Mod-29 Lec-31 Uniform Continuity and Absolute Continuity

# Mod-29 Lec-31 Uniform Continuity and Absolute Continuity

Today, we will discuss uniform continuity
and absolute continuity. Let us see the first uniform. We have already seen the continuity
definition. Let A, which is a subset of R and let f a mapping from A to R – set of
real numbers A to R; then the following two conditions, then following two statements
are equivalent. That is, the first statement says that, f is continuous at every point
u belongs to A; and second is given epsilon greater than 0 and u which is an element of
A; there is a delta, which depends on epsilon as well as the point u and greater than 0
such that for all x such that x belong to A and satisfy the
condition mod x minus u is less than delta, which is depends on epsilon, u; then the mod
of f x minus f u is less than epsilon. So, this we have discussed already, a function
is continuous at a point u if for a given epsilon, there exists a delta such that mod
of f x minus f u less than epsilon whenever mod of x minus u less than delta.
And here we have seen that, this delta depends on the point. If I change the point, correspondingly,
delta will change. So, what this shows? This shows the function f changes its behavior
when the point changes; maybe we have suppose at some point, the function is very slowly
changing their values; and, near to some point, it changes very rapidly. For example, if we
take the sin 1 by x when x is not equal to 0; if we look this function, then this function
is changing very rapidly when the point is very close to 0. It goes very up and down
from minus 1 to plus 1; and, very rapidly it goes. So, we are interested in such type
of the functions, where the change is smooth say. Or, we can say that, we are interested
in the delta, which is independent of u. And, that leads the concept of uniform continuity.
Though the function we say, it is continuous point wise here; but point wise means delta
will depend on the u. So, we wanted a definition in which the delta is independent of u. And,
that leads to the definition or concept of uniform continuity. So, let us see the definition
of uniform continuous function. Let A be a nonempty subset of R and let f is a mapping
from A to R. We say f is uniformly continuous. f is said to be uniformly continuous on the set A if
for each epsilon greater than 0, there is a delta, which depends only on epsilon – a
positive delta, which depends only on epsilon and independent of the point u of A such that
if x and u are any two arbitrary points of A, are any number
satisfying the condition – mod of x minus u is less than delta – depends on epsilon
only; then mod of f x minus f u is less than epsilon. So, what this shows is that, a function
is said to be uniformly continuous over the set A. Remember, when we say, the function
is continuous, then we can say function is continuous at a point. So, at a point, we
can identify a delta, which depends on epsilon and u. But when we say, the function is uniformly
continuous, then saying uniform continuous at a point is a meaningless. It will be continuous
over a set. So, a function is said to be uniform continuous over the set A, we mean that, if
for any epsilon greater than 0, if we are able to get a delta, which is independent
of the points of the set A such that whenever we pick up any two arbitrary points say in
the delta neighborhood of this, then corresponding fluctuation f x minus f u will remain less
than epsilon – the value of this. For example, if we take the function f x equal
to say 2 x; and, this f x equal to 2 x for x belongs to say the real number R. Now, if
I consider mod f x minus f u, where the u and v are the point… Let x and u – these
are the points in R; and, satisfies that condition. So, consider this. This is equal 2 x minus
2 u, which is equal to 2 times x minus u. So, if we choose delta, which
depends on say epsilon and u as epsilon by 2; obviously, this delta is independent of
u, because this is basically, we are taking delta to be epsilon by 2 whatever the u may
be. So, it is independent of this. So, if I take delta this, then obviously, this part
for all epsilon greater than 0, then this condition holds – less than epsilon for
all x and u belongs to R such that mod of this x minus u less than delta. And, this
is independent of… Delta is independent of u and positive quantity. So, this function
will be considered – f x equal to 2 x, we can say it is uniformly continuous over the
entire real line. However, there are the functions, which are
only continuous, but not uniformly. For example, we take the function. So, f x equal to 2 x
is uniformly continuous over any set A, which is subset of R or any subset of R or in entirely.
Now, take a function g x say 1 by x for x belonging to the set A, which is the set of
