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Math-661-A4 · revision r1 · on Math-661

gpt pro 5.2 attempt on Math-661, version 2

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Contents
  1. Problem 66 --- Round 2 (continuation & gap-closure mode)
  2. 1) ROUND-2 OBJECTIVE
  3. 2) ROUND-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL (EXACTLY ONE)
  9. 8) COMPLETION ESTIMATE (MANDATORY)
  10. 9) REFERENCES

Problem 66 --- Round 2 (continuation & gap-closure mode)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction direction. Starting from the Round 1 status UNRESOLVED, I pursue the "disproof/rigidity" track by extracting a strictly stronger necessary consequence of the existence of a nonzero limit \[ \frac{(1_A\ast 1_A)(n)}{\log n}\to L\in(0,\infty). \] Round 1 only obtained the order of growth $A(x)=\Theta(\sqrt{x\log x})$. In this round I close that gap: assuming the limit exists, I prove that the counting function in fact has a full asymptotic with an explicit constant depending on $L$: \[ A(x)\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}. \] This is a rigid constraint on any potential example and is a substantial strengthening of Round 1.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 items (as black boxes):
  • The formal definitions: $a_n:=1_A(n)$ and $r(n):=(1_A\ast 1_A)(n)=\sum_{k=1}^{n-1}a_ka_{n-k}$ counts ordered representations.
  • Corollary 66.1 only as context (I strengthen it).
No Round 1 probabilistic computations are used in the new proof.

3) NEW INSIGHT / TOOL (ROUND-2)

New tool: Abelian--Tauberian passage through generating functions/Laplace transforms.

Let \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}=\Big(\sum_{m\ge 1} a_m e^{-tm}\Big)^2\qquad(t>0), \] which is the Laplace transform of $r(n)$ and equals the square of the Laplace transform of $a_m$.

From the assumed pointwise asymptotic $r(n)\sim L\log n$ I first derive an Abelian asymptotic for $F(t)$ as $t\to 0^+$, then take a square root, and finally apply Karamata's Tauberian theorem (Laplace transform form) to recover an explicit asymptotic for the counting function $A(x)=\sum_{m\le x}a_m$.

4) ATTACK PLAN (ROUND-2)

Round 1 gap addressed: Corollary 66.1 only yields two-sided bounds $c\sqrt{x\log x}\le A(x)\le C\sqrt{x\log x}$.

Goal in this round: upgrade this to an explicit asymptotic with constant.

Plan:

  1. Prove an Abelian lemma: from $r(n)\sim L\log n$, show \[ F(t)=\sum_{n\ge 2} r(n)e^{-tn}\sim \frac{L}{t}\log\frac1t\qquad(t\to 0^+). \]
  2. Take square roots to obtain the Laplace transform asymptotic for $a_n$: \[ G(t):=\sum_{m\ge 1} a_m e^{-tm}=\sqrt{F(t)}\sim \sqrt{L}\,t^{-1/2}\,\sqrt{\log\tfrac1t}. \]
  3. Apply Karamata's Tauberian theorem to the nondecreasing function $A(x)$ to conclude \[ A(x)\sim \frac{\sqrt{L}}{\Gamma(3/2)}\,x^{1/2}\,\sqrt{\log x}=2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}. \]

5) WORK (ROUND-2)

Throughout, assume \begin{equation} \label{wsref-0-65-71-3a-6c-69-6d-69-74-2d-61-73-73-75-6d-70-74-69-6f-6e} \frac{r(n)}{\log n}\to L\in(0,\infty),\qquad r(n)=(1_A\ast 1_A)(n). \end{equation}

Lemma 66.3 (model Laplace asymptotic for $\sum \log n\,e^{-tn}$). Define \[ S(t):=\sum_{n\ge 2} (\log n)e^{-tn}\qquad(t>0). \] Then as $t\to 0^+$, \[ S(t)=\frac{1}{t}\log\frac{1}{t}+O\Big(\frac{1}{t}\Big). \] In particular $S(t)\sim \frac{1}{t}\log\frac{1}{t}$.

