# Missing digits points near manifolds

Han Yu  Han Yu, Mathematics Institute, Zeeman Building, University of Warwick,
Coventry CV4 7AL, UK<EMAIL_ADDRESS>
###### Abstract.

We consider a problem concerning the distribution of points with missing
digits coordinates that are close to non-degenerate analytic submanifolds. We
show that large enough (to be specified in the paper) sets of points with
missing digits coordinates distribute ’equally’ around non-degenerate
submanifolds. As a consequence, we show that intersecting those missing digits
sets with non-degenerate submanifolds always achieve the optimal dimension
reduction. On the other hand, we also prove that there is no lack of points
with missing digits that are contained in non-degenerate submanifolds. Among
the other results,

  * 1

we prove that the pinned distance sets of those missing digits sets contain
non-trivial intervals regardless of where the pin is.

  * 2

we prove that for each $\epsilon>0,$ for missing digits sets $K$ with large
bases, simple digit sets (to be specified in the paper), and
$\dim_{\mathrm{H}}K>3/4+\epsilon,$ the arithmetic product sets $K\cdot K$
contains non-trivial intervals.

###### Key words and phrases:

###### 2010 Mathematics Subject Classification:

11Z05, 11J83, 28A80

##  1\. Introduction

We discuss a problem concerning the distribution of missing digits points
around as well as on manifolds. Before we state the most general results, we
list three special cases with some mixtures of number theory and geometric
measure theory.

In what follows, let $K_{1}$ be the set of points on $[0,1]$ whose base
$10^{9000}$ expansions111Examples with much smaller bases can be found.
However, we did not carry out the most optimal numerical computations. contain
only digits in $\\{0,\dots,10^{8100}-1\\}.$ Let $K_{2}$ be the set of points
on $[0,1]$ whose base $10^{9000}$ expansions contain only digits in
$\\{0,\dots,10^{7000}-1\\}.$ We see that $\dim_{\mathrm{H}}K_{1}=9/10$ and
$\dim_{\mathrm{H}}K_{2}=7/9.$

######  Theorem A.

Let $(t,t^{2},t^{3})_{t\in\mathbb{R}}$ be the Veronese curve in
$\mathbb{R}^{3}.$ There is an integer $l\geq 0$ such that there are infinitely
many $t>0$ such that the fractional parts of

$10^{9000l}t,10^{9000l}t^{2},10^{9000l}t^{3}$

are contained in $K_{1}.$ Loosely speaking, there are many points on the
Veronese curve whose coordinates have special expansions in base $10^{9000}.$
Moreover, the upper box dimension of the set consisting such numbers
$t\in[0,1]$ is in $[1/30,7/10].$

######  Remark.

We expect that the (upper) box dimension of such $t$ should be exactly equal
to $7/10.$ This is obtained via

$7/10=3*9/10-(3-1).$

Here $3*9/10$ is the dimension of the missing digits set $K_{1}\times
K_{1}\times K_{1}$ in $\mathbb{R}^{3}$ considered in this theorem, $1$ is the
dimension of the Veronese curve. For more discussions, see Theorem 2.8.

######  Theorem B.

Let $K=K_{2}\times K_{2}\subset[0,1]^{2}$ be the twofold Cartesian product of
$K_{2}.$ Then for each $x\in\mathbb{R}^{2},$ the pinned distance set

$\Delta_{x}(K)=\\{|x-y|:y\in K\\}$

contains non-trivial interior and in particular, positive Lebesgue measure.
Moreover, for each circle, $C\subset\mathbb{R}^{2},$ we have

$\overline{\dim}_{\mathrm{B}}C\cap K\leq\dim_{\mathrm{H}}K-(2-\dim
C)=\frac{5}{9}.$

As a consequence, we can take $x=(0,0)$ and see that the set

$K^{2}_{2}+K^{2}_{2}=\\{x^{2}+y^{2}:x,y\in K_{2}\\}$

contains intervals.

######  Remark.

Previously, it is possible to show that:

1\. $\Delta_{x}(K)$ has positive Lebesgue measure for many choices of
$x\in\mathbb{R}^{2}.$ Notice that the set $K$ has Hausdorff dimension larger
than $1.5$. This statement follows from a classical result of Falconer. See
[31, Section 7].

2\. $\Delta_{x}(K)$ has full Hausdorff dimension for all $x\in\mathbb{R}^{2}.$
This follows by adapting the arguments in [19].

This theorem pushes the result further, i.e. $\Delta_{x}(K)$ has a positive
Lebesgue measure for all $x\in\mathbb{R}^{2}.$ This result holds in
$\mathbb{R}^{n},n\geq 3$ as well. Also, instead of a missing digits set in
$\mathbb{R}^{2},$ $K$ can be also chosen to be the Cartesian product of two
missing digits sets with possibly different bases on $\mathbb{R}.$ See Section
8.

######  Theorem C.

Consider the set $K_{2}.$ The arithmetic product set

$K_{2}\cdot K_{2}=\\{xy:(x,y)\in K_{2}\times K_{2}\\}$

has non-trivial interior.

######  Remark.

The sum set $K_{2}+K_{2}$ does not contain non-trivial intervals because it is
contained in the set of numbers whose base $10^{9000}$ expansions contain only
digits in $\\{0,\dots,2\times 10^{7000}-2\\}.$ Thus the nonlinearity of the
multiplication map makes a significant difference.

The role of missing digits sets in this paper is by no means restrictive. It
really just depends on Theorem 2.15. In fact, it is possible to extend Theorem
2.15 to homogeneous self-similar measures with rational scaling ratios, finite
rotation groups and rational translation vectors. However, in this paper, we
anyway restrict the discussions to missing digits measures because of their
intrinsic connections with number theory.

The roles of the specific missing digits sets $K_{1},K_{2}$ with base
$10^{9000}$ are also not restrictive. The precise conditions are on their
Fourier $l^{1}$-dimensions, see Section 2.1. We need that
$\dim_{l^{1}}K_{1}>8/9$ and $\dim_{l^{1}}K_{2}>3/4.$ These conditions are in
some sense optimal. See Theorem 2.12 and Remark 2.13.

For missing digits sets with simple digit sets (e.g. with consecutive digits),
their Fourier $l^{1}$-dimensions are almost equal to their Hausdorff
dimensions, see Theorem 2.15. Theorem C can be seen as a partial answer to the
following folklore conjecture.

######  Conjecture 1.1.

Let $K$ be a missing digits set with $\dim_{\mathrm{H}}K>1/2,$ then $K\cdot K$
contains intervals.

######  Remark 1.2.

Let’s see why this conjecture is plausible. By adapting the arguments in [19],
it can be shown that $\dim_{\mathrm{H}}K\cdot K=1$. Due to the nonlinearity of
the multiplication map, it is expected that $K\cdot K$ has a positive measure.
Finally, due to the results in [34], it is expected that $K\cdot K$ contains
intervals.

######  Remark 1.3.

If $K$ has Newhouse thickness (see [24, Definition 3.6]) $\tau(K)$ larger than
$1$, then $K\cdot K$ contains intervals. In fact, with the help of Newhouse
thickness of Cantor set ([22],[46],[24],[41]), it is possible to prove some of
the topological results in Theorems A, B, C for Cantor sets with thickness at
least one. To see where $K_{1},K_{2}$ are in the thickness story: $K_{1}$ at
the beginning has Newhouse thicknesses $1/(10^{900}-1-10^{-900});$ $K_{2}$ has
Newhouse thickness $1/(10^{2000}-1-10^{-2000}).$ Thus, in the sense of
Newhouse thickness, they are very thin.

######  Remark 1.4.

By [35], it is possible to prove that for each missing digits set $K$, there
is an integer $N$ so that the $N$-fold multiplication set $K\cdot
K\cdot\dots\cdot K$ contains intervals. In fact, it is even possible to show
that the corresponding measure $m_{*}((\mu_{K})^{N})$ is a smooth function
where $m:(x_{1},\dots,x_{N})\to x_{1}\dots x_{N}$ is the multiplication map
and $\mu_{K}$ is the natural missing digits measure supported on $K.$

To motivate the general results in this paper, we first introduce three open
problems (Questions 1.5, 1.10, 1.11). They are all linked together with a
generalized version of a difficult conjecture of É. Borel. See [8]. At the end
of this paper, we will provide some more specific applications of our general
results. See Sections 8, 9.

###  1.1. Missing digits solutions for algebraic equations

The first problem is related to the consideration of numbers of missing digits
satisfying algebraic equations. Arithmetic properties of numbers with
restricted digits have been studied for example in [7], [6], [13], [14], [32]
as well as [33].

Consider the equation

$x^{3}+y^{3}=1,$

and we want to ask whether or not there are irrational solutions $(x,y)$ such
that both $x$ and $y$ do not have digit $1$ in their ternary expansions. More
generally, we introduce the notion of missing digits sets: Let $n\geq 1$ be an
integer. Let $p\geq 3$ be an integer. Let $D\subset\\{0,\dots,p-1\\}^{n}.$
Consider the set

$K_{p,D}=\mathrm{cl}{\\{x\in\mathbb{R}^{n}:[p\\{p^{k}x\\}]\in D,k\geq 0\\}},$

where $\\{x\\},[x]$ are the component wise fractional part, integer part
respectively of $x\in\mathbb{R}^{n}$ and $\mathrm{cl}A$ is the closure of
$A\subset\mathbb{R}^{n}$ under the standard topology. More precisely,
$\\{x\\}$ is the unique $y$ point in $[0,1)^{n}$ with $y-x\in\mathbb{Z}^{n}$,
and $[x]=x-\\{x\\}\in\mathbb{Z}^{n}.$

For convenience, we use $\hat{D}$ for the complement
$\\{0,\dots,p-1\\}^{n}\setminus D.$ For example, if $n=2,$ then
$K_{3,\hat{\\{(1,1)\\}}}\cap[0,1]^{2}$ is the Sierpinski carpet with base $3$.
Later on, we will call such a set $K_{p,D}$ to be a missing digits set.

######  Question 1.5.

Let $n\geq 1$ be an integer. Let $M$ be a non-degenerate analytic manifold.
Let $p>2$ be an integer and $D\subset\\{0,\dots,p-1\\}^{n}$ be a choice of
digits. Determine whether or not $M\cap K_{p,D}$ is infinite.

To guide our intuitions, we formulate the following perhaps ambitious
conjecture.

######  Conjecture 1.6.

Let $n\geq 1$ be an integer. Let $M$ be a strongly non-degenerate analytic
manifold. Let $p>2$ be an integer and $D\subsetneq\\{0,\dots,p-1\\}^{n}$ be a
choice of at least two digits. For $\delta>0,$ let $M^{\delta}$ be the
$\delta$-neighbourhood of $M$ in $\mathbb{R}^{n}.$ Then $(M\cap
K_{p,D})^{\delta}$ can be covered by

$\ll\left(\frac{1}{\delta}\right)^{\max\\{0,\dim_{\mathrm{H}}K_{p,D}-(n-\dim_{\mathrm{H}}M)\\}}$

many $\delta$-balls.

######  Remark 1.7.

Strongly non-degeneracy is not a common notion and we will not use it anywhere
in this paper. If $M$ is strongly non-degenerate, it means that $M$ is non-
degenerate and for each affine subspace $L\subset\mathbb{R}^{n},$

$\dim(M\cap L)\leq\max\\{0,\dim M-(n-\dim L)\\}.$

Intuitively, this condition says that $M$ is ’sufficiently twisted’. This
condition is to avoid the cases when $K_{p,D}$ is contained in an affine
subspace $L$ and $M\cap L$ has a larger dimension than expected so that $M\cap
K_{p,D}$ can be larger than what is stated in the conjecture. Simple examples
of strongly non-degenerate manifolds include unit spheres, Veronese curves,
etc.

######  Remark 1.8.

If the exponent of $(1/\delta)$ is equal to zero, this conjecture says that
$M\cap K_{p,D}$ is a finite set. We can push a bit further. If in addition,
$M$ is also an algebraic variety over $\mathbb{Q}$ and
$\dim_{\mathrm{H}}K_{p,D}-(n-\dim_{\mathrm{H}}M)<0$, then we expect that
$M\cap K_{p,D}$ consists of only rational points. For example, when $n=1$,
this falls in the range of the aforementioned conjecture of É. Borel which
says that all algebraic irrational numbers cannot miss digits in any base.

######  Remark 1.9.

We can formulate a slightly weaker conjecture with the covering number being

$\ll(1/\delta)^{\max\\{0,\dim_{\mathrm{H}}K_{p,D}-(n-\dim_{\mathrm{H}}M)\\}+\epsilon}$

for each $\epsilon>0$. This weaker conjecture is also open in general.

One of the results we will prove is that Conjecture 1.6 holds when $K_{p,D}$
is large enough. In that case, we also provide a natural lower counting
estimate for missing digits points in $M$, see Theorem 2.8. It is of interest
to find the ’smallest possible’ missing digits set for the above conjecture to
hold. As long as $M$ is fixed, we are able to provide a positive number
$\sigma(M)>0$ and examples of $K_{p,D}$ with $\dim_{\mathrm{H}}K_{p,D}$ being
larger than and arbitrarily close to $n-\sigma(M)$. Thus the missing digits
sets need not be too large in the sense of Hausdorff dimension. In Section
5.8, we demonstrate a particularly subtle difficulty in Conjecture 1.6 for
small missing digits sets.

###  1.2. Intersecting manifolds with fractals

We now discuss the number theoretic problem from a slightly different point of
view. Let $n\geq 2$ and $M\subset\mathbb{R}^{n}$ be a manifold. Let
$F\subset\mathbb{R}^{n}$ be a fractal set, e.g. a Sierpinski sponge. We are
interested in considering the intersection $M\cap F.$ In view of the classical
Marstrand slicing theorem (see [31, Theorem 6.9 and Section 7]), we see that
for a ’generic’ translation vector $a\in\mathbb{R}^{n}$,

(Dimension Reduction) $\displaystyle\dim_{\mathrm{H}}((M+a)\cap
F)\leq\max\\{0,\dim_{\mathrm{H}}F-(n-\dim_{\mathrm{H}}M)\\}.$

Of course, it is possible to quantify the word ’generic’ in a precise way.
Loosely speaking, there is a small (in terms of the Lebesgue measure or
Hausdorff dimension) exceptional set $E$ such that the above holds for all
$a\notin E.$ In this direction, a great amount of efforts have been made to
discover the occasions in which one can reduce the exceptional set from a
small set to the empty set. For $M$ being affine subspaces, see [17], [20],
[38], [43].

Intuitively speaking, the only chance for the above (Dimension Reduction) not
to hold would be that $M$ and $F$ share similar structures in small scales.
For example, if $M$ is a line parallel to one of the coordinate axis, then it
is possible that

$\dim_{\mathrm{H}}(M\cap
F)>\max\\{0,\dim_{\mathrm{H}}F-(n-\dim_{\mathrm{H}}M)\\}.$

This phenomenon happens in $\mathbb{R}^{2}$ already: just consider $F$ being
the twofold Cartesian product of the middle third Cantor set and $M$ being the
$X$ coordinate line. In [38], Shmerkin showed that for $M$ being lines, those
are essentially the only cases for (Dimension Reduction) not to hold:
(Dimension Reduction) holds for all lines with irrational slopes.

