$\Delta^{\prime}_{B}=\Delta_{B}\cup\\{(q_{0},\Cat,0,q_{*})$, and

  * •

the final states $F^{\prime}_{B}=F_{B}\cup\\{q_{*}\\}$.

The OCN $\mathcal{M}$ is history-deterministic and the language accepted by
the net $\mathcal{M}$ is
$\mathcal{L}(\mathcal{M})=\mathcal{L}(\mathcal{H}_{B})\cup\\{\Cat\\}$. Note
that $\mathcal{L}(\mathcal{H}_{A})\subseteq\mathcal{L}(\mathcal{H}_{B})$ if
and only if $\mathcal{L}(\mathcal{H}_{A})\subsetneq\mathcal{L}(\mathcal{M})$.

Henceforth, we will only consider such history-deterministic OCNs
$\mathcal{H}_{A}$, $\mathcal{H}_{B}$ where $\mathcal{H}_{B}$ accepts a word
$\Cat$ which is not accepted by $\mathcal{H}_{A}$. We construct an OCN
$\mathcal{N}$, which is history-deterministic if and only if the language
inclusion $\mathcal{L}(\mathcal{H}_{A})\subset\mathcal{L}(\mathcal{H}_{B})$
holds. Let
$\mathcal{N}=(Q_{N},\Sigma\cup\\{\heartsuit\\},\Delta_{N},q^{0}_{N},F_{N})$,
where

  * •

the set of states $Q_{N}=Q_{A}\cup Q_{B}\cup\\{q^{0}_{N}\\}$,

  * •

the set of transitions
$\Delta_{N}=\Delta_{A}\cup\Delta^{\prime}_{B}\cup\\{q^{0}_{N},\heartsuit,0,q^{0}_{A}\\}\cup\\{(q^{0}_{N},\heartsuit,0,q^{0}_{B})\\}$,
and

  * •

the final states $F_{N}=F_{A}\cup F_{B}$.

Suppose $\mathcal{N}$ constructed as above is history-deterministic. Then,
there is a winning strategy for $\boldsymbol{\exists}$ in the letter game
which on reading $\heartsuit$ from $q_{N}^{0}$, chooses the transition
$q^{0}_{N}\xrightarrow{\heartsuit,0}q^{0}_{A}$ or
$q^{0}_{N}\xrightarrow{\heartsuit,0}q^{0}_{B}$. By our assumption, the
language of $\mathcal{H}_{B}$ contains a word which is not in the language of
$\mathcal{H}_{A}$. If $\boldsymbol{\exists}$ did not choose the transition to
$q^{0}_{B}$, then $\boldsymbol{\exists}$ looses the letter game as
$\boldsymbol{\forall}$ can give as input this word not in the language of
$\mathcal{H}_{A}$, but in $\mathcal{H}_{B}$. Therefore, any winning strategy
of $\boldsymbol{\exists}$ must choose the transition to the copy of
$\mathcal{H}_{B}$ from $q^{0}_{N}$. This implies that for any word accepted by
$\mathcal{H}_{A}$, the resolver has a strategy henceforth to produce a
sequence of transitions in $\mathcal{H}_{B}$ that leads to an accepting state,
implying that
$\mathcal{L}(\mathcal{H}_{A})\subset\mathcal{L}(\mathcal{H}_{B})$.

If $\mathcal{L}(\mathcal{H}_{A})\subset\mathcal{L}(\mathcal{H}_{B})$, the
resolver only needs to deal with non-determinism in the first step. Choosing
the transition $q^{0}_{N}\xrightarrow{\heartsuit,0}q^{0}_{B}$ at $q^{0}_{N}$
ensures $\boldsymbol{\exists}$ wins the letter game on $\mathcal{N}$, since
$\mathcal{H}_{B}$ is history-deterministic.

Since the obtained OCN $\mathcal{N}$ has size linear in $\mathcal{H}_{A}$ and
$\mathcal{H}_{B}$, we can check history-determinism of $\mathcal{N}$ to decide
whether the inclusion
$\mathcal{L}(\mathcal{H}_{A})\subseteq\mathcal{L}(\mathcal{H}_{B})$ holds, in
$\PSPACE$.

###  0.B.4 Proof of Lemma 9

###### Proof

Let us first show that given two history-deterministic one-counter nets
$\mathcal{M}$ and $\mathcal{N}$,

  * •

($\Rightarrow$) $\mathcal{N}$ simulates $\mathcal{M}$ then
$\mathcal{L}(\mathcal{M})\subseteq\mathcal{L}(\mathcal{N})$ and

  * •

($\Leftarrow$) $\mathcal{L}(\mathcal{M})\subseteq\mathcal{L}(\mathcal{N})$,
then $\mathcal{N}$ simulates $\mathcal{M}$.

