Chapter 3 – Limit and Continuity C ontinuity 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 , 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 . 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
x ∈ E for which d X ( x, p)
<
>0
such that dY ( f ( x), q )
<
ε
for all points
δ. k
If X and Y are replaced by a real line, complex plane or by Euclidean space ¡ , then the distances d X and d Y 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 .) ii) Even if p ∈ E , we may have f ( p) ≠ lim f ( x) .
q
x→ p
v Example
lim
2 x
1 + x 2x
x →∞
We have
x −1
Now if ε
=2
−2 =
2x − 2 − 2x 1+ x
> 0 is given we can find 2 x 1 + x
−2 <ε
δ
−2 < 1+ x
= =
2 ε
2 x
so that
whenever x > δ .
q
v Example
−1 . x − 1 It is to be noted that t hat f is not defined at x = 1 but if x ≠ 1 and is very close to 1 or Consider the function f ( x ) =
x 2
q
less then f ( x) equals to 2. 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 r ight limit l as x → c if for ε
f ( x) − l
<ε
> 0, ∃
>0
such that
> 0, ∃ a δ > 0
such that
δ
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
f ( x) − l
<ε
whenever x ∈ G and 0 < c − δ
And we write f ( c −) = lim f ( x) = l . x →c −
< x < c. q
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
pn
n→∞
pn = p . ≠ p , lim n→∞
Proof
Suppose lim f ( x) = q holds. x → p
Choose { pn } in E such that pn lim f ( pn ) = q
≠ p , lim p = p , we are to show that →∞ n n
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)
∃ 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
Also
Which shows that limit of the sequence seq uence lim f ( pn ) n→∞
Conversely, suppose that lim f ( pn ) n→∞
=
q
q is false.
Then ∃ some ε > 0 such that for every δ d y ( f ( x), q ) ≥ ε but 0 < d x ( x, p ) < δ .
> 0 , there is a point x ∈ E for which
1
, n = 1,2,3,...... n We find a sequence in E satisfied pn ≠ p , lim pn In particular, taking δ n
=
=
n→∞
=p
for which li li m f ( p n ) = q n→∞
q
is false. v Example
1 lim sin x →∞ x
does not exist.
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 and
sin
( 4n + 3)π 2
−l < 1
⇒ −1 − l < 1
or
1+ l
<1
3 So that
2 = 1 + l + 1− l
≤ 1 + l + 1− l < 1 + 1
⇒
2
<
2 q
This is impossible; hence limit of the function does d oes not exist. Alternative: Consider But
=
xn
2 ( 2n − 1)π
1 sin x n
{ f ( xn )} i.e.
then lim xn x →∞
=0
is an oscillatory sequence
i.e. .e. {1 {1, −1,1, −1,.... ,...... .......} ...} therefore sin Hence we conclude that lim sin x →∞
1 x
xn 1
diverges. q
does not exit.
v Example
Consider the function
x f ( x ) = 2 2 + ( x − 1)
;
x <1
;
x ≥1
We show that lim f ( x) does not exist. x →1
To prove this take xn But if we take xn
=1−
=1+
1
1 n
, then lim xn
then xn
n
x →∞
→1
=1
and lim f ( xn ) = 1 n→∞
as n → ∞
2
and lim f ( xn ) = lim x →∞
x →∞
This show that such that xn
1 2 + 1 + − 1 = 2 n
{ f ( xn )}
does not tend to a same limit as for all sequences
{ Sn }
→1 . q
Hence this limit does not exist. v Example
Consider the function f : [0,1] [0,1] → ¡ defined as
0 1
f ( x ) =
if x is rational if x is irratioanl
Show that lim f ( x) where p ∈[0,1] does not exist. x → p
Solution
> 0 we can find δ > 0 such that f ( x) − q < ε whenever x − p < δ . ( r − s, r + s) ⊂ [0,1] such that r is rational and
Let lim f ( x) = q , if given ε x → p
Consider the irrational
irrational. Then f (r ) = 0 & f (s ) = 1 Suppose lim f ( x) = q then x → p
=1 ⇒ 1 = f ( s) − q + q = ( f ( s) − q + q − 0 = f (s) − q + q − f ( r ) f ( s)
Q 0 = f (r )
s is
4
≤ f ( s) − q + f ( r ) − q < ε + ε i.e. 1 < ε + ε 1 1 1 if ε = ⇒ 1< + 4
4
4
Which is absurd. Hence the limit of the function does not n ot exist.
q
v Exercise
lim x sin
x→0
1 x
=0
We have x sin
1 x
1
⇒
x sin
⇒
x sin
⇒
x
⇒
x − 0
It shows that
−0 <ε
where ε
> 0 is a pre-assigned positive number.
