Kenneth Kuttler

1.11. SOME TOPICS FROM ANALYSIS 29

Corollary 1.11.4 If ai j ≥ 0, then ∑∞j=r ∑

∞i=r ai j = ∑i, j ai j the last symbol meaning for

Nr the integers larger than or equal to r,

sup

{∑

(i, j)∈Sai j where S is a finite subset of Nr×Nr

}

Proof: ∑(i, j)∈S ai j ≤∑∞j=r ∑

∞i=r ai j and so ∑i, j ai j ≤∑

∞j=r ∑

∞i=r ai j. Let λ <∑

∞j=r ∑

∞i=r ai j.

Then there exists M such that ∑Mj=r ∑

∞i=r ai j = ∑

∞i=r ∑

Mj=r ai j > λ . Now there is N such that

λ < ∑Ni=r ∑

Mj=r ai j < ∑i, j ai j. Since λ is arbitrary, it follows that ∑

∞j=r ∑

∞i=r ai j ≤ ∑i, j ai j ■

These theorems are special cases of Fubini’s theorem in Lebesgue integration as isshown later.

Corollary 1.11.5 If ∑i, j∣∣ai j∣∣ < ∞, then ∑

∞i=r ∑

∞j=r ai j = ∑

∞j=r ∑

∞i=r ai j. Here ai j are

complex numbers.

Proof: First note that ∑∞j=r ai j,∑

∞j=1 Reai j,∑

∞j=1 Imai j exist. This is because, for bi j =

ai j,Reai j, Imai j,∣∣∣∑q

j=p bi j

∣∣∣ ≤ ∑∞j=p∣∣ai j∣∣ which is small if p and q > p are large enough

because ∑ j∣∣ai j∣∣ 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 thatCauchy sequences converge. Now also, for the same bi j, if q > p, then∣∣∣∣∣ q

∑i=p

∞

∑j=r

bi j

∣∣∣∣∣≤ ∞

∑i=p

∣∣∣∣∣ ∞

∑j=r

bi j

∣∣∣∣∣≤ ∞

∑i=p

∞

∑j=r

∣∣ai j∣∣

which is small if p is large enough because ∑∞i=r ∑

∞j=r∣∣ai j∣∣ < ∞. Thus the partial sums

form a Cauchy sequence and so ∑∞i=r ∑

∞j=r bi j exists. Similarly ∑

∞i=r ∑

∞j=r bi j exists. Now∣∣ai j

∣∣+Reai j ≥ 0. Thus

∞

∑i=r

∞

∑j=r

∣∣ai j∣∣+Reai j =

∞

∑j=r

∞

∑i=r

∣∣ai j∣∣+Reai j

Thus, from the definition of the infinite sums,

∞

∑i=r

∞

∑j=r

∣∣ai j∣∣+ ∞

∑i=r

∞

∑j=r

Reai j =∞

∑j=r

∞

∑i=r

∣∣ai j∣∣+ ∞

∑j=r

∞

∑i=r

Reai j

Subtracting that which is known to be equal from both sides leads to the following equation:∑

∞i=r ∑

∞j=r Reai j = ∑

∞j=r ∑

∞i=r Reai j. A similar equation holds by the same reasoning for

Imai j in place of Reai j. Then this implies that ∑∞i=r ∑

∞j=r ai j = ∑

∞j=r ∑

∞i=r ai j. ■

1.11.1 lim sup and lim inf

Sometimes the limit of a sequence does not exist. For example, if an = (−1)n , thenlimn→∞ an does not exist. This is because the terms of the sequence are a distance of 1apart. Therefore there can’t exist a single number such that all the terms of the sequenceare ultimately within 1/4 of that number. The nice thing about limsup and liminf is thatthey always exist. First here is a simple lemma and definition.

1.11. SOME TOPICS FROM ANALYSIS 29Corollary 1.11.4 /f a;; > 0, then Y3_.Y2.,.4ij = Li,j aij the last symbol meaning forN,. the integers larger than or equal to r,sup y? aij where S is a finite subset of N; x ny}GesProof: Lies aij S Vier Lier Gj and so Yi aij S Vir Yi, Gij- Leta < Vir Yi, aij.Then there exists M such that yy Vier Gj = LE, yy, ajj > A. Now there is N such thatA< ye, yy aij < Yi aij. Since A is arbitrary, it follows that Y7_,. 2,4ij < Vij aijThese theorems are special cases of Fubini’s theorem in Lebesgue integration as isshown later.Corollary 1.11.5 Tf Mi j |ai;| < 00, then Vin, Vj, 4ij = Lia, Liz, ij. Here ajj arecomplex numbers.Proof: First note that = aj pLj- | Reajj, Yi- , Imaj; exist. This is because, for bj; =ajj,Rea;;,Imajj;, Yip bij <Vi=p |a;;| which is small if p and g > p are large enoughbecause )); |ai (| exists. Thus the partial sums form a Cauchy sequence and therefore, theseconverge. This follows from the assumption that R and C are complete which means thatCauchy sequences converge. Now also, for the same 5;;, if q > p, thenJ<¥Ii=piecoLdawhich is small if p is large enough because )’-,. V7_, |ai (| < co, Thus the partial sumsform a Cauchy sequence and so }/-,. bij exists. Similarly )j>,. D5, bij exists. Now|ai;| + Rea;; > 0. ThusYY Jaii|+Reaiy= VY Jaij| +Reaiji=r j=r j=ri=rThus, from the definition of the infinite sums,YL lai] +L LW Rea = Ye Yea] + VReaiji=r j=r i=r j=r j=ri=r j=ri=rSubtracting that which is known to be equal from both sides leads to the following equation:Ye rLj= ,Reajj = Li, Li-, Reaij. A similar equation holds by the same one forImajj; in place of Rea;j. Then this implies that )*, Y7—,.4ij = L5=, Lier Gij-1.11.1 lim sup and lim infSometimes the limit of a sequence does not exist. For example, if a, = (—1)”", thenlimy—s.0@, does not exist. This is because the terms of the sequence are a distance of |apart. Therefore there can’t exist a single number such that all the terms of the sequenceare ultimately within 1/4 of that number. The nice thing about limsup and liminf is thatthey always exist. First here is a simple lemma and definition.