Now, suppose that $M$ has some curved structures. Then intuitively, we can
think that $M$ cannot share any structures with $F$ in small scales and
(Dimension Reduction) should hold without exceptions. Towards this direction,
we pose the following question.

######  Question 1.10.

Let $M\subset\mathbb{R}^{n}$ be a ’curved’ submanifold. Let $F$ be the
Sierpinski sponge (based in 3, say). Then we have

$\dim_{\mathrm{H}}M\cap
F\leq\max\\{0,\dim_{\mathrm{H}}F-n+\dim_{\mathrm{H}}M\\}.$

Although both of the arguments in [43] and [38] do not work directly to gain
information about intersections between ’curved’ manifolds and fractals, it is
perhaps possible to adapt the arguments to work out the case when the manifold
is a hypersurface with non-vanishing curvatures.222We thank P. Shmerkin for
explaining the details. Here we do not fix the notion of a submanifold
(possibly with codimension larger than one) being ’curved’. We leave the
interpretation open. One of the results in this paper is to answer this
question with a specific meaning to the word ’curved’.

###  1.3. Counting missing digits points near manifolds

It turns out that the intersection problem above is closely related to a
special lattice counting problem that we now introduce.

Recall the notion of missing digits sets $K_{p,D}.$ It is possible to
introduce a natural Lebesgue measure $\lambda_{p,D}$ on $K_{p,D}$ which will
we discuss later. For now, we just think $\lambda=\lambda_{p,D}$ to be a
probability measure supported on $K=K_{p,D}.$

Let $M\subset\mathbb{R}^{n}$ be a submanifold. Let $\delta>0.$ Consider the
$\delta$-neighbourhood $M^{\delta}$ of $M.$ We want to study the quantity

$\lambda(M^{\delta})$

for $\delta\to 0$. Heuristically, if $\lambda$ and $M$ are somehow
’independent’, we expect that

(Independence) $\displaystyle\lambda(M^{\delta})\asymp\delta^{n-\dim M}.$

Here $\dim M$ is the standard notion of the dimension of the submanifold $M$.
In our situation, $\dim M=\dim_{\mathrm{H}}M.$ The value $n-\dim M$ is usually
called the codimension of $M.$ The value $\delta^{n-\dim M}$ is roughly the
Lebesgue measure of $M^{\delta}.$ Now, assuming (Independence), we see that
$M^{\delta}\cap K$ can be covered with

$\ll\delta^{n-\dim M}/\delta^{\dim_{\mathrm{H}}K}$

many balls with radius $\delta.$ For this, we need to assume that $\lambda$ is
AD-regular with exponent $\dim_{\mathrm{H}}K.$ See Section 4.2. From here, we
directly deduce that

$\dim_{\mathrm{H}}(M\cap K)\leq\dim_{\mathrm{H}}K-(n-\dim M)$

if $\dim_{\mathrm{H}}K>n-\dim M.$ Otherwise, we simply have

(Finiteness) $\displaystyle\\#M\cap K<\infty.$

The conclusion (Finiteness) is rather strong, it says that if $K$ is a small
enough missing digits set then $K\cap M$ has only finitely many points. Of
course, we would have to assume the asymptotic bound
$\lambda(M^{\delta})\asymp\delta^{n-\dim M}$ which is not easy to be tested. A
particular special case (of Question 1.5) in this direction can be formulated
as follows.

######  Question.

Consider the circle $x^{2}+y^{2}=1$ in $\mathbb{R}^{2}.$ Can we find
infinitely many points $(x,y)$ on the circle with $x,y\in K_{5,\\{0,4\\}}$?

This is an open problem. Methods in [38] or [43] can be probably used to
deduce that the points under consideration form a set with zero Hausdorff
dimension but this is not enough to deduce the finiteness. More generally, we
shall consider the following problem.

######  Question 1.11.

Find examples of $M$ and $K_{p,D}$ (and $\lambda_{p,D}$) for which the
asymptotic estimate holds

$\lambda_{p,D}(M^{\delta})\asymp\delta^{n-\dim M}.$

##  2\. results in this paper

To state the results, we first introduce some terminologies. The reader can
skip to Section 2.4 and return here later when it is necessary.

###  2.1. Fourier norm dimensions

We need some notions of Fourier norm dimensions. They are useful in e.g. [45]
where a problem of counting rational points near missing digits sets was
considered. In an ongoing project, [2], an algorithm for computing the Fourier
$l^{1}$ dimensions of missing digits sets is developed together with many
applications to metric Diophantine approximation. In this paper, we do not
need to consider the precise values of the Fourier $l^{1}$-dimensions. We only
provide some bounds which are rather crude but enough for the discussions in
this paper. See Theorem 2.15.

Let $n\geq 1$ be an integer. Let $\mu$ be a compactly supported Borel
probability measure on $\mathbb{R}^{n}.$ Consider the Fourier transform

$\hat{\mu}(\xi)=\int_{\mathbb{R}^{n}}e^{-2\pi i(x,\xi)}d\mu(x),$

where $(.,.)$ is the standard Euclidean bilinear form.

######  Definition 2.1.

Let $p>0.$ We define

$\dim_{l^{p}}\mu=\sup\left\\{s>0:\sup_{\theta\in[0,1]^{n}}\sum_{|\xi|\leq
R,\xi\in\mathbb{Z}^{n}}|\hat{\mu}(\xi+\theta)|^{p}\ll R^{n-s}\right\\}.$

With the help of the Cauchy-Schwarz inequality, it is possible to show that

$\frac{\dim_{l^{2}}\mu}{2}\leq\dim_{l^{1}}\mu.$

Moreover, we have for each AD-regular (see Section 4.2) measure $\mu$

$\dim_{l^{2}}\mu=\dim_{\mathrm{H}}\mu=\dim_{\mathrm{H}}\mathrm{supp}(\mu).$

Furthermore, let $n\geq 1$ be an integer. Let $\mu_{1},\dots,\mu_{n}$ be a
Borel probability measure on $\mathbb{R}.$ The $n$-fold Cartesian product
$\mu^{\prime}=\mu_{1}\times\dots\times\mu_{n}$ satisfies

$\dim_{l^{1}}\mu^{\prime}\geq\dim_{l^{1}}\mu_{1}+\dots+\dim_{l^{1}}\mu_{n}.$

In fact, the equality holds when $\mu_{1},\dots,\mu_{n}$ are missing digits
measures but we do not need this fact.

We have seen above that $\dim_{l^{2}}\mu$ is closely related to
$\dim_{\mathrm{H}}\mu.$ The reason that we also study $\dim_{l^{1}}\mu$ is
that it gauges, in some sense, how ’close’ is $\mu$ from being a continuous
function. Observe that if the exponent in the definition of $\dim_{l^{1}}\mu$
can be chosen to be negative, then $\mu$ has an absolutely integrable Fourier
transform. This says that $\mu$ can be chosen to be the distribution
associated with a continuous density function. In this case,
$\mathrm{supp}\mu$ can be seen as a topological manifold.

For computations, it is often not very convenient to have two $\sup$’s. For
this reason, we also introduce the following two additional definitions.

######  Definition 2.2 (Integral).

Let $p>0.$ We define

$\dim^{I}_{l^{p}}\mu=\sup\left\\{s>0:\int_{|\xi|\leq
R}|\hat{\mu}(\xi)|^{p}d\xi\ll R^{n-s}\right\\}.$

######  Definition 2.3 (Sum).

Let $p>0.$ We define

$\dim^{S}_{l^{p}}\mu=\sup\left\\{s>0:\sum_{|\xi|\leq
R,\xi\in\mathbb{Z}^{n}}|\hat{\mu}(\xi)|^{p}\ll R^{n-s}\right\\}.$

In most of the situations, $\dim^{I}_{l^{p}},\dim^{S}_{l^{p}}$ can provide
useful information individually. Notice that in general,

$\dim_{l^{p}}\mu\leq\min\\{\dim^{I}_{l^{p}}\mu,\dim^{S}_{l^{p}}\mu\\}.$

This is because

$\displaystyle\int_{|\xi|\leq
R}|\hat{\mu}(\xi)|^{p}d\xi\leq\sum_{\xi\in\mathbb{Z}^{n},|\xi|\leq
1.5R}\int_{\theta\in[0,1]^{n}}|\hat{\mu}(\xi+\theta)|^{p}d\theta$
$\displaystyle=\int_{\theta\in[0,1]^{n}}\sum_{\xi\in\mathbb{Z}^{n},|\xi|\leq
1.5R}|\hat{\mu}(\xi+\theta)|^{p}d\xi$
$\displaystyle\leq\sup_{\theta\in[0,1]^{n}}\sum_{|\xi|\leq
2R,\xi\in\mathbb{Z}^{n}}|\hat{\mu}(\xi+\theta)|^{p}$

and

$\sum_{|\xi|\leq
R,\xi\in\mathbb{Z}^{n}}|\hat{\mu}(\xi)|^{p}\leq\sup_{\theta\in[0,1]^{n}}\sum_{|\xi|\leq
R,\xi\in\mathbb{Z}^{n}}|\hat{\mu}(\xi+\theta)|^{p}.$

We also suspect that for missing digits measures, the three notions

$\dim_{l^{p}},\dim^{I}_{l^{p}},\dim^{S}_{l^{p}}$

are identical.

###  2.2. Manifolds of finite type

Let $n\geq 2$ be an integer. A smooth submanifold $M\subset\mathbb{R}^{n}$ is
of finite type if for each $x\in M,$ the manifold $M$ only has finite order
contacts with affine hyperplanes at $x$. For more details, see [40, Chapter
VIII, Section 3]. We do not repeat the definition here and only illustrate
some of the examples.

(1). Let $M$ be an analytic submanifold. Suppose that $M$ is not contained in
any affine hyperplane, then $M$ is of finite type.

(2). As a particular example, consider the Veronese curve
$(t,t^{2},t^{3},\dots,t^{n}),t\in\mathbb{R}.$ This curve is analytic and it is
not contained in any affine hyperplane. Therefore it is of finite type.

(3) If $M$ is a smooth hypersurface with non-vanishing Gaussian curvature,
then $M$ is of finite type.

###  2.3. Missing digits sets and measures

We recall the notion of missing digits sets. Let $n\geq 1$ be an integer. Let
$p\geq 3$ be an integer. Let $D\subset\\{0,\dots,p-1\\}^{n}.$ Consider the set

$K_{p,D}=\mathrm{cl}\\{x\in\mathbb{R}^{n}:[p\\{p^{k}x\\}]\in D,k\geq 0\\},$

where $\\{x\\},[x]$ are the component wise fractional part, integer part
respectively of $x\in\mathbb{R}^{n}.$ Let $p_{1},\dots,p_{\\#D}$ be a
probability vector, i.e. they are non-negative and sum to one. We can then
assign each element in $D$ a probability weight. To be specific, one can first
introduce an ordering on $D$ and assign the probabilities accordingly. We can
now construct the random sum

$S=\sum_{i\geq 1}p^{-i}\mathbf{d}_{i}$

where $\mathbf{d}_{i}\in D,i\geq 1$ are randomly and independently chosen from
the set $D$ with the assigned probabilities.

If $p_{1}=\dots=p_{\\#D}=1/\\#D,$ the distribution of $S$ is a Borel
probability measure supported on $[0,1]^{n}.$ We call this measure to be
$\lambda_{p,D}.$ It is a Borel probability measure supported on
$K_{p,D}\cap[0,1]^{n}.$ Moreover, it is AD-regular with exponent
$\dim_{\mathrm{H}}K_{p,D}.$ We also write

$\dim_{l^{1}}K_{p,D}=\dim_{l^{1}}\lambda_{p,D}.$

We provide some examples of missing digits sets. Recall that
$\hat{D}=\\{0,\dots,p-1\\}^{n}\setminus D.$

(1) If $n=1,$ then $K_{3,\\{0,2\\}}\cap[0,1]$ is the middle third Cantor set
and $\lambda_{3,\\{0,2\\}}$ is the natural middle third Cantor-Lebesgue
measure.

(2) If $n=2,$ then $K_{3,\hat{\\{(1,1)\\}}}\cap[0,1]^{2}$ is the Sierpinski
Carpet with base $3.$ In general, if $n\geq 3,$
$K_{3,\hat{\\{(1,\dots,1)\\}}}\cap[0,1]^{n}$ are Sierpinski sponges.

######  Remark 2.4.

It is often interesting to consider Cartesian products of missing-digit sets
(measures) which are not necessarily missing-digit sets (measures) themselves.
In fact, they are not self-similar in general. Although we do only consider
missing-digit sets (measures) in this paper, many results can be extended to
deal with Cartesian products of missing-digit sets (measures). We will discuss
this in Section 8. Before that, one can ignore this technical consideration.

###  2.4. Results

Towards Conjecture 1.6, we prove the following theorem. Later on, we will
provide examples that fall into the range of this theorem. The conditions are
related to the $l^{1}$-dimensions of missing digits sets. This is not a common
notion of dimension. Loosely speaking, missing digits sets with large bases
have almost equal Hausdorff and $l^{1}$-dimensions. So it is helpful to think
$\dim_{l^{1}}$ below just as $\dim_{\mathrm{H}}.$333In Theorem 2.15(2) we will
see a precise version of this ’loose’ statement.

######  Theorem 2.5.

Let $n\geq 2$ be an integer. Let $M\subset\mathbb{R}^{n}$ be a manifold of
finite type. Then there is a number $\sigma=\sigma(M)>0$ such that for each
compactly supported Borel probability measure $\lambda$ with
$\dim_{l^{1}}\lambda>n-\sigma,$

$\lambda(M^{\delta})\ll\delta^{n-\dim M}.$

This proves Conjecture 1.6 for large missing digits sets. The number
$\sigma(M)$ can be explicitly determined once we know $M.$ It is related to
the Fourier decay properties of smooth surface measures carried by $M.$ This
number is always $\leq\dim M/2.$ In the case when $M$ is a hypersurface with
non-vanishing Gaussian curvature, it can be chosen to be $\dim M/2=(n-1)/2.$
The condition we have in this theorem is in some sense sharp. We postpone the
discussion to Theorem 2.12.

We will provide some crude bounds for $\dim_{l^{1}}\lambda_{p,D}$, see Theorem
2.15. In particular, the following result can be deduced.

######  Corollary 2.6 (Missing digits points near manifolds).

Let $n\geq 2$ be an integer. Let $M\subset\mathbb{R}^{n}$ be a manifold of
finite type. Let $k\geq 1$ be an integer. There is a number $p_{0}(M,k)$ so
that for each integer $p>p_{0}(M,k)$, Theorem 2.5 holds for $\lambda_{p,D}$
where the digit set $D$ satisfies $\\#D\geq p^{n}-k.$

The number $p_{0}(M,k)$ can be explicitly determined once we know $\sigma(M)$
and $k.$ Theorem 2.5 leads to an intersection result.

