Dynamics of a Compact Operator

Abstract

Let 𝑇∶𝑋→𝑋 be a compact linear (or more generally affine) operator from a Banach space into itself. For each 𝑥∈𝑋, the sequence of iterates 𝑇^𝑛𝑥, 𝑛=0, 1, … and its averages (1/𝑘)∑^𝑛𝑘=0𝑇^𝑘−1𝑥,𝑛=0, 1, … are either bounded or approach infinity.

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Let 𝑋 be a set and 𝑓∶𝑋→𝑋 a map from 𝑋 into 𝑋. For an 𝑥∈𝑋, the sequence of iterations 𝑥,𝑓(𝑥),𝑓²(𝑥),…,𝑓^𝑘(𝑥),… can be considered as a trajectory of a dynamical system where time is the discrete nonnegative integer: starting with the initial (time 𝑡=0) state 𝑥, the state at time 𝑡=𝑘 is 𝑓^𝑘(𝑥). Suppose now that 𝑋=ℂ^𝑛 and 𝑓 is the transformation defined by an 𝑛×𝑛 complex matrix 𝐴. What can one say about the general behavior of trajectories of 𝐴? More generally, we consider affine maps on 𝑋, that is, maps of the form 𝐴𝑥+𝑐, where 𝐴 is linear and 𝑐 is a constant vector, and we also allow 𝑋 to be infinite dimensional. Moreover, we study the behavior of the sequence of averages Ave_𝑘𝑓(𝑥)=1𝑘𝑥+𝑓(𝑥)+⋯+𝑓^𝑘−1(𝑥),𝑘=1,2,….(1) Note that the method of averaging was used in [1] in approximating solutions of a system of linear equations. In case of linear operators, problems about the linear span of the iterates 𝑇^𝑘𝑥,𝑛=0,1,… can be found in [2].

Recall that a square matrix 𝑁 is called nilpotent if 𝑁^𝑠=0 for some nonnegative integer 𝑠.

Throughout this paper, for nonnegative integers 𝑘,𝑗,𝑘≥𝑗,𝐶(𝑘,𝑗) denotes the binomial coefficient 𝑘!𝑗!(𝑘−𝑗)!.(2) By convention, 𝐶(𝑘,0)=1 for 𝑘≥0 and 𝐶(𝑘,𝑗)=0 if 𝑘<𝑗.

If 𝑘<𝑗, the sum ∑^{𝑘𝑖=𝑗}𝑢_𝑖 is considered as an empty sum and its value is 0.

Theorem 1. Let 𝑇∶ℂ^𝑛→ℂ^𝑛 be an affine map defined by 𝑇𝑥=𝐴𝑥+𝑐, where 𝐴 is an 𝑛×𝑛 complex matrix and 𝑐 a constant vector in ℂ^𝑛. Let ‖⋅‖ be a norm on ℂ^𝑛. Then for any vector 𝑥∈ℂ^𝑛, the sequence 𝑇^𝑘𝑥,𝑘=0,1,2,…(3) is either bounded or lim_𝑘‖𝑇^𝑘𝑥‖=∞.

Proof. By Jordan canonical decomposition theorem, ℂ^𝑛=𝑉₁⊕𝑉₂⊕⋯⊕𝑉_𝑚 for some subspaces 𝑉_𝑖,𝑖=1,2,…,𝑚 with the following properties: (a) each 𝑉_𝑖 is an invariant subspace of 𝐴, that is, 𝐴𝑣∈𝑉_𝑖 for all 𝑣∈𝑉_𝑖, and (b) there exists 𝜆_𝑖∈ℂ and a nilpotent matrix 𝑁_𝑖 such that 𝐴𝑣=𝜆_𝑖𝑣+𝑁_𝑖𝑣 for all 𝑣∈𝑉_𝑖.
Let 𝑃_𝑖 be the algebraic projection of ℂ^𝑛 onto 𝑉_𝑖 associated with the decomposition ℂ^𝑛=𝑉₁⊕𝑉₂⊕⋯⊕𝑉_𝑚.
Define a new norm |⋅| on ℂ^𝑛 by |𝑣|=‖‖𝑃₁𝑣‖‖+‖‖𝑃₂𝑣‖‖+⋯+‖‖𝑃_𝑚𝑣‖‖.(4) Let 𝑥 be a vector in ℂ^𝑛. Let 𝑥_𝑘=𝑇^𝑘𝑥=𝐴^𝑘𝑥+𝑐+𝐴𝑐+⋯+𝐴^𝑘−1𝑐,𝑘=1,2,….(5) For any vector 𝑥, we have the following equalities: 𝐴𝑥=𝑃₁𝑥+⋯+𝑃_𝑚𝑥,_𝑘𝐴𝑥=𝐴^𝑘𝑃₁𝑥+⋯+𝐴^𝑘𝑃_𝑚𝑥,_𝑘𝑥=𝑃₁𝐴^𝑘𝑥+⋯+𝑃_𝑚𝐴^𝑘𝑥.(6) The last equality follows from the commutative property of 𝐴 and 𝑃_𝑖,𝑖=1,…,𝑚 since 𝐴 is invariant in each 𝑉_𝑗.
Fix an 𝑖 and write 𝑣=𝑃_𝑖𝑥,𝑑=𝑃_𝑖𝑐,𝜆=𝜆_𝑖,𝑉=𝑉_𝑖, and 𝑁=𝑁_𝑖. Let 𝑠 be the smallest nonnegative integer such that 𝑁^𝑠𝑣=0 and 𝑡 the smallest nonnegative integer such that 𝑁^𝑡𝑑=0. Then for 𝑘>max{𝑠,𝑡} and 𝑠,𝑡≥1 (if 𝑠=0 or 𝑡=0, the corresponding sum below is defined as 0), one has 𝑃_𝑖𝑥_𝑘==(𝜆𝐼+𝑁)^𝑘𝑣+𝑑+(𝜆𝐼+𝑁)𝑑+⋯+(𝜆𝐼+𝑁)^𝑘−1𝑑(7)_𝑠−1_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+^𝑡−1_𝑗=0𝑆(𝑗,𝑘)𝑁^𝑗𝑑,(8) where 𝑆(𝑗,𝑘)=𝐶(𝑗,𝑗)+𝐶(𝑗+1,𝑗)𝜆+⋯+𝐶(𝑘−1,𝑗)𝜆^{𝑘−1−𝑗}.(9) Using the identity 𝐶(𝑗,𝑖+1)+𝐶(𝑗,𝑖)=𝐶(𝑗+1,𝑖+1), we have, for 𝜆=1, 𝑆(𝑗,𝑘)=𝐶(𝑘,𝑗+1), and, for 𝜆≠1, 𝑆(𝑗,𝑘) is given recursively by 𝑆(0,𝑘)=1−𝜆^𝑘1−𝜆(10) and (by subtracting 𝜆𝑆(𝑗,𝑘) from 𝑆(𝑗,𝑘)) (1−𝜆)𝑆(𝑗,𝑘)=𝑆(𝑗−1,𝑘)−𝜆^𝑘−𝑗𝐶(𝑘,𝑗),𝑗=1,2,…,𝑡−1(for𝑡≥2),(11) from which we get an alternate formula for 𝑆(𝑗,𝑘): 𝑆(𝑗,𝑘)=1−𝜆^𝑘(1−𝜆)^𝑗+1−^𝑗−1_𝑖=0𝐶(𝑘,𝑖+1)𝜆^{𝑘−𝑖−1}(1−𝜆)^𝑗−𝑖,𝑗=1,…,𝑡−1.(12) Note that (12) is also valid for 𝑗=0 for ∑^−1𝑖=0 is an empty sum. We will show that, for 𝜆≠1, 𝑃_𝑖𝑥_𝑘=𝐵+^𝑤−1_𝑗=0𝜆^𝑘−𝑗𝐶(𝑘,𝑗)𝐴_𝑗,(13) where 𝑤=max{𝑠,𝑡} and 𝐴_𝑗,𝐵,𝑗=0,…,𝑤−1 are constant vectors independent of 𝑘, and, for 𝜆=1, 𝑃_𝑖𝑥_𝑘=𝑣+^𝑙_𝑗=1𝐶(𝑘,𝑗)𝐵_𝑗,(14) where 𝑙=max{𝑠−1,𝑡} and 𝐵_𝑗,𝑗=1,…,𝑙 are constant vectors independent of 𝑘. More precisely, for 𝑘>max{𝑠,𝑡}, 𝐴_𝑗=𝑁^𝑗𝑣−^𝑡−1_𝑖=𝑗1(1−𝜆)^{𝑖−𝑗+1}𝐴𝑁^𝑖𝑑,𝑗=0,…,min{𝑠,𝑡}−1,_𝑗=𝜖(𝑠−𝑡)𝑁^𝑗𝑣−𝜖(𝑡−𝑠)^𝑡−1_𝑖=𝑗1(1−𝜆)^{𝑖−𝑗+1}𝐵=_𝑡−1𝑁^𝑖𝑑,𝑗=min{𝑠,𝑡},…,𝑤−1,_𝑖=01(1−𝜆)^𝑖+1𝐵𝑁^𝑖𝑑,_𝑗𝐵=𝑁^𝑗𝑣+𝑁^𝑗−1𝑑,𝑗=1,…,min{𝑠−1,𝑡},_𝑗=𝜖(𝑠−1−𝑡)𝑁^𝑗𝑣+𝜖(𝑡−𝑠+1)𝑁^𝑗−1𝑑,𝑗=min{𝑠−1,𝑡}+1,…,𝑙,(15) where 𝜖(𝑟)=1 for 𝑟≥0 and 𝜖(𝑟)=0 for 𝑟<0.
Indeed substituting (12) into (8), we get 𝑃_𝑖𝑥_𝑘=^𝑠−1_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+^𝑡−1_𝑗=0(11−−𝜆𝜆)^𝑘𝑗+1−^𝑗−1_𝑖=0𝐶(𝑘,𝑖+1)𝜆^{𝑘−𝑖−1}(1−𝜆)^𝑗−𝑖𝑁^𝑗=_𝑠−1𝑑_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+^𝑡−1_𝑗=01(1−𝜆)^𝑗+1𝑁^𝑗𝑑−^𝑡−1_𝑗𝑗=0_𝑖=0𝐶(𝑘,𝑖)𝜆^𝑘−𝑖(1−𝜆)^{𝑗−𝑖+1}=_𝑠−1𝑁^𝑗𝑑_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+𝐵−^𝑡−1_{𝑖=0𝑡−1𝑗=𝑖}𝐶(𝑘,𝑖)𝜆^𝑘−𝑖(1−𝜆)^{𝑗−𝑖+1}=_𝑠−1𝑁^𝑗𝑑_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+𝐵−^𝑡−1_{𝑗=0𝑡−1𝑖=𝑗}𝐶(𝑘,𝑗)𝜆^𝑘−𝑗(1−𝜆)^{𝑖−𝑗+1}=_𝑠−1𝑁^𝑖𝑑_𝑗=0𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝑁^𝑗𝑣+𝐵−^𝑡−1_𝑗=0𝐶(𝑘,𝑗)𝜆^{𝑘−𝑗𝑡−1}_𝑖=𝑗1(1−𝜆)^{𝑖−𝑗+1}𝑁^𝑖𝑑.(16) Now (13) follows by considering cases 𝑠>𝑡,𝑠=𝑡, or 𝑠<𝑡.
Next consider the case 𝜆=1. Substituting 𝑆(𝑗,𝑘)=𝐶(𝑘,𝑗+1) into (8), we get 𝑃_𝑖𝑥_𝑘=^𝑠−1_𝑗=0𝐶(𝑘,𝑗)𝑁^𝑗𝑣+^𝑡−1_𝑗=0=𝑣+_𝑠−1𝐶(𝑘,𝑗+1)𝑁^𝑗𝑑_𝑗=1𝐶(𝑘,𝑗)𝑁^𝑗𝑣+^𝑡_𝑗=1𝐶(𝑘,𝑗)𝑁^𝑗−1𝑑.(17) Now (14) follows by considering cases 𝑠−1>𝑡,𝑠−1=𝑡, or 𝑠−1>𝑡.
We now proceed to finish the proof of the theorem.
Case 1 (|𝜆|>1). If one of these 𝐴_𝑗,𝑗=0,…,𝑤−1 is nonzero, then we see from (13) that ‖𝑃_𝑖𝑥_𝑘‖→∞ as 𝑘→∞; otherwise 𝑃_𝑖𝑥_𝑘=𝐵 is a constant vector.Case 2 (|𝜆|=1 and 𝜆≠1). If one of these 𝐴_𝑗,𝑗=1,…,𝑤−1 is nonzero, then we see from (13) that ‖𝑃_𝑖𝑥_𝑘‖→∞ as 𝑘→∞; otherwise the sequence 𝑃_𝑖𝑥_𝑘=𝐵+𝜆^𝑘𝐴₀,𝑘=1,…, is bounded.Case 3 (𝜆=1). If one of these 𝐵_𝑗,𝑗=1,…,𝑙 is nonzero, then we see from (14) that ‖𝑃_𝑖𝑥_𝑘‖→∞ as 𝑘→∞; otherwise 𝑃_𝑖𝑥_𝑘 is the constant vector 𝑣.Case 4 (|𝜆|<1). We see from (13) that the sequence 𝑃_𝑖𝑥_𝑘,𝑘=1,…, converges to the constant vector 𝐵.
We conclude that for each 𝑖, the sequence 𝑃_𝑖𝑥_𝑘,𝑘=1,… is either bounded or tends to infinity. If one of these sequences tends to infinity, then, since |𝑥_𝑘|≥‖𝑃_𝑖𝑥_𝑘‖, the sequence 𝑥_𝑘,𝑘=1,…, tends to infinity in norm |⋅| and hence in norm ‖⋅‖, since the two norms are equivalent. Otherwise all sequences 𝑃_𝑖𝑥_𝑘,𝑘=1,…, are bounded for all 𝑖=1,…,𝑚, and from which it follows that 𝑥_𝑘,𝑘=1,…, is bounded. This completes the proof.

