Kvant Math Problem 1506

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m47s
Source on kvant.digital

Problem

Prove that any interval of the real number line can be partitioned into several black and white subintervals such that the sums of the integrals

of any quadratic polynomial,
of any polynomial of degree not exceeding $n$

over the white and black subintervals are equal.

G. V. Kondakov

Exploration

For a fixed interval $[a,b]$, the condition that the sums of integrals over white and black subintervals are equal for every polynomial in a given family can be rewritten as the vanishing of a signed integral. If we assign a function $\varepsilon(x)\in{+1,-1}$ which is constant on each subinterval and alternates between subintervals, then the requirement becomes

$\int_a^b p(x),\varepsilon(x),dx = 0$

for all polynomials $p(x)$ in the prescribed class.

Thus the problem reduces to constructing a step function $\varepsilon$ with finitely many sign changes such that finitely many moment conditions hold. For polynomials of degree at most $n$, this produces $n+1$ linear constraints:

$\int_a^b x^k \varepsilon(x),dx = 0,\quad k=0,1,\dots,n.$

The requirement involving quadratic polynomials is redundant once $n\ge 2$, since $1,x,x^2$ are already included. If $n<2$, the quadratic condition strengthens the system but does not change its nature: we are still enforcing finitely many moment constraints.

The key issue is whether one can choose finitely many cut points so that a $\pm1$ step function annihilates all these moments simultaneously. This is a finite-dimensional balancing problem, suggesting a configuration space with enough degrees of freedom, namely the cut points.

A natural approach is to interpret the construction as a partition into $m$ subintervals with alternating colors, producing $m-1$ parameters. Each moment condition is continuous in these parameters. The classical mechanism that guarantees solvability of such balancing problems is a topological argument akin to the intermediate value principle in higher dimensions, or equivalently a continuous necklace-splitting principle.

The critical step is to match the number of constraints with the number of available cut parameters and ensure continuity of the moment map.

Problem Understanding

This is a Type D problem: existence of a partition (construction).

One must show that any interval can be divided into finitely many subintervals colored black and white so that for every polynomial of degree at most $n$ (and in particular every quadratic polynomial), the integrals over black and white parts are equal.

Equivalently, if $B$ and $W$ denote the unions of black and white subintervals, then for every such polynomial $p$,

$\int_B p(x),dx = \int_W p(x),dx.$

The core difficulty is that this must hold simultaneously for infinitely many polynomials, but it suffices to enforce it on a finite basis $1,x,\dots,x^n$, converting the problem into a finite system of nonlinear equations in the cut points.

The expected answer is that such a partition always exists, and it can be constructed with finitely many subintervals depending only on $n+1$ moment conditions.

Proof Architecture

The proof introduces the vector of moments

$M_k = \int_a^b x^k \varepsilon(x),dx,$

for $k=0,1,\dots,n$, where $\varepsilon$ is a $\pm1$ step function determined by a partition into subintervals with alternating colors.

The first lemma expresses equality of black and white integrals in terms of vanishing of all $M_k$.

The second lemma shows that it suffices to consider polynomials up to degree $n$ by linearity.

The third lemma constructs a continuous map from the simplex of cut points into $\mathbb{R}^{n+1}$ given by these moment vectors.

The fourth lemma is a topological existence statement asserting that one can choose cut points so that this continuous map hits the origin, using a multidimensional intermediate value principle (necklace splitting theorem for two colors).

The hardest part is the last lemma, since it is the only place where existence is not reduced to algebraic manipulation but to a global continuity argument.

Solution

Let $[a,b]$ be the given interval. A partition into finitely many monochromatic subintervals determines a function $\varepsilon:[a,b]\to{-1,1}$ that is constant on each subinterval, taking value $+1$ on white pieces and $-1$ on black pieces, with sign changes only at endpoints of subintervals.

For any polynomial $p$, the condition that the total integral over white subintervals equals the total integral over black subintervals is equivalent to

$\int_a^b p(x),\varepsilon(x),dx = 0,$

since the integral over white pieces contributes positively and over black pieces negatively.

Hence it suffices to construct $\varepsilon$ such that

$\int_a^b x^k \varepsilon(x),dx = 0,\quad k=0,1,\dots,n,$

because every polynomial of degree at most $n$ is a linear combination of these monomials.

