Chap 03 Real Analysis: Limit And Continuity

Chapter 3 – Limit and Continuity Subject: Real Analysis (Mathematics) Level: M.Sc. Source: Syyed Gul Shah (Chairman, Department of Mathematics, US Sargodha) Collected & Composed by: Atiq ur Rehman ([email protected]), http://www.mathcity.org

v Limit of the function Suppose (i) ( X , d X ) and (Y , dY ) be two metric spaces (ii) E ⊂ X (iii) f : E → Y i.e. f maps E into Y . (iv) p is the limit point of E . We write f ( x ) → q as x → p or lim f ( x ) = q , if there is a point q with the x→ p

following property; For every ε > 0 , there exists a δ > 0 such that dY ( f ( x), q ) < ε for all points x ∈ E for which d X ( x, p ) < δ .

If X and Y are replaced by a real line, complex plane or by Euclidean space ¡ k , then the distances d X and dY are replaced by absolute values or by appropriate norms. q Note: i) It is to be noted that p ∈ X but that p need not a point of E in the above definition ( p is a limit point of E which may or may not belong to E .) q ii) Even if p ∈ E , we may have f ( p ) ≠ lim f ( x) . x→ p

v Example 2x =2 x →∞ 1 + x 2x 2x − 2 − 2x −2 2 −2 = = < We have x −1 x 1+ x 1+ x 2 Now if ε > 0 is given we can find δ = so that ε 2x − 2 < ε whenever x > δ . 1+ x lim

q

v Example x2 − 1 . x −1 It is to be noted that f is not defined at x = 1 but if x ≠ 1 and is very close to 1 or less then f ( x) equals to 2. q Consider the function f ( x ) =

v Definitions i) Let X and Y be subsets of ¡ , a function f : X → Y is said to tend to limit l as x → ∞ , if for a real number ε > 0 however small, ∃ a positive number δ which depends upon ε such that distance f ( x ) − l < ε when x > δ and we write lim f ( x) = l . x →∞

ii) f is said to tend to a right limit l as x → c if for ε > 0 , ∃ δ > 0 such that f ( x ) − l < ε whenever x ∈ G and 0 < x < c + δ . And we write f (c + ) = lim f ( x) = l x →c +

iii) f is said to tend to a left limit l as x → c if for ε > 0 , ∃ a δ > 0 such that f ( x ) − l < ε whenever x ∈ G and 0 < c − δ < x < c . And we write f (c −) = lim f ( x ) = l . q x →c −

2 v Theorem Suppose (i) ( X , d x ) and (Y , d y ) be two metric spaces (ii) E ⊂ X (iii) f : E → Y i.e. f maps E into Y . (iv) p is the limit point of E . Then lim f ( x) = q iff lim f ( pn ) = q for every sequence { pn } in E such that x→ p

n→∞

pn ≠ p , lim pn = p . n→∞

Proof Suppose lim f ( x) = q holds. x→ p

Choose { pn } in E such that pn ≠ p , lim pn = p , we are to show that n→∞

lim f ( pn ) = q

n→∞

Then there exists a δ > 0 such that d y ( f ( x), q ) < ε if x ∈ E and 0 < d x ( x, p ) < δ ………. (i) Also ∃ a positive integer n0 such that n > n0 ⇒ d x ( pn , p ) < δ ………….. (ii) from (i) and (ii), we have for n > n0 d y ( f ( pn ), q ) < ε Which shows that limit of the sequence lim f ( pn ) = q n→∞

Conversely, suppose that lim f ( pn ) = q is false. n→∞

Then ∃ some ε > 0 such that for every δ > 0 , there is a point x ∈ E for which d y ( f ( x), q ) ≥ ε but 0 < d x ( x, p ) < δ . 1 , n = 1,2,3,...... n We find a sequence in E satisfied pn ≠ p , lim pn = p for which lim f ( pn ) = q In particular, taking δ n =

n→∞

n→∞

q

is false. v Example

1 does not exist. x →∞ x 1 Suppose that lim sin exists and take it to be l, then there exist a positive real x →∞ x number δ such that 1 sin − l < 1 when 0 < x − 0 < δ (we take ε = 1 > 0 here) x We can find a positive integer n such that 2 2 2 < δ then < δ and <δ nπ (4n + 1)π (4n + 3)π It thus follows (4n + 1)π sin −l < 1 ⇒ 1− l < 1 2 (4n + 3)π and sin − l < 1 ⇒ − 1 − l < 1 or 1 + l < 1 2 lim sin