<ε
x
1 x
<ε
<ε
Q sin
1 x
≤1
<ε =δ
lim x sin
x→0
1 x
= 0.
1 Same the case for function for f ( x) = x cos x Also we can derived the result that lim x 2 sin x →0
v
1 x
=0.
q
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 δ 1 and δ 2 such that
f ( x) − l1
Now
< ε whenever x − c < δ1 & f ( x) − l2 < ε whenever x − c < δ 2 l1 − l2 = ( f ( x ) − l1 ) − ( f ( x ) − l2 ) ≤ 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
>0 x − c < δ }.
ε , there is δ
{x :
f (t ) − f ( s)
such that
<ε
whenever s & t are in
Proof
Suppose lim f ( x) = l x→c
∴ for every
ε
&
⇒
> 0, ∃
> 0 such that 1 f ( s) − l < ε whenever
0< s−c
<δ
f (t ) − l
whenever
0< t −c
<δ
f ( s) − l
f (t ) − l
δ
f ( s) −
<
2 1
ε
2 f (t )
≤ ε
< +
ε
2
f (t ) −
+
whenever
s−c
2 f ( s) < ε whenever s & t are in
<δ
&
t−c
<δ
{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 → ∞ .
> 0 ∃ a natural number n0 such that ∀ m, n > n0 . xn − l < δ and xm − l < δ ε >0 f ( xn ) − f ( xm ) < ε whenever m, n > n0
Then for δ And for
⇒ { f ( xn )} is a Cauchy sequence and therefore it is convergent. v
q
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 li l im h( x) = l . x →c
Proof For ε
x →c
>0 ∃
x→c
δ 1 ,δ 2
>0
such that
f ( x) − l
⇒ &
⇒ ⇒ ⇒
< ε whenever 0 < x − c < δ1 & g ( x) − l < ε whenever 0 < x − c < δ 2 l − ε < f ( x ) < l + ε for 0 < x − c < δ1 l − ε < g ( x) < l + ε for 0 < x − c < δ 2 l − ε < f ( x ) ≤ h ( x ) ≤ g ( x) < l + ε l − ε < h( x ) < l + ε for 0 < x − c < min(δ1 , δ 2 ) lim h( x) = l 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 = B x → p g x ( )
iii- lim
provided B ≠ 0 .
Proof q
Do yourself v
Continuity
Suppose i) ( X , d X ) , (Y , d Y ) are two metric spaces ii) E ⊂ X iii) p ∈ E iv) f : E → Y
>0 ∃ a δ >0 d X ( x, p ) < δ .
Then f is said to be continuous at p if for every ε
dY ( f ( x), ), f ( p) )
<ε
for all points x ∈ E for which
such that
Note: (i) If f is continuous at every 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) = x2 is continuous
∀ x ∈ ¡ .
Here f ( x) = x2 , Take p ∈ ¡ Then
⇒ ⇒ ⇒
f ( x) − f ( p )
<ε
− p2 < ε ( x − p) ( x + p) < x − p < ε = δ x
2
ε
Q p is arbitrary real number
the fun functi ction f ( x) ∴ the
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 conti continu nuous ous at the the point 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
g
Y
X
Z
h
Q g is con continu tinuou ouss 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) )
< δ1 whenever
d X ( x, p )
<δ
………… ( ii)
Combin Combinin ing g (i ) and (ii ) , we have d Z ( g ( y), g ( f ( p) ) )
⇒
d Z ( h( x), h ( p ) )
<ε
<
whenever d X ( x, p )
ε
<δ
whenever d X ( x, p) < δ q
which shows that the function h is continuous at p . v Example
(i) f ( x) = (1 − x 2 ) is continuous
∀ x ∈ [0, ∞ ] , then
and g ( x) =
x is continuous
g ( f ( x) ) = 1 − x 2 is continuous x ∈ ( −1,1) .