######  Corollary 2.7 (Intersections).

Let $n\geq 2$ be an integer. Let $M\subset\mathbb{R}^{n}$ be a manifold of
finite type. Then there is a number $\sigma=\sigma(M)>0$ such that for each
compactly supported Borel probability measure $\mu$ which is AD-regular and
$\dim_{l^{1}}\mu>n-\sigma,$

$\overline{\dim}_{\mathrm{B}}M\cap
supp(\mu)\leq\max\\{0,\dim_{\mathrm{H}}\mathrm{supp}(\mu)-(n-\dim M)\\}.$

On the other hand, we are also interested in having a lower estimate for
$M\cap K_{p,D}.$ However, this problem is not always sensible to ask because
$K_{p,D}$ has holes and it can happen that $M$ is contained in one of the
holes. This phenomenon is certainly not capturing the real nature between $M$
and $K.$ In order to unveil the underlying truth. We propose two compromises
by modifying $K$ or $M$ in ’acceptable’ manners that will be clarified soon.

First, we can render this problem by relaxing our requirement for missing
digits sets. Let $l\geq 0$ be an integer. We define

$K_{p,D,l}=\mathrm{cl}\\{x\in\mathbb{R}^{n}:[p\\{p^{k}x\\}]\in D,k\geq l\\}.$

So $K_{p,D,l}$ fills some of the holes of $K_{p,D}.$ The purpose of doing this
is to get rid of the special cases when the manifold just happens to be
contained in the holes of the missing digits set. We do not relax the
situation too much by requiring that $l$ has to be a fixed integer. For
example, the digital representations of $x\in K_{p,D,l}\cap[0,1]^{n}$ are of
no restriction only for the first finitely many digits. This is the
’acceptable’ modification for $K$ we mentioned earlier.

######  Theorem 2.8.

[Compromise 1] Let $M$ be a manifold of finite type. Let $K=K_{p,D}$ be a
missing digits set. Suppose that $\dim_{l^{1}}K_{p,D}+\sigma(M)-n>0$. Then
there is an $l\geq 0$ such that the set

$M\cap K_{p,D,l}$

is infinite. In fact, there is an $l\geq 0$ such that for all small enough
$\delta>0,$ to cover $M^{\delta}\cap K_{p,D,l}^{\delta},$ at least
$\gg(1/\delta)^{\dim_{\mathrm{H}}K+\dim M-n}$ many $\delta$-balls are needed.
Moreover, we have

$\underline{\dim}_{\mathrm{B}}M\cap
K_{p,D,l}\geq\dim_{\mathrm{H}}K-(n-\sigma(M)).$

Finally, $\bigcup_{l\geq 0}K_{p,D,l}\cap M$ is dense in $M.$

######  Remark 2.9.

It is tempting to believe that

$\dim_{\mathrm{B}}K_{p,D,l}\cap M=\dim_{\mathrm{H}}K-(n-\dim M).$

So far, we can conclude that

$\dim_{\mathrm{H}}K-n+\sigma(M)\leq\underline{\dim}_{\mathrm{B}}M\cap
K_{p,D,l}\leq\overline{\dim}_{\mathrm{B}}M\cap
K_{p,D,l}\leq\dim_{\mathrm{H}}K-(n-\dim M).$

Since $\sigma(M)\leq\dim M/2,$ the above chain of inequalities will never be
close to being optimal. Some other ideas need to be used in order to fill in
this gap.

The last statement seems trivial. The set $\cup_{l\geq 0}K_{p,D,l}$ is dense
in $\mathbb{R}^{n}.$ However, without the previous statement, it is even
unclear whether or not $\cup_{l\geq 0}M\cap K_{p,D,l}$ is empty.

Another possible way to compromise the fact that $K=K_{p,D}$ has holes is to
consider a family of transformed manifolds of $M.$ In this paper, we deal with
the group $E_{n}=(0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n).$ For
$(t,v,g)\in E_{n},$ it acts on $\mathbb{R}^{n}$ by first applying the rotation
$g$, then applying the scaling $t$, then applying the translation $v$:

$x\in\mathbb{R}^{n}\to T_{t,v,g}(x)=tg(x)+v.$

Observe that $E_{n}$ also acts on the space of non-degenerate manifolds and it
will not distort Fourier transforms too much. Thus we expect that

(Dimension Conservation) $\displaystyle\dim_{\mathrm{B}}T_{t,v,g}(M)\cap
K=\dim_{\mathrm{H}}K-(n-\dim M)$

and in particular $T_{t,v,g}(M)\cap K\neq\emptyset$ for many choices of
$(t,v,g)\in E_{n}.$ In view of Mattila’s intersection theorem ([31, Section
7]), we already know that the above holds for ’many’ choices of $(t,v,g)$ in a
metric sense. We now upgrade the metric result at a cost of dropping the
(Dimension Conservation).

######  Theorem 2.10.

[Compromise 2] Let $n\geq 2$ be an integer. Let $M$ be a manifold of finite
type. Suppose further that $\lambda$ is a missing digits measure supported on
$K.$ The function

$f:(t,v,g)\to\limsup_{\delta\to
0}\frac{\lambda((T_{t,v,g}(M))^{\delta})}{\delta^{n-\dim M}}$

is well-defined and non-negative valued. If $M$ is compact, then there are a
constant $c>1$ and a continuous and non-vanishing function
$h:E_{n}\to[0,\infty)$ such that

$c^{-1}h\leq f\leq ch.$

######  Remark 2.11.

(a). The compactness of $M$ in the last statement is not strictly necessary.
In fact, if $M$ is not compact, one can replace $M$ with a compactly supported
and smooth surface measure on $M$.

(b). This result is likely to hold for other classes of possibly nonlinear
transformations of manifolds other than the scalings, translations and
rotations, for example, the evolution of $M$ under a smooth vector field with
some regularity conditions. As $E_{n}$ is already enough for most of our
arithmetic applications we do not pursue this degree of generality.444This
paragraph is not mathematically precise. We only want to illustrate an idea
rather than a rigorous definition. One possible way of seeing the idea in this
paragraph is to consider a light and soft tissue floating in the air.

(c). We are curious about whether or not the function $f$ can be defined via
$\mathcal{H}^{s}$ for $s=\dim_{\mathrm{H}}K-(n-\dim M).$ More precisely, let

$h:(t,v,g)\to\mathcal{H}^{s}(T_{t,v,g}(M)\cap K).$

Is $h$ continuous? From Theorem 2.5, it is not hard to show that $h$ takes
real values (not $\infty$). In this way, one can the understand the Hausdorff
dimension of $T_{t,v,g}(M)\cap K.$

(d). Notice that whenever $f(t,v,g)>0,$ it follows that

$T_{t,v,g}(M)\cap K\neq\emptyset.$

However, this is not enough to conclude that $\dim_{\mathrm{B}}$ or
$\dim_{\mathrm{H}}T_{t,v,g}(M)\cap K\geq s.$ In this case, we have the weaker
result

$\underline{\dim}_{\mathrm{B}}T_{t,v,g}(M)\cap
K\geq\dim_{\mathrm{H}}K-(n-\sigma(M))$

as in Theorem 2.8.

The above results also hold when $M$ is replaced with other sets supporting
measures with polynomial Fourier decay. We will discuss this part in Section
9.

Finally, we emphasize that the condition $\dim_{l^{1}}\lambda>n-\sigma(M)$
cannot be dropped although we believe that it might not be the optimal
condition to look at. See Section 10.

######  Theorem 2.12 (Sharpness).

Let $M\subset\mathbb{R}^{2}$ be the curve

$M:|x|^{k}+|y-1|^{k}=1$

where $k\geq 2$ is an integer. Let $p>3$ be an integer and let
$D_{1}=\\{0,\dots,l_{1}\\}$, $D_{2}=\\{0,\dots,l_{2}\\}$ for some integers
$0<l_{1},l_{2}<p-1.$ Let $K=K_{p,D_{1}}\times K_{p,D_{2}}\cap[0,1]^{2}.$ Then
for $\delta>0,$ $M^{\delta}\cap K$ cannot be covered with

$\ll\left(\frac{1}{\delta}\right)^{(k-1)s/k}$

many $\delta$-balls where
$s=\max\\{\dim_{\mathrm{H}}K_{p,D_{1}},\dim_{\mathrm{H}}K_{p,D_{2}}\\}.$

######  Remark 2.13.

In particular, if $k=2,$ then $M$ is a circle. In this case, the threshold for
$\dim_{l^{1}}$ in Theorem 2.5 is $n-\sigma(M)=2-1/2=3/2.$ It is possible to
choose $D_{1},D_{2}$ such that $\dim_{\mathrm{H}}K_{p,D_{1}}$ is close to one
and $\dim_{\mathrm{H}}K_{p,D_{2}}$ is close to $1/2.$ Then
$\dim_{\mathrm{H}}K_{p,D_{1}}\times K_{p,D_{2}}$ is close to $3/2.$ Moreover,
if $p$ is large enough, then $\dim_{l^{1}}K_{p,D_{1}}\times K_{p,D_{2}}$ is
also close to $3/2$. See Theorem 2.15. In this case we see that
$M^{\delta}\cap K$ cannot be covered with

$\ll\left(\frac{1}{\delta}\right)^{s/2}$

many $\delta$-balls. Notice that $s/2$ can be made arbitrarily close to $1/2$.
On the other hand, if the conclusion of Theorem 2.5 would hold, then
$M^{\delta}\cap K$ can be covered by

$\ll\left(\frac{1}{\delta}\right)^{\dim_{\mathrm{H}}K-1}$

many $\delta$-balls. Therefore if $\dim_{\mathrm{H}}K-1<s/2,$ then the
conclusion of Theorem 2.5 cannot hold. Since $s$ can be chosen to be close to
one, we are able to find examples of missing digits measures $\lambda$ with
$\dim_{l^{1}}\lambda$ smaller but arbitrarily close $3/2$ such that the
conclusion of Theorem 2.5 does not hold for $\lambda$ and $M.$

In general, the above discussion works for $k\geq 3$ as well. In this case,
$M$ is a ’flatter’ curve than the circle. We have $n-\sigma(M)=2-(1/k).$ This
is because $M$ has $k$th order contact with the $X$-axis. See Section 5.8 for
more details. On the other hand, as above we see that if
$\dim_{\mathrm{H}}K-1<(k-1)s/k,$ then Theorem 2.5 cannot hold for the missing
digit measure on $K.$ Thus we are able to find examples of missing digits
measures $\lambda$ with $\dim_{l^{1}}\lambda$ smaller but close to $2-(1/k)$
such that the conclusion of Theorem 2.5 does not hold for $\lambda$ and $M.$

######  Remark 2.14.

Theorem 2.12 does not disprove Conjecture 1.1. However, it does show that it
is perhaps a challenging task to reduce the $3/4$ threshold. Theorem 2.12 does
not disprove Conjecture 1.6 either, because in the statement of Conjecture
1.6, what we are interested in is to cover the set $(M\cap K)^{\delta},$ which
is in general smaller than $M^{\delta}\cap K^{\delta}.$

###  2.5. Fourier norm dimensions of missing digits measures

Missing digits sets and measures provide us with a large class of examples of
fractal sets and measures for the results in the previous section to be
applied. In this section, we list a few general results regarding the
$l^{1}$-dimensions of missing digits sets whose proofs are provided to make
this paper self-contained555In [2], a much more precise method is developed..

######  Theorem 2.15.

Let $n\geq 1$ be an integer. The following results hold.

  * •

1 Let $t\geq 1$ be an integer. We have

$\liminf_{p\to\infty,\\#D\geq p^{n}-t}\dim_{l^{1}}\lambda_{p,D}=n.$

In particular, for each number $\epsilon>0,$ as long as $p$ is large enough,
$\dim_{l^{1}}\lambda_{p,D}>n-\epsilon$ holds for each $D$ with $\\#D=p^{n}-1.$

  * •

2 For each integer $p\geq 4,$ let $D\subset\\{0,\dots,p-1\\}^{n}$ be a
’rectangle’, i.e. a set of form
$[a_{1},b_{1}]\times[a_{2},b_{2}]\dots[a_{n},b_{n}]\cap\\{0,\dots,p-1\\}^{n}.$
Then we have 666The base of $\log$ in this paper is $e$.

(T)
$\displaystyle\dim_{l^{1}}\lambda_{p,D}\geq\dim_{\mathrm{H}}\lambda_{p,D}-\frac{n\log\log
p^{2}}{\log p}.$

The significance of this result is to show that missing digits sets can have
almost equal Hausdorff and $l^{1}$-dimensions.

######  Remark 2.16.

The digit set structure is rather special for part 2. We emphasize that it can
be a little bit more complicated. More precisely, part 2 of this theorem also
works for rectangles with gaps larger than $1.$ For example, instead of being
a product set of sets of consecutive integers, it can be a product set of sets
of consecutive even numbers. Also, one can consider unions of rectangles as
long as the number of rectangles in the union is not too large compared with
$p,$ e.g. $O(p^{\epsilon})$. We are interested in finding the most general
condition for the digit sets so that a lower bound like (T) holds, i.e. as
$p\to\infty,$

$|\dim_{l^{1}}\lambda_{p,D}-\dim_{\mathrm{H}}\lambda_{p,D}|=o(1).$

Towards the question in this remark, we pose the following possibly too
optimistic conjecture.

######  Conjecture 2.17.

Let $p>2$ be a prime number. Let $\epsilon\in(0,1).$ Then we have

$|\dim_{l^{1}}\lambda_{p,D}-\dim_{\mathrm{H}}\lambda_{p,D}|=o(1),$

where the $o(1)$ term is uniform across all digit set $D$ with

$\\#D\geq p^{\epsilon}.$

The condition that $p$ is a prime number cannot be completely dropped. In
fact, if $p=3^{n}$ where $n$ is an integer. Then it is possible to choose $D$
such that $K_{p,D}$ is the middle third Cantor set. The statement of this
conjecture is obviously not true in this case.

##  3\. Other related works

The problem in this paper can be generally described as "understanding the
distribution of a special class of points near another set." Consider the
following three classes of sets:

1\. Manifolds

2\. Rational numbers with bounded heights (denominators).

3\. Missing digits sets.

We studied the distribution of 3 near 1 in this paper. On the other hand, it
has been of great interest in considering the distribution of 2 near 1 (see
[3], [5], [4], [27], [21], [36]) as well as 2 near 3 (see [1], [2], [9], [10],
[15], [25], [26], [28], [37], [39], [42], [45]).777Those are just a small
fraction of all the works on those topics.

The Fourier analysis method in this paper can be found in many of the above
references. Other than this Fourier analysis method, another major tool for
considering the above counting problems is by using the theory of dynamics on
homogeneous spaces, e.g. [27] (2 near 1), [25] (2 near 3). A dynamical
approach for the problems considered in this paper (3 near 1) is likely to be
found.

##  4\. Preliminaries

Before we prove Theorem 2.5, we need to introduce some more definitions as
well as some results which will be used without proofs. See [16], [30] for
more details on the notions of dimensions and regularities of measures.

