SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR

Pointwise and uniform convergence of sequences of functions Sanjay Gupta, Assistant Professor, Post Graduate Department of Mathematics, Dev Samaj College For Women, Ferozepur City, Email – [email protected]

INTRODUCTION Here, we shall define and study the convergence of sequences of functions. There are many different ways to define the convergence of a sequence of functions and different definitions lead to inequivalent types of convergence. We consider here two basic types : Pointwise and Uniform Convergence. DEFINITIONS AND EXAMPLES 1. Sequence of real valued functions. Let be a real valued function defined on an interval ⊆ for each ∈ . Then { , , , … , , … } is called a sequence of real valued functions on . It is denoted by { } or 〈 〉. Example 1. If {

( ), ( ),

( ), … } = { ,

( )=

,0 ≤

≤ 1,

, … } is a sequence of real valued functions on [0,1].

,

sanjay gupta

then

be a real valued function defined by

Example 2. If then

{

be a real valued function defined by ( ), ( ),

( ), … } =

,

,

( )=

,0 ≤

≤ 1,

, … is a sequence of real valued functions on [0,1].

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR 2. Sequence of real numbers. Let { } be a sequence of real valued functions defined on an interval , then for ∈ , { ( )} = { ( ), ( ), ( ), … , ( ), … } is a sequence of real numbers. Example 3. Let {

} be a sequence of real valued functions defined by

(i)

=

,

,

,…,

,… =

,

sanjay gupta

is a sequence of real numbers corresponding to the point

(ii)

{

(0) } = {

(0),

(0), (0), … ,

{

(1) } = {

(1),

(1), (1), … ,

,…,

≤ 1, then

,…

∈ [0,1].

=

= 0 ∈ [0,1].

(1), … } = { 1,1,1, … ,1, … }

is a sequence of real numbers corresponding to the point Thus to each

,

,0 ≤

(0), … } = { 0,0,0, … ,0, … }

is a sequence of real numbers corresponding to the point (ii)

( )=

= 1 ∈ [0,1].

∈ , we have a sequence of real numbers.

3. Convergence of sequence of real numbers. A sequence { } of real numbers is said to converge to a real number if for given > 0 ( however small ), there exist a positive integer ( depending upon ) such that |

− |<

− <

< +

− <

,

,

,

∀ ,

≥ ,………… < +

, … … ∈ ( − , + ) i.e. −

≥ .

Geometrically, it means that if we plot the graph of the sequence { } and draw an − band around = , say, = − to = + , then there exists a positive integer such that for all ≥ , the terms lies inside this − band.

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR

} defined by

Example 4. Consider a sequence { lim

Clearly, ⇒ upon

→ 1 as ) such that

= lim

=

→ ∞ i.e. for given

|

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= lim

−1| <

1− <

<1+

=

∀ =1

> 0 ( however small ), there exist a positive integer

≥ ∀

If we take = 0.3, then we have to find a positive integer |

−1| <

1 − 0.3 <

0.7 <

,

Here, Clearly,

= 0.5, ,

,

< 1 + 0.3 ∀

< 1.3 ∀

1− <

such that < 1+

, … … ∈ (0.7,1.3) i. e. 0.3 − = 0.67,

= 0.75,

∉ (0.7,1.3) and

,

= 0.8, ,

= 1. = 0.85

.

, … . ∈ (0.7,1.3)

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( depending

SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR ∴ For given such that

= 0.3 > 0, there exists a positive integer

|

−1| <

≥ 3 i.e.

,

,

= 3 ( as shown in convergent sequence diagram )

, … . ∈ (0.7,1.3) i. e. 0.3 −

} converges to 1.

Hence the sequence {

If we take = 0.1, then then we have to find a positive integer |

−1| <

1 − 0.1 <

⇒ ⇒

0.9 < ,

,

,

1− <

such that

< 1+

< 1.1 ∀ ≥ , … … ∈ (0.9,1.1) i. e. 0.1 −

,

,

= 0.75, = 0.923

,

,

,

,

= 0.8,

= 1. = 0.85,

= 0.86,

= 0.87,

= 0.89,

= 0.9,

. ∉ (0.9,1.1) and

,

,

, … . ∈ (0.9,1.1)

sanjay gupta

,

< 1 + 0.1 ∀

Here, = 0.5, = 0.67, = 0.91, = 0.917, Clearly,

= 1.