(ii) Let g ( x) = sin x and f ( x ) = Then
∀ x ∈ ¡
g ( f ( x) ) = − sin x
x − π , x ≤ 0 x + π , x > 0
{
∀ x
Then the function g ( f ( x) ) is continuous at x = 0 , although f is discontinuous at x = 0 . v
q
Theorem
Let f be defined on X . If f is continuous at c ∈ X then such that f is bounded on the open interval ( c − δ , c + δ ) .
∃ a number δ > 0
Proof
Since f is continuous at c ∈ X .
> 0, ∃ a real number δ > 0 f ( x) − f (c) < ε whenever x ∈ X and ⇒ f ( x) = f ( x) − f (c) + f (c)
Therefore for a real number ε
such that
x − c
<δ .
8
≤ f ( x ) − f ( c) − f ( c ) < ε + f (c) whenever
x − c
<δ . ]c − δ , c + δ [ .
It shows that f is bounded on the open interval v
q
Theorem
[ a, b] . If f (c ) > 0 for some c ∈[ a, b ] G ⊂ [ a, b] such that f ( x ) > 0 ∀ x ∈ G .
Suppose f is continuous on exist an open interval
then there
Proof Take ε
=
1
f (c ) 2 Q f is continuous on
∴
f ( x) − f (c)
<ε
[ a, b ]
Take G = { x ∈ [ a, b] : x − c
⇒
f ( x)
x−c
whenever
<δ }
= f ( x) − f (c) + f (c) ≤ f ( x) − f (c) + f (c) < ε + f (c) whenever
For x ∈ G , we have f ( x) = f ( c) − ( f (c) − f ( x) )
⇒
f ( x)
≥
f ( c ) − f ( x) − f ( c )
>
1 2
f (c )
>
< δ , x ∈ [ a, b ]
x − c
<δ
≥
f (c) − f (c) − f ( x)
>
f (c ) −
1 2
f (c ) q
0
v Example
Define a function f by
x cos x 0
f ( x ) =
; x≠o ; x = 0
This function is continuous at x = 0 because f ( x) − f (0) = x cos x ≤ x
> 0 , we can find δ > 0 f ( x) − f (0 ) < ε whenever
Which shows that for ε
(
Q cos x
≤1 )
such that
0
< x − c < δ = ε
q
v Example
f ( x) =
Let c be an arbitrary point such that 0 For ε > 0 , we have
⇒
f ( x ) − f ( c)
=
f ( x) − f (c)
<
<
x− c ε
<
=
whenever
c
<∞
x − c x
+
c x − c c
< <
x − c c ε
=δ ⇒ f is continuous for x = c . i.e.
x−c
[ 0, ∞ [ .
x is continuous on
cε
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 f ( x) = −1
, x is ratio nal , x is irrational
This function is discontinuous every where but f ( x) is continuous on ¡ . v
q
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
⇒ ∃
> 0 such that x ∈ f −1 (V ) as soon as d X ( x, p) < δ . that dY ( f ( x), x ∈ f −1 (V ) then f ( x ) ∈V so ), f ( y) ) < ε
δ
But if
q
Which proves that f is continuous.
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 . v
q
Theorem
Let f1 , f 2 , f3 ,...., f k be 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), f 3 ( x), ....., f k ( x) ) , x ∈ X then f is continuous on X if and only if f1, f 2 , f 3 , ....., f k are a re continuous on X .
Proof Let us suppose that the function f is continuous on X , we are to show that f1 , f 2 , f 3 ,......, f k are continuous on X .
If p ∈ X , then d ¡k ( f ( x), ), f ( p ) )
⇒
f ( x) − f ( p )
<
ε
<
whenever
ε
whenever d X ( x, p)
x − p
<δ
<δ
10
⇒
f1 ( x) − f1 ( p), f1( x) − f1 ( p), ...... f k ( x ) − f k ( p)
<
whenever x − p
ε
⇒ ( f1( x) − f1 ( p ) )2 , ( f 2 ( x) − f 2 ( p ) ) 2 ,......, ( f k ( x) − f k ( p ) )2
1
2
<
⇒ ∑ ( fi ( x) − fi ( p ) )2 i =1 ⇒ f1( x) − f1 ( p) < ε f 2 ( x ) − f 2 ( p) < ε
i.e.
1
2
<
ε
whenever
whenever
x − p
whenever
x − p
…………………… …………………… …………………… f k ( x) − fk ( x) < ε whenever
⇒
x − p
x − p
<
δ
ε
whenever x − p
k
<
<δ
δ
<δ <δ
<δ
all the functions f1, f 2 , f 3 , ....., f k are a re continuous at p .