###  4.1. Hausdorff dimension, box dimensions

Let $n\geq 1$ be an integer. Let $F\subset\mathbb{R}^{n}$ be a Borel set. Let
$g:[0,1)\to[0,\infty)$ be a continuous function such that $g(0)=0$. Then for
all $\delta>0$ we define the quantity

$\mathcal{H}^{g}_{\delta}(F)=\inf\left\\{\sum_{i=1}^{\infty}g(\mathrm{diam}(U_{i})):\bigcup_{i}U_{i}\supset
F,\mathrm{diam}(U_{i})<\delta\right\\}.$

The $g$-Hausdorff measure of $F$ is

$\mathcal{H}^{g}(F)=\lim_{\delta\to 0}\mathcal{H}^{g}_{\delta}(F).$

When $g(x)=x^{s}$ then $\mathcal{H}^{g}=\mathcal{H}^{s}$ is the $s$-Hausdorff
measure and Hausdorff dimension of $F$ is

$\dim_{\mathrm{H}}F=\inf\\{s\geq 0:\mathcal{H}^{s}(F)=0\\}=\sup\\{s\geq
0:\mathcal{H}^{s}(F)=\infty\\}.$

Next, let $F\subset\mathbb{R}^{n}$ be a Borel set. Let $\delta>0$ and
$N_{\delta}(F)$ be the minimum amount of $\delta$-balls needed to cover $F.$
Then the upper box dimension of $F$ is

$\overline{\dim}_{\mathrm{B}}F=\limsup_{\delta\to\infty}\frac{-\log
N_{\delta}(F)}{\log\delta}.$

The lower box dimension is

$\underline{\dim}_{\mathrm{B}}F=\liminf_{\delta\to\infty}\frac{-\log
N_{\delta}(F)}{\log\delta}.$

If the upper and lower box dimensions of $F$ are equal, we call the comment
value to be the box dimension of $F$

$\dim_{\mathrm{B}}F=\underline{\dim}_{\mathrm{B}}F=\overline{\dim}_{\mathrm{B}}F.$

If $F$ is compact, then we have the general result

$\dim_{\mathrm{H}}F\leq\underline{\dim}_{\mathrm{B}}F\leq\overline{\dim}_{\mathrm{B}}F.$

###  4.2. AD-regularity

Let $n\geq 1$ be an integer. Let $\mu$ be a Borel measure. Let $s>0$ be a
number. We say that $\mu$ is $s$-regular, or AD-regular with exponent $s$ if
there is a constant $C>1$ such that for all $x\in supp(\mu)$ and all small
enough $r>0$

$C^{-1}r^{s}\leq\mu(B_{r}(x))\leq Cr^{s},$

where $B_{r}(x)$ is the Euclidean ball of radius $r$ and centre $x$. For an
AD-regular measure $\mu$, the exponent can be seen as

$s=\dim_{\mathrm{H}}supp(\mu).$

For this reason, we simply define for AD-regular measure $\mu,$

$\dim_{\mathrm{H}}\mu=\dim_{\mathrm{H}}supp(\mu).$

Missing digits measure $\lambda_{p,D}$ in $\mathbb{R}^{n}$ are AD-regular
measures with exponent

$s=\dim_{\mathrm{H}}\lambda_{p,D}=\dim_{\mathrm{H}}K_{p,D}=\frac{\log\\#D}{\log
p^{n}}.$

###  4.3. Fourier transform of surface measures on manifolds of finite type

Discussions in this section follow [40, Chapter VIII, Section 3].

Let $n\geq 2$ be an integer and $M\subset\mathbb{R}^{n}$ be a manifold of
finite type. Let $\mu$ be the surface measure, i.e. the natural Lebesgue
measure carried by the manifold. If $M$ is compact, we can normalize $\mu$ to
be a probability measure. Otherwise, we shall truncate the measure to vanish
at infinity by choosing a smooth, compactly supported function $\phi$ and
denote

$\mu^{\prime}=c\phi\mu$

where $c$ is the normalization factor so that $\mu^{\prime}$ is a probability
measure.

We will always perform the truncating procedure. This will cause no loss of
generality for the problems considered in this paper.

A standard result of smooth surface measures carried by manifolds of finite
type is that the Fourier transforms decays polynomially,

$\hat{\phi\mu}(\xi)=O(|\xi|^{-\sigma}).$

Where $\phi$ is a smooth, compactly supported function on $\mathbb{R}^{n}.$ Of
course, if $M$ is compact, one can also let $\phi=1.$

Here lower bounds for $\sigma$ can be effectively determined. It is related to
the type of $M.$ Roughly speaking, $\sigma=1/k$ where $k$ is the smallest
integer such that $M$ does not have $k$-th order contact with affine
hyperplanes. For the Veronese curve, one can choose $\sigma=1/n.$ If $M$ is a
hypersurface with non-vanishing curvatures, then the Fourier transform has a
much better decay: one can choose $\sigma=(n-1)/2=2^{-1}\dim M.$

The choice of $\sigma(M)$ for non-degenerate manifolds is a challenging topic
in harmonic analysis and differential geometry. Apart from some general
results in Stein’s book [40], the article [11] contains some more technical
results that are useful to provide estimates of $\sigma(M)$ for some specially
defined $M$, e.g. joint graphs of analytic maps.

###  4.4. Fourier transform of missing digits measures

The discussion in this subsection works for all self-similar measures with a
uniform contraction ratio. We nonetheless focus only on missing digits
measures.

Let $n\geq 1$ be an integer. Let $p>2$ be an integer and
$D\subset\\{0,\dots,p-1\\}^{n}.$ Consider the missing digit measure
$\lambda_{p,D}.$ In this case, $\hat{\lambda}_{p,D}(\xi)$ can be computed with
the help of the formula

$\hat{\lambda}_{p,D}(\xi)=\prod_{j\geq 0}\frac{1}{\\#D}\sum_{d\in D}e^{-2\pi
i(d,\xi)/p^{j}}.$

###  4.5. Asymptotic notations

We use both the Vinogradov ($\ll,\gg,\asymp$) as well as Bachmann-Landau
($O(),o()$) notations:

Let $f(\delta),g(\delta)$ be two real valued quantities depending on
$\delta>0.$ Then

  * •

$f\ll g$ or $f=O(g)$ if $|f(\delta)|\leq C|g(\delta)|$ for a constant $C>0$
and all $\delta>0.$

  * •

$f=o(g)$ if for each $\epsilon>0,$ there is a $\delta_{0}>0$ such that for all
$\delta<\delta_{0},$

$|f(\delta)|\leq\epsilon|g(\delta)|.$

  * •

$f\asymp g$ if $f\ll g$ and $g\ll f$.

##  5\. Proof of the results

###  5.1. Thickening the surface measure

Let $n\geq 2$ be an integer. Let $M\subset\mathbb{R}^{n}$ be a smooth
submanifold of finite type. Let $\mu$ be a probability measure supported on
$M.$ We assume that $\mu$ is compactly supported and smooth. Those conditions
are not necessary for the following arguments, we only assume them for
convenience.

For later considerations, we need to thicken the measure $\mu.$ Let
$\delta>0.$ If $\delta$ is small enough, the neighbourhood $M^{\delta}$
containing points in $\mathbb{R}^{n}$ that are $\delta$ close to $M$ provides
us with a good approximation of $M.$ Let $\phi$ be a compactly supported and
smooth function which equals to one on $B_{1/2}(0)$ and vanish outside of
$B_{1}(0).$ We also arrange that $\phi$ and $\hat{\phi}$ are spherically
symmetric positive valued functions. See [44]. Furthermore, we shall assume
that $\int\phi=1.$ Let $\phi_{\delta}(x)=\delta^{-n}\phi(\delta x).$

We see that the convoluted measure $\mu_{\delta}=\phi_{\delta}*\mu$ is a
probability measure and it is also a smooth function supported on
$M^{\delta}.$ It can be also checked that on $M^{\delta/2},$
$\mu_{\delta}\asymp\delta^{-(n-\dim M)}$ and in general

$\mu_{\delta}(x)\ll\delta^{-(n-\dim M)}$

uniformly across $x\in\mathbb{R}^{n}.$

Observe that

$\hat{\phi_{\delta}*\mu}(\xi)=\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi).$

The function $\hat{\phi}$ decays fast at infinity. In fact, for each number
$N>0,$ we have

$|\hat{\phi}(\xi)|=O(|\xi|^{-N}).$

Intuitively (after scaling by $\delta$), this tells that $\hat{\phi}_{\delta}$
is a smooth function which is essentially supported on $B_{1/\delta}(0).$
Within the ball $B_{\delta}(0)$ we have

${\phi}_{\delta}\asymp\delta^{-n},$

that is, the value of $\phi_{\delta}/\delta^{-n}$ on $B_{\delta}(0)$ is
bounded and strictly positive (in a manner than does not depend on $\delta$).
Similarly, on $B_{1/\delta}(0),$ we have

$\hat{\phi}_{\delta}\asymp 1.$

###  5.2. Littlewood-Paley decomposition, proof of Theorem 2.5

Let $\lambda$ be a Borel probability measure, by Plancherel’s theorem, we have

$\int\mu_{\delta}(x)d\lambda(x)=\int\hat{\mu}_{\delta}(\xi)\overline{\hat{\lambda}(\xi)}d\xi.$

For each $r>0,$ let $B_{r}\subset\mathbb{R}^{n}$ be the metric ball with
radius $r$ centred at the origin. For each number $q\geq 0,$ let
$S_{q}=B_{2^{q+1}}\setminus B_{2^{q}}.$ Then we see that,

(I)
$\displaystyle\int|\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi)\overline{\hat{\lambda}(\xi)}|d\xi=\int_{B_{1}}|\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi)\overline{\hat{\lambda}(\xi)}|d\xi+\sum_{j\geq
0}\int_{S_{j}}|\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi)\overline{\hat{\lambda}(\xi)}|d\xi.$

Let $\kappa_{1}<\dim_{l^{1}}\lambda\leq\dim^{I}_{l^{1}}\lambda.$ Observe that

(II)
$\displaystyle\int_{S_{j}}|\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi)\overline{\hat{\lambda}(\xi)}|d\xi\ll\frac{1}{2^{j\sigma(M)}}\int_{S_{j}}|\overline{\hat{\lambda}(\xi)}|d\xi\ll\frac{2^{(j+1)(n-\kappa_{1})}}{2^{j\sigma(M)}},$

where $\sigma(M)>0$ can be chosen according to the discussions in Section 4.3.
Thus as long as $\sigma(M)+\dim_{l^{1}}\lambda>n$, we can choose $\kappa_{1}$
such that

$n-\kappa_{1}<\sigma(M).$

This implies that (we do not take the norm)

$\int\hat{\phi}_{\delta}(\xi)\hat{\mu}(\xi)\overline{\hat{\lambda}(\xi)}d\xi=O(1).$

Next, observe that

$\lambda(M^{\delta/2})=\int_{M^{\delta/2}}d\lambda(x)\ll\delta^{n-\dim
M}\int\mu_{\delta}(x)d\lambda(x)\ll\delta^{n-\dim M}.$

This finishes the proof of Theorem 2.5.

###  5.3. From the counting estimate to an intersection estimate

Now let $\lambda$ be an AD-regular measure. Suppose that $\dim_{l^{1}}\lambda$
is large enough so that Theorem 2.5 applies.

We then see that

$\lambda(M^{\delta})\ll\delta^{n-\dim M}.$

Since $\lambda$ is AD-regular, we see that
$M^{\delta}\cap\mathrm{supp}(\lambda)$ can be covered by

$\ll\delta^{n-\dim M}/\delta^{\dim_{\mathrm{H}}\lambda}$

many $\delta$-balls. By letting $\delta\to 0$ we see that

$\overline{\dim}_{\mathrm{B}}(M\cap\mathrm{supp}(\lambda))\leq\dim_{\mathrm{H}}\lambda-(n-\dim
M)$

if $\dim_{\mathrm{H}}\lambda-(n-\dim M)>0.$ Otherwise we have

$\overline{\dim}_{\mathrm{B}}(M\cap\mathrm{supp}(\lambda))=0.$

This proves Corollary 2.7.

###  5.4. Lower estimate, proof of Theorem 2.8

We first reduce the problem a little bit. First, we pick a compact subset of
$M$ by performing the smooth truncation discussed in Section 4.3. This step is
not necessary if $M$ is already compact. Let $\mu$ be the chosen smooth
compactly supported measure on $M$. Next, we choose an integer $l\geq 0.$ We
consider the measure

$\mu_{l}=\frac{1}{p^{nl}}\sum_{d\in\\{0,\dots,p^{l}-1\\}^{n}/p^{l}}\mu(.+d).$

It is an average of $p^{nl}$ many translated copies of $\mu.$ Next, we perform
the $\mod\mathbb{Z}^{n}$ action. We use $\mu_{l}$ to denote also the image of
$\mu_{l}$ under the action $\mod\mathbb{Z}^{n}.$ The $\mod\mathbb{Z}^{n}$ can
be performed because $K_{p,D}$ is already $\mathbb{Z}^{n}$ periodic and
$\mathrm{supp}({\mu}_{l})$ is compact.888It is important that
$\mathrm{supp}({\mu}_{l})$ is compact. Otherwise the image of $\mu_{l}$ under
$\mod\mathbb{Z}^{n}$ might be supported on a dense subset. This will let us
concentrate on the unit cube $[0,1]^{n}.$

Now we view the whole situation on
$\mathbb{R}^{n}/\mathbb{Z}^{n}\approx[0,1)^{n}.$ It can be checked that for
$\xi\in\mathbb{Z}^{n}$

$\hat{\mu}_{l}(\xi)=\Delta_{p^{l}|\xi}\hat{\mu}(\xi),$

where $\Delta_{p^{l}|\xi}=1$ if $p^{l}$ divides all the components of $\xi$ or
else $\Delta_{p^{l}|\xi}=0.$

Let $\delta>0$ be a small number. We consider the thickened measure
$\lambda_{\delta}=\phi_{\delta}*\lambda_{p,D}.$ Just as $\mu_{\delta},$ we see
that $\lambda_{\delta}\asymp\delta^{-(n-\dim_{\mathrm{H}}K_{p,D})}$ on
$K^{\delta/2}_{p,D}.$ We see that

(III)
$\displaystyle\mu_{l}(K^{\delta}_{p,D})\gg\delta^{n-\dim_{\mathrm{H}}K_{p,D}}\int_{\mathbb{R}^{n}/\mathbb{Z}^{n}}\lambda_{\delta}(x)d\mu_{l}(x)=\delta^{n-\dim_{\mathrm{H}}K_{p,D}}\sum_{\xi\in\mathbb{Z}^{n}}\hat{\lambda}_{\delta}(\xi)\hat{\mu}_{l}(-\xi).$

The above sum converges because $\lambda_{\delta}$ is a Schwartz function. We
can now perform the arguments in Section 5.2. First observe that

$\sum_{\xi\in\mathbb{Z}^{n}}\hat{\lambda}_{\delta}(\xi)\hat{\mu}_{l}(-\xi)=\hat{\lambda}_{\delta}(0)\hat{\mu}_{l}(0)+\sum_{p^{l}|\xi,\xi\neq
0}\hat{\lambda}_{\delta}(\xi)\hat{\mu}_{l}(-\xi).$

