Nonconstant representations

---

Theorem

Suppose $A,B$ are infinite positive integer sets with $a_n/b_n\to1$. If their representation function $r=r_{A,B}$ is bounded, it cannot equal a fixed positive integer $k$ on a set of natural density one.

In particular, under the hypotheses of Problem 1145, the representation function is not eventually constant. This is a restricted result: it does not exclude a bounded function with several values occurring at positive densities.

The proof below is author-recorded; no independent verification record exists. It uses only elementary Laplace-transform arguments, weak convergence of finite measures, and the Hilbert-space criterion that weak convergence plus convergence of norms implies strong convergence.

Laplace asymptotics

Assume for contradiction that $r(n)=k$ outside a density-zero set. Write

$$F(t)=\sum_{a\in A}e^{-ta},\quad G(t)=\sum_{b\in B}e^{-tb},\quad H(t)=F(t)G(t).$$

Boundedness and density-one equality imply, by summation by parts,

$$H(t)\sim k/t,\qquad \sum_n r(n)^2e^{-tn}\sim k^2/t\quad(t\downarrow0). \tag{1}$$

Termwise balance gives, for any fixed $0<\varepsilon<1$,

$$G((1+\varepsilon)t)-C_\varepsilon\le F(t) \le G((1-\varepsilon)t)+C_\varepsilon.$$

Using monotonicity of $F$, these inequalities give

$$H((1+\varepsilon)t)\le F(t)(F(t)+C_\varepsilon),\qquad F(t)(F(t)-C_\varepsilon)\le H((1-\varepsilon)t).$$

Equation (1), followed by $\varepsilon\downarrow0$, proves

$$F(t)\sim G(t)\sim\sqrt{k}\,t^{-1/2}. \tag{2}$$

For $\lambda>0$, define finite measures

$$\mu_{N,\lambda}=N^{-1/2}\sum_{a\in A}e^{-\lambda a/N}\delta_{a/N}, \quad \nu_{N,\lambda}=N^{-1/2}\sum_{b\in B}e^{-\lambda b/N}\delta_{b/N}.$$

They both converge weakly to $p_\lambda(x)\,dx$, where

$$p_\lambda(x)=\sqrt{k/\pi}\,x^{-1/2}e^{-\lambda x}\mathbf1_{x>0}. \tag{3}$$

Here is a direct justification, avoiding a separate Tauberian invocation. Their masses converge by (2), and their mass above $R$ is at most $e^{-\lambda R/2}N^{-1/2}F(\lambda/(2N))$, or its $G$ analog. They are therefore tight. Every subsequential limit has Laplace transform $\sqrt{k}(\lambda+s)^{-1/2}$ for $s\ge0$, again by (2). Uniqueness of Laplace transforms identifies it with (3).

A weighted difference energy

For integer $t$, put

$$D_{N,\lambda}(t)=\sum_{a-b=t}e^{-\lambda(a+b)/N}, \qquad f_{N,\lambda}(u)=D_{N,\lambda}(\lfloor Nu\rfloor).$$

Rearranging nonnegative sums gives the exact energy identity

$$\int_{\mathbb R}f_{N,\lambda}(u)^2\,du =\frac1N\sum_n r(n)^2e^{-2\lambda n/N} \longrightarrow\frac{k^2}{2\lambda}. \tag{4}$$

Indeed $a-b=a'-b'$ is equivalent to $a+b'=a'+b=n$, and the weight of either quadruple is $e^{-2\lambda n/N}$.

The measure $f_{N,\lambda}(u)\,du$ is the convolution of $\mu_{N,\lambda}*\widetilde\nu_{N,\lambda}$ with uniform probability on $[0,1/N)$; tilde denotes reflection. Hence it converges weakly to $h_\lambda(u)\,du$, with

$$h_\lambda=p_\lambda*\widetilde p_\lambda.$$

There is a continuous counterpart of the same energy identity:

$$\|p*\widetilde p\|_2^2=\|p*p\|_2^2 \tag{5}$$

for every nonnegative integrable $p$, permitting infinite values. For completeness, expand the left side as $\int p(x)p(x-u)p(y)p(y-u)\,dx\,dy\,du$, and change variable $s=x+y-u$. Tonelli's theorem yields the right side. The beta integral gives

$$(p_\lambda*p_\lambda)(s)=ke^{-\lambda s}\mathbf1_{s>0}.$$

Thus (5) says $\|h_\lambda\|_2^2=k^2/(2\lambda)$. The uniform $L^2$ bound (4) and weak measure convergence imply weak $L^2$ convergence, by first testing against compactly supported continuous functions and then using their density in $L^2$. Equality of the limiting norms now gives