3 So that

2 = 1+ l +1− l ≤ 1+ l + 1− l < 1+1 ⇒ 2 < 2 This is impossible; hence limit of the function does not exist.

q

Alternative: Consider But

{ f ( xn )}

2 then lim xn = 0 x →∞ (2n − 1)π  1 i.e. sin  is an oscillatory sequence  xn  xn =

 1 i.e. {1, −1,1, −1,..........} therefore sin  diverges.  xn  1 Hence we conclude that lim sin does not exit. x →∞ x

q

v Example Consider the function x ;  f ( x) =  2  2 + ( x − 1) ; We show that lim f ( x) does not exist.

x <1 x ≥1

x→1

1 To prove this take xn = 1 − , then lim xn = 1 and lim f ( xn ) = 1 x →∞ n→∞ n 1 But if we take xn = 1 + then xn → 1 as n → ∞ n 2  1  and lim f ( xn ) = lim 2 + 1 + − 1 = 2 x →∞ x→∞  n  This show that { f ( xn )} does not tend to a same limit as for all sequences such that xn → 1 . Hence this limit does not exist.

{ Sn }

v Example Consider the function f :[0,1] → ¡ defined as if x is rational 0 f ( x) =  if x is irratioanl 1 Show that lim f ( x) where p ∈ [0,1] does not exist. x→ p

Solution Let lim f ( x ) = q , if given ε > 0 we can find δ > 0 such that x→ p

f ( x ) − q < ε whenever x − p < δ . Consider the irrational (r − s, r + s ) ⊂ [0,1] such that r is rational and s is irrational. Then f (r ) = 0 & f (s ) = 1 Suppose lim f ( x ) = q then x→ p

f (s ) = 1 ⇒ 1 = f (s) − q + q = ( f (s ) − q + q − 0 = f (s ) − q + q − f (r )

Q 0 = f (r )

q

4

≤ f ( s) − q + f (r ) − q < ε + ε i.e. 1 < ε + ε 1 1 1 ⇒ 1< + if ε = 4 4 4 Which is absurd. Hence the limit of the function does not exist.

q

v Exercise lim x sin x →0

1 =0 x

We have 1 x sin − 0 < ε x 1 ⇒ x sin < ε x 1 ⇒ x sin < ε x

⇒ x <ε

where ε > 0 is a pre-assigned positive number.

Q sin

1 ≤1 x

⇒ x−0 <ε =δ 1 It shows that lim x sin = 0 . x →0 x Same the case for function for f ( x ) = x cos

1 x

Also we can derived the result that lim x 2 sin x →0

1 = 0. x

q

v Theorem If lim f ( x) exists then it is unique. x→c

Proof Suppose lim f ( x) is not unique. x→c

Take lim f ( x) = l1 and lim f ( x ) = l2 x →c

x →c

where l1 ≠ l2 .

⇒ ∃ real numbers δ1 and δ 2 such that f ( x ) − l1 < ε whenever & Now

l1 − l2 =

f ( x ) − l2 < ε

whenever

( f ( x) − l1 ) − ( f ( x) − l2 )

x − c < δ1 x − c < δ2

≤ f ( x ) − l1 + f ( x) − l2 < ε + ε whenever x − c < min(δ1 ,δ 2 ) ⇒ l1 = l2 …………………

q

5 v Theorem Suppose that a real valued function f is defined on an open interval G except possibly at c ∈ G . Then lim f ( x) = l if and only if for every positive real number x→c

ε , there is δ > 0 such that

{x :

x − c < δ }.