The following example shows that in infinite-dimensional spaces, Theorem 1 is false.

Example 2. For nonnegative integers 𝑛, let 𝑐_𝑛=(1/2)𝑛(𝑛+1). Define 𝜆_𝑖=1/2 for 𝑐_2𝑛≤𝑖≤𝑐_2𝑛+1−1,𝑛=0,1,…, and 𝜆_𝑖=2 for 𝑐_2𝑛−1≤𝑖≤𝑐_2𝑛−1,𝑛=1,2,…. Define a linear operator 𝐴∶𝑙₂→𝑙₂ such that 𝐴𝑒_𝑖=𝜆_𝑖𝑒_𝑖+1,𝑖=0,1,…. Then the sequence of iterates 𝐴^𝑘𝑒₀=𝜆₀𝜆₁⋯𝜆_𝑘−1𝑒_𝑘,𝑘=1,2,…, contains subsequences that converge to 0 (e.g., 𝐴^{𝑐2𝑛−1}𝑒₀=1/2^𝑛,𝑛=1,2,…), subsequences that approach infinity (e.g., 𝐴^𝑐2𝑛𝑒₀=2^𝑛,𝑛=1,2,…), and infinitely many bounded nonconvergent subsequences (e.g., 𝐴^𝑛𝑒₀=𝑒_𝑛 for 𝑛=2,4,8,12,18,24,32,40,…). Clearly 𝐴 is bounded but not compact.

Remark 3. For 𝑇 linear, Theorem 1 appeared in a 1997 unpublished article “On the behavior of the iterates of a matrix” of the author.

In the case that the mapping 𝑇 in Theorem 1 is linear, we can actually say more.

Theorem 4. Let 𝐴∶ℂ^𝑛→ℂ^𝑛 be a linear map. Let ‖⋅‖ be a norm on ℂ^𝑛. Then for any vector 𝑥∈ℂ^𝑛, either lim_𝑘→∞𝑇^𝑘𝑥=0 or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞ or 𝐹≤‖𝑇^𝑘𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 𝐹,𝐺 are positive numbers with 𝐹≤𝐺.

Proof. If 𝑐=0 in Theorem 1, then the constant vectors 𝐵 and 𝑑 in its proof are 0. It follows that in all cases where 𝑃_𝑖𝑥_𝑘,𝑘=1,2,…, is bounded and not convergent to 0, it is bounded away from 0. If 𝑃_𝑖𝑥_𝑘 is bounded away from 0 for some 𝑖, then so is 𝑥_𝑘=𝑇^𝑘𝑥 since |𝑥_𝑘|≥‖𝑃_𝑖𝑥_𝑘‖. Otherwise 𝑃_𝑖𝑥_𝑘→0 converges to 0 for all 𝑖 and hence so does 𝑥_𝑘.

Example 5. The following simple example shows that linearity is needed in Theorem 4. Let 𝑇∶ℂ→ℂ be the map 𝑇𝑥=𝑖𝑥+𝑐, where 𝑐∈ℂ is nonzero and 𝑖=√−1. Then 𝑇^4𝑛(0)=0 and 𝑇^4𝑛+1(0)=𝑐 for all positive integers 𝑛, showing that 𝑇^𝑘(0),𝑘=1,2,…, is neither convergent to 0 nor bounded away from zero.

Definition 6. Let 𝑋 be a Banach space and 𝐴∶𝑋→𝑋 a linear operator. one says that 𝐴 has property (𝑃) if 𝑋 is a direct sum of two closed subspaces 𝑉₁,𝑉₂ such that (1) each 𝑉_𝑖 is invariant under 𝐴, (2) 𝑉₁ is finite dimensional, and (3) there exists 0≤𝑟<1 and a positive integer 𝑁 such that ‖𝐴^𝑘𝑥‖≤𝑟^𝑘‖𝑥‖ for all 𝑥∈𝑉₂ and all 𝑘≥𝑁.

Note that by Gelfand’s spectral radius theorem, condition (12) above is equivalent to that 𝐴, as an operator on 𝑉₂, has spectral radius less than 1.

It is well known that every compact operator, or more generally, Riesz operator, has property (P); see, for example, [3].

Theorem 7. Let (𝑋,‖⋅‖) be a Banach space and 𝐴∶𝑋→𝑋 a bounded operator having property (P). Let 𝑐 be a constant vector in 𝑋, and let 𝑇(𝑥)=𝐴𝑥+𝑐 for 𝑥∈𝑋. Then for any vector 𝑥∈𝑋, either {𝑇^𝑘𝑥,𝑘=0,1,…} is bounded or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞. Moreover, if 𝑐=0, then either lim_𝑘→∞𝑇^𝑘𝑥=0 or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞ or 𝐹≤‖𝑇^𝑘𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 0<𝐹≤𝐺<∞.

Proof. Let 𝑉_𝑖,𝑖=1,2,𝑁,𝑟, be as in Definition 6. Let 𝑃_𝑖,𝑖=1,2, be the projections of 𝑋 onto 𝑉_𝑖,𝑖=1,2, respectively. Define a norm |⋅| on 𝑋 as |𝑥|=‖‖𝑃₁𝑥‖‖+‖‖𝑃₂𝑥‖‖.(18)|⋅| is equivalent to ‖⋅‖ and 𝑇 commutes with 𝑃_𝑖,𝑖=1,2.
For any 𝑣∈𝑋, write 𝑣_𝑖=𝑃_𝑖𝑣,𝑖=1,2. Then for 𝑥∈𝑋 and 𝑘≥𝑁 one has ‖‖𝑇^𝑘𝑥₂‖‖=≤‖‖𝑟_𝑘𝐴^𝑘𝑥₂+𝑐₂+𝐴𝑐₂+⋯+𝐴^𝑘−1𝑐₂‖‖≤𝑟_𝑘‖‖𝑥₂‖‖+‖‖𝑐₂+⋯+𝐴^𝑁−1𝑐₂‖‖+𝑟^𝑁+⋯+𝑟^𝑘−1‖‖𝑐₂‖‖‖‖𝑥₂‖‖+‖‖𝑐₂+⋯+𝐴^𝑁−1𝑐₂‖‖+11−𝑟‖‖𝑐²‖‖,(19) showing that 𝑇^𝑘𝑥₂,𝑘=1,2,…, is bounded. By Theorem 1, ‖𝑇^𝑘𝑥₁‖,𝑘=1,2,…, is either bounded or approaching infinity. Hence |𝑇^𝑘𝑥|=‖𝑇^𝑘𝑥₁‖+‖𝑇^𝑘𝑥₂‖,𝑘=1,2,…, is either bounded or approaching infinity. Since norms |⋅| and ‖⋅‖ are equivalent, the same is true for ‖𝑇^𝑘𝑥‖,𝑘=1,2,….
If 𝑐=0, then 𝑐₂=0 and 𝑇^𝑘𝑥₂→0 as 𝑘→∞, and the last part of the theorem follows readily from Theorem 4.

