MA3421 (Functional Analysis 1) Tutorial sheet 6
[November 13, 2014]
Name: Solutions
Let xn denote the sequence with j th term
(
xn,j =
1. Show that
P∞
n=1
1
n
0
if j = n
for j 6= n
xn is not absolutely convergent in `p for 1 ≤ p ≤ ∞.
Solution: If we write out xn in longhand, we find
1
xn = (0, 0, . . . , 0, , 0, . . .)
n
(or we could say that the sequence has all zero terms apart fron the nth term, which is 1/n).
So if we compute for 1 ≤ p < ∞, we find
kxn kp =
∞
X
j=1
!1/p
p
|xn,j |
1/p
p
1
p
p
p
p
+ 0 + ···
= 0 + 0 + ··· + 0 +
n
and so we get kxn kp = (1/np )1/p = 1/n. Even for p = ∞, where kxn k∞ = sup1≤j<∞ |xn,j |
we still get kxn k∞ = 1/n.
So
∞
X
n=1
kxn kp =
∞
X
1
n
n=1
and this is infinite (harmonic series does not converge).
P
p
So the series ∞
n=1 xn is not absolutrely convergent in ` (no matter what p we choose).
P
p
2. Show that ∞
n=1 xn is convergent in ` for 1 < p ≤ ∞ and also is convergent in c0 .
Solution: The partial sums Sn = x1 + x2 + · · · + xn work out as
x1 = (1, 0, 0, . . .)
1
x2 = (0, , 0, 0, . . .)
2
..
.
1
xn = (0, 0, . . . , 0, , 0, . . .)
n
1 1
1
Sn = (1, , , . . . , , 0, 0, . . .)
2 3
n
It seems reasonable to guess that the limit of these Sn must be the sequence
∞
1 1
1
S = (1, , , . . .) =
2 3
j j=1
and we can see if that is so by looking at limn→∞ kSn − Skp . We haver
S − Sn = (0, 0, . . . , 0, −
1
1
,−
, . . .),
n+1 n+2
aand for 1 < p < ∞
kSn − Skp =
=
p p
1/p
1 1 0 + 0 + · · · 0 + −
+ −
+ ···
n + 1
n + 2
!1/p
∞
X
1
.
jp
j=n+1
p
p
p
For p = ∞
kSn − Sk∞ =
X 1
1
=
j
n+1
j≥n+1
Now for p = ∞ we clearly have limn→∞P
kSn − Sk∞ = 0 while for 1 < p < ∞, the series
P
∞ 1
∞
1
j=1 j p converges, so that the tail sums
j=n+1 j p → 0 as n → ∞.
We have omitted one check
that S is in the right space. For 1 < p < ∞ we have
P here,
1
<
∞, while for the case p = ∞, we can see that S ∈ `∞
S ∈ `p because kSkpp = ∞
j=1 j p
since the absolute values of the terms are at most 1. In fact S ∈ c0 since limj→∞ 1j = 0.
The xn are clearly alsoPin c0 and so the convergence result limn→∞ kSn − Sk∞ = 0 shows
that Sn → S in c0 , or ∞
n=1 xn = S in c0 .
P
1
3. Show that ∞
n=1 xn is not convergent in ` .
Solution: Here the issue is more or less that S ∈
/ `1 , but we can’t then look at kSn − Sk1
because that is not defined (or finite).
What we can do is look at
kSn k1 =
n
X
1
j=1
j
and notice that limn→∞ kSn k1 = ∞. That means thre can be no limit in `1 for the sequence
S1 , S2 , . . . as convergent sequences are always bounded. (If the limit was s ∈ `1 say, then
for n big enough we would have kSn − sk1 < 1 and so kSn k1 = k(Sn − s) + sk1 ≤
kSn − sk1 + ksk1 < 1 + ksk1 . But that is not compatible with limn→∞ kSn k1 = ∞.)
P
n
1
Show that ∞
n=1 (−1) xn is also not convergent in ` .
Solution: If we take the partial sums in this case we get
1 1
1
Sn0 = −x1 + x2 − x3 + · · · + (−1)n xn = (−1, , − , . . . , (−1)n , 0, 0, . . .)
2 3
n
2
and the norm in `1 of Sn0 is the same as the norm of Sn , that is
kSn0 k1
n n
X
X
1
1
j
=
→∞
(−1) j =
j
j=1
j=1
(as n → ∞). So
there is no limit in `1 for limn→∞ Sn0 either (which is the same as saying
P∞
that the series n=1 (−1)n xn is not convergent in `1 ).
P
4. Show that if π : N → N is a bijection (orPa permutation of N) then ∞
n=1 xπ(n) converges
∞
p
in ` for 1 < p ≤ ∞ to the same sum as n=1 xn .
Solution: The point here is that the partial sums Sn00 = xπ(1) + xπ(2) + · · · + xπ(n) will still
be sequences with either 1j or 0 in the j th slot, but the slots get filled in in a diferent order
than from left to right.
However, if we take N big enough then π(1), π(2), . . . , π(N ) will include all of 1, 2, . . . , n.
To be precise, if we take
N ≥ max(π −1 (1), π −1 (2), . . . , π −1 (n))
00
will have all the slots 1, 2, . . . , n filled in (and some more perhaps). So if we
then SN
00
− S we will have all zero terms up to the nth term and probably some more
calculate SN
zeros.
00
− Skp ≤ kSn − Skp if N ≥ max1≤j≤n π −1 (j). Thus,
we can
So kSN
P∞if ε > 0 pis given,
p
00
guarantee that kSN −Skp < ε by choosing n big enough that j=n+1 1/j < ε (if p < ∞
and 1/(n + 1) < ε if p = ∞) and then N as above.
00
So limN →∞ SN
= S also (in `p ).
Richard M. Timoney
3