� IS

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� p

IS COMPLETE

Let 1

≤ p

≤ ∞

, and recall the definition of the metric space �

For 1

≤ p <

, � p

=

� p

: sequences a = ( a n

)

∞ n =1 in

R such that

| a n

| p

<

; n =1 whereas

∞ defined the p consists of all those sequences

-norm as the function � · � p

: � p a = ( a n

)

∞ n =1 such that sup n

N

[0 ,

)

, given by

| a n

|

<

∞ . We

� a

� p

=

� ∞

| a n

| p

1 /p

, for 1

≤ p <

, n =1 and � a

= sup n

N by d p

( a , b ) =

� a

| a b n

| . In class, we showed that the function d p

: � p

� p

×

� p

[0 ,

∞ is actually a metric. We now proceed to show that

( � p

)

, d p given

) is a complete metric space for

1

≤ p

≤ ∞ . For convenience, we will work with the case p <

∞ , as the case p =

∞ requires slightly different language (although the same ideas apply).

Suppose that a

1

, a

2

, a

3

, . . .

is a Cauchy sequence in � p

. Note, each term a k quence is a point in � p

, and so is itself a sequence: in the se­ a k

= ( a k

1

, a

2 k

, a

3 k

, . . .

) .

Now, to say that

( a k

)

∞ k =1 is a Cauchy sequence in

� p is precisely to say that

� > 0

K

N s.t.

∀ k, m

K,

� a k

− a m

� p

< �.

That is, for given � > 0 and sufficiently large k, m , we have

| a k n

− a m p n

|

=

� a k n =1

− a

� p

< � p

.

Now, the above series has all non-negative terms, and hence is an upper bound for any fixed term in the series. That is to say, for fixed n

0

N

,

| a k n

0

− a m n

0

| ≤

| a k n

− a m n

| p

< � p

, n =1 and so we see that the sequence

( a k n

0

)

∞ k =1 is a Cauchy sequence in is a complete metric space, and thus there is a limit a n

0

R

R

. But we know that

R to this sequence. This holds for each n

0

N

. The following diagram illustrates what’s going on. a

1 a

2 a

3 a

4

=

=

=

= a a a a

.

.

3

1

4

1

1

1

2

1 a

1 a a a a

3

2

4

2

1

2

2

2 a

.

.

2 a a a a

3

3

4

3

1

3

2

3 a

.

.

3 a a a a

3

4

4

4

1

4

2

4 a

.

.

4

·

·

· ·

·

·

· · ·

. ..

· · ·

·

·

·

So, we have shown that, in this � p

-Cauchy sequence of horizontal sequences, each vertical sequence actually converges. Hence, there is a sequence a = ( a

1

, a

2

, a

3

, a

4

, . . .

) to which

“ a k converges” in a vague sense. The sense is the “point-wise convergence” along vertical

1

2 lines in the above diagram. To be more precise, recall that a sequence a is a function a :

N

( a

1

, a

2

, a

3

R

, where we customarily write a ( n ) =

, . . .

) is a Cauchy sequence of such � p a n

. What we have shown is that, if functions, then there is a function a :

N

R such that a k converges to a point-wise ; i.e. a k

( n )

→ a ( n ) for each n

N

.

Now, our goal is to find a point b

� p that is, such that � a k

− b

� p

0 as k

→ ∞ such that a k

→ b as k

→ ∞

. The putative choice for this b in the sense of � is the sequence p a

; given above. In order to show that one works, we need to show first that it is actually an

� p sequence, and second that a k converges to a in � p sense, not just point-wise.

To do this, it is convenient to first pass to a family of subsequences of the ( a k n

) , as fol­ lows. Since

( a k

1

)

∞ k =1 converges to a

1

, we can choose

Having done so, and knowing that a k

2 we have

| a k

1

− a

1

|

<

1

4 and

| a k

2

− a

2

|

<

1 a

2 k

1 so that for ≥ k

1

, | a k

1

1

|

<

4

. Continuing this way iteratively, we can find an increasing sequence of integers k

1

< k

2

< k

3

<

· · · such that k

− a

1

1

2

.

, we can choose a larger k

2 so that for k

≥ k

2

, for each j

N

,

| a k n

− a n

|

< 2

− j for n = 1 , 2 , . . . , j and k

≥ k j

.

In particular, we have

| a k j n

− a n

|

< 2

− j for j

≥ n . That gives us the following.

(1)

Lemma 1.

The sequence a = ( a n

)

∞ n =1 of point-wise limits of ( a k

)

∞ k =1 is in � p

.

Proof. Fix N

N

, and recall that the finite-dimensional versions of the � p

-norms,

( a

1

, . . . , a

N

)

� p

=

� N

| a n

| p

1 /p n =1 also satisfy the triangle inequality (i.e. d p

( x , y ) =

� x

− y

� p can estimate the initial-segment of

N terms of a as follows: is a metric on

R

N

). Hence, we a n

= ( a n

− a k

N n

) + a k

N n

, and so

N

| a n

| p

1 /p

≤ n =1

N

| a n

− a k

N n

| p

1 /p

+

N

| a k

N n

| p

1 /p

.

n =1 n =1

(2)

Now, the last term in Equation 2 is bounded by the actual � p

-norm of the whole sequence a k

N

; that is, we can tack on the infinitely many more terms,

N

| a k n

N

| p

1 /p

≤ n =1

� ∞

| a k n

N

| p

1 /p n =1

=

� a k

N

� p

.