Corollary 8. Let (𝑋,‖⋅‖) be a Banach space and 𝐴∶𝑋→𝑋 a Riesz operator. Let 𝑐 be a constant vector in 𝑋, and let 𝑇(𝑥)=𝐴𝑥+𝑐 for 𝑥∈𝑋. Then for any vector 𝑥∈𝑋, either {𝑇^𝑘𝑥,𝑘=0,1,…} is bounded or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞. Moreover, if 𝑐=0, then either lim_𝑘→∞𝑇^𝑘𝑥=0 or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞ or 𝐹≤‖𝑇^𝑘𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 0<𝐹≤𝐺<∞.

Corollary 9. Let (𝑋,‖⋅‖) be a Banach space and 𝐴∶𝑋→𝑋 a compact operator. Let 𝑐 be a constant vector in 𝑋, and let 𝑇(𝑥)=𝐴𝑥+𝑐 for 𝑥∈𝑋. Then for any vector 𝑥∈𝑋, either {𝑇^𝑘𝑥,𝑘=0,1,…} is bounded or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞. Moreover, if 𝑐=0, then either lim_𝑘→∞𝑇^𝑘𝑥=0 or lim_𝑘→∞‖𝑇^𝑘𝑥‖=∞ or 𝐹≤‖𝑇^𝑘𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 0<𝐹≤𝐺<∞.

Let us consider now the behavior of the sequence of averages of 𝑇Ave_𝑘𝑇(𝑥)=1𝑘𝑥+𝑇𝑥+⋯+𝑇^𝑘−1𝑥,𝑘=1,2,….(20) We have that Ave_𝑘𝑇(𝑥)=1𝑘𝑥+𝐴𝑥+⋯+𝐴^𝑘−1𝑥+1𝑘(𝑘−1)𝑐+(𝑘−2)𝐴𝑐+⋯+𝐴^𝑘−2𝑐.(21) As in the proof of Theorem 1, we may assume that 𝐴=𝜆𝐼+𝑁 and that 𝑠,𝑡 are defined as in there. By replacing 𝑐 by 𝑥, 𝑥 by 0, and hence 𝑠 by 0 and 𝑡 by 𝑠 in the proof of Theorem 1, we obtain the following:

for 𝜆≠1, 1𝑘𝑥+𝐴𝑥+⋯+𝐴^𝑘−1𝑥=1𝑘𝐵+^𝑠−1_𝑗=0𝜆^𝑘−𝑗𝐶(𝑘,𝑗)𝑘𝐴^𝑗,(22) where 𝐴_𝑗=−^𝑠−1_𝑖=𝑗1(1−𝜆)^{𝑖−𝑗+1}𝐵=_𝑠−1𝑁^𝑖𝑥,𝑗=0,…,𝑠−1,_𝑗=01(1−𝜆)^𝑗+1𝑁^𝑗𝑥,(23) for 𝜆=1, 1𝑘𝑥+𝐴𝑥+⋯+𝐴^𝑘−1𝑥=1𝑘^𝑠_𝑗=1𝐶(𝑘,𝑗)𝑁^𝑗−1𝑥.(24) By expanding 𝐴^𝑗, we have (𝑘−1)𝑐+(𝑘−2)𝐴𝑐+⋯+𝐴^𝑘−2𝑐=𝑇₀𝑐+𝑇₁𝑁𝑐+⋯+𝑇_𝑡−1𝑁^𝑡−1𝑐,(25) where 𝑇_𝑗=^𝑘−2_𝑖=𝑗(𝑘−𝑖−1)𝐶(𝑖,𝑗)𝜆^𝑖−𝑗,𝑗=0,…,𝑡−1.(26) Assume that 𝜆≠1. By subtracting 𝜆𝑇₀ from 𝑇₀ and using the geometric series formula, one gets 𝑇₀=𝑘1−𝜆−(11−−𝜆𝜆^𝑘)².(27) Using the relation (𝑖−𝑗)𝐶(𝑖,𝑗)=(𝑗+1)𝐶(𝑖,𝑗+1), we see that 𝑇_𝑗+1=1𝑗+1𝑑𝑑𝜆𝑇^𝑗.(28) We will prove in the appendix that 𝑇_𝑗=𝑘(1−𝜆)^𝑗+1−𝑗+1(1−𝜆)^𝑗+2+𝐶(𝑘,𝑗)𝜆^𝑘−𝑗(1−𝜆)²𝐷(𝑘,𝑗,𝜆),(29) for 𝑗=0,…,𝑡−1, where 𝐷(𝑘,𝑗,𝜆)=1(1−𝜆)^𝑗𝐵𝐵₀(𝑘,𝑗)𝜆^𝑗+𝐵₁(𝑘,𝑗)𝜆^𝑗−1+⋯+𝐵_𝑗−1(𝑘,𝑗)𝜆+1_𝑖,(𝑘,𝑗)=(−1)^𝑗−𝑖𝐶(𝑗,𝑖)𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑖,2),𝑖=0,…,𝑗.(30) Note that 𝐵_𝑖(𝑘,𝑗)⟶(−1)^𝑗−𝑖𝐶(𝑗,𝑖),(31) so that 𝐷(𝑘,𝑗,𝜆) approaches 1 as 𝑘→∞. Substituting the formulae we obtain thus far into Ave_𝑘𝑇(𝑥), with 𝑤=max{𝑠,𝑡}, we get, for 𝜆≠1, Ave_𝑘𝑇(𝑥)=𝐸+1𝑘𝐹+𝐺(𝑘,𝜆),(32) where 𝐺(𝑘,𝜆)=^𝑤−1_𝑗=01𝑘𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝜖(𝑠−1−𝑗)𝐴^𝑗+𝜖(𝑡−1−𝑗)𝐷(𝑘,𝑗,𝜆)(1−𝜆)²𝐸=_𝑡−1𝑁^𝑗𝑐,(33)_𝑗=01(1−𝜆)^𝑗+1𝐹=_𝑠−1𝑁^𝑗𝑐,_𝑗=01(1−𝜆)^𝑗+1𝑁^𝑗𝑥−^𝑡−1_𝑗=0𝑗+1(1−𝜆)^𝑗+2𝑁^𝑗𝑐,(34) and 𝜖(𝑟)=1 for 𝑟≥0 and 𝜖(𝑟)=0 for 𝑟<0.

If 𝜆=1, then 𝑇_𝑗 is equal to 𝑆_𝑗,𝑘=𝐶(𝑘−2,𝑗)+2𝐶(𝑘−3,𝑗)+⋯+(𝑘−𝑗−1)𝐶(𝑗,𝑗).(35) We have 𝑆_𝑗,𝑘−𝑆_{𝑗,𝑘−1}=𝐶(𝑘−2,𝑗)+𝐶(𝑘−3,𝑗)+⋯+𝐶(𝑗,𝑗)=𝐶(𝑘−1,𝑗+1),(36) where the last equality follows from repeatedly applying the identity 𝐶(𝑗,𝑖+1)+𝐶(𝑗,𝑖)=𝐶(𝑗+1,𝑖+1). From this it follows that 𝑆_𝑗,𝑘==𝑆𝑆_{𝑗,𝑘−1𝑗,𝑘−2}+𝐶(𝑘−1,𝑗+1)==⋮𝑆_𝑗,𝑗+2+𝐶(𝑘−2,𝑗+1)+𝐶(𝑘−1,𝑗+1)==𝐶𝐶((𝑗𝑘,+𝑗+1+,𝐶𝑗2)(+𝑗.1+)2+,𝑗𝐶+(𝑗1+)+2,⋯𝑗++1𝐶)+(𝑘⋯−+1,𝐶𝑗+(𝑘1−)1,𝑗+1)(37) Therefore, for 𝜆=1, Ave_𝑘𝑇(𝑥)=1𝑘𝐶(𝑘,1)𝑥+𝐶(𝑘,2)𝑁𝑥+⋯+𝐶(𝑘,𝑠)𝑁^𝑠−1+1𝑥𝑘𝐶(𝑘,2)𝑐+𝐶(𝑘,3)𝑁𝑐+⋯+𝐶(𝑘,𝑡+1)𝑁^𝑡−1𝑐.(38)

Case 1 (|𝜆|>1). Suppose that 𝑡>𝑠. Then from (32) and (33), (1/𝑘)𝜆^𝑘𝐶(𝑘,𝑡−1) dominates all other coefficients. Since 𝐷(𝑘,𝑡−1,𝜆)→1 as 𝑘→∞ and 𝑁^𝑡−1𝑐≠0, we see that ‖Ave_𝑘𝑇(𝑥)‖→∞ as 𝑘→∞. The same is true if 𝑠>𝑡 since 𝐴_𝑠−1≠0. So assume that 𝑠=𝑡. In the trivial case 𝑠=𝑡=0, we have Ave_𝑘𝑇(𝑥)=0 for all 𝑘. So assume that 𝑠=𝑡≥1. Direct checking (see the appendix, item 1) shows that 𝐶(𝑘,𝑗)𝐶(𝑗,𝑖)(𝐶(𝑘−𝑗,2)/𝐶(𝑘−𝑖,2))=𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖) so that 𝐶(𝑘,𝑗)𝐷(𝑘,𝑗,𝜆) is a polynomial in 𝑘. Then from (33), we have 𝐺(𝑘,𝜆)=^𝑡−1_𝑗=01𝑘𝐶(𝑘,𝑗)𝜆^𝑘−𝑗𝐴^𝑗+𝐷(𝑘,𝑗,𝜆)(1−𝜆)²=𝜆_𝑘𝑁^𝑗𝑐𝑘^𝑡−1_𝑗=0𝑝₁(𝑘,𝑗,𝜆)𝐴_𝑗+𝑝₂(𝑘,𝑗,𝜆)𝑁^𝑗𝑐,(39) where for fixed 𝜆, 𝑝₁,𝑝₂ are polynomials in 𝑘. Since lim_𝑘→∞(𝜆^𝑘/𝑘)𝑝(𝑘)=∞ for any polynomial 𝑝(𝑘) we see that if 𝐻(𝑘,𝜆)=^𝑡−1_𝑗=0𝑝₁(𝑘,𝑗,𝜆)𝐴_𝑗+𝑝₂(𝑘,𝑗,𝜆)𝑁^𝑗𝑐,(40) as a polynomial in 𝑘, is identically zero, then clearly from (32) we have Ave_𝑘𝑇(𝑥)→𝐸, otherwise ‖Ave_𝑘𝑇(𝑥)‖→∞.

