Algebra of Continuous Functions PDF

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Chapter : 5 ( Theory of real function)
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Algebra of Continuous Functions

Lesson: Algebra of Continuous Functions Paper: Analysis - II Lesson Developer : Pragati Gautam and Sudha Gupta Department / College: Assistant Professor, Department of Mathematics, Kamala Nehru College/Lakshmibai College University of Delhi

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Algebra of Continuous Functions Table of Contents Chapter : Algebra of Continuous Functions 

1 : Learning Outcomes



2 : Introduction



3 : Continuous Functions o 3.1

Continuity of a Function at a Point

o 3.2

Another Definition of Continuity

o 3.3

Sequential Criteria for continuity



4 : Algebra of Continuous Functions



5: Composition of Continuous Functions



Exercise



References

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Algebra of Continuous Functions

1. Learning Outcomes After studying this chapter you should be able to  Have a deep understanding of continuous functions.  Understand Algebra of continuous functions.  Gain an understanding of Properties of algebra of continuous functions.  Will be able to solve questions related to them.

"Mathematicians do not study objects, but relations among objects; they are indifferent to the replacement of objects by others as long as relations do not change. Matter is not important, only form interests them....." Henri Poincare

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Algebra of Continuous Functions

2. Introduction The idea of a function has an irregular history. Until the time of L. Euler, functions were thought of largely in terms of formulas and thus were restricted to expressions that can be generated by elementary operations. The modern view of a function as a general assignment was first seen in Euler’s later work and particularly in the subsequent work of Cauchy and P. Dirichlet. A function f from A to B is a rule that assigns to each element in A one and only one element in B. We write f : A  B where A and B are nonempty sets. A graph is a function if it passes the vertical line test. i.e., if a vertical line is drawn anywhere on the co-ordinate plane, it crosses the graph only once. If a vertical line goes through the graph more than once, that means there is more than one y-value for the x-value. So if a graph fails the vertical line test, it is not a function. A continuous function is a function in which for “small” changes in the input result in “small” changes in the output. Continuity was first defined by Bernard Bolzano in 1817. Augustin-Louis Cauchy defined continuity for y = f (x) by saying for an infinitely small increment  of the independent variable x always produces an infinitely small change f (x + ) – f (x) of the dependent variable y. Cauchy called “infinitely small quantities” as variable quantities. He introduced the concept of continuous functions by requiring that indefinite small changes of x should produce indefinite small changes in y. But Bolzano (1817) and Weierstrass (1874) were more precise by saying the difference f (x) – f (x0) must be arbitrarily small for the difference x – x0 sufficiently small. Thus the concept of continuity is of great important in Analysis.

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Algebra of Continuous Functions Consider the graphs of the functions f and g. At the point x = c, there is a significant difference. The graph of g has a break at c, whereas the graph of f can be drawn without the pencil leaving the paper. This distinction, stated geometrically, can be made precise if we focus on the functions involved and not their graphs. The values of the function f are all near f (c) for x near c, i.e., lim f x c

(x) exists and equals f (c). However, this is not true for the function g. For some x near c (either for any x < c or x > c) the values of g are not near g (c). We do not have lim g (x) = g (c). In fact, lim g (x) does not exist. x c

x c

The function g is not continuous at the point c.

y = f (x)

y = g(x)

I.Q. 1 I.Q. 2 3.

Continuous Functions:

Definition: Let f be a function defined in a neighbourhood of the point c. Then f is continuous at c if

lim f  x  f  c x c

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Algebra of Continuous Functions In practice, most functions of a real variable have domains that are intervals or unions of separate intervals, thus we have to consider three type of points; interior points, left end points and right end points. In general, a function f is continuous at a left end point a of its domain if it is right continuous at a and continuous at a right end point b of its domain if it is left continuous at b. A function is continuous at an interior point c of its domain if and only if it is both right continuous and left continuous at c. Value addition : Example (1)

The function f  x 

1 is not continuous at x = 0, because x2

f  x  is not defined at x = 0. Moreover, this point of discontinuity cannot be removed, since we have

lim x c

(2)

The function

f  x 

1  x2

1 is not continuous at x = 0 and its x

discontinuity at x = 0

lim

x 0

cannot be removed, because

1 1    and lim    . x 0 x x

f ( x) 

1 x2

f ( x) 

I.Q. 3

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1 x

Algebra of Continuous Functions 3.1

Continuity of a Function at a Point The notion of continuity is one of the central concepts of

Mathematical Analysis. We will define what it means to say that a function is continuous at a point. It is known as an  –  criterion. Definition: Let A  R, f : A  R and c  A . Then function f is said to be continuous at c if for every  > 0, there exists  > 0 such that for x  A

and

xc   

f  x   f  c   .

