� p
Let 1 ≤ p ≤ ∞ , and recall the definition of the metric space �
�
For 1 ≤ p < ∞ , � p
= sequences a = ( a n
)
∞ n =1 p :
∞
� in
R such that | a n
| p
�
< ∞ ; n =1 whereas �
∞ consists of all those sequences a = ( a defined the p -norm as the function � · � p
: � p 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 by d p
�
( a a
�
,
∞ b
= sup n ∈
N
) = � 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
− a � p
< � p
.
n =1
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 ) = a n
. What we have shown is that, if
, . . .
) is a Cauchy sequence of such � p 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 we have | a k
1
− a
1
| < 1
4 and | a k
2
− a increasing sequence of integers k
1
2
| k
2
<
< k
2
→
1 a
2 k
1 so that for k ≥ k
1
, | a k
1
1
4
. Continuing this way iteratively, we can find an
< k
3
< · · · such that
− 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
�
| a n
− a k
N n
| p
�
1 /p
�
+
N
�
| a k
N n
| p
�
1 /p
.
n =1 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
�
≤
∞
�
| a k n
N | p
�
1 /p n =1 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
� ∞
�
≤ | a k n
| p
�
1 /p
�
+ n = N
� 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.

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Fall 2010
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