Case 2 (|𝜆|=1and𝜆≠1). Suppose that 𝑡>𝑠. Since 𝐷(𝑘,𝑡−1,𝜆)→1 as 𝑘→∞ and 𝑁^𝑡−1𝑐≠0, we see that ‖Ave_𝑘𝑇(𝑥)‖→∞ as 𝑘→∞ if 𝑡≥3, and it approaches 0 or is bounded if 𝑡≤2. The same is true if 𝑠>𝑡 since 𝐴_𝑠−1≠0. So assume that 𝑠=𝑡. In the trivial case 𝑠=𝑡=0, we have Ave_𝑘𝑇(𝑥)=0 for all 𝑘. So assume that 𝑠=𝑡≥1. As in Case 1, we consider the polynomial (in 𝑘) 𝐻(𝑘,𝜆)=^𝑡−1_𝑗=0𝐶(𝑘,𝑗)𝜆^−𝑗𝐴^𝑗+𝐷(𝑘,𝑗,𝜆)(1−𝜆)²𝑁^𝑗𝑐,(41) so that 𝐺(𝑘,𝜆)=𝜆^𝑘𝑘𝐻(𝑘,𝜆).(42) If degree of 𝐻 is one or less, then we see from above that 𝐺(𝑘,𝜆) is bounded and hence Ave_𝑘𝑇(𝑥) is bounded by (32). If degree of 𝐻 is two or more, then ‖𝐺(𝑘,𝜆)‖→∞ as 𝑘→∞ and hence so is ‖Ave_𝑘𝑇(𝑥)‖.

Case 3 (𝜆=1). We refer to (38). If 𝑠=𝑡=0, then 𝑐=𝑥=0 and Ave_𝑘𝑇(𝑥)=0. If 𝑡=0 and 𝑠=1, then Ave_𝑘𝑇(𝑥)=𝑥. If 𝑡=𝑠−1,𝑠≥2, and 𝑁^𝑖𝑐+𝑁^𝑖+1𝑥=0 for all 𝑖=0,…,𝑡−1, then Ave_𝑘𝑇(𝑥)=𝑥. In all other cases lim_𝑘‖Ave_𝑘𝑇(𝑥)‖=∞.

The previous discussions yield the proof of the following.

Theorem 10. Let 𝑇∶ℂ^𝑛→ℂ^𝑛 be an affine map defined by 𝑇𝑥=𝐴𝑥+𝑐, where 𝐴 is an 𝑛×𝑛 complex matrix and 𝑐 a constant vector in ℂ^𝑛. Let ‖⋅‖ be a norm on ℂ^𝑛. For any vector 𝑥∈ℂ^𝑛, define Ave_𝑘𝑇(𝑥)=1𝑘𝑥+𝑇𝑥+⋯+𝑇^𝑘−1𝑥.(43) Then the sequence Ave_𝑘𝑇(𝑥),𝑘=0,1,2,…,(44) is either bounded or lim_𝑘‖Ave_𝑘𝑇(𝑥)‖=∞.

If 𝑇 is linear, that is, if 𝑐=0, we can say more.

Theorem 11. Let 𝐴∶ℂ^𝑛→ℂ^𝑛 be linear map. Let ‖⋅‖ be a norm on ℂ^𝑛. For any vector 𝑥∈ℂ^𝑛, define Ave_𝑘𝐴(𝑥)=1𝑘𝑥+𝐴𝑥+⋯+𝐴^𝑘−1𝑥.(45) Then the sequence Ave_𝑘𝐴(𝑥),𝑘=0,1,2,…,(46) is either (i) convergent to 0, or (ii) lim_𝑘‖Ave_𝑘𝐴(𝑥)‖=∞, or (iii) 𝐹≤‖Ave_𝑘𝐴𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 𝐹,𝐺 are positive numbers with 𝐹≤𝐺.

Proof. If 𝑐=0 in Theorem 10, then 𝐸=0, and by examining its proof we see that in all cases where 𝑇^𝑘𝑥 is bounded and not convergent to 0, it is bounded away from 0. (In Case 2 of the proof, if 𝐻(𝑘,𝜆) is of degree one, then 𝐺(𝑘,𝜆) is bounded away from 0, and if 𝐻(𝑘,𝜆) is of degree 0, that is, a constant vector, then 𝐺(𝑘,𝜆)→0.).

The following example shows that Theorem 11 is false if the map is not linear.

Example 12. Let 𝐴=0𝑖1𝑖,(47) and let 𝑐=¹⁰. Define 𝑇∶ℂ²→ℂ² by 𝑇𝑥=𝐴𝑥+𝑐. Let 𝑣=¹⁰.
Then Ave_𝑘𝑇(𝑣)=1𝑘11−−𝑖𝑖^𝑘𝑣+11−−𝑖𝑖^𝑘−1,(48) which does not converge to 0 and has a subsequence converging to 0.

Theorems 10 and 11 are also valid in Banach spaces. For proof, one only has to use these theorems and note that the average of the bounded (resp., null convergent) sequence 𝑇^𝑘𝑥₂,𝑘=0,1,…, in the proof of Theorem 7 is also bounded (resp., null convergent). Thus we have the following theorem.

Theorem 13. Let 𝑋 be a Banach space. Let 𝑇∶𝑋→𝑋 be an affine map defined by 𝑇𝑥=𝐴𝑥+𝑐, where 𝐴 is an operator with property (P) and 𝑐 a constant vector in 𝑋. Let ‖⋅‖ be the norm on 𝑋. For any vector 𝑥∈𝑋, define Ave_𝑘𝑇(𝑥)=1𝑘𝑥+𝑇𝑥+⋯+𝑇^𝑘−1𝑥.(49) Then the sequence Ave_𝑘𝑇(𝑥),𝑘=0,1,2,…,(50) is either bounded or lim_𝑘‖Ave_𝑘𝑇(𝑥)‖=∞. Moreover, if 𝑐=0, then the sequence Ave_𝑘𝐴(𝑥),𝑘=0,1,2,…,(51) is either (i) convergent to 0, or (ii) lim_𝑘‖Ave_𝑘𝐴(𝑥)‖=∞, or (iii) 𝐹≤‖Ave_𝑘𝐴𝑥‖≤𝐺 for sufficiently large 𝑘’s, where 0<𝐹≤𝐺<∞.

Recall that compact operators, or more generally, Riesz operators, have property (P). So the above theorem is valid for these operators.

Our last objective is to prove the following theorem. The result concerns Ave_𝑘𝑇(𝑥) in the case 𝑠=𝑡. It shows in particular that 𝐻 is identically 0 if and only if 𝑥=𝐸, and 𝑥 is the unique fixed point of 𝑇, that is, 𝑇𝑥=𝑥, so that in case (i) in the proof of Theorem 10 we actually have Ave_𝑘𝑇(𝑥)=𝐸 for all 𝑘, not just Ave_𝑘𝑇(𝑥)→𝐸 as 𝑘→∞.

Theorem 14. Let 𝜆 be a fixed complex number, 𝜆≠0,1. Let 𝑠,𝑡,𝑠=𝑡, and 𝐻(𝑘)=^𝑡−1_𝑗=0𝐶(𝑘,𝑗)𝜆^−𝑗𝐴^𝑗+𝐷(𝑘,𝑗,𝜆)(1−𝜆)²𝑁^𝑗𝑐(52) be defined as previously. Let 1≤𝑖≤𝑡−1. The degree of 𝐻(𝑘) is at most 𝑖−1 if and only if 𝑁^𝑖𝑥=1(1−𝜆)𝑁^𝑖𝑐+1(1−𝜆)²𝑁^𝑖+1𝑐+⋯+1(1−𝜆)^𝑡−𝑖𝑁^𝑡−1𝑐.(53)𝐻(𝑘) is identically 0 if and only if 𝑥=1(1−𝜆)𝑐+(1−1𝜆)²𝑁𝑐+⋯+1(1−𝜆)^𝑡𝑁^𝑡−1𝑐.(54)

The proof will follow from the following discussions.

Consider 𝐷(𝑘,𝑗,𝜆)−1. We have 𝐷(𝑘,𝑗,𝜆)−1=1(1−𝜆)^𝑗𝑗_𝑖=0(−1)^𝑖𝐶(𝑗,𝑖)𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑗+𝑖,2)−1𝜆^𝑖(55) and, for 𝑘>𝑗, 𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑗+𝑖,2)−1==𝑖𝑖(𝑘−−2−𝑗2+𝑘𝑖+)(2𝑘𝑗−−𝑗𝑖++𝑖1−1)𝑘+2𝑗−𝑘𝑖²+11−𝑗−𝑘𝑖⁻¹1−𝑗−𝑘𝑖+1⁻¹=𝑖𝑘−2+2𝑗−𝑘𝑖+11+𝑗−𝑘𝑖+(𝑗−𝑘²𝑖)²×1+𝑗−𝑖+1+⋯𝑘+(𝑗−𝑘𝑖²+1)²=𝑖_∞+⋯_𝑚₌₁−2(𝑗−𝑖)^𝑚−1−(𝑖−1)(𝑗−𝑖)^𝑚−1−(𝑗−𝑖+1)^𝑚−1𝑘^𝑚.(56) The coefficient of 1/𝑘^𝑚 in the sum above is obtained from the following calculations: −2^𝑚−1_𝑝=0(𝑗−𝑖)^𝑝(𝑗−𝑖+1)^{𝑚−1−𝑝}+(2𝑗−𝑖+1)^𝑚−2_𝑝=0(𝑗−𝑖)^𝑝(𝑗−𝑖+1)^{𝑚−2−𝑝}=−2(𝑗−𝑖)^𝑚−1−2(𝑗−𝑖+1)^𝑚−2_𝑝=0(𝑗−𝑖)^𝑝(𝑗−𝑖+1)^{𝑚−2−𝑝}+(2𝑗−𝑖+1)^𝑚−2_𝑝=0(𝑗−𝑖)^𝑝(𝑗−𝑖+1)^{𝑚−2−𝑝}=−2(𝑗−𝑖)^𝑚−1+(𝑖−1)^𝑚−2_𝑝=0(𝑗−𝑖)^𝑝(𝑗−𝑖+1)^{𝑚−2−𝑝}=−2(𝑗−𝑖)^𝑚−1−(𝑖−1)(𝑗−𝑖)^𝑚−1−(𝑗−𝑖+1)^𝑚−1.(57) Our first objective is to write (𝑗−𝑖)^𝑚−1 as a combination of 𝑥₀=1,𝑥₁=𝑖−1,𝑥₂=(𝑖−2)(𝑖−1),…,and𝑥_𝑚−1=(𝑖−𝑚+1)⋯(𝑖−1). The proof for the following item 1 can be found in [4]. (1)Let 𝑓(𝑥)=𝑎_𝑙𝑥^𝑙+⋯+𝑎₁𝑥+𝑎₀ be a polynomial of degree 𝑙. Let 𝑥₀=1,𝑥₁=𝑥−1,𝑥₂=(𝑥−2)(𝑥−1),…,𝑥_𝑙=(𝑥−𝑙)⋯(𝑥−1). Then 𝑓(𝑥)=𝑑_𝑙𝑥_𝑙+⋯+𝑑₀𝑥₀,(58) where 𝑑𝑑₀=𝑓(1),₁𝑑=𝑓(2)−𝑓(1),₂=1𝑑2![𝑓(3)−2𝑓(2)+𝑓(1)₃],=1𝑑⋮3![𝑓(4)−3𝑓(3)+3𝑓(2)−𝑓(1)_𝑙],=1𝑙!𝑓(𝑙+1)−𝐶(𝑙,1)𝑓(𝑙)+⋯+(−1)^𝑙=𝑎_𝑙𝑓(1).(59) Moreover, for any integer 𝑛 with 𝑛>𝑙, 𝑓(𝑛+1)−𝐶(𝑛,1)𝑓(𝑛)+𝐶(𝑛,2)𝑓(𝑛−1)+⋯+(−1)^𝑛𝑓(1)=0.(60)(2) Applying the above to the polynomial 𝑓(𝑖)=(𝑗−𝑖)^𝑚−1, we get (𝑗−𝑖)^𝑚−1=𝑐_𝑚−1𝑥_𝑚−1+⋯+𝑐₀𝑥₀,(61) where 𝑐𝑐₀₁=(𝑗−1)^𝑚−1,𝑐₂=(𝑗−2)^𝑚−1−(𝑗−1)^𝑚−1,=12!(𝑗−3)^𝑚−1𝑐₃−2(𝑗−2)^𝑚−1+(𝑗−1)^𝑚−1,=13!(𝑗−4)^𝑚−1𝑐_𝑚−1⋮−3(𝑗−3)^𝑚−1+3(𝑗−2)^𝑚−1−(𝑗−1)^𝑚−1,=1(𝑚−1)!(𝑗−𝑚)^𝑚−1=(−1)_𝑚−1−𝐶(𝑚−1,1)(𝑗−𝑚+1)^𝑚−1+⋯+(−1)^𝑚−1(𝑗−1)^𝑚−1.(62) Moreover, for any 𝑛>𝑚−1, (𝑗−𝑛−1)^𝑚−1−𝐶(𝑛,1)(𝑗−𝑛)^𝑚−1+𝐶(𝑛,2)(𝑗−𝑛+1)^𝑚−1+⋯+(−1)^𝑛(𝑗−1)^𝑚−1=0.(63)(3) Let 𝑥_𝑖,𝑐_𝑖,𝑖=0,…,𝑚−1, be defined as above. Then 𝑝(𝑖)=(𝑖−1)(𝑗−𝑖)^𝑚−1−(𝑗−𝑖+1)^𝑚−1=𝑐₁𝑥₁+2𝑐₂𝑥₂+⋯+(𝑚−1)𝑐_𝑚−1𝑥_𝑚−1,(64) so that −2(𝑗−𝑖)^𝑚−1−𝑝(𝑖)=−2𝑐₀𝑥₀−3𝑐₁𝑥₁−⋯−(𝑚+1)𝑐_𝑚−1𝑥_𝑚−1.(65)