f (t ) − f (s ) < ε whenever s & t are in

Proof Suppose lim f ( x) = l x→c

∴ for every ε > 0 , ∃ δ > 0 such that 1 f ( s) − l < ε whenever 0 < s − c < δ 2 1 & f (t ) − l < ε whenever 0 < t − c < δ 2 ⇒ f ( s ) − f (t ) ≤ f ( s ) − l + f (t ) − l ε ε < + whenever s − c < δ & t − c < δ 2 2 f (t ) − f ( s ) < ε whenever s & t are in { x : x − c < δ } . Conversely, suppose that the given condition holds. Let { xn } be a sequence of distinct elements of G such that xn → c as n → ∞ . Then for δ > 0 ∃ a natural number n0 such that xn − l < δ and xm − l < δ ∀ m, n > n0 . And for ε > 0 f ( xn ) − f ( xm ) < ε whenever m, n > n0 ⇒ { f ( xn )} is a Cauchy sequence and therefore it is convergent. q v Theorem (Sandwiching Theorem) Suppose that f , g and h are functions defined on an open interval G except possibly at c ∈ G . Let f ≤ h ≤ g on G . If lim f ( x ) = lim g ( x) = l , then lim h( x ) = l . x →c

x→c

x →c

Proof For ε > 0 ∃ δ 1 ,δ 2 > 0 such that f ( x ) − l < ε whenever 0 < x − c < δ1 & g ( x) − l < ε

0 < x − c < δ1 l − ε < g ( x) < l + ε for 0 < x − c < δ2 l − ε < f ( x ) ≤ h( x ) ≤ g ( x) < l + ε l − ε < h( x ) < l + ε for 0 < x − c < min(δ 1 , δ 2 ) lim h( x) = l

⇒ l − ε < f ( x) < l + ε & ⇒ ⇒ ⇒

whenever 0 < x − c < δ 2 for

x →c

……………………….

q

6 v Theorem Let (i) ( X , d ) , (Y , d y ) be two metric spaces. (ii) E ⊂ X (iii) p is a limit point of E . (iv) f : E → Y . (v) g : E → Y and lim f ( x ) = A and lim g ( x) = B then x→ p

x→ p

i- lim ( f ( x) ± g ( x) ) = A ± B x→ p

ii- lim( fg )( x) = AB x→ p

 f ( x)  A iii- lim  provided B ≠ 0 . = x → p g ( x)   B Proof q

Do yourself

v Continuity Suppose i) ( X , d X ) , (Y , dY ) are two metric spaces ii) E ⊂ X iii) p ∈ E iv) f : E → Y Then f is said to be continuous at p if for every ε > 0 ∃ a δ > 0 such that dY ( f ( x), f ( p) ) < ε for all points x ∈ E for which d X ( x, p ) < δ . Note: (i) If f is continuous at every point of E . Then f is said to be continuous on E . (ii) It is to be noted that f has to be defined at p iff lim f ( x) = f ( p ) . q x→ p

v Examples f ( x) = x 2 is continuous ∀ x ∈ ¡ . Here f ( x) = x 2 , Take p ∈ ¡ Then f ( x) − f ( p) < ε ⇒ x2 − p 2 < ε

⇒ ( x − p)( x + p) < ε ⇒ x− p < ε =δ Q p is arbitrary real number ∴ the function f ( x) is continuous ∀ real numbers. ………………………..

q

7 v Theorem Let i) X ,Y , Z be metric spaces ii) E ⊂ X iii) f : E → Y , g : f ( E ) → Z and h : E → Z defined by h( x ) = g ( f ( x) ) If f is continuous at p ∈ E and if g is continuous at the point f ( p) , then h is continuous at p. Proof h f

g

x p E

f(x)

h(x) = g(f(x))

f(p)

h(p) = g(f(p))

f Y

X

g

Z

h

Q g is continuous at f ( p) ∴ for every ε > 0 , ∃ a δ > 0 such that d Z ( g ( y ), g ( f ( p ) ) ) < ε whenever dY ( y , f ( p) ) < δ1 ……….. (i) Q f is continuous at p ∈ E ∴ ∃ a δ > 0 such that dY ( f ( x ), f ( p) ) < δ 1 whenever d X ( x, p ) < δ ………… (ii) Combining (i ) and (ii ) , we have d Z ( g ( y ), g ( f ( p ) ) ) < ε whenever d X ( x, p ) < δ ⇒ d Z ( h( x), h( p) ) < ε whenever d X ( x, p ) < δ which shows that the function h is continuous at p .