∴ For given such that

= 0.1 > 0, there exists a positive integer |

−1| <

Hence the sequence {

≥ 10 i.e.

,

,

= 10 ( as shown in convergent sequence diagram ) , … . ∈ (0.9,1.1) i. e. 0.1 −

= 1.

} converges to 1.

This shows that if we change the value of , then the value of

also changes i.e.

depends upon .

4. Pointwise convergence. Pointwise convergence defines the convergence of functions in terms of the convergence of their values at each point of their domain. If { } is a sequence of real valued functions defined on an interval and for each ∈ , the corresponding sequence { ( ) } of real numbers is convergent, then we say that the sequence { } converges pointwise on .The limiting values of the sequences of real numbers corresponding to ∈ , define a function called the limit function or simply the limit of the sequence { } of functions on .

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR Example 5. Let { (i)

=

For

} be a sequence of real valued functions defined by

( )=

,0 ≤

≤ 1, then

; the corresponding sequence of real numbers (0), … } = { 0,0,0, … ,0, … }

(0), (0), (0), … ,

(0) } = {

{

is a constant sequence and hence convergent to 0

(ii)

<

For {

=0 ∀

= {

( ), … } = { ,

( ), ( ), … ,

( ),

( ) = lim

For

< 1 ; the corresponding sequence of real numbers

( )} = {

lim (iii)

(0) = 0

lim

i.e.

∈ (0,1).

,

, … } converges to 0 as

,…,

[ ∵ A sequence {

} converges to 0 if 0 <

<1]

; the corresponding sequence of real numbers (1), … } = { 1,1,1, … ,1, … }

(1), (1), (1), … ,

(1) } = {

is a constant sequence and hence convergent to 1.

sanjay gupta

(1) = 1

lim

i.e.

From (i), (ii) and (iii), we have ( ) = lim

( )=

⇒ For each

0, 1,

0≤

<1 =1

∈ [0,1], the corresponding sequence of real numbers is convergent.

Thus the sequence {

} converges pointwise to ( ) on [0,1] and ( ) is a limit function of the sequence {

}

5. Pointwise convergence of a sequence of functions. Let { } be a sequence of real valued functions defined on an interval . If to each ∈ and to each > 0, there exists a positive integer ( depending upon and each ∈ ) such that |

( ) − ( )| <

then we say that the sequence {

 The sequence {

,

} converges pointwise to the limit function

} converges pointwise to the limit function lim →

 The positive integer

depends on

( )=

∈ and given

( )∀

on

on .

if and only if

> 0 i.e.

=

( , ).

6.Geometrical Interpretation of pointwise convergence. For pointwise convergence, we first fix a value ∈ [ , ].Then we choose an arbitrary − neighbourhood around ( ), which corresponds to a vertical interval centered at ( ). Finally, we pick so that ( ) for all ≥ intersects the vertical line = inside the interval ( ( ) − , ( ) + ).

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR

Remark. For a sequence {

} of functions, an important question is :

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If each function of a sequence { } has a certain property such as boundedness, continuity, differentiability or integrability, then to what extent is this property transferred to the limit function ?. Infact, pointwise convergence is not strong enough to transfer any of the properties mentioned above from the terms of the sequence { } to the limit function . Let us consider the following few examples : Example 6. A sequence of bounded functions with an unbounded limit function. Consider a sequence { ( ) = lim

Then

⇒ The sequence { Now, 0 <

( )=

( ) = lim

, 0<

= lim

<1

=

} converges pointwise to the limit function ( ) on (0,1).

<1 ⇒0< |

∴ So each

} defined by

<

⇒1<

( )| =

=

+1 < <

+1 ⇒

for all

<

<1⇒

∈ (0,1)

is bounded on (0,1), but the limit function ( ) =

is not bounded on (0,1).