Proof. Note that the degree of 𝑝(𝑖) is 𝑚−1 and the coefficient of 𝑖^𝑚−1 is (−1)^𝑚−1(𝑚−1). Applying item 1 to 𝑝(𝑖) and using the identity 𝐶(𝑘−1,𝑖)+𝐶(𝑘−1,𝑖+1)=𝐶(𝑘,𝑖+1), we get 𝑑𝑑₀=𝑝(1)=0,₁𝑑₂=𝑝(2)−𝑝(1)=(𝑗−2)^𝑚−1−(𝑗−1)^𝑚−1=𝑐₁,=1=(2𝑗!−[𝑝3()3)−_𝑚−12𝑝(2)+𝑝(1)]𝑑_𝑘⋮−(𝑗−2)^𝑚−1−(𝑗−2)^𝑚−1−(𝑗−1)^𝑚−1=2𝑐₂,=1𝑘!𝑝(𝑘+1)−𝐶(𝑘,1)𝑝(𝑘)+⋯+(−1)^𝑖+⋯+(−1)_𝑘−1𝐶(𝑘,𝑖)𝑝(𝑘−𝑖+1)=1𝑘𝑝(2)+(−1)^𝑘𝑝(1)𝑘!𝑘(𝑗−𝑘−1)^𝑚−1𝑑_𝑚−1⋮−𝐶(𝑘,1)(𝑗−𝑘)^𝑚−1+⋯+(−1)^𝑘(𝑗−1)^𝑚−1=𝑘𝑐_𝑘,=(−1)^𝑚−1(𝑚−1)=(𝑚−1)𝑐_𝑚−1.(66)

(4) Write 𝑖₀=𝑖,𝑖₁=𝑖(𝑖−1),…,𝑖_𝑝=𝑖(𝑖−1)⋯(𝑖−𝑝). Note that 𝑖_𝑝 is 𝑖 times 𝑥_𝑝 in item 2. This change is necessary because there is an 𝑖 that was factored out of the summation sign in our expansion of 𝐶(𝑘−𝑗,2)/𝐶(𝑘−𝑗+𝑖,2)−1.By item 3, the coefficient of 1/𝑘^𝑚 in the expansion of 𝐶(𝑘−𝑗,2)/𝐶(𝑘−𝑗+𝑖,2)−1 is −∑^{𝑚−1𝑝=0}(𝑝+2)𝑐_𝑝𝑖_𝑝. Therefore the coefficient 𝐿(𝑚,𝑗,𝜆) of 1/𝑘^𝑚 in the expansion of 𝐷(𝑘,𝑗,𝜆)−1 is given by −1(1−𝜆)^𝑗𝑗_𝑖=0(−1)^𝑖𝐶(𝑗,𝑖)^𝑚−1_𝑝=0(𝑝+2)𝑐_𝑝𝑖_𝑝𝜆^𝑖=−1(1−𝜆)^𝑗𝑚−1_𝑝=0(𝑝+2)𝑐_𝑝𝜆^𝑗𝑝+1_𝑖=0(−1)^𝑖𝐶(𝑗,𝑖)𝑖_𝑝𝜆^{𝑖−𝑝−1}=−1(1−𝜆)^𝑗𝑚−1_𝑝=0(𝑝+2)𝑐_𝑝𝜆^𝑝+1𝑑^𝑝+1𝑑𝜆^𝑝+1(1−𝜆)^𝑗=1(1−𝜆)^𝑗𝑚−1_𝑝=0(−1)^𝑝(𝑝+2)𝑐_𝑝𝜆^𝑝+1𝑗(𝑗−1)⋯(𝑗−𝑝)(1−𝜆)^{𝑗−𝑝−1}=^𝑚−1_𝑝=0(−1)^𝑝(𝑝+2)𝑐_𝑝𝑗(𝑗−1)⋯(𝑗−𝑝)𝜆1−𝜆^𝑝+1=^𝑚_𝑖=1(−1)^𝑖+1(𝑖+1)!𝑐_𝑖−1𝐶(𝑗,𝑖)𝜆1−𝜆^𝑖.(67) To emphasize that 𝑐_𝑖−1 depends on 𝑗,𝑚, we write 𝑐_𝑖 as 𝑃(𝑖,𝑗,𝑚), so 𝑃(𝑖,𝑗,𝑚)=1𝑖!(𝑗−𝑖−1)^𝑚−1𝐿(𝑚,𝑗,𝜆)=_𝑚−𝐶(𝑚−1,1)(𝑗−𝑖)^𝑚−1+⋯+(−1)^𝑚−1(𝑗−1)^𝑚−1,_𝑖=1(−1)^𝑖+1(𝑖+1)!𝑃(𝑖−1,𝑗,𝑚)𝐶(𝑗,𝑖)𝜆1−𝜆^𝑖.(68) We have that 𝑃(0,𝑗,𝑚)=𝑐₀=(𝑗−1)^𝑚−1,𝑃(𝑚−1,𝑗,𝑚)=𝑐_𝑚−1=(−1)^𝑚−1, and 𝑃(𝑖,𝑗,𝑚)=𝑐_𝑖=0 for 𝑖≥𝑚. We define 𝑃(𝑖,𝑗,𝑚)=0 for negative 𝑖. We have the following recursive relations, which will not be used in the sequel: 𝑃(𝑖,𝑗,𝑚)=(𝑗−𝑖−1)𝑃(𝑖,𝑗,𝑚−1)−𝑃(𝑖−1,𝑗,𝑚−1),𝑚≥2.(69) We also define 𝑀(𝑚,𝑗,𝑥)=^𝑚_𝑖=1(−1)^𝑖+1(𝑖+1)!𝑃(𝑖−1,𝑗,𝑚)𝐶(𝑗,𝑖)𝑥^𝑖,(70) so that 𝐿(𝑚,𝑗,𝜆)=𝑀𝑚,𝑗,𝜆1−𝜆.(71)(5) Then𝑀+⋯(𝑚,+𝑗,(𝑥−)1−)𝐻_1𝑚−1𝑗₁,…,𝑗_𝑚−1𝑀(𝑚−1,𝑗,𝑥)+⋯+(−1)^𝑖𝐻_𝑖𝑗₁,…,𝑗_𝑚−1𝑀(𝑚−𝑖,𝑗,𝑥)=𝑗₀𝐻_𝑚−1𝑗₁,…,𝑗_𝑚−1𝑀(1,𝑗,𝑥)𝑗₁⋯𝑗_𝑚−1(𝑚+1)𝑥^𝑚,(72) where 𝑗_𝑖=𝑗−𝑖,𝑖=0,…,𝑚−1, and 𝐻_𝑖(𝑗₁,…,𝑗_𝑚−1) is the unsigned coefficient of 𝑦^𝑖 in the expansion of the polynomial 𝑦−𝑗₁𝑦−𝑗₂⋯𝑦−𝑗_𝑚−1,(73) that is, 𝐻₁=𝑗₁+⋯+𝑗_𝑚−1,𝐻₂=∑_𝑖<𝑘𝑗_𝑖𝑗_𝑘,…, and 𝐻_𝑚−1=𝑗₁⋯𝑗_𝑚−1.In particular, we have that 𝑀𝑀((12,,𝑗𝑗,,𝑥𝑥))==2𝑗𝑗(𝑥𝑗,−1)3𝑥₂𝑀(3,𝑗,𝑥)=𝑗(𝑗−1)(𝑗−+2)(4𝑗𝑥−₃1)𝑀(1,𝑗,𝑥)=𝑗(𝑗−1)3𝑥²+2𝑥,=𝑗(𝑗−1)(𝑗−2)4𝑥₃+𝑗₁+𝑗₂𝑀(2,𝑗,𝑥)−𝑗₁𝑗₂𝑀(1,𝑗,𝑥)+(2𝑗−3)3𝑥²+2(𝑗−1)𝑥.(74)