q

v Example (i) f ( x ) = (1 − x 2 ) is continuous ∀ x ∈ ¡ and g ( x) = x is continuous

∀ x ∈ [0, ∞ ] , then g ( f ( x) ) = 1 − x 2 is continuous x ∈ ( −1,1) .

{

(ii) Let g ( x) = sin x and f ( x) = x − π , x ≤ 0 x +π , x > 0 Then g ( f ( x) ) = − sin x ∀ x Then the function g ( f ( x) ) is continuous at x = 0 , although f is discontinuous at x = 0 . q v Theorem Let f be defined on X . If f is continuous at c ∈ X then ∃ a number δ > 0 such that f is bounded on the open interval (c − δ , c + δ ) . Proof Since f is continuous at c ∈ X . Therefore for a real number ε > 0, ∃ a real number δ > 0 such that f ( x ) − f (c) < ε whenever x ∈ X and x − c < δ . ⇒ f ( x ) = f ( x ) − f (c ) + f (c )

8

≤ f ( x ) − f (c ) − f (c ) < ε + f (c) whenever x − c < δ . It shows that f is bounded on the open interval ]c − δ , c + δ [ .

q

v Theorem Suppose f is continuous on [ a, b] . If f (c ) > 0 for some c ∈ [ a, b ] then there exist an open interval G ⊂ [ a, b] such that f ( x ) > 0 ∀ x ∈ G . Proof 1 f (c ) 2 Q f is continuous on [ a, b] ∴ f ( x) − f (c) < ε whenever x − c < δ , x ∈ [ a, b] Take ε =

Take G = { x ∈ [ a, b ] : x − c < δ }

⇒

f ( x ) = f ( x ) − f (c ) + f (c ) ≤ f ( x ) − f (c ) + f (c ) < ε + f (c) whenever x − c < δ For x ∈ G , we have f ( x ) = f ( c ) − ( f (c ) − f ( x ) ) ≥ f (c ) − f (c ) − f ( x ) 1 ≥ f (c ) − f ( x ) − f (c ) > f ( c ) − f ( c ) 2 1 ⇒ f ( x ) > f (c ) > 0 2 v Example Define a function f by  x cos x ; x ≠ o f ( x) =  ; x=0  0 This function is continuous at x = 0 because f ( x ) − f (0) = x cos x ≤ x ( Q cos x ≤ 1 Which shows that for ε > 0 , we can find δ > 0 such that f ( x ) − f (0) < ε whenever 0 < x − c < δ = ε v Example f ( x) = x is continuous on

q

) q

[ 0,∞ [ .

Let c be an arbitrary point such that 0 < c < ∞ For ε > 0 , we have x−c x−c f ( x ) − f (c ) = x − c = < x+ c c x−c ⇒ f ( x) − f (c) < ε whenever <ε c i.e. x − c < c ε = δ ⇒ f is continuous for x = c . Q c is an arbitrary point lying in [0,∞ [

∴ f ( x) = x is continuous on [0,∞ [ ………………………..

q

9 v Example Consider the function f defined on ¡ such that  1 , x is rational f ( x) =   −1 , x is irrational This function is discontinuous every where but f ( x) is continuous on ¡ .