Thus, pointwise convergence does not, in general, preserve boundedness. Example 7. A sequence of continuous functions with a discontinuous limit function. Consider a sequence {

( ) = lim

Then

⇒ The sequence { Clearly, each But ⇒

} defined by

( )=

( ) = lim →

=

, 0≤ 0, 1,

0≤

≤1 <1 =1

} converges pointwise to the limit function ( ) on [0,1].

, being a polynomial , is continuous on [0,1]. (1) = 1 and lim ( ) = 0 lim ( ) ≠ (1)

⇒ The limit function

is discontinuous at

<

= 1.

Thus, pointwise convergence does not, in general, preserve continuity.

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<

SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR Example 8. A sequence of differentiable functions in which the limit of the derivatives is not equal to the derivative of the limit function. Consider a sequence {

( ) = lim

( ) = lim

Then

( )=

} defined by

=0 ∀

,

[ ∵ lim →

The sequence {

= 0 and | sin

| ≤ 1∀

]

} converges pointwise to the limit function ( ) on .

Now,

( )=0 ∀

But

( )=

= √ cos

( )≠

lim →

( ) ∀

( ) = lim √ cos

lim

=∞

∈ } does not converge to the derivative of the limit function

i.e. a sequence of differentiable functions { on .

( )

Thus, pointwise convergence does not, in general, preserve differentiability.

sanjay gupta

Example 9. A sequence of integrable functions in which the limit of integrals is not equal to the integral of the limit function. Consider a sequence { When

< 1,

lim →

( ) = lim

) ,

∈ [0,1]

(

) = lim

(

)

(

)

[

)

= lim

(

)

=

< 1 and

form, using L’Hospital rule ]

(

→0

)

lim (1 − →

→∞

) =0 0<

< 1]

( )

∴ Also,

(1 −

<1 ⇒ 0< < 1 ⇒ 0 > − > −1 ⇒ 0 < 1 − ( ) = lim ( ) = 0 ∀ ∈ [0,1]

⇒ The sequence { Now, ∫

(1 −

= lim [∵ 0< ∴

( )=

( )=0

= 0 or 1,

When 0 <

} defined by

} converges pointwise to the limit function ( ) on [0,1]. =∫

(1 −

= −

[

lim ∫ →

( )

( ) =∫ 0

)

∫ (1 −

= −

1 0− ]= − 0 = lim ( ) = lim →

(

) (−2 )

= )

(

)

=

=0 lim ∫ →

( )

i.e. a sequence of integrable functions { ∫ on [0,1]. ∫ ( )

( )

≠∫

( )

} does not converge to the integral of the limit function

Thus, pointwise convergence does not, in general, preserve integrability. So, the above few examples show that we need to investigate under what supplementary conditions these or other properties of the terms of the sequence { } are transferred to the limit function . A concept of great importance in this respect is that known as uniform convergence.

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR 7. Uniform Convergence. In this section, we introduce a stronger notion of convergence of functions than pointwise convergence, called uniform convergence. The difference between pointwise convergence and uniform convergence is analogous to the difference between continuity and uniform continuity. We know that a sequence { } of real valued functions defined on an interval converges pointwise to a function if to each ∈ and to each > 0, there exists a positive integer ( depending upon and each ∈ ) such that |

( ) − ( )| <

The positive integer depends on ∈ and given positive integer which works for each ∈ . Example 10. Consider a sequence { ( ) = lim

Then

⇒ The sequence { Let

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0, 1,

=

0≤

∈ [0,1], there exists a positive integer

= 0, then

|

Thus (1) is true for = 1, then

≤1

<1 =1

|

∈ ) such that ............. (1)

∈ ∀

≥1

( )=1 ∀

( ) − ( )| = | 1 − 1 | = 0 <

≥1

<

= 1.

( ) = 0 and |

∴ Thus (1) is true for

( )=0 ∀

and each

( ) − ( )| = 0 <

( ) = 1 and

Thus (1) is true for

( depending upon

= 1.

= , then

( ) − ( )| <

( ) = 0 and

If

( ) = lim

, 0≤

} converges pointwise to the limit function ( ) on [0,1].

|

If

( )=

} defined by

( , ). It is not always possible to find a

=

= > 0 be given.