those points of real number such that x is positive. Now, we claim that, this function
g is continuous, but not uniformly over A; continuous at each point
of A, but not uniformly. Let us see how. Let us consider
g x minus g u. This is equal to what? 1 by x minus 1 by u which is the same as u minus
x over u x. Now, if u belongs to this – if u belongs
to A suppose; u belongs to A is given; where, we wanted to test the continuity is given.
So, if we take delta, which depends on u as the infimum of u by 2 and u square epsilon
by 2. So, let epsilon greater than 0 be given. Here we let us write, let epsilon greater
than 0 be given. And, let us pick up the point u at which the continuity is tested. So, now
choose the delta as this. So, when you take this delta, then if mod of x minus u is less
than delta and delta, which is depending on epsilon and u; then I can choose, suppose
first it is less than u by 2, so then take delta as u by 2. So, what happens is, this
shows that, x minus u is less than u by 2; or, this implies that, x lies between 3 by
2 u and half u, because x minus u is less than u by 2. So, it becomes less than 3 by
2. And then x minus u is u minus u by 2 is greater than this. So, it is greater. So,
it lies bound, therefore the bound for this… Therefore, 1 by x can be; x is greater than
u by 2. So, 1 by x will be greater than, sorry, x is less than, so 1 by x is less than, because
it will be; x is greater than this. So, 1 by x will remain less than 2 by u from here.
Once it is 2 by u, then in the condition, which we have taken as g x minus g u – in
this case, what we get? Mod of this; this is less than equal to u minus x mod over u
x, so u into x; x means 1 by x. So, it is 2 by u. So, it becomes the 2 by u square into
u minus x. Now, further choose delta to be this thing
– u square epsilon by 2. I have taken delta to be this. Another one I am taking this.
So, what we get is, from here, this shows that this part – mod of g x minus g u is
less than equal to. Delta means; mod of u minus x is less than delta, so this is less
than 2 by u square into u square by 2 epsilon that is epsilon. So, this holds that, if x
minus u is less than delta, for all such x, then g x minus g u will be less than epsilon.
And, this is true. So, this shows, that g is continuous at u belongs to A. But here
the delta which you are choosing is coming; which depends on u is positive quantity. But
what happens to this? This is not, each individual delta is positive when u is taken to be there.
But when you take the infimum value of this – when you take the delta as the infimum
of all such, then what happens to that? Infimum of delta… Here delta depends on epsilon u. So, g is
point wise continuous. But what is the infimum of all such deltas? Infimum of all such delta,
which depends on epsilon u; and, u is greater than 0. The infimum value of this is coming
to be 0. Why? Because each delta… Here is nothing but either u by 2 or u square by 2.
So, u is greater than 0. So, each delta is greater than 0. But when you take the infimum
value of this delta over u, then this infimum will be come out to be 0. So, we are not getting
a delta, which is greater than 0. So, for a given epsilon greater than 0, we cannot
get delta, which depends only on epsilon and greater than 0 such that condition – mod
of g x minus g u is less than epsilon provided mod of x minus u less than delta hold. This
we cannot get. Therefore, g is not uniformly continuous. So, we have seen the example where
the function is continuous, point wise and the function is uniformly continuous.
Now, to show the uniform continuity, we require the delta, where independent of point over
the entire set. So, that is not that easy. So, what we can develop that test which will
give at least sufficient criteria when the function is not uniformly continuous. So,
we just state those results without proof – the criteria for the non-uniform continuity.
So, the non-uniform continuity criteria –
this will be needed. So, proof – we just are dropping. But it can be easily done with
the help of previous knowledge. Let A be a nonempty subset of R and let f is a mapping
from A to R. Then the following statements are equivalent. The first statement
says, f is not uniformly continuous on A. Second statement says that, there exists an
epsilon naught greater than 0 such that for every delta greater than 0, there are points
say x depends on delta and then u depends on delta in A such that, the mod of x delta
minus u delta less than delta and mod of f x delta minus f u delta is greater than or
equal to epsilon naught. And, the third statement says, there exists an epsilon naught greater
than 0 and two sequences say x n and u n in A such that limit of x n minus u n over n
is 0 and mod of f of x n minus f of u n is greater than equal to epsilon naught for all