For the second sum above, we see that $|\xi|\geq p^{l}$ because at least one
of the components of $\xi$ is non-zero and divisible by $p^{l}.$

We can perform the summation version of the argument in (I), (II) in Section
5.2. The effect of considering $\mu_{l}$ instead of $\mu$ is to push the off-
zero $L^{2}$-sum in (III) away from the origin. Since
$\dim_{l^{1}}\lambda_{p,D}\leq\dim^{S}_{l^{1}}\lambda_{p,D}.$ We see that

$\dim^{S}_{l^{1}}\lambda_{p,D}+\sigma(M)\geq\dim_{l^{1}}\lambda_{p,D}+\sigma(M)>n.$

Then we see that (similar to (II)) as $l\to\infty$

$\displaystyle\sum_{p^{l}|\xi,\xi\neq
0}|\hat{\lambda}_{\delta}(\xi)\hat{\mu}_{l}(-\xi)|$
$\displaystyle\leq\sum_{\xi\in\mathbb{Z}^{n},|\xi|\geq
p^{l}}|\hat{\lambda}_{\delta}(\xi)\hat{\mu}(-\xi)|$
$\displaystyle\leq\sum_{j\geq
l}\sum_{\xi\in\mathbb{Z}^{n},p^{j-1}\leq|\xi|<p^{j}}|\hat{\lambda}_{\delta}(\xi)\hat{\mu}(-\xi)|$
$\displaystyle\ll\sum_{j\geq l}p^{-j\sigma(M)}p^{j(n-\kappa_{1})},$

where on the last line $\kappa_{1}$ is a positive number that can be chosen to
be arbitrarily close to $\dim_{l^{1}}\lambda_{p,D}.$ The implicit constant in
$\ll$ depends on $\kappa_{1}$ and does not depend on $l,\delta$. In
particular, $\kappa_{1}$ can be chosen such that

$\kappa_{1}+\sigma(M)>n.$

For this fixed $\kappa_{1},$ we see that

$\sum_{j\geq l}p^{-j\sigma(M)}p^{j(n-\kappa_{1})}\leq
p^{l(n-\sigma(M)-\kappa_{1})}\frac{1}{1-p^{n-\sigma(M)-\kappa_{1}}}.$

Thus we see that as $l\to\infty,$

$\sum_{p^{l}|\xi,\xi\neq 0}|\hat{\lambda}_{\delta}(\xi)\hat{\mu}(-\xi)|\to 0$

uniformly across $\delta>0.$ Now observe that

$\hat{\lambda}_{\delta}(0)\hat{\mu}_{l}(0)=1$

because $\lambda_{\delta},\mu_{l}$ are all probability measures. Thus as long
as $l$ is large enough we have for all small enough $\delta,$

(Lower)
$\displaystyle\mu_{l}(K^{\delta}_{p,D})>c\delta^{n-\dim_{\mathrm{H}}K_{p,D}}$

for a constant $c>0$ which depends on the choice of the bump function $\phi.$

Observe that $\mu_{l}$ is essentially a smooth surface measure carried by a
manifold. This is not exactly the case because $\mu_{l}$ is actually carried
by a finite union of manifolds. We denote the finite union of manifolds to be
$\tilde{M}.$ The estimate

$\mu_{l}(B_{\delta}(x))\asymp\delta^{\dim M}$

holds uniformly for all $x\in\mathrm{supp}(\mu_{l})$ and all small enough
$\delta>0.$ Of course, the implicit constants depend on $l.$ From here we see
that (Lower) implies that in order to cover $K^{\delta}_{p,D}\cap\tilde{M}$
with $\delta$-balls, one needs

$\gg\delta^{n-\dim_{\mathrm{H}}K_{p,D}-\dim M}=\delta^{-s}$

many of them where $s=\dim_{\mathrm{H}}K_{p,D}+\dim M-n>0.$

As a simple observation, from (Lower), we see that $\tilde{M}\cap K$ cannot be
empty. Indeed, if $\tilde{M}\cap K=\emptyset,$ then there is a $\delta_{0}>0$
such that $d(\tilde{M},K)>\delta_{0}.$ This is because $M$ and $K$ are closed
and compact (we already reduced the whole situation to $[0,1]^{n}$). This
means that $\mu_{l}(K^{\delta}_{p,D})=0$ as long as $\delta<\delta_{0}/10.$
This contradicts (Lower).

Having shown that $\tilde{M}\cap K$ is not empty, we now show that it has
positive dimension. Let $\epsilon>0.$ Let $\delta>0$ be a power of $p^{-1}$
and let $\mathcal{C}$ be any collection of $\delta$-balls with

$\\#\mathcal{C}\leq(1/\delta)^{\epsilon}.$

We want to show that $\mathcal{C}$ cannot cover $\tilde{M}\cap K$ as long as
$\epsilon$ is small enough. Let $C\in\mathcal{C}.$ Denote $10C$ by the ball
co-centered with $C$ of radius $10\delta.$ Then for each small enough
$\delta_{C}>0,$ there is a number $\eta>0$ such that

$\tilde{M}\cap K^{\delta_{C}}\cap 10C$

can be covered with at most

(Counting Down)
$\displaystyle\delta^{\eta}\left(\frac{1}{\delta_{C}}\right)^{s}$

many $\delta_{C}$-balls. This can be seen by rescaling the whole situation by
a factor $1/(10\delta).$ Such a zoom action does not change the Fourier decay
properties for $K$ or for $M$. Of course, some explicit constants are changed.
This results the additional scaling factor $\delta^{\eta}$ for some $\eta>0.$
We will discuss about this later.

We apply the above argument for each $C\in\mathcal{C}$ and find a small enough
number $\delta^{\prime}>0$ such that each $10C\cap\tilde{M}\cap K$ can be
covered by at most

$\delta^{\eta}\left(\frac{1}{\delta^{\prime}}\right)^{s}$

many $\delta^{\prime}$-balls. Thus in total, one needs at most

$\delta^{\eta}\left(\frac{1}{\delta}\right)^{\epsilon}\left(\frac{1}{\delta^{\prime}}\right)^{s}=\delta^{-\epsilon+\eta}\left(\frac{1}{\delta^{\prime}}\right)^{s}.$

many $\delta^{\prime}$-balls to cover
$\bigcup_{C\in\mathcal{C}}10C\cap\tilde{M}\cap K^{\delta^{\prime}}.$ If
$0<\epsilon<\eta$ we can make $\delta^{-\epsilon+\eta}$ arbitrarily small by
choosing $\delta$ to be small. Now we use (Lower) for $\delta^{\prime}.$ We
need at least

$\gg\left(\frac{1}{\delta^{\prime}}\right)^{s}$

many $\delta^{\prime}$-balls to cover

$\tilde{M}\cap K^{\delta^{\prime}}.$

We thus see that for each small enough $\delta>0,$ as long as
$\delta^{\prime}$ is small enough, there must exist possibly many
$\delta^{\prime}$-balls intersecting $\tilde{M}\cap K$ and they are not
intersecting $\cup_{C\in\mathcal{C}}2C.$ Thus, we apply the above result with
$\delta^{\prime}\to 0$. As a result, we find balls $B_{\delta^{\prime}}$ of
radius $\delta^{\prime}\to 0$,

$B_{\delta^{\prime}}\cap K\cap\tilde{M}\neq\emptyset$

and

$d\left(B_{\delta^{\prime}},\bigcup_{C\in\mathcal{C}}C\right)>\delta.$

Since $K$ and $\tilde{M}$ are compact, we see that there is a point $x\in
K\cap\tilde{M}$ with

$d\left(x,\bigcup_{C\in\mathcal{C}}C\right)\geq\delta>0.$

Thus $\mathcal{C}$ cannot cover $K\cap\tilde{M}.$ We have seen that
$K\cap\tilde{M}$ can not be covered by $(1/\delta)^{\epsilon}$ many
$\delta$-balls as long as $\delta$ is small enough. This shows that

$\underline{\dim}_{\mathrm{B}}K\cap\tilde{M}\geq\epsilon.$

Now we discuss how we can choose $\epsilon,\eta$ in (Counting Down). Let
$\delta>0$ be a power of $p^{-1}.$ We consider a $\delta$-branch of $K,$ say
$K^{\prime}.$ Notice that $K$ is self-similar and the $\delta$-branch
$K^{\prime}$ is just $K$ scaled down by the factor $\delta.$ This scaling
procedure affects the Fourier transform as follows. Let $\lambda^{\prime}$ be
the natural missing digits measure supported on $K^{\prime}$, i.e.
$\lambda^{\prime}=(\lambda(K^{\prime}))^{-1}\lambda_{|K^{\prime}}$. Denote
$\kappa_{2}=\dim_{\mathrm{H}}K_{p,D}$. Then we see that

(Rescale)
$\displaystyle|\hat{\lambda}^{\prime}(\xi)|=|\hat{\lambda}(\delta\xi)|.$

We can now argue as in Section 5.2 with $\lambda^{\prime}$ in the place of
$\lambda,$

$\displaystyle\int_{\mathbb{R}^{n}}|\hat{\mu}(\xi)\hat{\lambda^{\prime}}(\xi)|d\xi=\int_{\mathbb{R}^{n}}|\hat{\mu}(\xi)||\hat{\lambda}(\delta\xi)|d\xi$
$\displaystyle=\int_{\mathbb{R}^{n}}|\hat{\mu}(\xi/\delta)||\hat{\lambda}(\xi)|\delta^{-n}d\xi$
$\displaystyle\ll\delta^{-n}\int_{\mathbb{R}^{n}}|\xi/\delta|^{-\sigma(M)}|\hat{\lambda}(\xi)|d\xi$
$\displaystyle=\delta^{-n+\sigma(M)}\int_{\mathbb{R}^{n}}|\xi|^{-\sigma(M)}|\hat{\lambda}(\xi)|d\xi.$

From here we can use the inequalities in (II) to deduce that for
$\delta^{\prime}\to 0,$

$\lambda^{\prime}(M^{\delta^{\prime}})\ll(1/\delta)^{n-\sigma(M)}{\delta^{\prime}}^{n-\dim
M},$

where the multiplicative constant in $\ll$ is the same as at the end of
Section 5.2. Since

$\lambda(K^{\prime})\asymp\delta^{\kappa_{2}},$

we see that $K^{\prime}\cap M^{\delta^{\prime}}$ can be covered by at most

$\ll\frac{(1/\delta)^{n-\sigma}{\delta^{\prime}}^{n-\dim
M}}{\frac{{\delta^{\prime}}^{\kappa_{2}}}{\delta^{\kappa_{2}}}}\ll\delta^{\kappa_{2}-(n-\sigma)}\left(\frac{1}{\delta^{\prime}}\right)^{\kappa_{2}+\dim
M-n}$

many balls of radius $\delta^{\prime}.$ Thus we can choose
$\eta=\kappa_{2}-(n-\sigma)>0.$999Notice that $\kappa_{2}\geq\dim_{l^{1}}K$
and by assumption we already have $\dim_{l^{1}}K+\sigma>n.$ Finally, we can
choose any $\epsilon<\eta.$ This shows that

$\underline{\dim}_{\mathrm{B}}K\cap\tilde{M}\geq\kappa_{2}-(n-\sigma).$

Finally, if $x\in\tilde{M}\cap K$, then there is a translation vector
$d\in\\{0,\dots,p^{l}-1\\}^{n}/p^{l}$ such that

$x+d\in M\cap K_{p,D,l}.$

Thus $M\cap K_{p,D,l}\neq\emptyset$ and

$\underline{\dim}_{\mathrm{B}}K_{p,D,l}\cap M\geq\kappa_{2}-(n-\sigma).$

To see the latter, observe that

$\tilde{M}\cap K\subset
K_{p,D,l}\cap\bigcup_{d\in\\{0,\dots,p-1\\}^{n}/p^{l}}(M+d).$

For different $d,$ the sets

$M_{d}=K_{p,D,l}\cap(M+d)$

are translation copies of each other because $K_{p,D,l}$ is invariant with
respect to such translations $d.$ Thus $M_{d}$ for different $d$ all have the
same box covering properties. From here we conclude the lower bound for the
lower box dimension of $K_{p,D,l}\cap M.$

For the last statement of Theorem 2.8, it is enough to observe that for each
$\delta>0$ and $\delta$-cube $C$ with $C\cap M\neq\emptyset.$ From the
previous argument in this section. There is some possibly large integer $l\geq
0$ depending on $C$ such that

$C\cap M\cap K_{p,D,l}\neq\emptyset.$

###  5.5. A slightly more generalized argument for ’pushing away’ the non-zero
coefficients

This section is not needed outside of Section 8. The reader can skip it for
now and come back later. In the previous section, we replaced a measure $\mu$
with

$\mu_{l}=\frac{1}{p^{nl}}\sum_{d\in\\{0,\dots,p^{l}-1\\}^{n}/p^{l}}\mu(.+d).$

The effect of the above averaging process is that for $\xi\in\mathbb{Z}^{n},$
$\hat{\mu}_{l}(\xi)$ is not zero only when the components of $\xi$ are all
divisible by $p^{l}.$

Now, we can formulate a slightly more generalized way of performing the above
averaging process. Let $p_{1},\dots,p_{n}$ be integers larger than $1.$
Consider the measure

$\mu^{\prime}=\mu_{p_{1},\dots,p_{n}}=\frac{1}{p_{1}\dots
p_{n}}\sum_{d\in\\{0,\dots,p_{1}-1\\}p_{1}^{-1}\times\dots\times\\{0,\dots,p_{n}-1\\}p^{-1}_{n}}\mu(.+d).$

This measure $\mu^{\prime}$ is an average of $p_{1}\dots p_{n}$ translated
copies of $\mu.$ The Fourier coefficients $\hat{\mu}^{\prime}(\xi)$ at
$\xi\in\mathbb{Z}^{n}$ is not zero only when $\xi=(\xi_{1},\dots,\xi_{n})$
satisfies that for each $i\in\\{1,\dots,n\\},$ $p_{i}|\xi_{i}.$ By choosing
$p_{1},\dots,p_{n}$ to be all large enough, we again achieve the goal of
’pushing away’ the non-zero coefficients as in the previous section.

###  5.6. Proof of Theorem 2.10 part 1

Theorem 2.5 has another consequence. Let $M$ be a non-degenerate manifold (or
a manifold of finite type) and $K$ be a missing digits set with a large enough
$l^{1}$-dimension such that Theorem 2.5 applies. Let $\lambda$ be the
corresponding missing digits measure.

From Theorem 2.5, we know that

$\limsup_{\delta\to 0}\frac{\lambda(M^{\delta})}{\delta^{n-\dim M}}<\infty.$

In fact, with extra technical argument, it is likely that the above $\limsup$
can be replaced with $\lim$ but we do not need this. A much easier observation
is that one can replace $\limsup$ with $\liminf$ in all the later arguments.
We need to use this fact to prove the last assertion of this theorem.