Proof: Let \[ I(t):=\int_{1}^{\infty} (\log x)e^{-tx}\,dx. \] Substitute $u=tx$ to get \[ I(t)=\frac{1}{t}\int_{t}^{\infty} \big(\log u-\log t\big)e^{-u}\,du =\frac{1}{t}\Big(\log\frac1t\int_{t}^{\infty}e^{-u}\,du+\int_{t}^{\infty}(\log u)e^{-u}\,du\Big). \] As $t\to 0^+$, $\int_{t}^{\infty}e^{-u}\,du\to 1$ and $\int_{t}^{\infty}(\log u)e^{-u}\,du\to\int_{0}^{\infty}(\log u)e^{-u}\,du$, a finite constant. Hence \[ I(t)=\frac{1}{t}\log\frac1t+O\Big(\frac{1}{t}\Big). \]

It remains to compare $S(t)$ and $I(t)$. The function $g_t(x):=(\log x)e^{-tx}$ is smooth, nonnegative, and tends to $0$ as $x\to\infty$. A standard bounded-variation estimate (e.g. summation by parts) gives \[ \Big|\sum_{n\ge 2} g_t(n)-\int_{1}^{\infty} g_t(x)\,dx\Big|\le g_t(1)+\int_{1}^{\infty}|g_t'(x)|\,dx. \] Compute $g_t'(x)=(1/x-t\log x)e^{-tx}$, so \[ \int_{1}^{\infty}|g_t'(x)|\,dx\le \int_{1}^{\infty}\Big(\frac{1}{x}+t\log x\Big)e^{-tx}\,dx. \] After the substitution $u=tx$, both integrals on the right are $O(\log\tfrac1t)$ as $t\to 0^+$, hence the discretization error is $O(\log\tfrac1t)$. Therefore \[ S(t)=I(t)+O\Big(\log\frac1t\Big)=\frac{1}{t}\log\frac1t+O\Big(\frac{1}{t}\Big), \] as claimed. ∎

Lemma 66.4 (Abelian step: Laplace asymptotic for $r(n)$). Let \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}\qquad(t>0). \] Under (eq:limit-assumption), as $t\to 0^+$, \[ F(t)\sim \frac{L}{t}\log\frac1t. \]

Proof: Fix $\varepsilon\in(0,1)$. By (eq:limit-assumption) there exists $N_0(\varepsilon)$ such that for all $n\ge N_0$, \[ (L-\varepsilon)\log n\le r(n)\le (L+\varepsilon)\log n. \] Split \[ F(t)=\sum_{2\le n< N_0} r(n)e^{-tn}+\sum_{n\ge N_0} r(n)e^{-tn}. \] The finite initial sum is $O(1)$ uniformly in $t$. For the tail we get \[ (L-\varepsilon)\sum_{n\ge N_0}(\log n)e^{-tn}\le \sum_{n\ge N_0} r(n)e^{-tn}\le (L+\varepsilon)\sum_{n\ge N_0}(\log n)e^{-tn}. \] Adding back the $O(1)$ initial segment and using Lemma 66.3 (noting that removing finitely many terms does not change the $\sim$-asymptotic) gives \[ (L-\varepsilon)\Big(\frac{1}{t}\log\frac1t+O\Big(\frac{1}{t}\Big)\Big)+O(1) \le F(t)\le (L+\varepsilon)\Big(\frac{1}{t}\log\frac1t+O\Big(\frac{1}{t}\Big)\Big)+O(1). \] Divide by $(1/t)\log(1/t)$ and let $t\to 0^+$. The $O(1/t)$ term is smaller than $(1/t)\log(1/t)$ by a factor $1/\log(1/t)\to 0$, and $O(1)$ is negligible. Hence \[ L-\varepsilon\le \liminf_{t\to 0^+}\frac{F(t)}{(1/t)\log(1/t)}\le \limsup_{t\to 0^+}\frac{F(t)}{(1/t)\log(1/t)}\le L+\varepsilon. \] Since $\varepsilon$ is arbitrary, $F(t)/((1/t)\log(1/t))\to L$. ∎

Lemma 66.5 (square-root transfer to the Laplace transform of $A$). Define \[ G(t):=\sum_{m\ge 1} a_m e^{-tm}=\sum_{m\in A} e^{-tm} \qquad(t>0). \] Then $F(t)=G(t)^2$ and, under (eq:limit-assumption), \[ G(t)\sim \sqrt{L}\,t^{-1/2}\,\sqrt{\log\tfrac1t}\qquad(t\to 0^+). \]

Proof: Since $r(n)=\sum_{k=1}^{n-1}a_ka_{n-k}$, we have the identity of absolutely convergent series for $t>0$: \[ \sum_{n\ge 2} r(n)e^{-tn}=\sum_{n\ge 2}\Big(\sum_{k=1}^{n-1}a_ka_{n-k}\Big)e^{-tn} =\Big(\sum_{m\ge 1}a_m e^{-tm}\Big)^2, \] i.e. $F(t)=G(t)^2$. For $t>0$, $G(t)>0$ (as soon as $A\neq\emptyset$) and $F(t)>0$, so taking square roots is legitimate. Lemma 66.4 yields \[ G(t)=\sqrt{F(t)}\sim \sqrt{\frac{L}{t}\log\frac1t}=\sqrt{L}\,t^{-1/2}\,\sqrt{\log\tfrac1t}. \] ∎