n and belongs to capital N. Let us see what is this?
Uniform continuity criteria says, if suppose function is not uniform; then by the definition
of not uniform means a function is said to be uniform continuous over the set A if for
each epsilon, there exists a delta, which depends only on epsilon, not on the delta
such that the difference of f x minus f u can be made less than epsilon provided the
points are in the delta neighborhood. So, if the function is not uniformly continuous,
it means this condition will be violated. If we choose the point in the neighborhood
of delta, the images of this, the fluctuation may not be less than epsilon; it can exceed
to any arbitrary number epsilon naught. So, that is why, what he says is that, if f is
not uniformly continuous, then there exists an epsilon naught such that, whenever the
point x and u are in the delta neighborhood, the corresponding images exceed that bound
epsilon naught greater than… Similarly, this is Cauchy’s definition;
this is Heine’s definition. Instead of choosing the two arbitrary points, if we picked up
the two sequences x n, u n, which are tending to 0, the difference of this is tending 0.
This means x n and u n are very close to each other as n is sufficiently large. But the
corresponding image is not close, is greater than equal to some positive number epsilon
naught. Then we say, the function f is not uniformly continuous. We got this. Now, this
criteria can be applied very directly. Suppose I apply this function to show g x,
which is 1 by x, is not uniformly continuous on the set A, where x
is greater than 0 – set of all real number n. So, what we do is, we have to pick up the
two arbitrary sequences. So, let x n – I choose 1 by n; and, u n – I take to be 1
over n plus 1. Both are in A, and the difference of these two sequences – obviously, x n
minus u n. This goes to 0 as n tends to infinity. But what is g of x n minus g of u n? This
mod is nothing but what? g of x n is nothing but n plus 1 minus n, that is, 1, which does
not go to 0; in fact, it is greater than equal to any number epsilon naught. Therefore, g
is not uniformly continuous. That is what. Now, uniform continuity – when its function
is defined over a closed interval and the function is continuous, then it will be uniform
continuous. This result is known as the uniform continuity
theorem. The theorem says, let I be a closed bounded interval and let f, which is a mapping
from I to R be continuous on I. Then f is uniformly continuous on I. So, what is said,
if the function f, which is continuous over a closed and bounded interval; then the function
must be uniformly continuous. Suppose f is not uniformly continuous
on I. So, I can use one of the criteria, which I listed earlier. I will take in the form
of the sequence. So, by the previous results or previous theorems, where the criteria are
there, we can choose; then there exists an epsilon naught greater than 0 and two sequences
x n’s and u n’s in I such that the difference between these two… that is, the limit of
this is going to 0 means difference is very very small – say 1 by n. But the mod of
f x n minus f u n – this difference exceeds by this epsilon naught for all n. So, this
is by the criteria when non-uniformly continuous criteria. From here we are getting this one.
Now, since I is bounded and the sequence x n and u n’s – both are the sequences belong
to I. So, by Bolzano-Weierstrass theorem, every bounded sequence has a convergence subsequence.
So, there is convergent subsequences say x n k. If there is convergent subsequence, let
us take first, this part; then we can take belongs to this. So, there exists a convergent
subsequence x n k of x n that converges to an element
say z belongs to I; that converges to z. Now, we wanted to show, z is a point in I; which
follows, because I is closed. Since I is closed… So, all the limit points of a sequence in
I must be point in there. Also, all these sequences x n k must be lies between the lower
and the upper bound of this interval. So, by the Sandwich theorem, the limit of this
sequence x n k must be the point in I. So, this implies, since I is closed, the limit
point z belongs to I. Similarly, we also claim that, the sequence
u n will have a subsequence u n k in I, whose limit point belongs to I. But the limit point
of u n k and x n k will be the same. The reason is, because if I consider
the u n k minus z, then this can be written as the u n k minus x n k plus x n k minus
z. Now, this term is less than equal to 1 by k by condition, which have chosen, because
both are in A and we are choosing the interval and neighborhood in such a way, so that this
is less than equal to 1 by k and this converges to z. This is the limit point of this. So,
it goes to 0. So, as tends to total tends to 0. Therefore, both will have the same limit