Now we can define a function from
$[0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n)$ to $[0,\infty),$

$f_{K,M}(t,v,g)=\limsup_{\delta\to
0}\frac{\lambda(T_{t,v,g}(M)^{\delta})}{\delta^{n-\dim M}}$

where $T_{t,v,g}(M)=t\times g(M)+v,$ i.e., it is the image of $M$ under the
rotation $g$, then scaled by $t$ and then translated by $v.$ From here the
first part of Theorem 2.10 follows.

Let $\mu$ be a smooth and compactly supported surface measure on $M.$ Then we
replace $M^{\delta}$ with the Schwartz function $\mu_{\delta}$ (as in Section
5.2) and define the function

$\displaystyle f_{K,\mu}(t,v,g)=\limsup_{\delta\to 0}\int
T_{t,v,g}(\mu_{\delta})(x)d\lambda(x).$

For each fixed $\delta>0,$ the quantity in the $\lim$ symbol is continuous
viewed as a function with variables $t,v,g.$ Observe that $\mu_{\delta}$ is a
smooth function with
$\mathrm{supp}(\mu_{\delta})\subset\mathrm{supp}(\mu)^{\delta}$ and
$\mu_{\delta}\asymp\delta^{-(n-\dim M)}$ on $\mathrm{supp}(\mu)^{\delta/2}.$
From here, we see that

(*) $\displaystyle T_{t,v,g}(\mu_{\delta})(x)\asymp\delta^{-(n-\dim M)}$

uniformly for $x\in T_{t,v,g}(\mathrm{supp}(\mu))^{t\delta/2}.$ From here we
see that for all $t,v,g$ and all small enough $\delta>0,$

$\frac{\lambda((T_{t,v,g}(M))^{t\delta/2})}{\delta^{n-\dim M}}\ll\int
T_{t,v,g}(\mu_{\delta})(x)d\lambda(x)\ll\frac{\lambda((T_{t,v,g}(M))^{t\delta})}{\delta^{n-\dim
M}}.$

Thus there is a constant $c>1$ such that

$c^{-1}t^{n-\dim M}f_{K,M}(t,v,g)\leq f_{K,\mu}(t,v,g)\leq ct^{n-\dim
M}f_{K,M}(t,v,g).$

We want to show that $f_{K,\mu}$ is continuous. Then $h=t^{-(n-\dim
M)}f_{K,\mu}$ is continuous. In the next section, we will show that $h$ is
non-vanishing and this will conclude Theorem 2.10.

Observe that

$\int
T_{t,v,g}(\mu_{\delta})(x)d\lambda(x)=\int\hat{T_{t,v,g}(\mu_{\delta})}(\xi)\overline{\hat{\lambda}(\xi)}d\xi.$

From (I), (II) we see that the above integrals converge absolutely in a manner
that is uniform across $\delta>0$. This is because under the action
$T_{t,v,g},$ the Fourier transform is changed accordingly in a manner that
preserves the polynomial decay with the same exponent. Thus we can apply (I),
(II). Moreover, if we restrict $t$ within a compact interval away from $0$,
then there are constants $c_{1},c_{2},C_{1},C_{2}>0$ with

$C_{1}\min_{|\xi^{\prime}|\in[c_{1}|\xi|,c_{2}|\xi|]|}|\hat{\mu_{\delta}}(\xi^{\prime})|\leq|\hat{T_{t,v,g}(\mu_{\delta})}(\xi)|\leq
C_{2}\max_{|\xi^{\prime}|\in[c_{1}|\xi|,c_{2}|\xi|]}|\hat{\mu_{\delta}}(\xi^{\prime})|$

for all $\xi.$ The above discussion does not depend on $\delta.$ From here we
see that

$\limsup_{\delta\to 0}\int
T_{t,v,g}(\mu_{\delta})d\lambda(x)=\limsup_{\delta\to
0}\int\hat{T_{t,v,g}(\mu_{\delta})}(\xi)\overline{\hat{\lambda}(\xi)}d\xi.$

The RHS above is continuous viewed as a function with variable $(t,v,g).$ This
follows from the following two facts:

a. $T_{t,v,g}$ acts on each of the Fourier coefficients of $\mu_{\delta}$
continuously in a manner that is uniform across $\delta>0$. Here we emphasize
that $T_{t,v,g}$ does not act continuously on the Fourier coefficients in a
manner that is uniform across all the frequencies. The continuity here is only
pointwise. Thus the role of the cut-off function $\hat{\phi}_{\delta}$ is
important.

b. $\xi\to|\hat{T_{t,v,g}(\mu_{\delta})}(\xi)\hat{\lambda}(\xi)|$ is
integrable in a manner that is uniform across $\delta>0$ and $t,v,g$ inside
any (fixed) compact set
$U\subset(0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n).$

From here we deduce that

$(t,v,g)\to\limsup_{\delta\to 0}\int T_{t,v,g}(\mu_{\delta})(x)d\lambda(x)$

is continuous. Thus $f_{K,\mu}$ is continuous. Moreover, the convergence and
the continuity is uniform when we restrict $(t,v,g)$ within a compact set
$U\subset[0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n)$ away from the set
$\\{t=0\\}\subset[0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n).$

###  5.7. The integral over group actions: Proof of Theorem 2.10 part 2

In order to show that $h=t^{-n+\dim M}f_{K,\mu}$ is non-vanishing, it is
enough to show that $f_{K,\mu}$ is non-vanishing (i.e. not identically zero).
To show that $f_{K,\mu}$ is non-vanishing, we show that the integral of
$f_{K,\mu}$ over a large enough region is positive. Following the previous
section, for each $\delta>0,$ consider the integral

$\int_{U}\int T_{t,v,g}(\mu_{\delta})(x)d\lambda(x)d(t,v,g),$

where $t,v,g$ is integrated over a compact subset $U$ with respect to the Haar
measure on $U$. We can exchange the order of the double integral. Observe that
for each $x\in\mathbb{R}^{n}$

$\int_{U}T_{t,v,g}(\mu_{\delta})(x)d(t,v,g)\asymp|\\{(t,v,g)\in
U:T^{-1}_{t,v,g}(x)\in supp(\mu_{\delta})\\}|/\delta^{n-\dim M}.$

where $|.|$ is with respect to the Haar measure restricted on $U$. This
measure is not always the Haar measure on
$[0,\infty)\times\mathbb{R}^{n}\times\mathbb{O}(n).$ In fact, it can also be
the Haar measure on a subgroup. As $\mu_{\delta}$ is a compactly supported
Schwartz function,

$x\to\int_{U}T_{t,v,g}(\mu_{\delta})(x)d(t,v,g)$

is non-negative and continuous. Our requirement for $U$ is that
$U\cap\\{t=0\\}=\emptyset$ and for each ball $B\subset\mathbb{R}^{n}$, each
$(t,v,g)\in U$ we have

(Positive) $\displaystyle|\\{(t,v,g)\in U:T^{-1}_{t,v,g}(x)\in
supp(\mu_{\delta})\\}|/\delta^{n-\dim M}>\epsilon>0$

for some $\epsilon>0$ and all $x\in B$ in a manner than does not depend on
$\delta.$

There are many possible choices of $U.$

### Case 1:

For example, we can fix $t>0,g\in\mathbb{O}(n)$ and let $v$ range over a
sufficiently large ball $B^{\prime}\subset\mathbb{R}^{n}$ centred at the
origin (i.e. $U=\\{t\\}\times B^{\prime}\times\\{g\\}$). In this way, the Haar
measure restricted to $U$ is the Lebesgue measure on the $v$ component
restricted to $B^{\prime}.$

### Case 2:

Another choice is to fix $v$ and then fix an interval $[a,b]\subset(0,\infty)$
for $t$ with a small enough $a>0$ and a large enough $b>a.$ Finally, we do not
restrict $g.$ In this way, the Haar measure restricted to $U$ is the Haar
measure of the scaling and rotation group $[0,\infty)\times\mathbb{O}(n).$ In
this case, the Haar measure on $U$ is equivalent to the $n$-dimensional
Lebesgue measure.

In both of the cases illustrated above, (Positive) is satisfied. In the first
example,

$\displaystyle|\\{(t,v,g)\in U:T^{-1}_{t,v,g}(x)\in supp(\mu_{\delta})\\}|$

reduces to

$|\\{v\in B^{\prime}:x+v\in supp(\mu_{\delta})\\}|=|supp(\mu_{\delta})|.$

The last term is the Lebesgue measure of $\mathrm{supp}(\mu_{\delta})$ which
is the $\delta$-neighbourhood of a compact piece of $M.$ It is of order
$\delta^{n-\dim M}.$ The second example can be tested via a similar method.

### Case 3:

We now introduce the third case. We restrict the discussion to
$\mathbb{R}^{2}.$ Let $r>0$ and consider the hyperbola
$H_{r}=\\{xy=r\\}_{x,y>0}.$ We can choose a piece of $H_{r}$ by considering
$\mu$ to be any compactly support smooth measure on $H_{r}.$ Let
$x\in(0,\infty)^{2}.$ Consider the line between $x$ and $(0,0)$: $l_{x}.$ If
$l_{x}\cap\mathrm{supp}(\mu)\neq\emptyset,$ then

$|\\{c>0:cx\in supp(\mu_{\delta})\\}|\asymp\delta,$

where $|.|$ is the one dimensional Lebesgue measure. The implicit constants in
$\asymp$ depend on $|x|.$ Furthermore, let $r>0$ and $P\subset\mathbb{R}^{2}$
be compact such that $d(P,(0,0))=r.$ Then the implicit constants in $\asymp$
can be chosen to be the same for all $x\in P.$

We have mentioned a few scenarios in which (Positive) holds. There are far
more situations and we do not provide further examples. Notice that under the
Condition (Positive),

$\int_{U}f_{K,\mu}(t,v,g)d(t,v,g)=\limsup_{\delta\to 0}\int_{U}\int
T_{t,v,g}(\mu_{\delta})(x)d\lambda(x)d(t,v,g)>0.$

Here the order of $\lim_{\delta\to 0}$ and the integral $\int_{U}$ can be
changed because that $U$ is compact and $U\cap\\{t=0\\}=\emptyset$ which
implies that the limit is uniform on $U$. Since $f_{k,\mu}$ is continuous, we
see that there exists a non-trivial ball $E\subset U$ such that
$f_{K,\mu}(t,v,g)>0$ for all $(t,v,g)\in E.$ This concludes the non-vanishing
part of Theorem 2.10.

Finally, for the lower box dimension
$\underline{\dim}_{\mathrm{B}}T_{t,v,g}(M)\cap K,$ observe that an estimate
like (Lower) holds. However, there is a slight difference. We need to replace
the above arguments with $\limsup$ being replaced by $\liminf$. Then we have

$\lambda((T_{t,v,g}(\mathrm{supp}(\mu)))^{\delta})\gg\delta^{n-\dim M}.$

This means that $T_{t,v,g}(supp(\mu)^{\delta})\cap K$ cannot be covered with
$o(\delta^{n-\dim M-\dim_{\mathrm{H}}K})$ many $\delta$-balls. Thus we see
that

$\mu((T_{t,v,g}(\mathrm{supp}(\mu)))^{\delta})\gg\delta^{n-\dim_{\mathrm{H}}K}.$

From here the rest of the arguments follow as in Section 5.4 and we have

$\underline{\dim}_{\mathrm{B}}T_{t,v,g}(M)\cap
K\geq\dim_{\mathrm{H}}K-(n-\sigma(M)).$

###  5.8. Sharpness of Theorem 2.5 for small missing digits sets: Proof of
Theorem 2.12

The curve $M$ under consideration is

$|x|^{k}+|y-1|^{k}=1.$

This curve passes through the origin and has an order $k$ contact with the
$X$-axis at $(0,0).$ This is the highest order of contact of $M$ with
hyperplanes (lines). Loosely speaking, this is because around $(0,0)$ the
curve looks like $y=|x|^{k}$ which has vanishing $(k-1)$th derivative and non-
vanishing $k$th derivative at $x=0.$

Let $\delta_{0}=p^{-l}>0$ for a large integer $l>0.$ Consider the square
$[0,\delta_{0}]^{2}.$ We decompose this square into smaller squares with side
length $p^{-kl}.$ Consider the ’first row’ of those smaller squares, i.e.
those that intersect the $X$-axis. We see that $M$ intersects each of those
smaller squares. Because the digit set $D$ contains $0.$ This implies that
$M^{p^{-kl}}\cap K$ must intersect
$\gg(p^{-l}/p^{-kl})^{\dim_{\mathrm{H}}K_{p,D_{1}}}$ many of those squares.

Let $\delta=p^{-kl}.$ We see that $M^{\delta}\cap K$ cannot be covered with

$\ll\left(\frac{1}{\delta^{(k-1)/k}}\right)^{\dim_{\mathrm{H}}K_{p,D_{1}}}$

many $\delta$-balls. Similarly, consider the point $(1,0)$ where $M$ has a
$k$th order contact with a vertical line. We see that $M^{\delta}\cap K$
cannot be covered with

$\ll\left(\frac{1}{\delta^{(k-1)/k}}\right)^{\dim_{\mathrm{H}}K_{p,D_{2}}}$

many $\delta$-balls. This proves Theorem 2.12.

##  6\. Fourier norm dimensions for missing digits measures

###  6.1. General Algorithm

For the case when $n=1,$ this matter has been discussed in [2]. See also [33].
For $n\geq 2$, the arguments change very little. In this paper, we provide
some details for being self-contained.