q

v Theorem A mapping of a metric space X into a metric space Y is continuous on X iff f −1 (V ) is open in X for every open set V in Y . Proof Suppose f is continuous on X and V is open in Y . We are to show that f −1 (V ) is open in X i.e. every point of f −1 (V ) is an interior point of f −1 (V ) . Let p ∈ X and f ( p ) ∈V Q V is open ∴ ∃ ε > 0 such that y ∈V if dY ( y, f ( p) ) < ε …….. (i) Q f is continuous at p ∴ ∃ δ > 0 such that dY ( f ( x), f ( p) ) < ε when d X ( x, p ) < δ ……… (ii) From (i) and (ii), we conclude that x ∈ f −1 (V ) as soon as d X ( x, p ) < δ Which shows that f −1 (V ) is open in X . Conversely, suppose f −1 (V ) is open in X for every open set V in Y . We are to prove that f is continuous for this. Fix p ∈ X and ε > 0 . Let V be the set of all y ∈ Y such that dY ( y, f ( p) ) < ε

V is open, f −1 (V ) is open ⇒ ∃ δ > 0 such that x ∈ f −1 (V ) as soon as d X ( x, p ) < δ . But if x ∈ f −1 (V ) then f ( x ) ∈V so that dY ( f ( x ), f ( y ) ) < ε Which proves that f is continuous.

q

Note The above theorem can also be stated as a mapping f : X → Y is continuous iff

f −1 (C ) is closed in X for every closed set C in Y .

q

v Theorem Let f1 , f 2 , f3 ,...., f k be real valued functions on a metric space X and f be a mapping from X on to ¡ k defined by f ( x ) = ( f1 ( x), f2 ( x ), f3 ( x),....., f k ( x) ) , x ∈ X then f is continuous on X if and only if f1, f 2 , f 3 ,....., f k are continuous on X . Proof Let us suppose that the function f is continuous on X , we are to show that f1 , f 2 , f3 ,......, f k are continuous on X .

(

)

If p ∈ X , then d ¡k f ( x), f ( p ) < ε

whenever d X ( x, p ) < δ

⇒

x− p < δ

f ( x) − f ( p ) < ε

whenever

10

⇒

f1 ( x) − f1 ( p), f1 ( x) − f1 ( p),...... f k ( x) − f k ( p) < ε whenever

x− p < δ

1

2 2 2 2 ⇒ ( f1 ( x) − f1 ( p ) ) , ( f 2 ( x) − f 2 ( p ) ) ,......, ( f k ( x) − f k ( p ) )  < ε   whenever x − p < δ

1

2  k 2 ⇒ ∑ ( f i ( x) − f i ( p ) )  < ε whenever x − p < δ  i =1  ⇒ f1 ( x) − f1 ( p) < ε whenever x − p < δ

i.e.

f 2 ( x) − f 2 ( p) < ε whenever x − p < δ …………………… …………………… …………………… f k ( x) − fk ( x ) < ε whenever x − p < δ ⇒ all the functions f1, f 2 , f 3 ,....., f k are continuous at p . Q p is arbitrary point of x , therefore f1, f 2 , f 3 ,....., f k are continuous on X . Conversely, suppose that the function f1, f 2 , f 3 ,....., f k are continuous on X , we are to show that f is continuous on X . For p ∈ X and given ε i > 0 , i = 1,2,.....k ∃ δ i > 0 , i = 1,2,...., k Such that f1 ( x) − f1 ( p) < ε1 whenever x − p < δ1 f 2 ( x) − f2 ( p) < ε 2 whenever x − p < δ2 …………………… …………………… …………………… f k ( x) − f k ( x) < ε k whenever x − p < δk Take δ = min (δ 1, δ 2 ,δ 3 ,...., δ k ) then fi ( x) − fi ( p) < ε i whenever x− p < δ 2 2 2 ⇒ ( f1 ( x) − f1 ( p ) ) + ( f 2 ( x) − f 2 ( p ) ) + .... + ( f k ( x) − f k ( p ) ) 





1

2

(

< ε12 + ε 22 + .... + ε k2

)2

1

2 2 2 2 i.e. ⇒ ( f1 ( x) − f1 ( p ) ) + ( f 2 ( x) − f 2 ( p ) ) + ..... + ( f k ( x) − f k ( p ) )  < ε   whenever x − p < δ

where

(ε

2 1

(

+ ε 22 + ..... + ε k2

)

) 2= ε 1

Then d ¡k f ( x), f ( p ) < ε

whenever d X ( x, p ) < δ

⇒ f ( x) is continuous at p .