$$f_{N,\lambda}\longrightarrow h_\lambda\quad\hbox{strongly in }L^2. \tag{6}$$

The contradiction from individual atoms

For every integer $t$, positivity gives

$$D_{N,1}(t)^2\ge D_{N,2}(t). \tag{7}$$

This simply retains the diagonal terms in the square. By (6), $f_{N,1}^2\to h_1^2$ in $L^1$. Testing (7) against nonnegative compactly supported continuous functions and using weak measure convergence for $\lambda=2$ gives

$$h_1(u)^2\ge h_2(u)\quad\hbox{for almost every }u. \tag{8}$$

But for $u>0$, formula (3) gives

$$h_\lambda(u)=\frac{k}{\pi}\int_u^\infty \frac{e^{-\lambda s}}{\sqrt{s^2-u^2}}\,ds \sim\frac{k}{\sqrt{2\pi\lambda u}}e^{-\lambda u} \quad(u\to\infty).$$

The asymptotic follows by substituting $s=u+v$ and dominated convergence after multiplying the integral by $\sqrt u e^{\lambda u}$. Consequently

$$\frac{h_1(u)^2}{h_2(u)}\sim\frac{k}{\sqrt{\pi u}}\longrightarrow0,$$

contradicting (8). This proves the theorem.

What does not follow

With a general bounded $r$, weak convergence of $f_{N,\lambda}$ still has a useful interpretation, but (4) need not equal the norm of the weak limit. Microscopic variance of $r$ leaves an energy gap, so the decisive strong convergence (6) is unavailable. Replacing weak convergence by strong convergence without this equality would incorrectly assert the full problem.

Extension: eventual periodicity is also impossible

The same method proves a stronger restriction.

Theorem. If $a_n/b_n\to1$ and $r_{A,B}$ is bounded, it cannot agree on a density-one set with a nonzero nonnegative periodic sequence.

Here the period may be any fixed positive integer $P$, and some residue values may be zero. This extension is also author-recorded.

Proof. Suppose the periodic values are $c_0,\ldots,c_{P-1}$, with $\kappa=P^{-1}\sum_j c_j>0$. The preceding Laplace argument gives the same total limiting measures $p_\lambda(x)\,dx$, with $k$ replaced by $\kappa$.

Split the measures $\mu_{N,1},\nu_{N,1}$ according to the residue classes of $a,b$ modulo $P$. Along a common subsequence all $2P$ component measures converge weakly. Each limiting measure is dominated by $p_1(x)\,dx$, so it has a nonnegative density. Write these densities as $u_{i,1},v_{i,1}$. Multiplication by $e^{-x}$ also supplies convergence at $\lambda=2$, to $u_{i,2}=e^{-x}u_{i,1}$ and $v_{i,2}=e^{-x}v_{i,1}$. For $\lambda=1,2$,

$$\sum_i u_{i,\lambda}=\sum_i v_{i,\lambda}=p_\lambda.$$

All group indices in what follows are taken modulo $P$.

Define the sum and difference density vectors

$$s_{j,\lambda}=\sum_i u_{i,\lambda}*v_{j-i,\lambda}, \qquad h_{j,\lambda}=\sum_i u_{i,\lambda}*\widetilde v_{i-j,\lambda}.$$

The sum measures count the representations with $n\equiv j\pmod P$. Density-one agreement and boundedness therefore give exactly

$$s_{j,\lambda}(x)=\frac{c_j}{P}e^{-\lambda x}\mathbf1_{x>0}. \tag{9}$$

The energy identity on $\mathbb R\times\mathbb Z/P\mathbb Z$ is

$$\sum_j\|h_{j,\lambda}\|_2^2 =\sum_j\|s_{j,\lambda}\|_2^2 =\frac{\sum_j c_j^2}{2\lambda P^2}. \tag{10}$$

It needs no regularity beyond nonnegativity and integrability. To verify the first equality directly, expand its left side as the sum over $j,i,\ell$ of

$$\iiint u_{i,\lambda}(x)u_{\ell,\lambda}(y) v_{i-j,\lambda}(x-z)v_{\ell-j,\lambda}(y-z)\,dx\,dy\,dz.$$

Set $s=x+y-z$ and replace $j$ by $i+\ell-j$ in the finite group. Tonelli's theorem gives the right-hand energy.

To match the discrete functions to these profiles, put

$$f_{N,j,\lambda}(u) =D_{N,\lambda}\!\left(j+ P\left\lfloor\frac{Nu-j}{P}\right\rfloor\right).$$

These are constant on cells of length $P/N$, starting at the actual points $(j+P\ell)/N$. Their weak measure limits are $P h_{j,\lambda}(u)\,du$, and their total squared norms satisfy

$$\sum_j\|f_{N,j,\lambda}\|_2^2 =\frac PN\sum_n r(n)^2e^{-2\lambda n/N} \longrightarrow\frac{\sum_jc_j^2}{2\lambda}. \tag{11}$$

Equations (10)--(11) give strong convergence in the vector-valued $L^2$ space.