Then for each

If

> 0 i.e.

( )=

( ) − ( )| =

∈ −0 =

≥3

= 3. = 7 when

Similarly, (1) is true for

Hence there is no single value of

=

.

for which (1) holds for all

If we can find ∈ which depends only on uniformly convergent to on .

∈ [0,1] i.e.

and not on

depends on both

and .

∈ , then we say that the sequence {

} is

8. Uniform convergence of a sequence of functions. A sequence { } of real valued functions defined on an interval is said to be uniformly convergent to a function on if to each > 0, there exists a positive integer ( depending only on ) such that | The function

( ) − ( )| <

is called uniform limit of the sequence {

and ∀

∈ .

} on .

The crucial point in this definition is that depends only on and not on ∈ , whereas for a pointwise convergent sequence may depend on both and . A uniformly convergent sequence is always pointwise

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR convergent ( to the same limit ), but the converse is not true. If a sequence converges pointwise, it may happen that for some > 0 one needs to choose arbitrarily large ′ for different points ∈ , meaning that the sequences of values converge arbitrarily slowly on . In that case a pointwise convergent sequence of functions is not uniformly convergent. Example 11. Consider a sequence { ( ) = lim

Then

( )=

} defined by

( ) = lim →

=

0, 1,

0≤

, 0≤

≤1

<1 =1

⇒ The sequence { } converges pointwise to the limit function ( ) on [0,1] but { uniformly on [0,1]. Example 12. Consider a sequence { ( ) = lim

Then Let

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( ) = lim →

, 0≤

≤1

∈ [0,1]

= 0 for all

> 0 be given. |

Then ∵

( )=

} defined by

} does not converge

0≤

( ) − ( )| = ≤1 ⇒ 0≤

−0 = ≤ 1∀

is

.

Maximum value of

=

If we choose a positive integer |

( ) − ( )| <

Hence the sequence {

<

>

⇒ 0 ≤

+ 1, then ∀

and ∀

∈ [0,1].

} converges uniformly to the limit function ( ) on [0,1].

9. Geometrical Interpretation of uniform convergence. A sequence { } of real valued functions defined on an interval is said to be uniformly convergent to a function on if to each > 0, there exists a positive integer ( depending only on ) such that | i.e.

( ) − ( )| <

( )− <

( )< ( )+

≥ ∀

and ∀ ≥

∈ .

and ∀

∈ .

This shows that for uniform convergence, we draw an − neighbourhood around the limit function ( ), which results in an " −strip" i.e. ( ( ) − , ( ) + ) with ( ) in the middle. Now, we pick so that the graphs of ( ) for all ≥ lie completely inside that strip for all ∈ .

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR In Example 12., if we take = , then we have to find a positive integer |

( ) − ( )| <

|

( )−0|<

and ∀

∈ [0,1]

− <

( )<

and ∀

∈ [0,1]

( ),

( ),

Here, we have

( )=

( )= ,

0<

<1 ⇒ 0<

Also,

0<

<1 ⇒ 0<

0<

∴ For given

<1 ⇒ 0<

,

<

,

( )=

etc.

( )∉ − ,

⇒ 0< ⇒ 0<

∈ [0,1]

( )=

( )<1 ⇒ <

only) such that

∈ [0,1]

and ∀

( ), … … … ∈ − ,

Now,

and

(depending upon

( )< ⇒ ( )< ⇒

( )∈ − , ( )∈ − ,

= = 0.5 > 0, there exists a positive integer

and so on.

= 2 (as shown in convergent sequence diagram)

sanjay gupta

(depending upon only) such that |

( )−0|<

≥ 2 i.e.

( ),

( ) ,….∈ − ,

i. e. 0.5 −

( ) = 0.

Hence the sequence { } converges uniformly to 0.

10. Test for uniform convergence of sequence of functions. In order to test whether a given sequence { } is uniformly convergent or not in a given interval, so far we have been using the definition of uniform convergence. Accordingly, we tried to get a positive integer , independent of x, which is not easy in practice. This method can be replaced by an easy method given in the following theorem.

