that by the same by mean value
theorem we can say that, by mean value theorem
there exist a zi between x n t such that such
that f n x minus f m t. Then minus f m x,
sorry f n x minus f n t minus f n x minus
f n t that is nothing but t minus x into f
n prime at zi minus f m prime at zi, we are
applying mean value theorem to this function
f n minus f m.
Remember, here we have f n minus this whole
thing can be written as this numerator is
nothing but f n minus f m at x minus f n minus
f m at t what we are saying this will be x
minus t fine. This will be x minus t fine,
it will not matter because we are going to
take absolute values, so what, so finally
what we will get is this t minus x will cancel
with that x minus t with a negative sign.
So, since we are going to take absolute value
what we will get is this.
Mod phi n x minus phi n, sorry mod phi n t
minus phi n t this will be equal to mod f
m prime zi minus f n prime zi for some zi
lying between t and x. Do you agree after
applying mean value theorem and after cancelling
this t minus x and x minus t etcetera. Now,
what is to be done after this is clear we
already know that f n prime converges to g
uniformly, we already that f n prime converges
to g uniformly. So, we can always find given
epsilon, we can always find some let me just
write since f n prime converges 
uniformly there exist n 0 in n such that m
and n bigger not equal to n 0.
This implies mod f n prime zi minus f m prime
zi is less than epsilon for zi in a b that
is important for all zi in a b. So, going
back here since this is less than epsilon
it will mean that this is also less than epsilon
for since this happens for every zi this is
less than epsilon for every t in a b, for
every t in a b. It is basically an application
of mean value theorem once again and this
shows that this sequence f n is uniformly
Cauchy, the sequence f n is uniformly Cauchy.
Hence, it follows again that phi n converges
to phi uniformly we already know that phi
n converges to phi point wise, now we got
phi n converges uniformly. So, the limit function
must be same as phi and as we have already
observed once this claim is prove the proof
of the theorem is this over. So, this says
about the relationship between uniform convergence
and the differentiation and as in the earlier
case we can also ask similar questions about
the series that is till, now we have taken
f n converges to f.
Suppose we look at the series like this sigma
f n converges to f sigma n going from about
to infinity suppose, here the convergence
is uniform. But, then also you cannot say
immediately that you can differentiate the
series term by term it does not follows at
sigma f n convergence to sigma f n prime converges
to f n. We need additional condition, namely
that sigma f n prime should also converge
uniformly and this original series should
converge at least at one point then we can
differentiate the series term by term. So,
let us take record those observation, so corollary
is this if sigma f n prime converges to f
uniformly let us say not f.
Now, let us call this g uniformly and sigma
f n is equal to f point wise, let us say actually
it is enough to say at least at one point
by using that theorem then f is differentiable,
then f is differentiable. Its derivative is
nothing but g, its derivative is nothing but
g and f prime is equal to g in other words
sigma f n prime is equal to f if sigma f n
is equal to, f sigma f n prime is equal to
f prime. Which basically means that you can
interchange the operations, summation and
differentiation that you can do provided this
happens sigma f n prime equals to g?
That is the series obtain by differentiating
each term must be uniformly convergent, then
only you can interchange the summation and
differentiation. Now, let us see how all these
theorems which we have considered about the
uniform convergence and various uniform convergence
and continuity. Uniform convergence and integration
and differentiation can be used, there are
several uses of this there are two major things
that are coming into picture those are certain
kind of series and in proving that those series
are uniformly convergent take new continuous
functions, new differential functions etcetera.
One set is what is called power series something
that you have heard of already power series
and other type of series are what are called
Fourier series. Now, these are also meting
like power series, but here the function involved
are polynomials and, here the functions involved
are what are called trigonometric polynomials.
So, these are the positive results that will
follow from the discussion of uniform convergence
there is also, some one very interesting result
which in some sense is negative. But, it is
interesting in the sense that it says something
different from our usual intuitions says and
that is a following, that is if you look at
by if you say that something is a continuous
function you are expected to think.
That it is graph will be something like a
curve, so if you have that kind of idea then
it is then it will appear that whenever a
function is continuous it has to be differentiable
also, it has to be differentiable also. But,
you already know that that is not true the
graph can be something like this, so its function
can be continuous, but not differentiable.