Q p is arbitrary point of x , therefore f1, f 2 , f 3 , ....., f k are a re continuous on X .
Conversely, suppose that the function f1, f 2 , f 3 , ....., f k are a re 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 < δ 1
f 2 ( x ) − f 2 ( p)
ε2
whenever
x − p
< δ2
…………………… …………………… …………………… f k ( x) − f k ( x) < ε k
whenever
x − p
<
Take δ
= min (δ1,δ 2 ,δ 3 , ....,δ k ) then fi ( x) − f i ( p ) < ε i whenever
⇒ ( f1 ( x) − i.e.
<
f1 ( p ) )
2
⇒ ( f1( x) −
+ ( f2 ( x) − f1 ( p ) )
2
f2 ( p) )
2
x − p
+ .... + ( f k ( x) −
δ k
<δ f k ( p ))
2
1
2
< (ε12 + ε 22 + .... + ε k 2 )
+ ( f 2 ( x ) − f 2 ( p ) ) + ..... + ( f k ( x) − f k ( p ) ) 2
2
whenever where Then
⇒
2 ε2
2 ε k
2
<ε x − p < δ
1
( + + ..... + ) 2 = ε ) , f ( p ) ) < ε whenever d ( f ( x), 2 ε1
1
¡ k
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
2
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 n
Then
I I 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 hich can not happen ppen.. Henc ence f ( c) = 0
ii) Do yourself as above v
q
Example
Show that x 3 − 2 x 2
− 3x + 1 = 0
has a solution c ∈ [ −1,1]
Solution Let f ( x) = x3 − 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 q
i.e. c is the root of the equation. 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 betw betwee een n 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 )
>
≥ 1.
n for all n
⇒ { f ( xn )} diverges. But a ≤ xn ≤ b ; n ≥ 1 ⇒ ∃ a subsequence { xn } k
such that { xnk } converges to λ .
⇒ { f ( xn )} also converges to ⇒ { f ( xn )} converges to λ . k
λ.
Which is contradiction Hence our supposition is wrong. v
q
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 is a global property but continuity can be defined at a single single point i.e. it is 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
< 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 a ll point p of X . δ > 0 which will do for all 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 d X ( x, p )
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) = x2 . It is uniformly continuous on S . To prove this observe that we have f ( x) − f ( y)
If x − y
<δ
then
= = <
− y2 x− y x+ y 2 x − y f ( x) − f ( y) < 2δ = ε x
2
Hence if ε is given we need only to take δ
f ( x) − f ( y)
<
=
ε
to guarantee that
2 ε for every pair x, y with x − y
Thus f is uniformly continuous on the set S .
<δ q
13 v Example
f ( x) = xn , n
≥
0 is uniformly continuous of [0,1]
Solution
[0,1] we have = ( x1 − x2 ) ( x1n−1 + x1n−2 x2 + x1n−3x 22 + ..... + x 2n−1 )
For any two values x1, x2 in
x1n
− x2n
≤ Given ε 2
x1
n x1 − x2
> 0 , we can find − x22 <
δ
n x1 − x2
ε
=
<
independent of x1 and x2 such that
n
ε
whenever x1, x2 ∈ [0,1] and x1 − x2
<δ =
ε
n
Hence the function f is uniformly continuous on [0,1] .
q
v Example
Let S be the half open interval 0 < x ≤ 1 and let a function f be defined for each
x in S by the formula f ( x ) =
1
. 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 condition of the definition. δ
Taking x = δ , y =
11
< 1, to satisfy the
, we obtain
x − y
=
10δ 11
<
δ
and f (x) − f ( y)
=
1
−
δ
11 δ
=
10 δ
f ( x) − f ( y)
Hence for these two points we have
> 10
> 10
(always)
Which contradict the definition of uniform continuity. Hence the given function being continuous on a set S is not uniformly continuous on S .
q
v Example
f ( x ) = sin
1 x
; x ≠ 0 . is not uniformly uniformly continuous on 0 < x ≤ 1 i.e (0,1].
Proof Suppose that f is uniformly continuous on the given interval then for ε there is δ > 0 such that f ( x1 ) − f ( x2 ) < 1 whenever x1 − x2 < δ Take x1 = So that But
1
and x2
( n − 12 )π
x1 − x2
<
δ
f ( x1 ) − f ( x2 )
= =
=
1 3( n −
1 2
)π
= 1,
, n ≥ 1.
