� p
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 nonnegative 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 “pointwise 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 pointwise ; 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 pointwise.
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 pointwise limits of ( a k
)
∞ k =1 is in � p
.
Proof. Fix N
∈
N
, and recall that the ﬁnitedimensional 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 initialsegment 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 righthandside 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 pointwise 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 pointwise 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 nonnegative 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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