Recall from calculus that if A is a nonempty set, sup_{a∈A}f
(a)
denotes the least upper bound of f
(A )
or if
this set is not bounded above, it equals ∞. Also inf _{a∈A}f
(a)
denotes the greatest lower bound of f
(A)
if
this set is bounded below and it equals −∞ if f
(A)
is not bounded below. Thus to say sup_{a∈A}f
(a)
= ∞ is
just a way to say that A is not bounded above. The existence of these quantities is what we mean when we
say that ℝ is complete.
Definition 1.13.1Let f
(a,b)
∈
[− ∞, ∞ ]
for a ∈ A and b ∈ B where A,B are sets which meansthat f
(a,b)
is either a number, ∞, or −∞. The symbol, +∞ is interpreted as a point out at the endof the number line which is larger than every real number. Of course there is no such number. Thatis why it is called ∞. The symbol, −∞ is interpreted similarly. Then sup_{a∈A}f
(a,b)
means sup
(S )
b
where S_{b}≡
{f (a,b) : a ∈ A}
.
Unlike limits, you can take the sup in different orders.
Lemma 1.13.2Let f
(a,b)
∈
[− ∞,∞ ]
for a ∈ A and b ∈ B where A,B are sets. Then
sau∈pA sub∈pB f (a,b) = sbu∈pBsau∈pA f (a,b).
Proof:Note that for all a,b,
f (a,b) ≤ supsupf (a,b)
b∈Ba∈A
and therefore, for all a, sup_{b∈B}f
(a,b)
≤ sup_{b∈B} sup_{a∈A}f
(a,b)
. Therefore,
sup sup f (a,b) ≤ supsupf (a,b).
a∈A b∈B b∈Ba∈A
Repeat the same argument interchanging a and b, to get the conclusion of the lemma. ■
Theorem 1.13.3Let a_{ij}≥ 0. Then
∑∞ ∑∞ ∑∞ ∑∞
aij = aij.
i=1j=1 j=1i=1
Proof: First note there is no trouble in defining these sums because the a_{ij} are all nonnegative. If a
sum diverges, it only diverges to ∞ and so ∞ is the value of the sum. Next note that
∞ ∞ ∞ n
∑ ∑ aij ≥ sup∑ ∑ aij
j=r i=r n j=ri=r
because for all j,
∑∞ ∑n
aij ≥ aij.
i=r i=r
Therefore,
∞ ∞ ∞ n m n
∑ ∑ a ≥ sup∑ ∑ a = sup lim ∑ ∑ a
j=ri=r ij n j=r i=r ij n m→ ∞j=r i=r ij
Interchanging the i and j in the above argument proves the theorem. ■
Corollary 1.13.4If a_{ij}≥ 0, then
∑∞ ∑∞ ∑
aij = aij
j=ri=r i,j
the last symbol meaning for ℕ_{r}the integers larger than or equal to r,
({ )}
sup ∑ aij where S is a finite subset of ℕr × ℕr
( (i,j)∈S )
Proof:Obviously ∑_{}
(i,j)
∈Sa_{ij}≤∑_{j=r}^{∞}∑_{i=r}^{∞}a_{ij} and so ∑_{i,j}a_{ij}≤∑_{j=r}^{∞}∑_{i=r}^{∞}a_{ij}. Now let
λ <∑_{j=r}^{∞}∑_{i=r}^{∞}a_{ij}. Then there exists M such that
∑M ∞∑ ∞∑ M∑
aij = aij > λ
j=r i=r i=r j=r
Now there is N such that
N∑ M∑ ∑
λ < aij < aij
i=r j=r i,j
Since λ is arbitrary, it follows that
∞∑ ∞∑ ∑
aij ≤ aij ■
j=ri=r i,j
Corollary 1.13.5If∑_{i,j}
|aij|
< ∞, then∑_{i=r}^{∞}∑_{j=r}^{∞}a_{ij} = ∑_{j=r}^{∞}∑_{i=r}^{∞}a_{ij}. Here a_{ij}arecomplex numbers.
Proof: First note that ∑_{j=r}^{∞}a_{ij},∑_{j=1}^{∞}Rea_{ij},∑_{j=1}^{∞}Ima_{ij} exist. This is because, for
b_{ij} = a_{ij},Rea_{ij},Ima_{ij},
||∑q || ∑∞
|| bij||≤ |aij|
||j=p || j=p
which is small if p and q > p are large enough because ∑_{j}
|a |
ij
exists. Thus the partial sums form a
Cauchy sequence and therefore, these converge. This follows from the assumption that ℝ and ℂ are
complete which means that Cauchy sequences converge. Now also, for the same b_{ij}, if q > p,
then
Subtracting that which is known to be equal from both sides,
∞∑ ∞∑ ∑∞ ∑∞
Reaij = Re aij
i=r j=r j=r i=r
A similar equation holds by the same reasoning for Ima_{ij} in place of Rea_{ij}. Then this implies that
∑_{i=r}^{∞}∑_{j=r}^{∞}a_{ij} = ∑_{j=r}^{∞}∑_{i=r}^{∞}a_{ij}.■