Proposition 66.6 (Tauberian sharpening of Round 1 Corollary 66.1). Assume (eq:limit-assumption). Then the counting function satisfies the full asymptotic \[ A(x):=|A\cap[1,x]|\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}\qquad(x\to\infty). \] Equivalently, \[ \lim_{x\to\infty}\frac{A(x)^2}{x\log x}=\frac{4L}{\pi}. \]

Proof: We invoke Karamata's Tauberian theorem in Laplace-transform form (for monotone functions):

Karamata (Laplace form, special case). Let $A(x)$ be nondecreasing, right-continuous, with $A(x)=0$ for $x<0$, and let \[ \mathcal L_A(t):=\int_{[0,\infty)} e^{-tx}\,dA(x)\qquad(t>0) \] be its Laplace--Stieltjes transform. Suppose that for some $\rho>0$ and slowly varying $\ell$, \[ \mathcal L_A(t)\sim C\,t^{-\rho}\,\ell(1/t)\qquad(t\to 0^+). \] Then \[ A(x)\sim \frac{C}{\Gamma(\rho+1)}\,x^{\rho}\,\ell(x)\qquad(x\to\infty). \] (See, e.g., Bingham--Goldie--Teugels, Regular Variation, or Feller, vol. II, Tauberian theorems chapter.)

In our setting, $A(x)=\sum_{m\le x}a_m$ is nondecreasing and \[ G(t)=\sum_{m\ge 1}a_m e^{-tm}=\int_{[0,\infty)} e^{-tx}\,dA(x), \] so $G(t)$ is exactly the Laplace--Stieltjes transform of $A$. Lemma 66.5 gives \[ G(t)\sim \sqrt{L}\,t^{-1/2}\,\ell(1/t),\qquad \ell(u):=\sqrt{\log u}, \] with $\rho=1/2$ and slowly varying $\ell$. Applying Karamata yields \[ A(x)\sim \frac{\sqrt{L}}{\Gamma(3/2)}\,x^{1/2}\,\sqrt{\log x}. \] Since $\Gamma(3/2)=\tfrac{\sqrt\pi}{2}$, this becomes \[ A(x)\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}. \] Squaring and dividing by $x\log x$ gives the equivalent limit $A(x)^2/(x\log x)\to 4L/\pi$. ∎

6) ADVERSARIAL VERIFICATION

  • Ordered vs unordered representations. The identity $F(t)=G(t)^2$ uses ordered convolution. If one used unordered representations, $F(t)$ would instead correspond to $(G(t)^2+G(2t))/2$; constants would change. Round 1 and this round consistently use ordered pairs.
  • Asymptotic of $\sum (\log n)e^{-tn}$. Lemma 66.3 only needs a leading term. The discretization error is at most $O(\log(1/t))$, far smaller than the main term $(1/t)\log(1/t)$, so the $\sim$ claim is stable.
  • Taking square roots. For $t>0$, $G(t)\ge 0$ and $F(t)=G(t)^2$, so $G(t)=\sqrt{F(t)}$ with no sign ambiguity.
  • Karamata hypotheses. The function $A(x)$ is nondecreasing and right-continuous; $\ell(u)=\sqrt{\log u}$ is slowly varying. Also $\rho=1/2>0$, so the theorem applies.
  • Boundary cases. If $L=0$ the same method gives $G(t)=o(t^{-1/2}\sqrt{\log(1/t)})$, hence $A(x)=o(\sqrt{x\log x})$, consistent with Round 1. If $L=\infty$ the method is inapplicable; the statement of the problem asks $L$ finite.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 closes a concrete Round 1 gap: under the existence of a finite nonzero limit $r(n)/\log n\to L$, the counting function must satisfy the precise asymptotic \[ A(x)\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}, \] which is substantially stronger than $A(x)=\Theta(\sqrt{x\log x})$.

The core existence/nonexistence question remains open: I did not derive a contradiction nor construct such an $A$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

  • N. H. Bingham, C. M. Goldie, J. L. Teugels, Regular Variation, Cambridge Univ. Press (for Karamata Tauberian theorem, Laplace-transform form).
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. II (Tauberian theorems chapter).