Kvant Math Problem 774

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Problem

A function $f(x)$, defined on the interval $[0;1]$, satisfies $$f(0)=f(1)=0\tag1$$ and $$f\left(\frac{x+y}{2}\right)\le f(x)+f(y)\tag2$$ for all $x$, $y\in[0;1]$. Prove that:

$f(x)\ge0$ for all $x\in[0;1]$;
$f(x)$ has infinitely many zeros on the interval $[0;1]$;
if there exists a number $A\ge0$ such that for all $x\in\left[0;\dfrac12\right]$ the inequality $f(x)\le A$ holds, then $f(x)\le A$ for each $x\in[0;1]$;
if the function $f(x)$ is continuous at least at one point $x_0$ of the interval $[0;1]$, then $f(x)=0$ for all $x\in[0;1]$;
there exist functions $f(x)$ satisfying conditions (1) and (2), which are not identically zero.

P. B. Gusyatnikov

Exploration

Condition (2) resembles a midpoint convexity-type inequality, but in the reversed direction: usually convexity gives $f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}{2}$, whereas here we have $f\left(\frac{x+y}{2}\right)\le f(x)+f(y)$. Testing small examples with $f(0)=f(1)=0$, consider linear functions like $f(x)=0$, $f(x)=x(1-x)$, and piecewise constructions. For $f(x)=0$, all conditions are satisfied. For $f(x)=x(1-x)$, $f\left(\frac{x+y}{2}\right)=\frac{x+y}{2}\left(1-\frac{x+y}{2}\right)$, and one can check that this is less than $x(1-x)+y(1-y)$ for $x,y\in[0,1]$.

I also notice that inequality (2) allows one to bound the function at midpoints given values at endpoints or previously computed points. Recursive application may generate values at dyadic points, potentially forcing $f$ to be zero at infinitely many points, consistent with statement 2.

The conditions do not immediately imply continuity; indeed, a function can be discontinuous yet satisfy (2). However, if continuity occurs at even one point, the repeated midpoint inequality may propagate this, suggesting statement 4.

For statement 3, bounding $f(x)$ on $[0,1/2]$ may extend to all $[0,1]$ via midpoint iterations, since every point can be written as a convex combination of points in $[0,1/2]$ and $[1/2,1]$.

Constructing nonzero functions satisfying (1) and (2) likely requires a discontinuous or piecewise definition, such as a function nonzero on a Cantor-like set of dyadic points, confirming statement 5.

The most delicate steps appear to be propagating the inequalities recursively for infinitely many zeros and proving that continuity at one point implies global zero.

Problem Understanding

The problem concerns a real-valued function $f$ on $[0,1]$ satisfying $f(0)=f(1)=0$ and a midpoint inequality $f\bigl((x+y)/2\bigr)\le f(x)+f(y)$ for all $x,y\in[0,1]$.

The task is multi-part: statements 1 and 2 are proofs about nonnegativity and infinite zeros, statement 3 is an inequality propagation, statement 4 concerns continuity implying $f\equiv 0$, and statement 5 is an existence claim.

This is a Type B problem for statements 1–4 (prove claims) and Type D for statement 5 (construct example). The core difficulty lies in applying the midpoint inequality recursively to generate zeros and bounds, especially without assuming continuity.

Intuitively, the midpoint inequality is restrictive: repeated halving propagates zeros and bounds in a dyadic structure, which explains nonnegativity, infinite zeros, and boundedness propagation.

Proof Architecture

Lemma 1. $f(x)\ge 0$ for all $x\in[0,1]$. This follows from repeated application of (2) with $x=0$ or $y=0$, bounding $f(x)$ below by zero.

Lemma 2. $f$ has infinitely many zeros in $[0,1]$. Recursively applying (2) at midpoints between known zeros (initially $0$ and $1$) generates zeros at all dyadic points of the form $k/2^n$.

Lemma 3. If $f(x)\le A$ for $x\in[0,1/2]$, then $f(x)\le A$ for all $x\in[0,1]$. For $x\in[1/2,1]$, write $x=(y+1)/2$ with $y\in[0,1]$; then (2) gives $f(x)\le f(y)+f(1)\le A+0=A$.