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR ) Let { } be a sequence of real valued functions defined on an interval [ , ] such ∈ [ , ] and let = sup{ | ( ) − ( )| ∶ ∈ [ , ] }. Then { } converges

Theorem 1. ( − that lim ( ) = ( ) ∀ →

uniformly on [ , ] if and only if

→ 0 as

Proof. Let { } converge uniformly to ( depending only on ) such that | sup{ |

⇔ ⇔

|

∈[ , ]

and ∀

∈[ , ] }<

> 0, there exists a positive integer

−0| <

} converges to 0 as

→ 0 as

i.e.

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( ) − ( )| ∶

The sequence {

on [ , ]. Then, for given

( ) − ( )| <

<

→ ∞.

→ ∞

→ ∞.

} does not converge to 0 . .  If the sequence { ↛ 0 as → ∞, then the sequence { } is not uniformly convergent.  = max { | ( ) − ( )| ∶ ∈ [ , ] } = Maximum value of { | ( ) − ( )| } for fixed and for all ∈ [ , ].  A function ( ) is maximum at = if ( ) = 0 and ( ) < 0.  Example 13. Consider a sequence { ( ) = lim

Then

} defined by

( ) = lim

( )=

[ ∵ lim →

|

( ) − ( )| =

−0 =

where

=

and

= − √ ( sin

=

so that

At ⇒

=

,

is maximum at

=

=

= sup{ |

i.e.

= −

= 0 and |sin

| ≤ 1∀

= , (say)

, the sequence {

∈ [0, ] ]

[ ∵

.

, we have =

=

= − n sin

=− =

<0 . √

=

→∞

( ) − ( )|: ∈ [0, ] } = max

→0

∈ [0, ] ]

= √ cos

and maximum value of

→ 0 as

Moreover

By

=0 ⇒

= − n sin

) = −n sin

For maximum and minimum value of = 0 ⇒ cos

∈ [0, ]

∈ [0, ]

=0 ∀

,

=

∈ [0, ]

→∞ } is uniformly convergent on [0, ].

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=

SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR 11. Properties of uniform convergence In this section we prove that, unlike pointwise convergence, uniform convergence preserves boundedness,continuity and integrability. Uniform convergence does not preserve differentiability any better than pointwise convergence. Nevertheless, we give a result that allows us to differentiate a convergent sequence , the key assumption is that the derivatives converge uniformly. 12. Boundedness. First, we consider the uniform convergence of bounded functions. Theorem 1. If a sequence of bounded functions { bounded on [ , ]. But the converse is not true. Proof. Since { } is uniformly convergent to integer ( depending only on ) such that |

( ) − ( )| <

Choose some positive integer > 0 such that |

( )| <

Now, | ( )| = | ( ) − =|

( )+

( ) − ( )| + |

} is uniformly convergent to a function

on [ , ], therefore for given ∀

is

> 0, there exists a positive

∈[ , ]

......... (1)

is bounded on [ , ], therefore there exists a real number

. Then, since ∀

and ∀

on [ , ], then

∈[ , ]

.........(2) ( )| + |

( )| ≤ | ( ) − ( )| < +

( )|

∈[ , ]

sanjay gupta

[ using (1) & (2)] is bounded on [ , ].

In particular, it follows that if a sequence of bounded functions converges pointwise to an unbounded function, then the convergence is not uniform. Thus it provides a good negative test for uniform convergence. 13. Continuity. One of the most important properties of uniform convergence is that it preserves continuity. Theorem 2. If a sequence of continuous functions { is continuous on [ , ]. But the converse is not true. Proof. Since { } is uniformly convergent to integer ( depending only on ) such that | Let

( ) − ( )| <

} is uniformly convergent to a function

on [ , ], therefore for given

and ∀

> 0, there exists a positive

∈[ , ]

......... (1)

be any point of [ , ]. Then in particular, from (1), we have |

( ) − ( )| <

.......... (2)

Since is continuous on [ , ], therefore is continuous at ∈ [ , ] ∀ ∴ for given > 0, there exists > 0 (depending upon ) such that |

≤| ( )− =| < is continuous at

whenever | − | <

( ) − ( )| <

Now, | ( ) − ( ) | = | ( ) −

on [ , ], then

( )+

( )−

.............(3)

( )+

( )− ( )|

( )|+|

( )−

( )|+|

( )− ( )|

( )− ( )|+|

( )−

( )|+|

( )− ( )|

+ + =

whenever | − | <

= .