If f fails to be continuous at c, then we say that f is discontinuous at c. Value Addition: Note (1)

In the above definition c must be in A but need not be a cluster point of A. This lim f  x need not exist, even when f is x c

continuous at c. Consider the function

f : A  R defined by f ( x) 

x 3  x2

where A  {0}  [1, ) is the domain of f. f is continuous at 0, but lim f ( x) does not exist. x 0

(2)

In case c is a cluster point of A, then definition 2.2 is

equivalent to (i)

f  c exists

(ii)

lim f  x  exists and

(iii)

lim f  x   f  c 

x c

x c

I.Q. 4 I.Q. 5 In fact we say that If c is a cluster point of A, then f is continuous at c iff

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Algebra of Continuous Functions

lim f  x   f  c  x c

3.2

Another Definition of Continuity

The definition of continuity at a point can be reformulated in terms of neighbourhoods. The function f is continuous at c if for every  > 0, there exists  > 0 such that x  A and x  c    f  x  f  c   i.e.,

x  A and c    x  c    f c     f  x   f c   

i.e.,

x  A and x ]c  , c  [ f  x  ] f c   , f c    [

i.e.,

xA

] c  , c  [ f  x ] f  c   , f  c   [

Thus in other words, Let A  R , f : A  R and c  A. Then function f is continuous at c iff for every   nbd V  f c  of f  c , there exists   nbd V c  of c such that

i.e.,

x A

 f x  V  f c 

fA

 V  f  c  

Value Addition : Note If c  A is not a cluster point of A, then there exists a neighbourhood V  c  of c such that

A

=

 c .

Then the

function will be automatically continuous at c  A. Such points are called Isolated Points. For the function f : A  R defined by f ( x) 

x 3  x2 .

where A  {0}  [1, ) is the domain of f. Here, 0 is an isolated point of the domain of f.

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Algebra of Continuous Functions

Example

1:

Consider

the

function

f

:

R



R,

defined

by

1  x sin , x  0 f  x   x  0 , x 0  Solution: Here lim f  x  lim x sin x0



x0

1  0  f  0 x

the function f is continuous at 0.

Example 2: Let f  x   sgn  x  x  R . Where sgn is the signum function Show that f is not continuous at 0. Solution: We have

1 , x  0  f  x   0 , x  0  1 , x  0  Here lim f  x    1 x 0

lim f  x   1

x 0



lim f  x   lim f  x 

x 0

x 0



lim f  x does not exist.



f is not continuous at x = 0.

x 0

Example 3: The function f : R  R, defined by f (x) = [x] is called the greatest integer function. Check the continuity of f (x). Solution: We define [x] to be the greatest integer n  Z such that n  x. Let z be any integer.

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Algebra of Continuous Functions

lim f  x   z  1

x z 

Also lim f x   z x z

lim f  x  lim f  x 



x z 

x z



lim f  x does not exist



f is not continuous at any integer. But when x is not an integer, i.e.,

x z

x is a real number then lim f  x  lim f  x  f  c . x c

x c

I.Q. 6 Example

4:

Determine

the

points

of

discontinuity

of

f  x   sin x ,0  x  2 Solution:

0 , 1 ,  Here f  x  0 ,  1 ,   1 ,

0  x  / 2 x  / 2 / 2  x     x  3 / 2 3 / 2  x  2 

This function is discontinuity at 0, /2, , 2, 5/2, …… Example 5: Determine the points of continuity of the following functions:

(i)

f  x 

x2  2 x  1  x R x2  1

=

 x 1

2

x 2 1

It is continuous everywhere on R.

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Algebra of Continuous Functions

(ii)

hx  1

Solution:

| sin x | , x 0 x

hx 

1  sin x x



f x where x  0. g x 

sin x ; x  0 sin x    sin x ; x  0 Here g  x  0 for x = 0 and g(x)  0 ; x  0

f  x

g x

is defined  x  0

sin x is continuous  |sin x| is continuous 

1  sin x is continuous.

Also g x   x is continuous. Therefore, h  x  

(iii)

f x  is continuous  x  0. g  x

k  x   cos 1  x2  x  R 1  x 2 are continuous functions.

Solution: cos and Therefore, cos 1  x

2

is also a continuous function.

Example 6: Show that if f : A  R is continuous on A  R and if n  N, n then the function f defined by f n  x    f  x  for x  A, is continuous n

on A. Solution: Here f is continuous on A For

f 2  x    f f  x

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Algebra of Continuous Functions = f  x f  x   f  x 

2

If f is continuous so f 2 will also be continuous. 3

4

Similarly, f , f ,... will also be continuous.

f n  x   f  x  will also be continuous. n

 I.Q. 7 3.3

Sequential Criteria for continuity:

Theorem 1: A function f : A  R is continuous at a point c  A iff for every sequence

 xn 

in A that converges to c, the sequence

f  xn 

converges to f  c . Proof: Necessary part Suppose f is continuous at c and let the sequence

 xn 

in A be such that it

converges to c. Since f is continuous at c, therefore for every  > 0, there exists  > 0 such that x  A and x  c    f  x  f  c   (1) Now the sequence  xn



converges to c, therefore for  > 0, there exists a

+ve integer m such that xn  c    n  m (2) From (1) and (2) we have

f  xn   f  c    n  m (replacing x by xn in (1)) 