Q p is an arbitrary point therefore we conclude that f is continuous on X .

…………………………

q

1

11 v Theorem Suppose f is continuous on [ a, b] i) If f (a ) < 0 and f (b ) > 0 then there is a point c , a < c < b such that f (c ) = 0 . ii) If f (a ) > 0 and f (b ) < 0 , then there is a point c , a < c < b such that f (c ) = 0 . Proof i) Bisect [ a, b] then f must satisfy the given condition on at least one of the sub-interval so obtained. Denote this interval by [ a2 , b2 ] If f satisfies the condition on both sub-interval then choose the right hand one [ a2 , b2 ] . It is obvious that a ≤ a2 ≤ b2 ≤ b . By repeated bisection we can find nested intervals {I n } , I n+1 ⊆ I n , I n = [ an , bn ] so that f satisfies the given condition on

[ an , bn ] , n = 1,2,.......

And a = a1 ≤ a2 ≤ a3 ≤ ..... ≤ an ≤ bn ≤ ..... ≤ b2 ≤ b1 = b n

1 Where bn − an =   ( b − a ) 2

II n

Then

n

contain one and only one point. Let that point be c such that

i =1

f (c ) = 0 If f (c ) ≠ 0 , let f (c ) > 0 then there is a subinterval [ am , bm ] such that am < bm < c Which can not happen. Hence f (c ) = 0 ii) Do yourself as above

q

v Example Show that x 3 − 2 x 2 − 3 x + 1 = 0 has a solution c∈ [ −1,1] Solution Let f ( x) = x 3 − 2 x 2 − 3x + 1 Q f ( x) is polynomial ∴ it is continuous everywhere. (for being a polynomial continuous everywhere) Now f (−1) = (−1)3 − 2(−1)2 − 3(−1) + 1 = −1 − 2 + 3 + 1 = 1 > 0 f (1) = (1)3 − 2(1)2 − 3(1) + 1 = 1 − 2 − 3 + 1 = −3 < 0 Therefore there is a point c∈ [ −1,1] such that f (c ) = 0 i.e. c is the root of the equation. q v Theorem (The intermediate value theorem) Suppose f is continuous on [ a, b] and f (a ) ≠ f (b ) , then given a number λ that lies between f (a ) and f (b) , ∃ a point c , a < c < b with f (c) = λ . Proof Let f (a ) < f (b) and f (a ) < λ < f (b) . Suppose g ( x) = f ( x ) − λ Then g (a ) = f (a ) − λ < 0 and g (b) = f (b) − λ > 0 ⇒ ∃ a point c between a and b such that g (c) = 0 ⇒ f (c ) − λ = 0 ⇒ f (c ) = λ If f (a ) > f (b) then take g ( x) = λ − f ( x) to obtain the required result.

q

12 v Theorem Suppose f is continuous on [ a, b] , then f is bounded on [ a, b] (Continuity implies boundedness) Proof Suppose that f is not bounded on [ a, b] , We can, therefore, find a sequence { xn } in the interval [ a, b] such that f ( xn ) > n for all n ≥ 1 . ⇒ { f ( xn )} diverges. But a ≤ xn ≤ b ; n ≥ 1 ⇒ ∃ a subsequence ⇒

{ f ( x )} nk

{ x } such that { x } converges to λ . nk

nk

also converges to λ .