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[ using (1), (2) & (3)]

SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR ∈ [ , ] is arbitrary, therefore

Since

is continuous on [ , ].

In particular, it follows that if a sequence of continuous functions converges pointwise to a discontinuous function, then the convergence is not uniform. Thus it provides a good negative test for uniform convergence. 14. Differentiability. The uniform convergence of differentiable functions does not, in general, imply anything about the convergence of their derivatives or the differentiability of their limit. This is because the values of two functions may be close together while the values of their derivatives are far apart ( for example, if one function varies slowly while the other oscillates rapidly). Thus, we have to impose strong conditions on a sequence of → . functions and their derivatives if we hope to prove that → imples The following example shows that the limit of the derivatives need not equal the derivative of the limit even if a sequence of differentiable functions converges uniformly and their derivatives converge pointwise. Example 14. Consider a sequence {

( ) = lim

( ) = lim

Then |

( ) − ( )| = =

where

sanjay gupta

=0 ⇒1− (

At

=

,

=−

is maximum at

=

=

= sup{ |

i.e.

= −

Clearly, each It follows that

=(

)

√ (

)

=

)

(

)

<0

→ 0 as

and maximum value of

=

=

.

→∞

( ) − ( )| ∶

→0

, the sequence {

} = max{ | | ∶

} =

→∞ } is uniformly convergent on

is differentiable on →

|, (say)

)

) (

)

)

)

(

Moreover

By

(

=

(

, we have

=0 ⇒

=

(

=

so that

,

∈ =|

−0 =

For maximum and minimum value of

and

=0 ∀

( )=

} defined by

pointwise as

and

( )=(

→ ∞ where ( ) =

. )

0, 1,

≠0 =0

The convergence is not uniform , since is discontinuous at 0. Thus, limit of the derivatives is not the derivative of the limit.

→ 0 uniformly,but

(0) → 1, so the

However, we do get a useful result if we strengthen the assumptions and require that the derivatives converge uniformly, not just pointwise. The proof involves a slightly tricky application of the Mean value Theorem.

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SANJAY GUPTA, DEV SAMAJ COLLEGE FOR WOMEN, FEROZEPUR Theorem 3. Let { } be a sequence of differentiable functions with pointwise limit . If the sequence { uniformly convergent with as uniform limit on [ , ], then is differentiable and derivative is equal to i.e.

= lim →

( )

} is

= lim →

i.e the derivative of the limit of the sequence is equal to the limit of the sequence of derivatives. In particular, it follows that if the derivative of the limit of the sequence is not equal to the limit of the sequence of the derivative, then the sequence { } cannot be uniformly convergent. Thus it provides a good negative test for the uniform convergence the sequence { }. are not assumed to be continuous. If they are It is worth noting that in theorem 3. the derivatives continuous, then one can use Riemann Integration and the fundamental theorem of calculus to give a simpler proof. 15. Integrability. Another one of the most important properties of uniform convergence is that it preserves integrability. Theorem 4. If { } is a uniformly convergent sequence with uniform limit [ , ] ∀ ∈ , then is integrable on [ , ]. But the converse is not true. lim ∫

Moreover

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In particular, it follows that if {

( )

=∫

on [ , ] and

( )

} is a sequence of integrable functions converging to →

then { } cannot converge uniformly to convergence of the sequence { }.

( )

≠∫

is integrable on

on [ , ] and

( )

on [ , ]. Thus it provides a good negative test for the uniform

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## POINTWISE AND UNIFORM CONVERGENCE OF SEQUENCES OF ...

Sanjay Gupta, Assistant Professor, Post Graduate Department of .... POINTWISE AND UNIFORM CONVERGENCE OF SEQUENCES OF FUNCTIONS.pdf.

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