The sequence

 f  x  converges to f  c . n

Sufficient part

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Algebra of Continuous Functions Suppose that for every sequence

 xn 

in A that converges to c, the

sequence f  xn  converges to f  c . We have to prove that f is continuous. On contrary, assume that f is not continuous at c. Then there exist an  > 0 for which no  works. i.e., for n 

1 , n  N ,  yn  A s.t n

yn  c 

1 but f  yn   f  c    n n

This implies that the sequence

1    0 but the sequence aslim n n 

 yn 

 f  y  n

in A is such that

lim yn  c n

does not converge to f  c which

is a contradiction. Therefore f is continuous. Example 7: Show that the Dirichlet's functions is discontinuous at every real number.

Solution: Let

1 if x is rational f  x   0 if x is irrational

Let c be a real number. Case-I: Let c is a rational number then

f c   1 By density theorem, there exists a sequence such that Here

 xn 

of irrational numbers

lim xn  c n

f  xn   0

n

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Algebra of Continuous Functions Thus,  a sequence  xn



such that

lim xn  c n

but

lim f  xn   f  c  n

By sequential criteria, we have that the function f is not continuous at c. Case-II: Let c be an irrational number. Then f (c) = 0 By density theorem,  a sequence

 yn  of rational numbers such that

lim yn  c n

Here f ( yn )  1  n

lim f ( yn )  1  f (c)  n

Thus,  a sequence  yn  s.t

lim yn  c n

but

lim f  yn   f  c n

By sequential criteria, the function is not continuous at c. Thus from Case I and Case II, we have that function f is not continuous at any real number.

Example 8:

Let

 2 1  x sin   , x  0 f  x    x  0 , x 0 

Use the  –  definition to show that f is continuous at 0.

Solution: Let  > 0

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Algebra of Continuous Functions Now f  x  f  0  f  x  x2  x We want this to be less than , so we let    . Then x  0   implies x2  2   

x  0    f  x  f  0  

Hence  –  property holds and f is continuous at 0. We can also show that if lim  xn   0 for every seq  xn



then lim f (xn) = 0

Thus sequential criteria can also be used to prove that f is continuous at 0. Example 9: Prove that the function

x is continuous on its domain [0, )

. Solution: Let f  x 

x , x 0

The function f is continuous at c  c  0. Let  > 0 be arbitrary. Case-I: When c = 0, then f  x   f  c  

x 0

x  0   if 0  x  2 Let  = 2 Then 0  x  2 

x 0 

This shows that the function f is continuous at 0.

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Algebra of Continuous Functions Case-II: When c  0 then f  x  f  c 



x c 

x c xc  x c c

f  x  f  c   if x  c  c  c

take  = then x  c   

x c 

This implies that the function

x is continuous at c.

4. Algebra of Continuous Functions Let A  R and let f and g be functions defined on A to R. We define the sum f + g, the difference f – g, and the product fg on A to R to be the functions given by

f

 g x  f  x  g  x , for all x  A

f

 g x  f  x  g  x , for all x  A

 f g  x  f  x g  x,

for all x  A

Further, if b  R, we define the multiple bf to be the function given by

 bf  x   bf  x  for all

x  A.

Also, if h  x  0 for x  A, we define the quotient

f to be the function h

f x  f  for all x  A x    h  x h 

given by 

Value addition If f and g are functions, both bounded on A and c is any real

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Algebra of Continuous Functions number, then the function f + g, f – g, cf and f g are each bounded on A.

I.Q. 8 Theorem 2: Suppose the function f and g are continuous at the point c. Then the functions f + g, f – g, f g are continuous at c. Proof: If c  A is not a cluster point of A, then the conclusion is obvious. Hence we assume that c is a cluster point of A. Since f and g are continuous at c, then

f  c  lim f  x x c

g c   lim g  x 

and

x c

We have lim  f  g  x   lim  f  x   g  x  x c

x c

= lim f  x  lim g x x c

x c

= f (c) + g(c) = 

 f  g  c 

lim  f  g  x =  f  g  c  x c

Thus f + g is continuous at c. In a similar manner we can show that f – g is continuous at c. Again, lim  f g  x   lim f  x  g x  x c

x c

= lim f  x lim g  x  x c

x c

= f (c) g (c)

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Algebra of Continuous Functions = f g (c ) Thus we have lim  f g  x   f g  c  x c



The function f g is continuous at c.

Converse of the above theorem need not be true. For Example:

(i)

Let

 1x  e  1 , f  x   1x  e 1  0,

x0 x 0

and 1  x  1 e  , g x    1 x e  1  0,

x 0 x 0

then

f

 g  x   0 for all x



f + g is continuous at 0 (being a constant function)

but the functions f and g are discontinuous at x = 0

(ii)

Let
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