Reading Eight Erdős Problems Together

Nullity at and below the critical exponent

Here only eventual evenness is needed. Fix a cutoff $K$ and a prefix before $K$, and allow every continuation with $0\le e_n\le n^c$ and $2\mid e_n$ for $n\ge K$. The derivative vectors form a compact set $V$. For all sufficiently large $n$, there are at most

choices at position $n$. Thus the number of prefixes through $N$ is at most $C(3/4{)}^N(N!{)}^c$. Every remaining derivative tail has norm at most $C_{c,d}{N}^{c+d-1}/(N+1)!$, by comparison with its first term. Balls of this radius, one for each prefix, cover $V$ with total $d$-dimensional volume at most

For $c\le d$ this tends to zero. Hence $V$ is null and, being compact, nowhere dense. Countably many cutoffs and integer prefixes cover the attainable set. This proves both conclusions, including the critical case.

Lifting a scalar interval to an open set

First allow signed integer coefficients. For every $\eta >0$ and cutoff $N\ge 1$, consider series supported on $n\ge N$ with $|e_n|\le \eta n^c$ and the same eventual divisibilities. We prove by induction on $d$ that their derivative vectors contain an open set whenever $c>d$. For $d=1$, apply Theorem 3.1 to $F_n=\lfloor \eta n^c\rfloor$ and $Q_n=n!$, starting after $N$. The first tail term makes $U_N\to \mathrm{\infty }$ when $c>1$; the all-position version gives an interval with common congruence cutoffs.

For the induction step, use dimension $d-1$ and exponent $c-1$ for $B(z)=\sum b_n{z}^n/n!$ with $|b_n|\le (\eta /4)n^{c-1}$ and $b_n=0$ before $N$. The higher derivative coordinates of $(z-1)B$ form an open set $W\subseteq {\mathbb{R}}^{d-1}$, because

Choose $y_0+[-2\rho ,2\rho {]}^{d-1}\subset W$. Choose a later cutoff $M$ so that every $A$ supported on $n\ge M$ with $0\le a_n\le (\eta /2)n^c$ has $|A^{(r)}(1)|<\rho$ for $1\le r<d$. This follows uniformly from the convergent derivative tails. The scalar theorem gives a nondegenerate interval $I$ of values $A(1)$. For each $x\in I$ choose one such $A$. For every $y\in y_0+(-\rho ,\rho {)}^{d-1}$, a suitable $B$ cancels the higher coordinates of $A$ and gives $y$. Hence the attainable vectors contain

No continuous choice of $A$ is assumed: uniform smallness of all its higher derivatives makes the same cube work for every $x$. The coefficients of $A+(z-1)B$ obey the bound, since $n|b_{n-1}|+|b_n|\le (\eta /2)n^c$; divisibility holds term by term. The finitely many families in the induction have common cutoffs for each $q$.

To restore nonnegativity, choose nested positive integers $m_n\to \mathrm{\infty }$ such that every fixed integer eventually divides $m_n$ and $\log m_n=o(\log n)$. One choice is the largest factorial at most $\log (n+3)$. For large $n$,

Set earlier $h_n$ to zero and start the signed construction after that point. Take $\eta =1/4$. Adding $\sum h_n{z}^n/n!$ translates the open set and puts every sufficiently late coefficient between $n^c/12$ and $3n^c/4$. All coefficients lie between $0$ and $n^c$, and each function is nonpolynomial.

The upper bound for Hausdorff dimension

Fix a finite prefix and an allowance cutoff. Through index $N$ there are at most $\prod _{n\le N}(\lfloor n^c\rfloor +1)$ continuations, whose logarithm is $c\log (N!)+O(N)$. Each derivative tail has norm at most $O_{c,d}(N^{c+d-1}/(N+1)!)$. For any $s>c$, the sum of the $s$-th powers of the diameters of these covering balls tends to zero. Countably many prefixes and cutoffs preserve this bound; the ambient bound is $d$.

A separated construction attaining the dimension

For $0\le k<d$ with $k<c$, set ${\alpha }_k=min(1,c-k)$. Use only these indices. Their sum is $s=min(c,d)$. Let $\epsilon =2^{-d-6}$ and use the same $m_n$. Independently for every $n$ and $k$, choose

All digits vanish before a large cutoff. Put

The coefficient added at position $n$ is

with negative indices interpreted as zero. Its absolute value is at most $\epsilon \sum _{k<d}{2}^k{n}^c<n^c/4$, since $j+{\alpha }_k\le c$. Thus $f\in {\mathcal{H}}_c$, with the allowance everywhere and common congruence cutoffs. Local uniform convergence makes the image of this product of finite digit sets compact.

The crucial step is to rule out cancellation between different codes. Suppose two codes first differ at index $n$, and let $g$ be their difference. For $0\le r<d$, take the Taylor coefficients of $z^{-n}g(z)$ at $1$:

Unused coordinates are zero. The binomial coefficient now depends on $m-n$, not on $m$. Since $|\mathrm{\Delta }{b}_m^{(k)}|\le \epsilon m$, the absolute tail, multiplied by $n!$, is at most

Here $n!/(n+t-1)!\le 1/(t-1)!$ and $\sum _{k\le r}(\genfrac{}{}{0ex}{}{t}{r-k})\le 2^t$. Some $\mathrm{\Delta }{b}_n^{(r)}$ is a nonzero integer. Therefore one coordinate on the left of (T) has absolute value at least $1/(2n!)$. The linear map from $J_d(g)$ to those coordinates has norm at most $C_d{n}^{d-1}$: its entries are $(-1{)}^j(\genfrac{}{}{0ex}{}{n+j-1}{j})/(r-j)!$, $0\le j\le r<d$. Distinct codes first differing at $n$ consequently have separation at least

in the supremum norm of ${\mathbb{R}}^d$.

Give all choices in (D) uniform independent probabilities. If $A_N$ counts prefixes through $N$, then

because $\log m_n=o(\log n)$ and each used ${\alpha }_k$ is positive. The separation makes the coding injective. A ball of radius less than ${\delta }_N/2$ meets at most one length-$N$ cylinder and has measure at most $A_N^{-1}$. For a small radius $r$, choose $N$ with ${\delta }_{N+1}/2\le r<{\delta }_N/2$. Since $\log (1/{\delta }_{N+1})/\log (N!)\to 1$, for every $t<s$ this implies $\mu (B(x,r))\le C_t{r}^t$. Summing this inequality over any ball cover gives positive $t$-dimensional Hausdorff content. The dimension is at least $s$, completing Theorem 2.1.