⇒ { f ( xn )} converges to λ . Which is contradiction Hence our supposition is wrong.

q

v Uniform continuity Let f be a mapping of a metric space X into a metric space Y . We say that f is uniformly continuous on X if for every ε > 0 there exists δ > 0 such that dY ( f ( p), f (q) ) < ε ∀ p, q ∈ X for which d x ( p, q ) < δ The uniform continuity is a property of a function on a set i.e. it is a global property but continuity can be defined at a single point i.e. it is a local property. Uniform continuity of a function at a point has no meaning. If f is continuous on X then it is possible to find for each ε > 0 and for each point p of X , a number δ > 0 such that dY ( f ( x), f ( p) ) < ε whenever d X ( x, p ) < δ . Then number δ depends upon ε and on p in this case but if f is uniformly continuous on X then it is possible for each ε > 0 to find one number δ > 0 which will do for all point p of X . It is evident that every uniformly continuous function is continuous. To emphasize a difference between continuity and uniform continuity on set S , we consider the following examples. q v Example Let S be a half open interval 0 < x ≤ 1 and let f be defined for each x in S by the formula f ( x) = x 2 . It is uniformly continuous on S . To prove this observe that we have f ( x) − f ( y) = x 2 − y 2

= x− y x+ y < 2 x− y If x − y < δ then f ( x ) − f ( y ) < 2δ = ε ε Hence if ε is given we need only to take δ = to guarantee that 2 f ( x ) − f ( y ) < ε for every pair x, y with x − y < δ Thus f is uniformly continuous on the set S .

q

13 v Example f ( x) = x n , n ≥ 0 is uniformly continuous of [0,1] Solution For any two values x1, x2 in [0,1] we have

x1n − x2n =

( x1 − x2 ) ( x1n−1 + x1n−2 x2 + x1n−3 x22 + ..... + x2n−1 )

≤ n x1 − x2 ε Given ε > 0 , we can find δ = independent of x1 and x2 such that n x12 − x22 < n x1 − x2 < ε whenever x1, x2 ∈ [0,1] and x1 − x2 < δ = Hence the function f is uniformly continuous on [0,1] .

ε n q

v Example Let S be the half open interval 0 < x ≤ 1 and let a function f be defined for each 1 x in S by the formula f ( x ) = . This function is continuous on the set S , x however we shall prove that this function is not uniformly continuous on S . Solution Let suppose ε = 10 and suppose we can find a δ , 0 < δ < 1 , to satisfy the condition of the definition. δ Taking x = δ , y = , we obtain 11 10δ x− y = < δ 11 and 1 11 10 f ( x) − f ( y) = − = > 10 δ δ δ Hence for these two points we have f ( x ) − f ( y ) > 10 (always) Which contradict the definition of uniform continuity. Hence the given function being continuous on a set S is not uniformly q continuous on S . v Example 1 f ( x ) = sin ; x ≠ 0 . is not uniformly continuous on 0 < x ≤ 1 i.e (0,1]. x Proof Suppose that f is uniformly continuous on the given interval then for ε = 1 , there is δ > 0 such that f ( x1 ) − f ( x2 ) < 1 whenever x1 − x2 < δ 1 1 Take x1 = and x2 = , n ≥1. 1 3 ( n − 12 )π ( n − 2 )π 2 So that x1 − x2 < δ = 3 ( n − 12 )π But

f ( x1 ) − f ( x2 ) = sin ( n − 12 )π − sin 3 ( n − 12 )π

Which contradict the assumption. Hence f is not uniformly continuous on the interval.

= 2 >1 q

14 v Example Prove that f ( x) = x

is uniformly continuous on [0,1] .

Solution Suppose ε = 1 and suppose we can find δ , 0 < δ < 1 to satisfy the condition of the definition. δ2 2 Taking x = δ , y = 4 δ2 3δ 2 2 = <δ Then x − y = δ − 4 4 And

δ2 δ − 4

f ( x) − f ( y) =

2

δ δ = < 1=ε 2 2 Hence f is uniformly continuous on [0,1] . = δ−

q

v Theorem If f is continuous on a closed and bounded interval [ a, b] , then f is uniformly continuous on [ a, b] . Proof Suppose that f is not uniformly continuous on [ a, b] then ∃ a real number ε > 0 such that for every real number δ > 0 . We can find a pair u , v satisfying u − v < δ but f (u ) − f (v) ≥ ε > 0 1 If δ = , n = 1,2,3,.... n We can determine two sequence {un } and {vn } such that 1 u n − vn < but f (un ) − f (vn ) ≥ ε n Q a ≤ un ≤ b ∀ n = 1,2,3.......