point. Once they are having same limit point, f is continuous.
Now, further, since f is continuous at z, then both the sequences: f of x n k
and f of u n k must converge to f z, because x n k goes to z. So, f of x n k will go to
f of z; u n k goes to z. So, f of u n k will also go to f of z because f is continuous.
So, once they are continuous… But the given hypothesis is that, mod of f x n minus f of
u n is greater than equal to epsilon naught. This is given, so the contradiction. And contradiction
is because of our wrong assumption that, function is not uniformly continuous. Therefore, f
is uniformly continuous over A. That shows the result. So, this proves that uniform continuity.
Now, there is one more result, which we say. Then we come to that. Theorem is, if f is a mapping from A to R,
is uniformly continuous on a subset A of R on a subset
A of R and if x n is a Cauchy sequence in A, then f of x n – this
sequence is a Cauchy sequence in R. It means if f is a uniformly continuous function, then
it will transfer the Cauchy sequence to the Cauchy sequence. The proof is let x n be a
Cauchy sequence in A and f is given to be uniform continuous
over A. So, by the definition of continuity, let for a given epsilon greater than 0, we
can identify a delta. There exists a delta, which depends only on epsilon greater than
0 such that the mod x minus u is less than delta such that, if x comma u belongs to A
satisfies this condition; then f of x minus f of u – this is less than epsilon. So,
let it be 1. Now, it has given the sequence x n is a Cauchy
sequence. Since x n is a Cauchy sequence. By definition of Cauchy, for a given delta
greater than 0, there exists an H, which depends on delta such that the difference between
any two arbitrary terms of the sequence after a certain stage can be made less than delta
for all m, n, which are greater than equal to H. This is true. Now, by the same choice
of delta, since mod of x n minus x m is less than this, if we take x equal to x n, u equal
to x m; from 1, it follows that f of x n minus f of x m – mod of this will be less than
epsilon for all n, m greater than equal to H. This shows the sequence f x n is a Cauchy
sequence in R. So, that proves the result. Now, we come to the thing, which is say absolute
continuity. That is the new concept. Absolute continuity means a function f x is said to be absolutely
continuous in the interval say a, b if corresponding
to an arbitrarily chosen positive number epsilon. Another positive number delta can
be determined such that, for a sequence of non-overlapping intervals – open interval h r, k r defined
in a, b. The sigma of mod f of h r minus f of k r is less than epsilon provided sigma
of k r minus h r – this length is less than delta. So, what we
say, it means we say, that is, f x is absolutely continuous function. The meaning of this is
absolute continuous over the interval a, b if corresponding to a given epsilon a
number delta exists such that in a countable set of
non-overlapping intervals of total length – I am not using the measure –
less than delta, the sum of the fluctuations of the function
is less than epsilon. Let us see what the meaning of this is. Suppose we have an interval a, b and a function
f is given. We say this function is absolutely continuous in the interval a, b if for a given
epsilon greater than 0, if we have a non-overlapping intervals means divide this one. Say here
this is like this – if I take this non-overlapping intervals, and so on like this. So, if we
take a countable number on non-overlapping interval of this, whose length is less than
delta. So, we have to take a small portion. Suppose I take this small portion. Now, this
small portion total length is delta; total length of this is delta. So, this small length
over this small interval – we can find a non-overlapping intervals, such that sigma
of this length k r minus h r – this length is less than delta. These are countable – 1
to infinity; R is 1 to infinity. Total length is less than delta.
For a given epsilon – say this epsilon is there; for this given epsilon, we can find
a delta such that whenever we have countable number of non-overlapping intervals, whose
length is less than delta, then corresponding fluctuation over these subintervals – the
total sum the corresponding fluctuation should not exceed by epsilon. If so then we say,
that is, sigma of mod f of h r minus f of k r – this should be less than epsilon 1
to infinity. So, if the total fluctuation of the function over this subintervals is
less than epsilon whenever the points are in the countable number of intervals, whose
length total sum is less than delta, then function is said to be an absolutely continuous
function. So, obviously, every absolutely continuous is continuous.
As a note, we can say, function, which is absolutely continuous
