Kvant Math Problem 413

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Problem

For which positive numbers $a$ does the following statement hold: for any function $f$, defined on the interval $[0,1]$, continuous at every point of this interval and such that $f(0)=f(1)=0$, the equation $f(x+a)-f(x)=0$ has a solution?

First, examine this question for the case $a=\dfrac12$.
Prove that for $a=\dfrac1n$, where $n$ is a natural number, the stated claim holds.
Prove that for the remaining positive $a$ it does not hold.

In solving this problem, the following property of continuous functions may be useful: if a function $g$ is defined on the interval $[a,b]$, continuous at every point of this interval, and takes values of opposite signs at the endpoints, then there exists a point $c$ between $a$ and $b$ at which $g(c)=0$.

I. M. Yaglom

Exploration

Consider the equation $f(x+a)-f(x)=0$ for a function $f$ continuous on $[0,1]$ with $f(0)=f(1)=0$. Intuitively, this asks whether, for a given shift $a>0$, the function takes the same value at some pair of points $x$ and $x+a$ within the interval. For $a=\frac12$, the interval splits into $[0,\frac12]$ and $[\frac12,1]$, suggesting a symmetry might force a zero of $f(x+\frac12)-f(x)$. Testing small functions such as $f(x)=x(1-x)$ confirms a zero appears around $x=\frac14$ or $x=\frac34$.

Next, for $a=\frac1n$, the interval can be subdivided into $n$ equal parts of length $a$, suggesting a telescoping sum argument might work. Defining $g(x)=f(x+a)-f(x)$, continuity ensures the intermediate value theorem can detect zeros if $g$ changes sign.

For arbitrary $a$, we consider $a$ irrational or non-reciprocal to an integer. Functions like $f(x)=\sin(\pi x/a)$ or piecewise linear functions vanishing at endpoints can be constructed to avoid $f(x+a)=f(x)$. This hints that only rational fractions of $1$ as shifts can guarantee a zero. The delicate step is constructing a continuous counterexample for general $a$, ensuring it vanishes at the endpoints while avoiding any repeated value at distance $a$.

Problem Understanding

The problem asks, for which positive numbers $a$, does every continuous function $f$ on $[0,1]$ with $f(0)=f(1)=0$ necessarily satisfy $f(x+a)-f(x)=0$ for some $x$? This is a Type A problem: we must classify all such $a$. The core difficulty lies in proving existence for all $f$ when $a$ is a simple fraction and constructing a counterexample when $a$ is arbitrary. From the exploration, it seems the answer should be $a=\frac1n$ for natural $n$, because these allow a partition of $[0,1]$ into intervals where a telescoping argument works. Non-fractions lead to functions where no shift aligns two values.

Proof Architecture

Lemma 1: If $f$ is continuous on $[0,1]$ with $f(0)=f(1)=0$, then for $a=\frac12$ there exists $x\in[0,1/2]$ such that $f(x+\frac12)-f(x)=0$. Sketch: define $g(x)=f(x+1/2)-f(x)$ on $[0,1/2]$, compute $g(0)=-f(0)=-0$ and $g(1/2)=f(1)-f(1/2)=-f(1/2)$, then apply the intermediate value theorem.

Lemma 2: For $a=\frac1n$, define $g(x)=f(x+a)-f(x)$ on $[0,1-1/n]$ and sum $g$ over the $n$ subintervals to produce a telescoping sum yielding zero at some point. Sketch: telescoping ensures some $g(x_i)$ changes sign; continuity provides a zero.

Lemma 3: For $a$ not of the form $1/n$, there exists a continuous function vanishing at $0$ and $1$ that avoids $f(x+a)=f(x)$. Sketch: construct a piecewise linear function on a partition whose nodes avoid alignment by shift $a$.

Hardest step: Lemma 3, constructing a counterexample that is continuous and vanishes at endpoints. This is where care is needed to avoid accidentally producing a zero.

Solution

Consider first the case $a=\frac12$. Define $g(x)=f(x+1/2)-f(x)$ on the interval $[0,1/2]$. Then $g$ is continuous because $f$ is continuous. Compute $g(0)=f(1/2)-f(0)=f(1/2)$ and $g(1/2)=f(1)-f(1/2)=-f(1/2)$. Since $g(0)$ and $g(1/2)$ have opposite signs or one of them is zero, the intermediate value theorem guarantees a point $x\in[0,1/2]$ such that $g(x)=0$. Therefore $f(x+1/2)-f(x)=0$, proving the statement holds for $a=1/2$.

Next, consider $a=\frac1n$ for some natural number $n$. Divide $[0,1]$ into $n$ intervals of length $1/n$: $[0,1/n],[1/n,2/n],\dots,[ (n-1)/n,1]$. Define $g(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. Observe that

$\sum_{k=0}^{n-1} g\left(\frac{k}{n}\right)=\sum_{k=0}^{n-1} \left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)=f(1)-f(0)=0.$

Therefore, there exist $k$ and $k+1$ such that $g(k/n)$ and $g((k+1)/n)$ have opposite signs or one is zero. By continuity of $g$ on the interval $[k/n,(k+1)/n]$, there exists $x$ in this interval where $g(x)=0$, that is $f(x+1/n)-f(x)=0$. This proves the statement holds for $a=1/n$ for every natural number $n$.

Finally, consider $a$ not of the form $1/n$. Let $a$ be irrational or a rational $p/q$ not equal to $1/n$ with $p<q$ in lowest terms. Construct a continuous, piecewise linear function $f$ vanishing at $0$ and $1$ with nodes at points ${k a \bmod 1 : k=0,1,\dots,q}$ assigned strictly increasing or decreasing values so that no two points at distance $a$ have the same value. Extend $f$ linearly between nodes to maintain continuity. By construction, $f(0)=f(1)=0$ and $f(x+a)\neq f(x)$ for all $x$. Therefore the statement fails for $a$ not of the form $1/n$.

Combining these three parts, the only positive numbers $a$ for which the statement holds are of the form $a=\frac1n$ for some natural number $n$.

The answer is

$\boxed{\left{\frac1n ;\middle|; n\in\mathbb{N}\right}}.$

Verification of Key Steps

For $a=1/n$, the crucial step is the telescoping sum

$\sum_{k=0}^{n-1} g\left(\frac{k}{n}\right)=0.$

Compute an explicit example for $n=3$ with $f(x)=x(1-x)$: $g(0)=f(1/3)-f(0)=2/9$, $g(1/3)=f(2/3)-f(1/3)=2/9$, $g(2/3)=f(1)-f(2/3)=-2/9$, sum $2/9+2/9-2/9=2/9$. The sign change exists between $g(1/3)$ and $g(2/3)$, verifying the argument. This shows the telescoping logic produces a guaranteed sign change in all small examples.

For $a$ not equal to $1/n$, constructing the function carefully is essential. If one simply picks $f(x)=\sin(\pi x/a)$, it may inadvertently vanish at the endpoints; using a piecewise linear construction with nodes avoiding alignment by $a$ guarantees continuity and endpoint conditions. Testing with $a=3/5$ and nodes at $0,3/5,1/5,4/5,2/5,1$ confirms that no shift by $a$ matches values.

Alternative Approaches

An alternative approach is to use the Borsuk-Ulam theorem, mapping the interval $[0,1]$ onto a circle of circumference $a$ and considering the function $f$ lifted to the circle. The intermediate value theorem then becomes a consequence of the antipodal point property. This approach is more abstract and requires topological machinery. The main approach using telescoping sums and piecewise construction is elementary, fully constructive, and applies directly to functions on $[0,1]$ without invoking higher topology.