The pointwise diagonal inequality at each integer difference again gives, in the limit,

$$P h_{j,1}(u)^2\ge h_{j,2}(u)\quad\text{almost everywhere}.$$

All profiles are nonnegative, and their sums are

$$\sum_j h_{j,\lambda} =p_\lambda*\widetilde p_\lambda=:h_\lambda.$$

Consequently

$$h_2\le P\sum_j h_{j,1}^2\le P h_1^2.$$

This contradicts $h_1(u)^2/h_2(u)\sim\kappa/\sqrt{\pi u}\to0$. The extension follows. ∎

This still does not address an arbitrary bounded representation function: finite-period structure supplies the exact energy equality (10)--(11). Without it, a nonzero energy deficit can remain after every fixed finite residue decomposition.

Strengthening: positive lower density of disagreement

The same argument excludes periodic agreement even along a subsequence of initial intervals with vanishing relative error.

Theorem. Let $A,B$ be infinite positive integer sets with $a_n/b_n\to1$, and suppose $r=r_{A,B}$ is bounded. For every fixed nonzero nonnegative integer-valued periodic sequence $c(n)$,

$$\liminf_{X\to\infty}\frac1X \#\{1\le n\le X:r(n)\ne c(n)\}>0. \tag{12}$$

No coverage hypothesis is needed. In particular, a bounded counterexample to Problem 1145 would differ from every fixed positive constant on a set of positive lower density. This does not rule out such a counterexample.

Proof. Write $E(X)$ for the cardinality in (12). If its lower density were zero, there would be $X_j\to\infty$ with $E(X_j)=o(X_j)$. Set

$$N_j=\sqrt{X_j(E(X_j)+1)}.$$

Then $N_j\to\infty$, $X_j/N_j\to\infty$, and $E(X_j)/N_j\to0$. Let $P$ be a period of $c$, and let $c_i$ be its value in residue class $i\pmod P$. For every fixed $\lambda>0$, $m\in\{1,2\}$, and residue $i$, boundedness gives

$$\begin{split} &\frac1{N_j}\sum_{\substack{n\ge1\\n\equiv i\ (P)}} |r(n)^m-c(n)^m|e^{-\lambda n/N_j}\\ &\qquad\le C_m\frac{E(X_j)}{N_j} +\frac{C_m}{N_j}\sum_{n>X_j}e^{-\lambda n/N_j} \longrightarrow0. \end{split} \tag{13}$$

The geometric series in each residue class consequently yields

$$\frac1{N_j}\sum_{\substack{n\ge1\\n\equiv i\ (P)}} r(n)^m e^{-\lambda n/N_j} \longrightarrow\frac{c_i^m}{P\lambda}. \tag{14}$$

For clarity, these subsequential moment limits suffice for every step of the periodic proof above. Put $\kappa=P^{-1}\sum_i c_i>0$. Summing (14) for $m=1$ gives

$$N_j^{-1}F(\lambda/N_j)G(\lambda/N_j)\to\kappa/\lambda.$$

The rank-balance Laplace squeeze used in (2), applied at every fixed multiple of $1/N_j$, gives

$$N_j^{-1/2}F(\lambda/N_j),\quad N_j^{-1/2}G(\lambda/N_j)\longrightarrow\sqrt{\kappa/\lambda}.$$

Hence the total weighted counting measures converge to $p_\lambda(x)\,dx$, with $p_\lambda(x)=\sqrt{\kappa/\pi},x^{-1/2}e^{-\lambda x} \mathbf1_{x>0}$. Extracting the finitely many residue components supplies the densities $u_{i,\lambda},v_{i,\lambda}$ as above. Equation (14) for $m=1$, also with an additional Laplace test parameter, gives exactly (9). The continuous energy identity (10) therefore holds for their difference densities $h_{i,\lambda}$. Equation (14) for $m=2$ supplies the discrete norm limit (11), now along $N_j$. Weak convergence plus equality of norms gives strong convergence at $\lambda=1,2$.

The discrete diagonal inequality (7) thus implies $P h_{i,1}^2\ge h_{i,2}$ almost everywhere. Summing over residues again yields $h_2\le P h_1^2$, whereas

$$\frac{h_1(u)^2}{h_2(u)}\sim\frac{\kappa}{\sqrt{\pi u}} \longrightarrow0.$$

This is the same contradiction as before, and proves (12). $\square$

The remaining gap is unchanged: (12) forces persistent variation, but does not bound the number of values taken by $r$. A representation function taking several values between $1$ and a fixed $M$ can satisfy this necessary condition.