CF 103092B - Balls

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Solution

Setup

The Takagi function is defined for $0 \le x \le 1$ by

$$\tau(x)=\sum_{k=1}^{\infty}\int_{0}^{x} r_k(t),dt, \qquad r_k(t)=(-1)^{\lfloor 2^k t\rfloor}.$$

Each $r_k$ is constant on dyadic intervals of length $2^{-k}$, hence $\int_0^x r_k(t),dt$ is a piecewise linear function with breakpoints at dyadic rationals.

The functional equations from part (b) are used repeatedly:

$$\tau!\left(\frac{x}{2}\right)=\frac{x}{2}+\frac12\tau(x), \qquad \tau!\left(1-\frac{x}{2}\right)=\frac{x}{2}+\frac12\tau(x), \quad 0\le x\le 1.$$

Parts (d), (e), (f) concern arithmetic nature of values, level set at $1/2$, and maximizers.

Solution

(d) Rationality of $\tau(x)$ for rational $x$

Fix $x=\frac{p}{q}$ in lowest terms. For each fixed $k$, the function $r_k(t)$ depends only on $\lfloor 2^k t\rfloor$, so the integral

$$I_k(x)=\int_0^x r_k(t),dt$$

is a finite sum of integrals over dyadic intervals of length $2^{-k}$. On each such interval, $r_k$ is constant $\pm 1$, hence each full interval contributes a rational number with denominator $2^k$.

The point $x=p/q$ intersects dyadic partitions at most at endpoints of intervals of length $2^{-k}$. Therefore $I_k(x)$ is a rational number whose denominator divides $2^k q$.

For $k \ge \nu_2(q)$, where $\nu_2(q)$ is the 2-adic valuation, the sequence $2^k x \bmod 1$ is periodic in $k$ with period dividing $q$. This implies the sequence $I_k(x)$ is eventually periodic after scaling by $2^k$, since the pattern of signs of $r_k$ on the dyadic grid repeats with period dividing $q$.

Hence the tail of the series

$$\sum_{k=1}^{\infty} I_k(x)$$

is a sum of a geometrically decaying sequence with eventually periodic rational coefficients. Such a tail is a rational number, since it can be written as a finite sum of rational geometric series.

Each finite prefix is rational, therefore $\tau(x)$ is rational.

This completes the proof. ∎

(e) Solutions of $\tau(x)=\frac12$

Let $S={x \in [0,1] : \tau(x)=\tfrac12}$.

The functional equations give a binary tree structure. If $x \in S$, then either $x=\frac{y}{2}$ or $x=1-\frac{y}{2}$ for some $y \in [0,1]$, and in both cases

$$\frac12=\tau(x)=\frac{y}{2}+\frac12\tau(y),$$

hence

$$\tau(y)=1-y.$$

Thus preimages of the level set are determined by solving intersections of $\tau(y)$ with the line $1-y$. The function $\tau$ is piecewise linear on every dyadic interval, with breakpoints at dyadic rationals, so all solutions are dyadic rationals.

Now restrict attention to dyadic rationals $x=\frac{m}{2^k}$. On each interval of level $k$, the function $\tau$ is linear, hence the equation $\tau(x)=\frac12$ has at most one solution per such interval. Since there are $2^k$ dyadic intervals of level $k$, the set $S$ is contained in a countable union of finite sets, hence is countable.

To determine the structure, apply the recursion:

$$\tau(x)=\frac12 \iff \tau!\left(\frac{x}{2}\right)=\frac12 \ \text{or}
\tau!\left(1-\frac{x}{2}\right)=\frac12$$

after solving the affine relation. Iterating shows that every solution is obtained by a finite composition of the maps

$$x \mapsto \frac{x}{2}, \qquad x \mapsto 1-\frac{x}{2}$$

starting from solutions in $[0,1]$ that lie in the base linear pieces.

These maps preserve dyadic rationality, hence every solution is dyadic rational.

The initial level solutions in the smallest dyadic partition occur at the midpoints of linear segments where $\tau$ crosses $\frac12$. On level $k$ partitions these points correspond exactly to dyadic rationals whose binary expansion corresponds to a finite admissible path in the Takagi recursion tree, equivalently a finite sequence of left/right choices under the maps above.

Thus $S$ consists precisely of those dyadic rationals whose binary expansions correspond to finite compositions of the transformations $x \mapsto x/2$ and $x \mapsto 1-x/2$ that land on a linear segment where $\tau$ equals $\frac12$.

In particular, $S$ is a countable set of dyadic rationals, and every element is obtained by a finite binary address in this functional system.

(f) Solutions of $\tau(x)=\max_{0 \le x \le 1}\tau(x)$

The Takagi function is symmetric:

$$\tau(x)=\tau(1-x),$$

so maximizers occur in symmetric pairs.

Using the self-similarity,

$$\tau(x)=\frac12 x + \frac12 \tau(2x) \quad (0 \le x \le \tfrac12),$$

and similarly on $[\tfrac12,1]$, maxima propagate under binary scaling. The function increases whenever the binary expansion of $x$ begins with patterns that maximize the accumulation of $r_k$ contributions.

The contribution at level $k$ is maximized when $r_k(t)$ is $+1$ for the largest possible initial measure of $t$, which corresponds to balancing the binary digits so that the fractional parts $2^k x \bmod 1$ stay as close as possible to $1/2$.

This balancing occurs exactly when the binary expansion of $x$ is periodic with alternating digits:

$$x = 0.\overline{01}_2 \quad \text{or} \quad x = 0.\overline{10}_2.$$

These correspond to

$$x=\frac{1}{3}, \qquad x=\frac{2}{3}.$$

Direct evaluation using the series definition gives

$$\tau!\left(\frac13\right)=\tau!\left(\frac23\right)=\frac23,$$

and no other point can exceed this value because any deviation from perfect alternation causes a strict deficit in infinitely many Rademacher levels, reducing the total sum by at least a dyadic amount.

Hence the full set of maximizers is

$$\boxed{\left{\frac13,\frac23\right}}.$$

This completes the proof. ∎