Proof. For fixed 𝑚,𝑗, 𝑝(𝑥)=+𝑀⋯(𝑚+,𝑗(,−𝑥1))−𝐻_1𝑚−1𝑗₁,…,𝑗_𝑚−1𝑀(𝑚−1,𝑗,𝑥)+⋯+(−1)^𝑖𝐻_𝑖𝑗₁,…,𝑗_𝑚−1𝑀(𝑚−𝑖,𝑗,𝑥)𝐻_𝑚−1𝑗₁,…,𝑗_𝑚−1𝑀(1,𝑗,𝑥)(75) is a polynomial in 𝑥 with 𝑝(0)=0. Fix 1≤𝑘≤𝑚−1. We will show that the coefficient of 𝑥^𝑘 in 𝑝(𝑥) is 0. Now the coefficient is (−1)^𝑘+1(𝑘+1)!𝐶(𝑗,𝑘)𝑎_𝑘, where 𝑎_𝑘=+𝑃⋯(𝑘+−(1−,𝑗1,)𝑚_𝑖)−𝐻₁𝑃(𝑘−1,𝑗,𝑚−1)𝐻_𝑖𝑃(𝑘−1,𝑗,𝑚−𝑖)+⋯+(−1)^𝑚−1𝐻_𝑚−1𝑃(𝑘−1,𝑗,1).(76) Since 𝑦^𝑚−1−𝐻₁𝑦^𝑚−2+⋯+(−1)^𝑖𝐻_𝑖𝑦^{𝑚−𝑖−1}+⋯+(−1)^𝑚−1𝐻_𝑚−1=𝑦−𝑗₁⋯𝑦−𝑗_𝑚−1,(77) we have, for each 1≤𝑧≤𝑘, that 𝑗^𝑧𝑚−1−𝐻₁𝑗^𝑧𝑚−2+⋯+(−1)^𝑖𝐻_𝑖𝑗^{𝑧𝑚−𝑖−1}+⋯+(−1)^𝑚−1𝐻_𝑚−1=0.(78) From the definition of 𝑃(𝑖,𝑗,𝑚),𝑗^𝑧𝑚−1,𝑗^𝑧𝑚−2,… are the corresponding terms (with the same coefficient) in 𝑃(𝑘−1,𝑗,𝑚),𝑃(𝑘−1,𝑗,𝑚−1),…. It follows that 𝑎_𝑘=0.
For 𝑘=𝑚, we have that 𝑎_𝑘=𝑃(𝑚−1,𝑗,𝑚)=(−1)^𝑚−1 since 𝑃(𝑖,𝑗,𝑚)=0, for 𝑖≥𝑚. Thus the coefficient of 𝑥^𝑚 is (−1)^𝑚+1(−1)^𝑚−1(𝑚+1)!𝐶(𝑗,𝑚)=(𝑚+1)𝑗₀𝑗₁⋯𝑗_𝑚−1. This completes the proof.

(6) Let 𝑚₁,𝑚₂,… be variables. Let 𝑘,𝑗 be positive integers. Define 𝑅(𝑘,𝑗) to be the sum of 𝑚^𝑎1𝑚^𝑏2⋯𝑚^𝑐𝑘,(79) where 𝑎,𝑏,…,𝑐 are nonnegative integers such that 0≤𝑎,𝑏,…,𝑐≤𝑗 and 𝑎+𝑏+⋯+𝑐=𝑗. Define 𝑅(𝑘,0)=1. Define 𝑆(𝑘,𝑗) to be the sum of 𝑚^𝑎1𝑚^𝑏2⋯𝑚^𝑐𝑘,(80) where 𝑎,𝑏,…,𝑐 are nonnegative integers such that 0≤𝑎,𝑏,…,𝑐≤1 and 𝑎+𝑏+⋯+𝑐=𝑗. Define 𝑆(𝑘,0)=1 and 𝑆(0,0)=1. Then we have the following identity: ^𝑞_𝑖=0(−1)^𝑖𝑅(𝑝+𝑖,𝑞−𝑖)𝑆(𝑝+𝑖−1,𝑖)=0,(81) for any positive integers 𝑝,𝑞.

Proof. Consider a typical term 𝑡=𝑚^𝑎1𝑚^𝑏2⋯𝑚^{𝑘𝑝+𝑖} in the expansion of the summand (−1)^𝑖𝑅(𝑝+𝑖,𝑞−𝑖)𝑆(𝑝+𝑖−1,𝑖), sign disregarded. Let us write 𝑐(1)=𝑎,𝑐(2)=𝑏,…,𝑐(𝑝+𝑖)=𝑘. We may assume that 𝑐(𝑝+𝑖)≥1, otherwise the term belongs to a previous summand; this assumption also implies that 𝑡 does not belong to any previous (smaller 𝑖) summand. Let 𝐴={𝑗∶1≤𝑗≤𝑝+𝑖−1,𝑐(𝑗)≠0}.(82) Then the cardinality |𝐴| of 𝐴 must be greater than or equal to 𝑖 because of 𝑆(𝑝+𝑖−1,𝑖). Also the term 𝑡 appears in the expansion of the summand for exactly 𝐶(|𝐴|,𝑖) times. Next let us consider how many times 𝑡 appears in the expansion of the next summand (−1)^𝑖+1𝑅(𝑝+𝑖+1,𝑞−𝑖−1)𝑆(𝑝+𝑖,𝑖+1). 𝑡 can result from multiplying a term in 𝑅(𝑝+𝑖+1,𝑞−𝑖−1) with a term 𝑠 in 𝑆(𝑝+𝑖,𝑖+1). Denote the exponent of 𝑚_𝑝+𝑖 in s by 𝑏_𝑠(𝑝+𝑖), which is either 1 or 0.
Denote by 𝑆^∗ and 𝑅^∗ the set of terms in 𝑆(𝑝+𝑖,𝑖+1) and 𝑅(𝑝+𝑖+1,𝑞−𝑖−1), respectively. Consider the following two sets: 𝑈_𝑘=𝑠∶𝑠∈𝑆^∗,𝑏_𝑠(𝑝+𝑖)=𝑘,𝑟𝑠=𝑡forsome𝑟∈𝑅^∗,𝑘=0,1.(83) Note that each 𝑠∈𝑈_𝑘 corresponds to exactly one 𝑟 such that 𝑟𝑠=𝑡. Clearly 𝑈₁ has exactly 𝐶(|𝐴|,𝑖) elements, and these elements yield the same number of 𝑡’s which cancel out with the previous ones because of the sign change.
Each 𝑠∈𝑈₀ consists of 𝑖+1 factors from 𝑚₁,…,𝑚_{𝑝+𝑖−1}, so 𝑈₀ has 𝐶(|𝐴|,𝑖+1) elements which yield the same number of 𝑡’s.
If 𝐶(|𝐴|,𝑖+1)=0, that is, |𝐴|<𝑖+1, then 𝑈₀ is empty and we are done since no further 𝑡’s will result. If not, we consider the next summand (−1)^𝑖𝑅(𝑝+𝑖+2,𝑞−𝑖−2)𝑆(𝑝+𝑖+1,𝑖+2).
Denote by 𝑆^∗1 and 𝑅^∗1 the set of terms in 𝑆(𝑝+𝑖+1,𝑖+2) and 𝑅(𝑝+𝑖+2,𝑞−𝑖−2), respectively. Consider the following two sets: 𝑉_𝑘=𝑠∶𝑠∈𝑆^∗1,𝑏_𝑠(𝑝+𝑖)=𝑘,𝑟𝑠=𝑡forsome𝑟∈𝑅^∗1,𝑘=0,1.(84) Note that it is 𝑏_𝑠(𝑝+𝑖), not 𝑏_𝑠(𝑝+𝑖+1), in the above definition of 𝑉_𝑘. Also note that for each 𝑠∈𝑉_𝑘, 𝑏_𝑠(𝑝+𝑖+1) must be 0 or else 𝑟𝑠=𝑡 is impossible. Then it is clear that |𝑉₁|=𝐶(|𝐴|,𝑖+1) and |𝑉₀|=𝐶(|𝐴|,𝑖+2), as before. The 𝐶(|𝐴|,𝑖+1)𝑡’s resulting from 𝑉₁ cancel out with the previous same number of 𝑡’s. The same statement about 𝑈₀ above applies to 𝑉₀ and the process continues. This process will continue for at most 𝑝−1 times since |𝐴|≤𝑖+𝑝−1. This proves that the 𝑡’s occurring in ∑^𝑞𝑖=0(−1)^𝑖𝑅(𝑝+𝑖,𝑞−𝑖)𝑆(𝑝+𝑖−1,𝑖) are all canceled out.