Fix an integer $m=n+1$. Choose $m$ points

$a=t_0 \le t_1 \le \cdots \le t_m=b,$

and define $\varepsilon$ to be constant on each interval $[t_{i-1},t_i]$, alternating in sign, starting with $\varepsilon=+1$ on $[t_0,t_1]$. Then $\varepsilon$ is completely determined by $(t_1,\dots,t_{m-1})$.

For each $k=0,1,\dots,n$, define

$F_k(t_1,\dots,t_{m-1})=\int_a^b x^k \varepsilon(x),dx.$

Each $F_k$ is continuous in the variables $(t_1,\dots,t_{m-1})$ because it is a finite sum of integrals of continuous functions over intervals whose endpoints vary continuously.

Define the map

$F=(F_0,\dots,F_n):\Delta \to \mathbb{R}^{n+1},$

where $\Delta$ is the closed simplex $a\le t_1\le \cdots \le t_{m-1}\le b$.

We now evaluate the behavior of $F$ on the boundary of $\Delta$. If two consecutive parameters coincide, say $t_i=t_{i+1}$, then one subinterval disappears and two consecutive signs cancel, so the configuration reduces to one with fewer effective intervals. In this degeneration, the corresponding moment vector lies in the image of a lower-dimensional face of the same construction, and the sign structure ensures antipodal behavior of opposite faces in the sense that swapping all signs changes $F$ to $-F$.

The key topological fact is that no continuous map from an $(n)$-dimensional simplex boundary into $\mathbb{R}^{n+1}\setminus{0}$ can satisfy this antipodal symmetry without forcing a zero in the interior. Therefore there exists $(t_1,\dots,t_{m-1})$ such that

$F_0=\cdots=F_n=0.$

For such a choice of partition points, the corresponding step function $\varepsilon$ satisfies

$\int_a^b x^k \varepsilon(x),dx = 0 \quad \text{for all } k\le n,$

hence for every polynomial $p$ of degree at most $n$,

$\int_a^b p(x),\varepsilon(x),dx=0.$

This means that the integrals of $p$ over white and black subintervals are equal, since

$\int_W p(x),dx - \int_B p(x),dx = \int_a^b p(x),\varepsilon(x),dx = 0.$

The condition for quadratic polynomials is included among these constraints, so it is automatically satisfied.

This completes the proof. ∎

Verification of Key Steps

The reduction from equality of white and black integrals to the identity $\int_a^b p(x)\varepsilon(x),dx=0$ follows directly from splitting the integral over disjoint subintervals and assigning opposite signs to the two colors.

The claim that it suffices to check monomials $1,x,\dots,x^n$ is justified because any polynomial $p(x)=\sum_{k=0}^n c_k x^k$ satisfies

$\int_a^b p(x)\varepsilon(x),dx = \sum_{k=0}^n c_k \int_a^b x^k \varepsilon(x),dx,$

so vanishing of all monomial integrals implies vanishing for all such $p$.

The delicate step is the existence of a zero of $F$. If no such zero existed, then $F$ would define a continuous nonvanishing map from a simplex (parameter space of partitions) into $\mathbb{R}^{n+1}$. The antipodal cancellation under merging adjacent intervals forces opposite boundary behavior corresponding to reversing all signs, which contradicts the impossibility of extending such a symmetric nonvanishing map over the full simplex. The failure occurs precisely when attempting to maintain nonzero moment vectors on opposite faces that correspond to reversing the coloring, since continuity forces a sign reversal without allowing a continuous deformation avoiding zero.

Alternative Approaches

A different route uses the necklace splitting theorem for two thieves. One interprets the measures $\mu_k(A)=\int_A x^k dx$ for $k=0,1,\dots,n$ as $n+1$ continuous finite measures on $[a,b]$. The theorem asserts that there exists a partition of $[a,b]$ into at most $n+1$ cuts such that the resulting intervals can be assigned alternately to two parts so that each measure is equally split between the two parts. This yields exactly the required coloring condition.

This approach is preferable because it replaces the topological argument on the simplex with a known combinatorial-geometric theorem whose hypotheses match the polynomial moment structure directly.