in the interval a, b is also continuous in accordance with Cauchy definition, because
what we do, we replace it that, total set k r – we consider single interval; some
of these, we can replace by a small interval; total sum is less than delta. And correspondingly,
here we get the total fluctuation is less than epsilon. So, this will be obviously true
– consists of this; less than that. Then the condition of uniform continuity. Similarly,
we can also say that, we may take this interval – single interval delta; and, the definition
of uniform continuity is also satisfied over this. So, function, which is absolutely continuous
is a continuous. And, for b also, we say it is uniformly continuous.
And, uniform continuity – uniform continuity condition is also satisfied if we choose the
k r to consider single interval length delta. Converse is not true; that is, a function,
which is continuous, may not be absolutely continuous in that interval; that is, a function,
which is continuous in the interval a, b may not be
absolutely continuous in that interval. A function, which
is continuous, may not be absolutely continuous. For example, if we take the function f x,
which is defined as x sin 1 by x when x is different from 0 and 0 when x is 0. We know
this function is a continuous function. This is a continuous. This we have seen. This function f x is continuous.
This we have already seen. So, because the limit of f x – continuous at 0. And, otherwise
also, it is continuous in the total limit as x tends to 0 is the value of this function
coming to 0, which is f 0. And, for other point at x belongs to 0, 1 interval, in fact,
entire interval it is continuous, because at the point 0, the value is coming to be
this and continuity follows. Now, it is not absolutely continuous we claim. But this function
f x is not absolutely continuous on the interval 0, 1. Why? It means the condition of the absolute
condition is not satisfied; that is, if we choose the infinite number or countable number
of subintervals, whose total length is less than delta, but the fluctuations may not be
less than epsilon. So, that is what. Suppose I divide the interval 0, 1 – in
each of these subintervals, we get into subintervals say 1 by r pi, 1 over r plus 1 pi – these
subintervals, where r is 1, 2, 3 and so on – into subintervals. And, in each intervals
in the subintervals 1 upon r pi and 1 upon r plus 1 pi; r is 1, 2, 3. The fluctuation
of the function f x, which is x sin 1 by x exceed by this number 1 by r plus 1 pi. Let
us see, why? Over this function the reason is. Over this
function, this is the interval 1 by r pi; this is the interval 1 by r plus 1 pi. The
function x sin 1 by x… If we write the function f x, which is x sin 1 by x, at the end point,
the sin of this is 0, because it is an integral multiple of pi. So, basically, the function
will go like this, because at this end point, r pi or r plus 1 pi, this value will give
the value 0. So, the fluctuation of the function – the value this minus the value this will
be something, which is greater than this one. Now, here is the maximum value. Suppose I
take x equal to say… Let us choose 1 by x to be 2 r plus 1 pi by
2. So, x will be 1 upon this lying say here. Then what will be the sin of this value? Sin
of 1 by x is 1 becase odd multiple of this 1. And, this value will give the f of x 2
r plus 1 pi by 2 – 1 upon 2 r plus 1 pi by 2. So, this will be greater than the value
at this point r plus 1. So, fluctuation will exceed by this number, which is greater than
this – by this. So, we get from here is that, the fluctuation is exceeding by this
number. So, sum of the fluctuation
in the sequence of intervals say 1 by r pi and 1 over r plus 1 pi – this sequence of
intervals corresponding to r is equal to n, n plus 1 and so on, is greater than is greater
than – if I take the sum, is greater than 1 by pi 1 over n plus 1 plus 1 over n plus
2 and so on. But this series is a divergent series. So, this cannot be made as small as
we please. So, it is greater than by any assigned positive number A for the series. Therefore,
this large number cannot be… So, the function f x, which is x sin 1 by x; x is different
from 0 and 0 when x is 0, is not absolute continuous. So, this shows. Results are just I state one result. The result
states, the sum and the product of two absolutely continuous functions
are absolutely continuous. So, this proof follows by the definition; and, one can go
with it. Thank you very much.

## 4 Replies to “Mod-29 Lec-31 Uniform Continuity and Absolute Continuity”

1. deepak singh rawat says:

Sir kya aap mujhe koi book suggest ker sakte hai real analysis ki.

2. Ray ll says:

It is so clear! Thank you so much for this quality video

3. Dineshkumar Kanesan says:

Absolute Continuity 36:10 , thank you sir!

4. Chaudhary aman Patel says:

epsinol,epsilon