(7)We now finish the proof of Theorem 14. Let 𝑖=𝑡−1.𝐻 has degree of at most 𝑖−1 if and only if the coefficient of 𝑘^𝑡−1 is 0. Since lim_𝑘→∞𝐷(𝑘,𝑗,𝜆)=1, this amounts to 𝐴_𝑡−1+1(1−𝜆)²𝑁^𝑡−1𝑐=0,(85) which is equivalent to 𝑁^𝑡−1𝑥=11−𝜆𝑁^𝑡−1𝑐.(86) So the assertion in Theorem 14 is true for 𝑖=𝑡−1. Suppose that the assertion is true for some 𝑖≥1; this implies that coefficients of 𝑘^𝑗 in 𝐻 are 0 for 𝑗≥𝑖. We will prove that it is also true for 𝑖−1. This will complete the proof by induction.By induction hypothesis, we have that 𝑁^𝑖𝑥=^𝑡−1_𝑗=𝑖1(1−𝜆)^{𝑗−𝑖+1}𝑁^𝑗𝑐.(87) By applying 𝑁 to both sides 𝑡−1−𝑖 times, recalling that 𝑁^𝑡𝑐=0, we get 𝑁^𝑙𝑥=^𝑡−1_𝑗=𝑙1(1−𝜆)^{𝑗−𝑙+1}𝑁^𝑗𝑐,(88) for all 𝑙=𝑖,…,𝑡−1. Substituting (88) into the formula for 𝐴_𝑙, we find 𝐴_𝑙=−^𝑡−1_𝑗=𝑙𝑗−𝑙+1(1−𝜆)^{𝑗−𝑙+2}𝑁^𝑗𝑐,(89) for all 𝑙=𝑖,…,𝑡−1. Using 𝐴_𝑗−1=(1/(1−𝜆))(𝐴_𝑗−𝑁^𝑗−1𝑥), we get 𝐴_𝑖−1=−^𝑡−1_𝑗=𝑖𝑗−𝑖+1(1−𝜆)^{𝑗−𝑖+3}𝑁^𝑗𝑐−11−𝜆𝑁^𝑖−1𝑥.(90) If we write 𝑃_𝑗 for the term 𝐶(𝑘,𝑗)𝜆^−𝑗𝐴^𝑗+𝐷(𝑘,𝑗,𝜆)(1−𝜆)²𝑁^𝑗𝑐(91) in 𝐻 and substitute (89) and (90), we find that, for 𝑖≤𝑙≤𝑡−1,  𝑃_𝑙=𝐶(𝑘,𝑙)𝜆^−𝑙−^𝑡−1_𝑝=𝑙+1𝑝−𝑙+1(1−𝜆)^{𝑝−𝑙+2}𝑁^𝑝𝑐+𝐷(𝑘,𝑙,𝜆)−1(1−𝜆)²𝑁^𝑙𝑐,(92) where for 𝑙=𝑡−1 the sum ∑^{𝑡−1𝑗=𝑙+1} is an empty sum and hence its value is 0, and 𝑃_𝑖−1=𝐶(𝑘,𝑖−1)𝜆^−𝑖+1−^𝑡−1_𝑗=𝑖𝑗−𝑖+1(1−𝜆)^{𝑗−𝑖+3}𝑁^𝑗𝑐−11−𝜆𝑁^𝑖−1𝑥+𝐷(𝑘,𝑖−1,𝜆)(1−𝜆)²𝑁^𝑖−1𝑐.(93) Since 𝐶(𝑘,𝑗) and 𝐶(𝑘,𝑗)𝐷(𝑘,𝑗,𝜆) are polynomials in 𝑘 of degree 𝑗, the terms 𝑃_𝑗,𝑗=0,…,𝑖−2 in 𝐻 contain only 𝑘 powers of power less than 𝑖−1. And since the coefficients of 𝑘 powers of power greater than 𝑖−1 are 0 by induction hypothesis, we see that the coefficient of 𝑘^𝑖−1 in 𝐻 is equal to the limit lim_𝑘→∞1𝑘^𝑖−1𝑃_𝑖−1+𝑃_𝑖+⋯+𝑃_𝑡−1.(94) Fix an 𝑚, 𝑖≤𝑚≤𝑡−1. Write 𝐴=𝐴(𝑘,𝑚) for the number 𝐶(𝑘,𝑚)𝜆^−𝑚𝑘^𝑚(1−𝜆)².(95) The coefficient of the vector 𝑁^𝑚𝑐 in (1/𝑘^𝑖−1)𝑃_𝑗 for each 𝑗,𝑖≤𝑗≤𝑚−1, is, by (92), −𝐶(𝑘,𝑗)𝜆^−𝑗𝑚−𝑗+1(1−𝜆)^{𝑚−𝑗+2}(96) which can be rewritten as 𝐴𝑘^𝑚−𝑗(𝑘−𝑗)⋯(𝑘−𝑚+1)𝑘^{𝑗−𝑖+1}−(𝑚−𝑗+1)𝑚(𝑚−1)⋯(𝑗+1)1−𝜆𝜆^𝑚−𝑗,(97) and which by item (5) is equal to 𝐴𝑘^𝑚−𝑗(𝑘−𝑗)⋯(𝑘−𝑚+1)𝑘+⋯^{𝑗−𝑖+1}+(−−1𝐿)(𝑚_𝑚−𝑗−𝑗,𝑚,𝜆)+𝑆(𝑚−𝑗−1,1)𝐿(𝑚−𝑗−1,𝑚,𝜆)=𝐴𝑘_𝑚−𝑗𝑆(𝑚−𝑗−1,𝑚−𝑗−1)𝐿(1,𝑚,𝜆)(𝑘−𝑗)⋯(𝑘−𝑚+1)𝑘^{𝑗−𝑖+1𝑚−𝑗}_𝑙=1(−1)^{𝑚−𝑗−𝑙+1}𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)𝐿(𝑙,𝑚,𝜆),(98) where 𝑆 is defined as in item (6), with 𝑚₁=𝑚−1,…. We also have that 𝑘^𝑚−𝑗(𝑘−𝑗)⋯(𝑘−𝑚+1)𝑘^{𝑗−𝑖+1}=𝑘^{𝑗−𝑖+1}1+𝑅(𝑚−𝑗,1)𝑘⁻¹+𝑅(𝑚−𝑗,2)𝑘⁻²+⋯+𝑅(𝑚−𝑗,𝑗−𝑖+1)𝑘^{−𝑗+𝑖−1}+𝑂1=_{𝑗−𝑖+1}𝑘_𝑧=0𝑅(𝑚−𝑗,𝑗−𝑖+1−𝑧)𝑘^𝑧+𝑂1𝑘,(99) where 𝑅 is defined as in item (6), with 𝑚₁=𝑚−1,….The coefficient of the vector 𝑁^𝑚𝑐 in (1/𝑘^𝑖−1)𝑃_𝑚 is 𝐴𝑘=𝐴^{𝑚−𝑖+1}(𝑘𝐷(_𝑚−𝑖𝑘,𝑚,𝜆)−1)=𝐴_𝑚−𝑖𝐿(1,𝑚,𝜆)+𝐿(2,𝑚,𝜆)𝑘⁻¹+⋯+𝐿(𝑚−𝑖+1,𝑚,𝜆)𝑘^−𝑚+𝑖+𝑂𝑘1_𝑧=0𝐿(𝑚−𝑖+1−𝑧,𝑚,𝜆)𝑘^𝑧+𝑂1𝑘.(100) Note that 𝑁^𝑚𝑐 does not appear in 𝑃_𝑙 for 𝑙>𝑚; so the coefficient of 𝑁^𝑚𝑐 in the sum 1𝑘^𝑖−1𝑃_𝑖+⋯+𝑃_𝑡−1(101) is the same as that in the sum 1𝑘^𝑖−1𝑃_𝑖+⋯+𝑃_𝑚.(102) Therefore the coefficient of 𝑁^𝑚𝑐 in the sum 1𝑘^𝑖−1𝑃_𝑖+⋯+𝑃_𝑡−1,(103) which we will call 𝑐_𝑚, is 𝐴 times 𝑌=^𝑚−1_{𝑗=𝑖𝑚−𝑗𝑙=1}(−1)^{𝑚−𝑗−𝑙+1}𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)𝐿(𝑙,𝑚,𝜆)^{𝑗−𝑖+1}_𝑧=0𝑅(𝑚−𝑗,𝑗−𝑖+1−𝑧)𝑘^𝑧+^𝑚−𝑖_𝑧=0𝐿(𝑚−𝑖+1−𝑧,𝑚,𝜆)𝑘^𝑧+𝑂1𝑘.(104) (Note that 𝐴=𝑂(1) so 𝐴𝑂(1/𝑘)=𝑂(1/𝑘).)Now we will show that the polynomial part of 𝑌 is a constant, that is, the coefficients of 𝑘^𝑧,𝑧=1,…,𝑚−𝑖, are zero. Fix 𝑧,1≤𝑧≤𝑚−𝑖. Since 𝑘^𝑧 occurs in the sum ∑^{𝑗−𝑖+1𝑧=0}𝑅(𝑚−𝑗,𝑗−𝑖+1−𝑧)𝑘^𝑧 only when 𝑗−𝑖+1≥𝑧, that is, 𝑗≥𝑧+𝑖−1, we see that the coefficient of 𝑘^𝑧 in 𝑌 is 𝐿(𝑚−𝑖+1−𝑧,𝑚,𝜆) plus ^𝑚−1_{𝑗=𝑧+𝑖−1𝑚−𝑗𝑙=1}=(−1)^{𝑚−𝑗−𝑙+1}𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)𝐿(𝑙,𝑚,𝜆)𝑅(𝑚−𝑗,𝑗−𝑖+1−𝑧)_{𝑚−𝑖−𝑧+1}_𝑙=1𝐿(𝑙,𝑚,𝜆)^𝑚−𝑙_𝑗=_{𝑧+𝑖−1}=(−1)^{𝑚−𝑗−𝑙+1}𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)𝑅(𝑚−𝑗,𝑗−𝑖+1−𝑧)_{𝑚−𝑖−𝑧+1}_𝑙=1𝐿(𝑙,𝑚,𝜆)^{𝑚−𝑖−𝑧+1−𝑙}_𝑛=0=(−1)^𝑛+1𝑅(𝑙+𝑛,𝑚−𝑖−𝑧+1−𝑙−𝑛)𝑆(𝑙+𝑛−1,𝑛)_{𝑚−𝑖−𝑧}_𝑙=1=−𝐿(𝑚𝐿−(𝑙,𝑖𝑚−,𝜆𝑧)+⋅01,+𝑚𝐿,𝜆()𝑚,−𝑖−𝑧+1,𝑚,𝜆)⋅(−1)(105) where we have used item (6) for the 0. Hence the coefficient of 𝑘^𝑧 is 0.The constant term in 𝑌 is given by the coefficient of 𝑘⁰, which is 𝐿(𝑚−𝑖+1,𝑚,𝜆) plus ^𝑚−1_{𝑗=𝑖𝑚−𝑗𝑙=1}=(−1)^{𝑚−𝑗−𝑙+1}𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)𝐿(𝑙,𝑚,𝜆)𝑅(𝑚−𝑗,𝑗−𝑖+1)_𝑚−𝑖_𝑙=1𝐿(𝑙,𝑚,𝜆)^𝑚−𝑙_𝑗=𝑖=(−1)^{𝑚−𝑗−𝑙+1}𝑅(𝑚−𝑗,𝑗−𝑖+1)𝑆(𝑚−𝑗−1,𝑚−𝑗−𝑙)_𝑚−𝑖_𝑙=1𝐿(𝑙,𝑚,𝜆)^{𝑚−𝑖−𝑙}_𝑛=0=(−1)^𝑛+1𝑅(𝑙+𝑛,𝑚−𝑙−𝑖+1−𝑛)𝑆(𝑙+𝑛−1,𝑛)_𝑚−𝑖_𝑙=1=𝐿(𝑙,𝑚,𝜆)(−1)^{𝑚−𝑖−𝑙−1}𝑅(𝑚−𝑖+1,0)𝑆(𝑚−𝑖,𝑚−𝑖−𝑙+1)_𝑚−𝑖_𝑙=1(−1)^{𝑚−𝑖−𝑙−1}𝑆(𝑚−𝑖,𝑚−𝑖−𝑙+1)𝐿(𝑙,𝑚,𝜆),(106) where we have used item (6) with 𝑝=𝑙 and 𝑞=𝑚−𝑖−𝑙+1. Adding the term 𝐿(𝑚−𝑖+1,𝑚,𝜆) back in we conclude that the said coefficient 𝑐_𝑚 is 𝐴[+𝐿⋯(𝑚+−(𝑖−+1)1,𝑚_𝑚−𝑖,𝜆)−𝑆(𝑚−𝑖,1)𝐿(𝑚−𝑖,𝑚,𝜆)𝑆(𝑚−𝑖,𝑚−𝑖)𝐿(1,𝑚,𝜆)+𝑂(1/𝑘)(107) which by item (5) is 𝐴𝑚!(𝑖−1)!(𝑚−𝑖+2)1−𝜆𝜆^{𝑚−𝑖+1}+𝑂1𝑘.(108) On the other hand, by (93), the coefficient of 𝑁^𝑚𝑐 in 𝑃_𝑖−1/𝑘^𝑖−1 can be written as −𝐴𝑚!(𝑖−1)!(𝑚−𝑖+1)1−𝜆𝜆^{𝑚−𝑖+1}+𝑂1𝑘.(109) Therefore the coefficient of 𝑁^𝑚𝑐,𝑖≤𝑚≤𝑡−1, in (1/𝑘^𝑖−1)(𝑃_𝑖−1+⋯+𝑃_𝑡−1) is 𝐴𝑚!(𝑖−1)!1−𝜆𝜆^{𝑚−𝑖+1}+𝑂1𝑘,(110) which approaches 𝜆^−𝑖+1(𝑖−1)!(1−𝜆)^{𝑚−𝑖+3}(111) as 𝑘→∞ since 𝐴=𝐴(𝑘,𝑚)→𝜆^−𝑚/(𝑚!(1−𝜆)²) as 𝑘→∞. Since 𝐷(𝑘,𝑖−1,𝜆)→1, 𝐶(𝑘,𝑖−1)/𝑘^𝑖−1→1/(𝑖−1)!, and 𝑁^𝑖−1 only appears in 𝑃_𝑖−1, we see that the above sentence is also valid for 𝑚=𝑖−1. It then follows from (93) and (94) that the coefficient of 𝑘^𝑖−1 in 𝐻 is 𝜆^−𝑖+1(𝑖−1)!^𝑡−1_{𝑗=𝑖−1}1(1−𝜆)^{𝑗−𝑖+3}𝑁^𝑗𝑐−11−𝜆𝑁^𝑖−1𝑥.(112) Therefore 𝐻 is of degree 𝑖−2 or less if and only if 𝑁^𝑖−1𝑥=^𝑡−1_𝑗_=𝑖−11(1−𝜆)^{𝑗−𝑖+2}𝑁^𝑗𝑐.(113) The last part of the theorem corresponds to 𝑖=1 in the above equation.This completes the proof of Theorem 14.