This is a well know example of modulus, suppose
you want a function which is continuous, but
not differentiable at two points when one
can have a function like this.
Suppose more than two points it can be a function
something like this that is, so it is fairly
easy to construct a function which is continuous
everywhere let us say. But, not differentiable
at two points, three points etcetera, but
suppose one asks a question does there exist
a function which is continuous everywhere
but, not differentiable. Anywhere that does,
there exist a is that, is that question clear
does there exist a function which is continuous
for all x in R, but not differentiable for
any x in R.
Then we will say that our usual intuitions
does not help us in answering that question
and one feels that such a function may not
exist at all. But, the answer is yes there
is a function does there exist a function
which is continuous everywhere, but not differentiable
at any point and the construction of such
function requires everything. We have done,
so far about 
make some preparation by the way it is not
very difficult constructing that function
is not very difficult what is involve is little
bit of calculations. Some use of all these
theorems which we have met. Now, to begin
with let me first make a few observations
one observation which we have already made,
let me repeat it once again.
You know that saying that some limit of f
x as x tends to a suppose this is equal to
n we have already seen that this is equivalent
of saying that if you take any sequence x
n converging to a then f of x n must converges
to l. If you take any sequence x n converging
to a f of x n must converge to l suppose I
have applied this concept to derivatives,
suppose I apply this concept to derivatives
suppose f is differentiable at x.
What is the meaning of that, it means that
limit of let us say again we will say let
the limit of t minus f x at t minus x limit
as t tends to x exist suppose f is differentiable
then this is same as this limit is same as
f prime x, this limit is same as f prime x.
How do I convert this using this earlier idea,
this will mean that if I take any sequence
suppose I call sequence alpha n, suppose I
take any sequence alpha n converging to x
then f of alpha n minus f x divided by alpha
n minus x limit of that as n tends to infinity.
That should be same as f prime x that means
let us, let us suppose if that means this
is, so this is equivalent to say that for
every sequence alpha n alpha n converging
to x 
limit of f at alpha n minus f x divided by
alpha n minus x limit of this as n tends to
infinity. This limit should exist and that
limit should be same as f prime x, now this
very easy that follows immediately by applying
this principle. But, what we want is, now
this a slight modification of this what I
want is that suppose instead of one sequence
alpha n, suppose we take two sequences alpha
n and beta n and suppose both of them are
converging to x.
I look at this ratios f at beta n minus f
at alpha n divided by beta n minus alpha n
and take the limit of that as n tends to infinity.
Then what are beta n and alpha n, beta n and
alpha n are two sequences converging to x
beta n and alpha n are two sequences converging
to x. Now, the question is whether the limit
of such thing is also same as f prime x that
is the question and the answer is in general
this is not true if you take alpha n and beta
n as totally arbitrary sequences converging
to x this may or may not happen. But, if you
choose alpha n and beta n in a particular
way this is true and what is that particular
way let me just explain it.
Now, let us say this for theorem let us say
that let f suppose f is defined on some interval
a b open interval because we are talking of
differentiability we will take open interval
f from a b to R is differentiable 
at let us say some x in a b. We take two sequences
alpha n and beta n in the following manner,
let a less not equal to alpha n less not equal
to x, less not equal to beta n and less not
equal to b. That is all alpha n are less not
equal to x and all beta n are bigger than
or equal to x for all n and both of them converse
to x and limit of alpha n as n tends to infinity.
This is x and limit of beta n as enters to
infinity is also x then this is true then
limit as n tends to infinity of beta n minus
f at alpha n divided by beta n minus alpha
n this is equal to f prime as and we shall
first prove this. Then we shall use this to
construct that function which is continuous
everywhere, but differentiable nowhere, in
fact that was a very famous function first
constructed by a Weirstrass.
Who is famous for several things and this
function construction of this function because
this said something totally opposite to whatever
is the usual intuition of everybody. That
led to deeper thinking about what how the
function as to be defined etcetera because
till that time everybody use to think that
function means something defined by a formula.
Functions means 2 x cube plus 3 x square of
f x is equal to sine x or functions means
something defined by that kind of formula.
But, after this example came everybody thought
that the concept of function has to be looked
at more carefully. Then we have the modern
definition of functions after several people
contributed to that definition we shall not
go to that kind of history. But, we shall,
we shall look at construction of this function
in tomorrow class because today the time is
already over.