∴ there is a subsequence

{u } which converges to some number nk

u0 in [ a, b]

⇒ for some λ > 0 , we can find an integer n0 such that unk − u0 < λ ∀ n ≥ n0 ⇒ vnk − u0 ≤ vnk − unk + unk − u0 <

{ } ⇒ { f ( u )} and { f ( v )} converge to f (u ) . Consequently, f ( u ) − f ( v ) < ε whenever

1 +λ n

⇒ vnk also converges to u0 . nk

nk

nk

0

nk

unk − vnk < ε

Which contradict our supposition. Hence we conclude that f is uniformly continuous on [ a, b] .

…………………………….

q

15 v Theorem Let f and g be two continuous mappings from a metric space X into ¡ k , then the mappings f + g and f ⋅ g are also continuous on X . i.e. the sum and product of two continuous vector valued function are also continuous. Proof i) Q f & g are continuous on X .

∴ by the definition of continuity, we have for a point p ∈ X . ε f ( x) − f ( p ) < whenever x − p < δ 1 2 ε and g ( x) − g ( p) < whenever x − p < δ2 2 Now consider f ( x) + g ( x) − f ( x) − g ( p ) = f ( x ) − f ( p ) + g ( x) − g ( p ) ≤

f ( x) − f ( p ) + g ( x) − g ( p)

ε ε + = ε whenever x − p < δ where δ = min (δ 1, δ 2 ) 2 2 which shows that the vector valued function f + g is continuous at x = p and hence on X . <

ii)

f ⋅g =

k

∑ f ⋅g i =1

i

i

= f1g1 + f 2 g 2 + f 3 g 3 + ..... + f k g k Q the function f and g are continuous on X

∴ their components f i and g i are continuous on X .

q

v Question Suppose f is a real valued function define on ¡ which satisfies lim [ f ( x + h) − f ( x − h)] = 0 ∀ x ∈ ¡ h→0

Does this imply that the function f is continuous on ¡ .

Solution Q lim [ f ( x + h) − f ( x − h) ] = 0 h→0

∀ x∈¡

⇒ lim f ( x + h) = lim f ( x − h) h→0

h→0

⇒ f ( x + 0) = f ( x − 0) ∀ x ∈ ¡ Also it is given that f ( x) = f ( x + 0) = f ( x − 0) It means f is continuous on x ∈ ¡ . ……………………………

q

16 v Discontinuities If x is a point in the domain of definition of the function f at which f is not continuous, we say that f is discontinuous at x or that f has a discontinuity at x. If the function f is defined on an interval, the discontinuity is divided into two types 1. Let f be defined on ( a, b ) . If f is discontinuous at a point x and if f ( x+ ) and f ( x−) exist then f is said to have a discontinuity of first kind or a simple discontinuity at x . 2. Otherwise the discontinuity is said to be second kind. For simple discontinuity i. either f ( x +) ≠ f ( x −) [ f ( x) is immaterial] ii. or f ( x + ) = f ( x −) ≠ f ( x) q v Example 1 , x is rational i) Define f ( x) =  0 , x is irrational The function f has discontinuity of second kind on every point x because neither f ( x+ ) nor f ( x−) exists.

q

 x , x is rational ii) Define f ( x) =   0 , x is irrational Then f is continuous at x = 0 and has a discontinuity of the second kind at every other point. q (−3 < x < −2)  x+2 iii) Define f ( x ) =  − x − 2 (−2 < x < 0)  x+2 (0 < x < 1)  The function has simple discontinuity at x = 0 and it is continuous at every other point of the interval (−3,1) q  1 , x≠0 sin iv) Define f ( x ) =  x , x=0  0  Q neither f (0+ ) nor f (0−) exists, therefore the function f has discontinuity of second kind. f is continuous at every point except x = 0 . q References:

(1) Lectures (2003-04) Prof. Syyed Gull Shah

Chairman, Department of Mathematics. University of Sargodha, Sargodha.

(2) Book Principles of Mathematical Analysis Walter Rudin (McGraw-Hill, Inc.)

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