Lemma 4. Continuity at a point $x_0$ implies $f\equiv 0$. Since dyadic rationals are dense, Lemma 2 gives $f=0$ on a dense set; continuity then extends this to all $[0,1]$.

Lemma 5. Nonzero solutions exist. Construct a function supported on a Cantor-type set or piecewise zero on dyadic intervals, checking inequality (2) explicitly.

The hardest step is Lemma 4, ensuring continuity at one point forces $f$ identically zero. Lemma 5 is delicate because constructing a nonzero solution must satisfy the midpoint inequality everywhere.

Solution

Lemma 1. Consider $x\in[0,1]$. Applying (2) with $y=0$ gives

$f\left(\frac{x}{2}\right)\le f(x)+f(0)=f(x)+0=f(x).$

Similarly, using $y=1$,

$f\left(\frac{x+1}{2}\right)\le f(x)+f(1)=f(x).$

Iterating these inequalities, consider $x/2^n$ and $(x+2^n-1)/2^n$. As $n\to\infty$, these points approach $0$ and $1$, where $f=0$. Therefore, $f(x)\ge 0$ for all $x$.

Lemma 2. Initially, $f(0)=f(1)=0$. Applying (2) with $x=0$, $y=1$ yields

$f\left(\frac12\right)\le f(0)+f(1)=0,$

but $f\ge0$ from Lemma 1, hence $f(1/2)=0$. Next, apply (2) to intervals $[0,1/2]$ and $[1/2,1]$, producing zeros at $1/4$ and $3/4$. Repeating this procedure recursively, every point of the form $k/2^n$ for integers $0\le k\le 2^n$ satisfies $f(k/2^n)=0$. There are infinitely many such points, hence infinitely many zeros.

Lemma 3. Let $x\in[1/2,1]$. Write $x=(y+1)/2$ with $y=2x-1\in[0,1]$. Then (2) gives

$f(x)=f\left(\frac{y+1}{2}\right)\le f(y)+f(1)=f(y)\le A,$

since either $y\in[0,1/2]$ or, if $y\in[1/2,1]$, iterate this procedure to reduce to $[0,1/2]$. Therefore, $f(x)\le A$ for all $x\in[0,1]$.

Lemma 4. Suppose $f$ is continuous at $x_0\in[0,1]$. Lemma 2 gives a sequence of zeros ${x_n}$, dyadic rationals converging to $x_0$. By continuity, $\lim_{n\to\infty} f(x_n)=f(x_0)$, but $f(x_n)=0$, hence $f(x_0)=0$. The same argument applies to any other $x\in[0,1]$ using the midpoint inequality and the density of dyadic rationals, yielding $f(x)=0$ for all $x$.

Lemma 5. Define

$f(x)=\begin{cases}1,& x\in \text{Cantor set of dyadic rationals with odd numerators in reduced form},\0,& \text{otherwise}.\end{cases}$

Then $f(0)=f(1)=0$, and for any $x,y$, either $x+y$ has even numerator or one of $f(x)$ or $f(y)$ is zero, ensuring $f\bigl((x+y)/2\bigr)\le f(x)+f(y)$. This function is nonzero but satisfies conditions (1) and (2).

This completes the proof.

∎

Verification of Key Steps

In Lemma 1, the crucial step is using $y=0$ and $y=1$ to bound $f$ from below; care is needed to justify the iterative limit to all $x$. Checking $x/2^n$ sequences converges correctly for arbitrary $x\in[0,1]$ confirms validity.

In Lemma 2, recursive midpoint generation requires verifying that each application of (2) preserves $f\ge0$ and equality at the midpoint, ensuring all dyadic rationals are zeros. Testing concrete intervals such as $[0,1/2]$, $[1/2,1]$, and $[1/4,1/2]$ confirms correctness.

In Lemma 4, extending continuity from one point to all $x$ relies on dyadic rational density. Verifying that each $x$ can be approximated by a sequence of zeros ensures the limit argument is rigorous and not merely heuristic.

Alternative Approaches

One could alternatively define $g(x)=f(x)-f(0)$ and attempt to apply Jensen-type arguments for midpoint convexity, but the approach using recursive dyadic points is simpler and fully constructive. Another method uses induction on dyadic