limit_interchange-1
PDF · 1 pages · 50.6 KB
Open PDF file
Mathematical note on interchange of limits and uniform convergence. It defines uniform convergence in one variable and proves that if one limit is uniform, the iterated limits agree, using an epsilon/3 argument. It then applies the result to show that for a double series with absolutely summable rows and a finite sum of absolute row sums, the order of summation can be swapped.
AI-written summary; may contain errors.
Extracted text (machine-read; may contain errors)
Interchange of limits and uniform convergence
Lets→aandt→bbe two limiting processes. (For example, these could be a real variable
tending to a number, or towards infinity, a natural number tending towards infinity, a
partition tending towards ‘infinity’ in the sense of refinement, etc.) Let f(s, t) be a function
(real- or complex-valued now, though this is not necessary). We will give a general theorem
justifying the interchange of the order of these two limiting processes if a certain one of
them occurs uniformly, in the sense of the following definition.
Definition. LetL(t) be a function of t. We say that lim s→af(s, t) =L(t)uniformly in tif
for every /epsilon1 >0 there is s0such that for any s≥s0, and for all t, we have/vextendsingle/vextendsinglef(s, t)−L(t)/vextendsingle/vextendsingle< /epsilon1.
Theorem. Suppose that
(i) For each s, lim t→bf(s, t) =R(s).
(ii) For each t, lim s→af(s, t) =L(t).
(iii) lim s→a/parenleftbig
lim t→bf(s, t)/parenrightbig
=C.
(iv) lim s→af(s, t) =L(t) uniformly in t.
Then lim t→b/parenleftbig
lim s→af(s, t)/parenrightbig
=C.
Proof. Let/epsilon1 > 0. By (iv) we may choose s0so that if s≥s0then for all twe have/vextendsingle/vextendsinglef(s, t)−L(t)/vextendsingle/vextendsingle< /epsilon1/ 3. By (iii) we may choose s1so that if s≥s1then/vextendsingle/vextendsingleR(s)−C/vextendsingle/vextendsingle<
/epsilon1/3. Choose s2≥max{s0, s1}. We may use (i) to choose t0so that if t≥t0then/vextendsingle/vextendsinglef(s2, t)−R(s2)/vextendsingle/vextendsingle< /epsilon1/ 3. Now for t≥t0we have
/vextendsingle/vextendsingleL(t)−C| ≤/vextendsingle/vextendsingleL(t)−f(s2, t)/vextendsingle/vextendsingle+/vextendsingle/vextendsinglef(s2, t)−R(s2)/vextendsingle/vextendsingle+/vextendsingle/vextendsingleR(s2)−C/vextendsingle/vextendsingle
< /epsilon1/ 3 +/epsilon1/3 +/epsilon1/3
=/epsilon1.
The standard theorems stating that the (appropriately uniform) limit of continu-
ous/integrable/differentiable functions is continuous/integrable/differentiable, and that
the limit/integral/derivative of the limit is the limit of the limit/integral/derivative, can
all be derived from the above theorem. We will use it to prove a result on double series.
Theorem. Letaijbe complex numbers, for i,j∈ {1, 2, 3, . . .}. Suppose that
(i) For all i,/summationtext
j|aij|<∞.
(ii)/summationtext
i/parenleftbig/summationtext
j|aij|/parenrightbig
<∞.
Then
(iii) For all j,/summationtext
i|aij|<∞.
(iv)/summationtext
j/parenleftbig/summationtext
iaij/parenrightbig
=/summationtext
i/parenleftbig/summationtext
jaij/parenrightbig
.
Proof. Fixj. For any iwe have that |aij| ≤/summationtext
/lscript|ai/lscript|. Hence/summationtext
i|aij|converges by (ii) and
the comparison test. Therefore (iii) holds. In particular,/summationtext
iaijconverges (absolutely). By
(ii) and the comparison test again,/summationtext
i/vextendsingle/vextendsingle/summationtext
jaij/vextendsingle/vextendsingleconverges, so that/summationtext
i/parenleftbig/summationtext
jaij/parenrightbig
converges
(absolutely). We claim that/summationtext
iaijconverges uniformly in j. Let /epsilon1 > 0. By (ii) there
isi0such that/summationtext∞
i=i0/parenleftbig/summationtext
/lscript|ai/lscript|/parenrightbig
< /epsilon1. For any j, and for all i≥i0,|aij| ≤/summationtext
/lscript|ai/lscript|.
Therefore/summationtext∞
i=i0|aij| ≤/summationtext∞
i=i0/parenleftbig/summationtext
/lscript|ai/lscript|/parenrightbig
< /epsilon1. The theorem now follows by using the
previous theorem.