Let $n\geq 1$ be an integer. Let $p>2$ be an integer and
$D\subset\\{0,\dots,p-1\\}^{n}.$ Consider the missing digit measure
$\lambda_{p,D}.$ In this case, $\hat{\lambda}(\xi)$ can be computed with the
help of the formula,

$\hat{\lambda}_{p,D}(\xi)=\prod_{j\geq 0}\frac{1}{\\#D}\sum_{d\in D}e^{-2\pi
i(d,\xi)/p^{j}}.$

For convenience, let

$g(\xi)=\frac{1}{\\#D}\sum_{d\in D}e^{-2\pi i(d,\xi)}.$

Then we have

$\hat{\lambda}_{p,D}(\xi)=\prod_{j\geq 0}g(\xi/p^{j}).$

We want to estimate for large integers $k\geq 1,$

$S_{k}=\sup_{\theta\in[0,1]^{n}}\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k}}|\hat{\lambda}_{p,D}(\xi+\theta)|.$

Notice that in the sum, we conditioned on the max norm
$|\xi|_{\infty}=\max\\{|\xi_{1}|,\dots,|\xi_{n}|\\}.$ We now estimate $S_{k}.$
Let $\theta\in(0,1)^{n}$ be a vector. Consider the function

$f(\theta)=\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}}|g((\mathbf{i}+\theta)/p)|.$

Clearly we have for all $\theta,$

$0\leq f(\theta)\leq p^{n}.$

Observe that for each $\theta\in[0,1]^{n}$

$\displaystyle
S_{k}(\theta)=\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k}}|\hat{\lambda}_{p,D}(\xi+\theta)|$
$\displaystyle=\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k}}\left|\prod_{j\geq
0}g((\xi+\theta)/p^{j})\right|.$

Let $\xi^{\prime}=\xi+\mathbf{i}$ for some $\mathbf{i}\in
p^{k-1}\mathbb{Z}^{n}.$ Then we have

$g((\xi^{\prime}+\theta)/p^{j})=g((\xi+\theta)/p^{j})$

for all $j=0,1,\dots,k-1.$ From here we see that (recall that $|g|\leq 1$)

$\displaystyle S_{k}(\theta)$
$\displaystyle=\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k}}\left|\prod_{j\geq
0}g((\xi+\theta)/p^{j})\right|$
$\displaystyle\leq\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k}}\left|\prod^{k}_{j=0}g((\xi+\theta)/p^{j})\right|$
$\displaystyle=\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k-1}}\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}p^{k-1}}\left|\prod^{k}_{j=0}g((\xi+\mathbf{i}+\theta)/p^{j})\right|$
$\displaystyle=\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k-1}}\left|\prod_{j=0}^{k-1}g((\xi+\theta)/p^{j})\right|\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}}\left|g(\mathbf{i}p^{-1}+\theta
p^{-k}+\xi p^{-k})\right|$
$\displaystyle\leq\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k-1}}\left|\prod_{j=0}^{k-1}g((\xi+\theta)/p^{j})\right|\sup_{\theta^{\prime}}f(\theta^{\prime})$
$\displaystyle\overset{\text{Continue
inductively}}{\leq}\sum_{\xi\in\mathbb{Z}^{n},|\xi|_{\infty}<p^{k-2}}\left|\prod_{j=0}^{k-2}g((\xi+\theta)/p^{j})\right|(\sup_{\theta^{\prime}}f(\theta^{\prime}))^{2}$
$\displaystyle\dots$
$\displaystyle\leq(\sup_{\theta^{\prime}}f(\theta^{\prime}))^{k}.$

Thus we have

$S_{k}(\theta)\leq(\sup_{\theta^{\prime}}f(\theta^{\prime}))^{k}.$

Therefore we see that

$S_{k}\leq(\sup_{\theta^{\prime}}f(\theta^{\prime}))^{k}.$

This implies that (we take the Euclidean norm $|\xi|$)

$\sup_{\theta\in[0,1]^{n}}\sum_{\xi\in\mathbb{Z}^{n},|\xi|\leq
p^{k}}|\hat{\lambda}_{p,D}(\xi+\theta)|\leq
S_{k}\leq(\sup_{\theta^{\prime}}f(\theta^{\prime}))^{k}.$

From here one can see that

$n-\frac{\log\sup_{\theta}f(\theta)}{\log p}\leq\dim_{l^{1}}\lambda_{p,D}.$

###  6.2. A crude bound: proof of Theorem 2.15 part 1

Although it is in principle possible to compute $\dim_{l^{1}}\lambda_{p,D}$
within any precision, we are not too obsessed with the exact computations.
Instead, we provide a rather crude but still useful upper bound for the value

$\sup_{\theta}f(\theta)$

when $D$ is a large set. This will give us a lower bound for
$\dim_{l^{1}}\lambda_{p,D}.$

First, observe that

$\\#Dg(\xi)=\prod_{j=1}^{n}\frac{1-e^{2\pi ip\xi_{j}}}{1-e^{2\pi
i\xi_{j}}}-\sum_{d\notin D}e^{-2\pi i(d,\xi)}.$

Let $\\#D=p^{n}-t$ for some integer $t>0.$ Then we have

$-t\leq|(p^{n}-t)g(\xi)|-\left|\prod_{j=1}^{n}\frac{1-e^{2\pi
ip\xi_{j}}}{1-e^{2\pi i\xi_{j}}}\right|\leq t.$

Now we want to consider the sum

$f_{1}(\theta)=\sum_{\mathbf{i}\in\\{0,p-1\\}^{n}}|g_{1}((\mathbf{i}+\theta)/p)|,$

where

$g_{1}(\xi)=\prod_{j=1}^{n}\frac{1-e^{2\pi ip\xi_{j}}}{1-e^{2\pi i\xi_{j}}}.$

To do this, consider the function $h:\mathbb{R}\to\mathbb{R},$

$h(x)=\left|\frac{1-e^{2\pi ipx}}{1-e^{2\pi ix}}\right|.$

We want to provide an estimate for

$H(\theta)=\sum_{j\in\\{0,\dots,p-1\\}}h((j+\theta)/p).$

Notice that ($|e^{ix}-1|\leq 2,\forall x\in\mathbb{R}$)

(*) $\displaystyle H(\theta)\leq 2\sum_{j=0,\\{(j+\theta)/p\\}\geq
1/p}^{p-1}\frac{1}{|1-e^{2\pi i(j/p)}e^{2\pi
i(\theta/p)}|}+\sum_{j:\\{(j+\theta)/p\\}<1/p}h((j+\theta)/p).$

For the first sum in (* ‣ 6.2), we see that ($|1-e^{2\pi
ix}|^{2}=2(1-\cos(2\pi x))\geq 16x^{2}$, for $x\in[0,1/2]$)

$\displaystyle 2\sum_{j=0,\\{(j+\theta)/p\\}\geq 1/p}^{p-1}\frac{1}{|1-e^{2\pi
i(j/p)}e^{2\pi i(\theta/p)}|}\leq 8\sum_{k=1}^{[(p-1)/2]}\frac{1}{|1-e^{2\pi
i(k/p)}|}$
$\displaystyle\leq\frac{1}{\pi}\sum_{k=1}^{[(p-1)/2]}\frac{k}{p}\leq\frac{p}{\pi}(\log
p+1).$

For the second sum in (* ‣ 6.2), notice that there are at most two $j$’s in
the sum. As $|h|\leq p$, we have

$\sum_{j:\\{(j+\theta)/p\\}<1/p}h((j+\theta)/p)\leq 2p.$

From here we see that for $p\geq 4,$

$H(\theta)\leq 2p\log p.$

We can then use this estimate to see that

$\sup_{\theta}f_{1}(\theta)\leq(2p\log p)^{n}.$

Thus we have

$(p^{n}-t)\sup_{\theta}f(\theta)\leq tp^{n}+(2p\log p)^{n}.$

This implies that

(Crude bound) $\displaystyle\dim_{l^{1}}\lambda_{p,D}\geq
n-\frac{\log\frac{tp^{n}+(2p\log p)^{n}}{p^{n}-t}}{\log p}.$

This (Crude bound) tells us, for example, if we fix $t>0,$ then as long as $p$
is large enough, $\dim_{l^{1}}\lambda_{p,D}$ can be arbitrarily close to $n.$

###  6.3. Missing digits measures with large bases and rectangular digit sets:
proof of Theorem 2.15 part 2

Let $K_{p,D}$ be a missing digits set. In general, we have

$\frac{1}{2}\dim_{\mathrm{H}}K_{p,D}\leq\dim_{l^{1}}K_{p,D}\leq\dim_{\mathrm{H}}K_{p,D}.$

For missing digits sets, we expect the rightmost side should be closer to the
truth. We now make this point clearer.

Let $n\geq 1$ be an integer. Let $p>1$ be an integer. For a missing digits
set, we choose a subset $D\subset\\{0,\dots,p-1\\}^{n}$ and construct
$K_{p,D}.$ We have seen that it is in principle possible to find the value of
$\dim_{l^{1}}\lambda_{p,D}$ with arbitrarily small errors. We can also find a
lower bound for $\dim_{l^{1}}$ with the help of (Crude bound). It turns out
that if the digit sets are chosen to be well structured, then we can have a
much better estimate than the (Crude bound).

Let $D\subset\\{0,\dots,p-1\\}^{n}$ be an rectangle, i.e. it is of form

$[a_{1},b_{1}]\times\dots\times[a_{n},b_{n}]\cap\\{0,\dots,p-1\\}^{n}$

with integers $a_{1}\leq b_{1},\dots,a_{n}\leq b_{n}.$ With this special digit
set $D$, we see that the corresponding function $g_{b,D}$ is of form

$g_{b,D}(\theta)=(\\#D)^{-1}\sum_{z\in D}e^{-2\pi
i((z,\theta))}=(\\#D)^{-1}\prod_{j=1}^{n}e^{-2\pi
ia_{j}\theta_{j}}\frac{1-e^{-2\pi i(b_{j}-a_{j}+1)\theta_{j}}}{1-e^{-2\pi
i\theta_{j}}}.$

Next, we estimate the sum

$(\\#D)f_{b,D}(\theta)=(\\#D)\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}}|g_{p,D}(\mathbf{i}+\theta)/p|.$

For each $j\in\\{1,\dots,n\\},$ define

$S_{j}(\theta_{j})=\left|\frac{1-e^{-2\pi
i(b_{j}-a_{j}+1)\theta_{j}}}{1-e^{-2\pi i\theta_{j}}}\right|.$

Then we see that

$\displaystyle(\\#D)\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}}|g_{p,D}(\mathbf{i}+\theta)/p|$
$\displaystyle=\sum_{\mathbf{i}\in\\{0,\dots,p-1\\}^{n}}\prod_{j=1}^{n}S_{j}((\mathbf{i}_{j}+\theta_{j})/p)$
$\displaystyle=\prod_{j=1}^{n}\sum_{i\in\\{0,\dots,p-1\\}}S_{j}((i+\theta_{j})/p).$

Now we need to estimate for each $j\in\\{1,\dots,n\\},$

$\sum_{i=0}^{p-1}S_{j}((i+\theta_{j})/p).$

We have already considered this type of sums before, see (* ‣ 6.2). Notice
that $S_{j}(\theta_{j})\leq b_{j}-a_{j}+1.$ As a result, we have for $p\geq
4,$

$\sup_{\theta_{j}}\sum_{i=0}^{p-1}S_{j}((i+\theta_{j})/p)\leq\frac{p}{\pi}(\log
p+1)+2(b_{j}-a_{j}+1)\leq 2p\log p.$

Thus we see that

$\sup_{\theta}f_{b,D}(\theta)\leq(\\#D)^{-1}(2p\log p)^{n}.$

From here we see that

$\displaystyle\dim_{l^{1}}\lambda_{p,D}\geq$ $\displaystyle
n-\frac{\log((\\#D)^{-1}(2p\log p)^{n})}{\log p}$
$\displaystyle=n+\frac{\log\\#D}{\log p}-n-\frac{n\log\log p^{2}}{\log p}$
$\displaystyle=\dim_{\mathrm{H}}K_{p,D}-\frac{n\log\log p^{2}}{\log p},$

where have used the fact that $\dim_{\mathrm{H}}K_{p,D}=\log\\#D/\log p.$

##  7\. Proofs of Theorems A, B, C

We now prove the special results at the beginning.

###  7.1. Theorem A

For Theorem A, one can use Theorems 2.5, 2.8, 2.15(2) together with the fact
that for the Veronese curve in $\mathbb{R}^{3},$ the positive number $\sigma$
can be chosen to be $1/3.$ Notice that the missing digits sets in Theorem A is
the threefold Cartesian product $K^{3}$ where $K$ is the set of numbers on
$\mathbb{R}$ whose base $10^{9000}$ expansions only contain digits
$\\{0,\dots,10^{8100}-1\\}.$ With the help of Theorem 2.15(2), it can be
checked that $\dim_{l^{1}}K>8/9.$ Thus we have $\dim_{l^{1}}K^{3}>3-3^{-1}.$
The lower bound $1/30$ for box dimension can be derived from the numbers given
above.

###  7.2. Theorem B

For Theorem B, there is still one more step to perform. The missing digits set
in consideration is $K\times K$ where $K$ is the set of numbers in
$\mathbb{R}$ whose base $10^{9000}$ expansions only have digits in
$\\{0,\dots,10^{7000}-1\\}.$ By Theorem 2.15(2), it can be checked that
$\dim_{l^{1}}$ and then $\dim_{l^{1}}K\times K>1.5.$ Let $\lambda$ be the
missing digits measures on $K.$

We first illustrate how to show that pinned distance sets have a positive
Lebesgue measure. Then we upgrade the result to the non-trivial interval
version with the help of Section 5.7. In this way, we hope to first illustrate
clearly the idea behind the proof before the need for too many technical
details (zipped in Section 5.7).

For each $r>0,$ consider the circle $C_{r}:x^{2}+y^{2}=r^{2}.$ For circles in
$\mathbb{R}^{2}$ (or spheres in $\mathbb{R}^{n}$), the Fourier decay exponent
$\sigma$ can be chosen to be $1/2$ (or $(n-1)/2$ for spheres in
$\mathbb{R}^{n}$). Theorem 2.5 tells us that
$\lambda(C^{\delta}_{r})\ll_{r}\delta.$ Moreover, a further insight of Section
5.2 tells us that the $\ll_{r}$ estimate is uniform for $r$ ranging over a
bounded interval which is away from zero, i.e. for positive numbers $b>a>0$

$\lambda(C^{\delta}_{r})\ll_{a,b}\delta$

for $r\in[a,b].$ Now we can choose $a>0$ be sufficiently small and $b>0$ being
sufficiently large such that

$\lambda(B_{b}(0)\setminus B_{a}(0))>0.$

Consider the set

$\Sigma_{a,b}=\\{r\in[a,b]:C_{r}\cap K\neq\emptyset\\}.$

Suppose that $\Sigma_{a,b}$ has zero Lebesgue measure. Then we can cover
$\Sigma_{a,b}$ with small intervals. More precisely, for each $\epsilon>0,$
there is a countable collection of intervals $I_{j},j\geq 1$ of total length
at most $\epsilon$ and they cover $\Sigma_{a,b}.$ Let $I_{j}$ be one of those
intervals. Let $r_{j}$ be the centre Then we have

$\lambda(C_{r_{j}}^{|I_{j}|/2})\leq c|I_{j}|,$

where $c>0$ is a constant depending on $a,b.$ We can sum up the above
estimates for all $j,$

$\lambda(B_{b}(0)\setminus B_{a}(0))\leq c\sum_{i}|I|_{j}\leq c\epsilon.$

As $\epsilon>0$ can be arbitrarily chosen, this contradicts the statement
(Positive). Thus we see that

$\Delta_{(0,0)}(K)\supset\Sigma_{a,b}$

has positive Lebesgue measure. Of course, the choice $(0,0)$ is of no
speciality. One can replace it with any other point in $\mathbb{R}^{2}.$ So
that we showed that the pinned distance sets have positive Lebesgue measure.

Now we want to show that the pinned distance sets in fact have non-trivial
intervals. This is not very straightforward to show. We use the method
introduced in Section 5.7. Since the circles are compact, we do not need to
choose compactly supported smooth surface measures for them. Thus
$\mu,\mu_{\delta}$ in Section 5.7 can be simply taken to be the natural
($\delta$-thickened) Lebesgue measures on the circles. For the range of group
actions, we take $U=(R^{-1},R)\times\\{(0,0)\\}\times\mathbb{O}(2)$ for a
large enough number $R>0.$ Observe that the circles are invariant under
rotations. The arguments in Section 5.7 provide us with a non-vanishing
continuous function $f:[R^{-1},R]\to[0,\infty)$ so that whenever $f(r)>0,$
$C_{r}\cap K\neq\emptyset.$ This shows that $\Delta_{(0,0)}(K)$ contains non-
trivial intervals. Again, the point $(0,0)$ is of no significance. One can
replace it with any $x\in\mathbb{R}^{2}.$ However, the value $R$ needs to be
changed accordingly. From here the proof of Theorem B concludes.