2 3( n − 12 )π 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
Solution
Suppose ε = 1 and suppose we can find δ , 0 < δ the definition. Taking x = δ
, y =
2
Then
x − y
=
And
f (x) − f ( y)
δ
δ
−
2
[0,1] .
is uniformly continuous on
2
4 δ
2
4
= =
< 1 to satisfy the condition of
= δ
δ
3δ
2
<δ
4
2
−
δ
4
−δ = 2
2
δ
< 1=ε
2
Hence f is uniformly continuous on [0,1] . v
q
Theorem
[ a, b] , then f is uniformly
If f is continuous on a closed and bounded interval continuous on [ a, b] .
Proof Suppose that f is not uniformly continuous on [ a, b] then ε
such that for every real number δ > 0 . We can find a pair u , v satisfying u − v < δ but f (u ) − f (v)
∃ a real number
>0
=
>
ε
0
1
, n = 1,2,3,.... n We can determine two sequence
If δ
≥
un
Q a ≤ un
≤b ∀
− vn <
1
n n = 1,2,3.......
{un }
and
{vn } such that
f (un ) − f (vn )
but
≥
ε
∴ there is a subsequence {un } which converges to some number ⇒ for some λ > 0 , we can find an integer n0 such that un − u0 < λ ∀ n ≥ n0 k
u0 in
[ a, b ]
k
⇒
vnk
− u0 ≤
vnk
− un + k
unk
− u0 <
1 n
+λ
⇒ {vn } also converges to u0 . ⇒ { f ( un )} and { f ( vn )} converge to f (u0 ) . k
k
Consequently,
k
f (unk ) − f ( vnk )
<
ε
whenever
unk
− vn <
Which contradict our supposition. Hence we conclude that f is uniformly continuous on [ a, b] .
…………………………….
ε
k
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 .
and
f ( x) − f ( p )
<
g ( x) − g ( p)
<
ε
whenever
x − p
< δ1
whenever
x − p
<
2 ε
2
δ2
Now consider f ( x) + g ( x) − f ( x) − g ( p )
= ≤ <
f ( x ) − f ( p ) + g ( x) − g ( p ) f ( x) − f ( p ) ε
ε
+ =ε
+
g ( x) − g ( p)
whenever
x − p
2 2 which shows that the vector valued function
< δ where δ = min (δ1,δ 2 ) f + g is continuous at x = p
and
hence on X . ii)
f ⋅g
=
k
∑= f ⋅ g i
i
i 1
=
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 v
f i and g i are continuous on X .
q
Question
Suppose f is a real valued function define on ¡ which satisfies lim [ f ( x + h) − f ( x − h) ] h →0
=
0
∀ x ∈ ¡
Does this imply that the function f is continuous on ¡ .
Solution Q lim [ f ( x + h) − f ( x − h) ] = 0
⇒ ⇒
h →0
lim f ( x + h) h →0
f ( x + 0)
=
=
∀ x ∈ ¡
lim f ( x − h) h→0
f ( x − 0)
∀ x ∈ ¡ Also it is given that f ( x) = f ( x + 0) = It means f is continuous on x ∈ ¡ .
f ( x − 0)
……………………………
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 0
i) Define f ( x ) =
, x is rational , x is irrational
The function f has discontinuity of second kind on every point x because neither f ( x +) nor f ( x −) exists.
x ii) Define f ( x ) = 0
q
, x is rational , x is irrational
Then f is continuous at x = 0 every other point. x + 2 iii) Define f ( x ) = − x − 2 x + 2
and has a discontinuity of the second kind at q
(−3 < x < −2) ( −2 < x < 0 ) (0 < x < 1)
The function has simple discontinuity at x = 0 and it is continuous at every other point point of the interv interval al ( −3,1 3,1) q
sin 1 , x ≠ 0 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 .
References:
(1) Lectures (2003-04) Prof. Syyed Gull Shah Chairman, Chairman, Department of Mathematics. University of Sargodha, Sargodha.
(2) Book Principles of Mathematical Analysis Walter Rudin (McGraw-Hill, Inc.)
Collected and composed by: Atiq ur Rehman (
[email protected]) http://www.mathcity.org in PDF Format. Available online at http://www.mathcity.org Page Setup: Legal ( 8′′ 1 2 × 14′′ ) Printed: October 20, 2004. Updated: November 03, 2005
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