Recall that ( a k

)

∞ k =1 is a Cauchy sequence in the metric space � p

. We have proved that any

Cauchy sequence in a metric space is bounded . Thus, there is a constant R independent of

N such that

� a k

N

� p

R . Combining this with Equation 1, we can therefore estimate the right-hand-side of Equation 2 by

N

| a n

| p

1 /p

N

(2

N

) p

1 /p

+ R =

N 2

N p �

1 /p

+ R.

n =1 n =1

3

Finally, the term

N 2

N p

1 /p

= N

1 /p

2

N converges to 0 as N

→ ∞

(remember your calculus!), and hence this sequence is also bounded by some constant

S

. In total, then, we have

N

| a n

| p

1 /p

R + S for all N

N

.

n =1

In other words,

N n =1

| a n

| ≤

( R + S ) p

. The constant on the right does not depend on it is an upper bound for the increasing sequence of partial sums of the series

� a

� p p

. Thus, we have

� a

� p

R + S , and so a

� p

.

∞ n =1

| a n

| p

N

=

;

So, we have shown that the putative limit a (the point-wise limit of the sequence ( a k

)

∞ k =1 of points in � p

) is actually an element of the metric space � p

. But we have yet to show that it is the limit of the sequence

( a k

) in

� p

. That somewhat involved proof now follows.

Proposition

in

� p

2.

Let ( a k

)

∞ k =1 be a Cauchy sequence in

, by Lemma 1). Then � a k

− a

� p

0 as k

→ ∞ .

� p

, and let a be its point-wise limit (which is

Proof. Let

� > 0

. Lemma 1 shows that by the Cauchy criterion, there is an

N

1 a

� p

, which means that

∞ n =1

N so that

| a n

| p

<

∞ . Hence,

| a n

| p

< � p

.

n = N

1

In addition, we know that

� a k

− a m

� p

( a k

)

∞ k =1 is

< � . Letting N = max

{

N

1 p

-Cauchy, so there is

, N

2

}

N

2 so that, whenever

, we therefore have k, m

N

2

,

| a n

| p

< � p and

� a

N

− a k

� p

< �

∀ k

N.

n = N

(3)

Now, the sequence a

N large enough so that is in

� p

, and so we can apply the Cauchy criterion again: select

N

| a

N n

| p

< � p

.

n = N

(4)

Note, we can always increase

N

N

N

.

� and still maintain this estimate, so we are free to chose

We now use the constant

N

� we defined above in the bounds we will need later. Since a k n

→ a n for each fixed we can choose

K

2 n

, we can choose so that | a k

2 we can take K = max

{

K

1

, K

2

− a

2

|

< �

, . . . , K

N

� p

K

/N

1

� so that | a k

1

− a

1

} and then we have

|

< � p

/N

� for k

K

1

. Likewise, for k

K

2

. Continuing this way for

N

� steps,

| a k n

− a n

|

<

� p

N

, for k

K and n

N

.

(5)

For good measure, we will also (increasing it if necessary) make sure that

K

N

. Now, for any k

K , break up b = a k

− a as follows:

( b n

)

∞ n =1

= ( b n

)

N

− 1 n =1

+ ( b n

)

∞ n = N

.

4

(To be a little more pedantic, we are expressing b n

= x n

+ y n where x n

= b n when n < N and = 0 when n

N

, and y n

= 0 when n < N

� and = b n when n

N

.) The triangle inequality for the p -norm then gives

� a k

− a

� p

� N

− 1

| a k n

− a n

| p

1 /p

+

� ∞

| a k n

− a n

| p

1 /p

.

(6) n =1 n = N

Equation 5 shows that, for k

K

, the first term here is

N

1

| a k n

− a n

| p

1 /p

N

1

� p

N

1 /p

=

N

1

1 /p

� < �.

N

� n =1 n =1

For the second term in Equation 6, we use the triangle inequality for the � to the range n

N

� to get p

-norm restricted

� ∞

| a k n

− a n

| p

1 /p

≤ n = N

� ∞

| a k n

| p

1 /p

+

� n = N

� n = N

| a n

| p

1 /p

.

Since

N

N

, Equation 3 shows that the second term here is

< �

. So, summing up the last two estimates, we have

� a k

− a

� p

2 � +

� ∞

| a k n

| p

1 /p

, n = N

(7) whenever k

K . So we need only show this final term is small. Here we make one more decomposition: a k n

= a k n

− a

N n

+ a

N n

, and so once again applying the triangle inequality,

� ∞

� n = N

| a k p n

|

1 /p

� ∞

| a k n

− a

N p n

|

1 /p

+

� n = N

| a n

| n = N

1 /p

.

The first of these terms is a sum of non-negative terms over n

N

, and so it is bounded above by the sum over n

1 which is equal to

� a k

(since k

K

N

N

). And the second term is also a

N

< � p

, which is < � by Equation 3

, by Equation 4. Whence, the last term in Equation 7 is also < 2 � , and so we have shown that

� > 0 ,

K

N such that ∀ k

K

� a k

− a

� p

< 4 �.

Of course, we should have been more clever and chosen all our constants in terms of

�/ 4 to get a clean

� in the end, but such tidying is not really necessary;

4 � is also arbitrarily small, and so we have shown that ( a k

)

∞ k =1 does converge to a in � p

. This concludes the proof that � p is complete. Whew! �

Let us conclude by remarking that a very similar (though somewhat simpler) proof works for p =

∞ ; the details are left to the reader.

MIT OpenCourseWare http://ocw.mit.edu

18.100B Analysis I

Fall 2010

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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