###  7.3. Theorem C

Finally, we prove Theorem C. Consider the class of hyperbola

$H_{r}=\\{(x,y):xy=r\\},r>0.$

For each $r>0,$ let $(x_{1},x_{2})\in K\times K\cap(0,1]^{2}.$ We see that the
line connecting $(0,0)$ and $(x_{1},x_{2})$ will intersect $H_{r}.$ However,
the intersection might be too close to the origin. To overcome this issue, we
can consider a branch of $K\times K$ that is away from both of the coordinate
lines. Such branches certainly exist, e.g. the image $Y=T_{t,v,g}(K\times K)$
with

$g=\mathbb{I},t=10^{-9000},v=((10^{7000}-1)10^{-9000},(10^{7000}-1)10^{-9000}).$

Now $Y\subset K\times K$ and $Y$ is far away from the coordinate lines. Let
$C_{1},C_{2}>0$ be large enough numbers such that for each $x\in Y,$ the line
connecting $(0,0)$ and $x$ intersects $H_{r},r\in[C_{1}^{-1},C_{1}]$ in
$[0,C_{2}]^{2}.$

Notice that the $H_{r},r>0$ is a class of curves that can be obtained from
each via scalings. Now we can apply Section 5.7 (more specifically, the third
case) to deduce that

$\\{r\in[C_{1}^{-1},C_{1}]:H_{r}\cap Y\\}$

contains intervals. From here, the proof of Theorem C concludes.

##  8\. More examples and a question on linear forms

We explain more applications in this section. For convenience, we fix three
missing digits sets on $\mathbb{R}:$

$K_{1}$: numbers whose base $10^{9000}$ expansions only contain digits in

$\\{0,\dots,10^{8100}-1\\}.$

$K_{2}$: numbers whose base $11^{9000}$ expansions only contain digits in

$\\{0,\dots,11^{8100}-1\\}.$

$K_{3}$: numbers whose base $12^{9000}$ expansions only contain digits in

$\\{0,\dots,12^{8100}-1\\}.$

The Hausdorff dimensions of $K_{1},\dots,K_{3}$ are equal to $9/10.$ The
$l^{1}$ dimensions of $K_{1},\dots,K_{3}$ are all very close to $9/10$ and in
fact they are all larger than $8/9$ by using Theorem 2.15(2).

######  Example 8.1.

Consider the hyperbola $\\{xy=1\\}$ in $\mathbb{R}^{2}.$ We can apply Theorems
2.5, 2.8 to see that there is an integer $l\geq 0$ and there are infinitely
many numbers $t>0$ with

$10^{l}t,\frac{10^{l}}{t}$

both in $K_{1}.$

We want to list more results that go slightly beyond the scope of Theorem 2.8.
Notice that $K=K_{1}\times K_{2}\times K_{3}$ is not a missing-digit nor a
self-similar set by our definition. It is nonetheless AD-regular. Theorem 2.5
still applies to this set. However, Theorem 2.8 does not apply to the set $K$.
One can review the proof of Theorem 2.8 and find two places where it stops
working.

The first place is at the beginning of Section 5.4 where we constructed
$\mu_{l}.$ We need to replace it with the argument in Section 5.5. In our
case, we can set $p_{1}=10^{l_{1}},p_{2}=11^{l_{2}},p_{3}=12^{l_{3}}$ for
suitable numbers $l_{1},l_{2},l_{3}\to\infty$.

The second place is at the end of Section 5.4 where we used (Rescale). It
depends on the self-similarity of the underlying set (measure). As our current
$K$ is not self-similar, we cannot follow the proof without further
modification.

We now make this modification. Let $\delta>0$ be a small number. We can choose
integers $l_{1},l_{2},l_{3}$ such that

$10^{l_{1}}\leq\delta^{-1}<10^{l_{1}+1},$
$11{l_{2}}\leq\delta^{-1}<11^{l_{2}+1},$
$12^{l_{3}}\leq\delta^{-1}<12^{l_{3}+1}.$

Then we can decompose $[0,1]^{3}$ into rectangles of dimension
$10^{-l_{1}}\times 11^{-l_{2}}\times 12^{-l_{3}}.$ In this way, we can
decompose $K$ into small pieces of the above size. Let $K^{\prime}$ be one of
such pieces. Notice that $K^{\prime}$ is roughly a box of size $\delta$ up to
some bounded multiplicative error. However, $K^{\prime}$ is not a rescaled
copy of $K.$ We can now find the corresponding restricted and normalized
measure $\lambda^{\prime}$ on $K^{\prime}$ given the original measure
$\lambda$ on $K$ which is the product measure of missing-digit measures on
$K_{1},K_{2},K_{3}.$ Then one can continue the argument in Section 5.4.

After this modification, one can obtain results on products of missing-digit
sets (measures). We list three results.

######  Example 8.2.

Again consider the hyperbola $\\{xy=1\\}$ in $\mathbb{R}^{2}.$ We see that
there is an integer $l\geq 0$ and there are infinitely many numbers $t>0$ with
$10^{l}t\in K_{1}$ and $11^{l}/t\in K_{2}.$

######  Example 8.3.

Consider the Veronese curve $(t,t^{2},t^{3})_{t\in\mathbb{R}}$ in
$\mathbb{R}^{3}.$ We see that there is an integer $l\geq 0$ and there are
infinitely many numbers $t>0$ such that

$10^{t}t\in K_{1},11^{t}t^{2}\in K_{2},12^{t}t^{3}\in K_{3}.$

######  Example 8.4.

Consider the curve $\\{x^{3}+y^{3}=1\\}$ on $\mathbb{R}^{2}.$ For this curve,
we can choose $\sigma=1/3.$ There are points $(t_{1},t_{2})$ on this curve
with $10^{l}t_{1}\in K_{1}$ and $11^{l}t_{2}\in K_{2}.$

In those examples, it is possible to study the lower box dimension of the set
of points in consideration. This can be done with the same method as in the
proof of Theorem A.

In this paper, we require that the manifold in consideration is of finite
type. This excludes the case when it is a line. In fact, despite the simple
geometry of lines, their intersections with fractals still remain mysterious:

######  Question 8.5.

In Theorems 2.5,2.8 can the manifold $M$ be taken to be irrational
hyperplanes?

Here, the irrationality condition is crucial. It says that the normal
direction of $M$ has rationally independent coordinates. If this condition is
not satisfied, then the intersection can be larger than expected, see Section
1.2. If one allows an $\epsilon$ uncertainty on the exponents then there are
satisfactory results. See [38], [43].

##  9\. Measures with polynomial Fourier decay

So far, we have only considered problems regarding manifolds intersecting
missing digits sets. From the proofs of Theorems 2.5, 2.8, we see that the
property we need for manifolds of finite type is that they support natural
surface measures $\mu$ with whose Fourier transforms have polynomial decay,
i.e. for some $\sigma>0$

$|\hat{\mu}(\xi)|\ll|\xi|^{-\sigma}.$

The proofs of Theorems 2.5, 2.8 can be done with all measures satisfying the
above properties. There is no lack of such measures other than surface
measures of finite type. We list two special examples. Note that unlike for
manifolds with finite type, here the decay exponent $\sigma$ may not be easy
to determine and $\mu$ may not be AD-regular. The following example shows that
digit expansion and continued fraction expansion are in some sense
’independent’.

######  Example 9.1.

Gibbs measures from non-linear interval maps. See [23], [12]. In particular,
from [23, Corollary 1], we can deduce the following result:

Let $A\subset\mathbb{N}$ be a finite set. Consider the set $B(A)$ of numbers
whose continued fractions only contain digits in $A$. Suppose that
$s=\dim_{\mathrm{H}}B(A)>1/2.$ Then there is a number $\sigma>0$ such that for
missing digits measure $\lambda$ with $\dim_{l^{1}}\lambda>1-\sigma,$

$\lambda(B(A)^{\delta})\ll\delta^{1-s}.$

To prove this result, from [23] (or [12]) we know that there is a natural
measure $\mu$ supported on $B(A)$ such that
$|\hat{\mu}(\xi)|\ll|\xi|^{-\sigma}$ for some $\sigma>0.$ Loosely speaking,
this example shows that the continued fraction expansions and digit expansions
of numbers are somehow independent in a quantitative way. For example, since
$s<1,$ we deduce that $\lambda(B(A)=0.$ This means that $\lambda.a.e$ points
$x\in K$ are not in $B(A)$ (This can be also deduced from a much stronger
result in [39]).

Next, we want to find points in $B(A)$ that also has missing digits in base
$p$ expansion. From the facts that $\dim_{l^{1}}\lambda>1-\sigma$ and
$|\hat{\mu}(\xi)|\ll|\xi|^{-\sigma}$ we deduce that

$\int|\hat{\mu}(\xi)\hat{\lambda}(-\xi)|d\xi<\infty.$

This implies that $\mu*\lambda$ is absolutely continuous with respect to the
Lebesgue measure and moreover the density function is continuous. This implies
that the arithmetic sumset $-\mathrm{supp}(\lambda)+\mathrm{supp}(\mu)$
contains intervals. Let $K_{p,D}$ be a missing digits set whose missing digits
measure $\lambda$ satisfies $\dim_{l^{1}}\lambda>1-\sigma.$ Let $I$ be one of
those intervals. Let $a\in I$ be any number with terminating base $p$
expansion. Then we see that there are $x\in K_{p,D}$ and $y\in B(A)$ with

$-x+y=a.$

This implies that

$y=a+x.$

Since $a$ has terminating base $p$-expansion and $x\in K_{p,D},$ $y$ has
missing $p$-adic digits eventually (i.e. there is an integer $l\geq 1$ such
that $\\{p^{l}a\\}\in K_{p,D}$).

######  Example 9.2.

Patterson-Sullivan measures. See [29]. A counting result can be deduced as in
Example 9.1. We omit further details.

##  10\. Further discussions

We do not have a complete picture of the distribution of missing digits points
around the manifold yet. However, results in this paper provide us with rather
satisfactory information in the following situation:

  * •

The manifold is sufficiently curved and the missing digits set is sufficiently
large with respect to the Fourier $l^{1}$-dimension.

There are several directions for further study:

  * •

First, the largeness of missing digits sets are quantified with the help of
$\dim_{l^{1}}.$ We believe this is not the optimal condition to look at. In
fact, we believe that the results in this paper can be proved with
$\dim_{\mathrm{H}}$ in the place of $\dim_{l^{1}}.$ For example, we formulate
the following conjecture.

######  Conjecture 10.1.

Let $n\geq 2$ be an integer. Let $M\subset\mathbb{R}^{n}$ be a manifold of
finite type. Then there is a number $\sigma=\sigma(M)>0$ such that for each
missing digits measure $\lambda$ with $\dim_{\mathrm{H}}\lambda>\sigma,$

$\lambda(M^{\delta})\ll\delta^{n-\dim M}.$

Part 2 of Theorem 2.15 provides us with examples of missing digits measures
that satisfy the conclusion of this conjecture. However, the base of those
missing digits measures are all large and the digit sets have to be chosen to
be specially structured. Thus the task is to reduce the size of the bases and
the structural requirement of the digit sets.

  * •

Second, what happens if the size of the missing digits set is small? Our
theory so far can only be applied when the size of the missing digits set is
large. Then we have obtained an optimal intersection result by combining
Theorems 2.5, 2.8.

We expect that if the missing digits set has a small enough Hausdorff
dimension then it should be rather rare to find those points inside the
manifold. We mentioned this point briefly at the beginning of this paper. We
formulate the following concrete problem.

######  Conjecture 10.2.

Consider the circle $C:x^{2}+y^{2}=1.$ For large enough integers $p,$ suppose
that $D\subset\\{0,\dots,p-1\\}^{2}$ is small enough, say, $\\#D\leq 100,$
then $C\cap K_{p,D,l}$ only contains rational points for each $l\geq 0.$

  * •

Third, for the method we used in this paper, there are two important factors.
We have two sets $A,B$ and we want to study $A\cap B.$ To do this, we find
nicely supported measures $\mu_{A},\mu_{B}$ on $A,B$ respectively. Then we
need one of the measures, $\mu_{A},$ say, to have a power Fourier decay, i.e.
for some $\sigma>0,$

$|\hat{\mu}(\xi)|\ll|\xi|^{-\sigma}.$

For $\mu_{B},$ we need that $\dim_{l^{1}}\mu_{B}$ is sufficiently large.

There is no lack of studies of the power Fourier decay property for various
measures, e.g. as we have mentioned surface measures carried by manifolds,
Gibbs measures, Patterson-Sullivan measures. On the other hand, the study of
$\dim_{l^{1}}$ is relatively new. So far, the best knowledge we have for
$\dim_{l^{1}}\mu_{B}$ is when $\mu_{B}$ is a missing digits measure. See also
[2] and [45]. In particular, in [45] a numerical method was proposed to treat
self-similar measures which are not missing digits measures. This numerical
method does not provide accurate estimates. Apart from these results, almost
nothing is known in the case when $\mu_{B}$ is a general Borel probability
measure. We want to ask the following particular problems.

######  Question 10.3.

Estimate $\dim_{l^{1}}\mu$ for $\mu$ being:

1\. A smooth surface measure carried by non-degenerate manifolds.

2\. A self-similar measure with the open set condition.

3\. A self-affine measure with the strong irreducibility condition.

4\. A self-conformal measure with the open set condition.

As in [45], answers to this question can help us gain more insights about how
rational points are distributed around a specific set, e.g. a self-conformal
set, e.g. a Julia set. More generally, as we have discussed in this paper, it
is possible to study intersections between different sets from the above list.

  * •

Theorem 2.8 is not satisfactory because we are only able to find a possibly
not sharp lower bound for $\underline{\dim}_{\mathrm{B}}M\cap K_{p,D,l}.$

In fact, as mentioned earlier, we believe that under the hypothesis of Theorem
2.8, it should be that

$\dim_{\mathrm{B}}M\cap K_{p,D,l}=\dim_{\mathrm{H}}K_{p,D}-(n-\dim M).$

We are in fact not too far away from such a result because Theorem 2.8 also
tells us that

$M^{\delta}\cap K_{p,D,l}$

have the ’correct’ size for small enough $\delta>0.$ Thus we see that there
are ’enough’ points in $K$ which are also close to $M$ but we are not yet able
to say that there are ’enough’ points in $K_{p,D,l}\cap M.$

##  11\. Acknowledgement

HY was financially supported by the University of Cambridge and the Corpus
Christi College, Cambridge. HY has received funding from the European Research
Council (ERC) under the European Union’s Horizon 2020 research and innovation
programme (grant agreement No. 803711). HY has received funding from the
Leverhulme Trust (ECF-2023-186). HY thank P.Varjú and P. Shmerkin for various
comments.

### Rights

For the purpose of open access, the authors have applied a Creative Commons
Attribution (CC-BY) licence to any Author Accepted Manuscript version arising
from this submission.

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