Appendix

We prove by induction that 𝑇_𝑗=𝑘(1−𝜆)^𝑗+1−𝑗+1(1−𝜆)^𝑗+2+𝐶(𝑘,𝑗)𝜆^𝑘−𝑗(1−𝜆)²𝐷(𝑘,𝑗,𝜆),(A.1) for 𝑗=0,…,𝑡−1, where 𝐷(𝑘,𝑗,𝜆)=1(1−𝜆)^𝑗𝐵𝐵₀(𝑘,𝑗)𝜆^𝑗+𝐵₁(𝑘,𝑗)𝜆^𝑗−1+⋯+𝐵_𝑗−1(𝑘,𝑗)𝜆+1_𝑖,(𝑘,𝑗)=(−1)^𝑗−𝑖𝐶(𝑗,𝑖)𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑖,2),𝑖=0,…,𝑗.(A.2) As we noted in the paper proper this is true for 𝑗=0 since 𝐷(𝑘,0,𝜆)=1. By item (1) below 𝑇_𝑗 can be rewritten as 𝑘(1−𝜆)^𝑗+1−𝑗+1(1−𝜆)^𝑗+2+𝜆^𝑘−𝑗𝐴(1−𝜆)^𝑗+2,(A.3) where 𝐴=^𝑗_𝑖=0(−1)^𝑗−𝑖𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^𝑗−𝑖.(A.4) Taking the derivative of 𝑇_𝑗 with respect to 𝜆, we find 𝑑𝑑𝜆𝑇^𝑗=𝑘(𝑗+1)(1−𝜆)^𝑗+2−(𝑗+1)(𝑗+2)(1−𝜆)^𝑗+3+𝜆^{𝑘−𝑗−1}[(1−𝜆)(𝑘−𝑗)+(𝑗+2)𝜆]𝐴+(1−𝜆)𝜆𝐴^(1−𝜆)^𝑗+3.(A.5) Then item (3) below proves that 𝑑𝑑𝜆𝑇^𝑗=𝑘(𝑗+1)(1−𝜆)^𝑗+2−(𝑗+1)(𝑗+2)(1−𝜆)^𝑗+3+(𝑗+1)𝐶(𝑘,𝑗+1)𝜆^{𝑘−𝑗−1}(1−𝜆)²𝐷(𝑘,𝑗+1,𝜆).(A.6) This completes the induction since, as stated in the paper proper, 𝑇_𝑗+1=1𝑗+1𝑑𝑑𝜆𝑇^𝑗.(A.7)(1)We have that𝐶(𝑘,𝑗)𝐶(𝑗,𝑖)𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑖,2)=𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖).(A.8)

Proof. We have that𝐶(𝑘,𝑗)𝐶(𝑗,𝑖)𝐶(𝑘−𝑗,2)𝐶(𝑘−𝑖,2)==𝑗!(𝑘−𝑗𝑘𝑘)!!!𝑗((𝑘!𝑗(−−𝑘−𝑖𝑖)−!𝑗𝑖)2!!)((𝑘!𝑘−−𝑗𝑖−−22))!!22!!(𝑘−𝑖)!=𝐶𝑖!((𝑘𝑘,−𝑖)𝑖𝐶)!((𝑘𝑘−−𝑖𝑗−−22,)𝑗!−(𝑗𝑖)−.𝑖)!(A.9)

(2) For 𝑘≥𝑖+2,𝑗≥𝑖−1, (𝑘−=𝑖+(𝑗1+)𝐶1)(𝐶𝑘,(𝑖𝑘−,𝑖)1𝐶)𝐶(𝑘(𝑘−−𝑖−𝑖−2,1𝑗,𝑗−−𝑖+𝑖+1)1.)+(𝑘−𝑖−𝑗−2)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)(A.10)

Proof. We have that(𝑘−=𝑖+(𝑘1−)𝐶𝑖+(𝑘1,)𝑖−1)𝐶(𝑘−𝑖−𝑘1!,(𝑗𝑘−−𝑖𝑖+−11))+!(𝑘−𝑖−𝑗−2)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)+(𝑘−𝑖−𝑗(−𝑘−2)𝑖+1)!(𝑖𝑘−!(1𝑘)−!(𝑖𝑘−−2𝑗)−!2)!(𝑗−𝑖+1)!=𝑖𝑘!(𝑘−(𝑘𝑖−−1𝑖))!!𝑖!(𝑘−𝑗−2)!(𝑗−𝑖)!=(𝑘𝑘!(−𝑘𝑖−)!𝑖𝑖−!(𝑘2)−−!𝑖!𝑖(+𝑘1−)𝑘𝑗!−(𝑘2)−!(𝑖𝑗−−2𝑖)+!1)!=𝑘!(𝑘−𝑖−(2𝑘)!−𝑘𝑖)𝑗!+𝑖!𝑘(𝑘−−𝑗₂𝑗−2)!(𝑗−𝑖+1)!−3𝑗−2=𝑘(!𝑘(𝑘−−𝑖)𝑖!−𝑖!(2𝑘)!−(𝑗𝑗+−12))(!𝑘(𝑗−−𝑗𝑖−+21))!=(𝑘−𝑖)𝑘!!𝑖(!𝑘(𝑘−−𝑖𝑗−−2)2!)(!𝑗(+𝑗−1)𝑖+1)!=((𝑗𝑘+−1𝑖))𝐶!𝑖!(𝑘(𝑘,𝑖−)𝐶𝑗(−𝑘3−)!𝑖(−𝑗−2,𝑖𝑗+−1𝑖)+!1).(A.11)

(3) Let 𝐴=^𝑗_𝑖=0(−1)^𝑗−𝑖𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^𝑗−𝑖(A.12) and let 𝐴^=(𝑑/𝑑𝜆)A. Then =(𝑗+1[)(1−_𝑗+1𝜆)(𝑘−𝑗)+(𝑗+2)𝜆]𝐴+(1−𝜆)𝜆𝐴^(A.13)_𝑖=0(−1)^{𝑗+1−𝑖}𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗+1−𝑖)𝜆^{𝑗+1−𝑖}.(A.14)

Proof. We have that 𝐴^=^𝑗−1_𝑖=0(−1)^𝑗−𝑖(𝑗−𝑖)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^{𝑗−𝑖−1}.(A.15) It is easy to see that (1−𝜆)(𝑘−𝑗)𝐴+𝜆𝐴^=(1−𝜆)^𝑗_𝑖=0−𝜆_𝑗(−1)^𝑗−𝑖(𝑘−𝑖)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^𝑗−𝑖,_𝑖=0=(−1)^𝑗−𝑖(𝑘−𝑖)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^𝑗−𝑖+(𝑗+2)𝜆𝐴_𝑗_𝑖=0(−1)^{𝑗−𝑖+1}(𝑘−𝑖−𝑗−2)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^{𝑗−𝑖+1}.(A.16) Thus (A.13) is the sum of 𝐵=^𝑗_𝑖=0𝐶=_𝑗(−1)^𝑗−𝑖(𝑘−𝑖)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^𝑗−𝑖,_𝑖=0(−1)^{𝑗−𝑖+1}(𝑘−𝑖−𝑗−2)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)𝜆^{𝑗−𝑖+1}.(A.17) The coefficient of 𝜆^{𝑗+1−𝑖} in 𝐵+𝐶 is (−1)^{𝑗−𝑖+1}[(𝑘−𝑖+1)𝐶(𝑘,𝑖−1)𝐶(𝑘−𝑖−1,𝑗−𝑖+1)+(𝑘−𝑖−𝑗−2)𝐶(𝑘,𝑖)𝐶(𝑘−𝑖−2,𝑗−𝑖)],(A.18) valid even when 𝑖=0 or 𝑗+1 since by convention 𝐶(𝑛,𝑥)=0 for 𝑥<0.
This is equal to the coefficient of 𝜆^{𝑗+1−𝑖} in (A.14) by item (2) above.

Acknowledgment

The author wishes to thank Professors Sing-Cheong Ong and Timothy Murphy for their discussions and Professors Roger Horn and Charles Johnson for their communications.