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limit_interchange-1

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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.

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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.