$\Rightarrow:$ Suppose $\mathcal{N}$ simulates $\mathcal{M}$. Then, over any
accepting word $w\in\mathcal{L}(\mathcal{M})$, there is an accepting run
$\rho$ of it in $\mathcal{M}$, and as $\mathcal{N}$ simulates $\mathcal{M}$,
the run corresponding to $\rho$ in $\mathcal{N}$ must be accepting as well.
Thus, $w\in\mathcal{L}(\mathcal{N})$.

$\Leftarrow:$ Suppose
$\mathcal{L}(\mathcal{M})\subseteq\mathcal{L}(\mathcal{N})$. Then, the player
$\boldsymbol{\exists}$ wins the simulation game
$\mathcal{G}(\mathcal{M}\longhookrightarrow\mathcal{N})$: The player
$\boldsymbol{\exists}$ can simply ignore $\boldsymbol{\forall}$’s run in
$\mathcal{M}$, and play according to her letter game strategy in
$\mathcal{N}$. If $\boldsymbol{\forall}$’s token is at an accepting state at
the end of any round in the game after having read $w$, then
$w\in\mathcal{L}(\mathcal{M})\subseteq\mathcal{L}(\mathcal{N})$, which implies
$\boldsymbol{\exists}$’s token must be on an accepting state as well, as
$\boldsymbol{\exists}$ was playing according to her letter game strategy in
$\mathcal{N}$.

Note that $\Sigma^{*}$ can be given by a one state finite state automata.
Thus, the problem of universality reduces to checking for simulation between
the input net $\mathcal{M}$ and a one-state finite automaton, which is in $\P$
from the results of Kucera (Lemma 2, [24]).

##  Appendix 0.C Appendix for Section 5

###  0.C.1 Proof of Lemma 10

###### Proof

We first describe what reachability games on succinct one-counter nets (SOCN)
are [20, 22]. The arena of a SOCN-reachability game $\mathcal{G}(\mathcal{N})$
consists of a one-counter net $\mathcal{N}=(Q,\\{a\\},\Delta,q_{0},F)$ over an
unary alphabet. However, the states are partitioned as $Q=Q_{\lor}\uplus
Q_{\land}$ among the players $\lor$ and $\land$ respectively, such that any
transition is only between a $\land$ and $\lor$ state or a $\lor$ and $\land$
state. If the play is at a $\lor$ ($\land$) state, then the player $\lor$
($\land$) chooses a transition at that state to go to the next configuration.
Moreover, these transitions are allowed to increment and decrement the counter
more than 1, and can be any arbitrary value $d\in\mathbb{N}_{\geq 0}$, where
$d$ is given in binary. The starting state of the game is the configuration
$(q_{0},0)$. We consider, for the $\lor$ player. the problem of reachability
to a configuration $(q_{F},0)$ for some $q_{F}\in F$.

The version of SOCN-reachability game we have defined is slightly different
from the version considered in the theorem statement of Hunter [20], but one
can see that these can be shown to be inter-reducible [22].

The decision problem, which we call Socn-Game is

> Given: A SOCN-reachability game $\mathcal{G}(\mathcal{N})$ such that the
> counter change in transitions are encoded in binary.
> Question: Does there exist a winning strategy for $\lor$ in
> $\mathcal{G}(\mathcal{N})$.

The above problem is known to be $\EXPSPACE$-complete [20, 22], and we shall
show the $\EXPSPACE$-hardness for deciding history-determinism by reducing
from the above problem.

Given a SOCN-reachability game $\mathcal{G}(\mathcal{N})$, we construct a net
$\mathcal{M}$ which is history-deterministic if and only if the $\land$ player
wins the reachability game. The net $\mathcal{M}$ is designed such that a
transcription of any play on the succinct one-counter game is an accepting
word.

The following introduction to the construction of $\mathcal{M}$ is best read
referring to the rigorous construction that follows it.

Describing the states of the automaton:  To be able to recreate the transcript
of a play of the SOCN-reachability game in the letter game of $\mathcal{M}$,
the state space contains

  1. 1.

A ‘main’ copy of the states of $\mathcal{N}$, in which the game would stay if
$\boldsymbol{\forall}$ had a strategy to win. These are used to maintain a run
on $\mathcal{N}$.

  2. 2.

A resolution copy of states of $Q_{\land}$ for $\boldsymbol{\exists}$ to stay
in, until the non-determinism chosen by $\boldsymbol{\exists}$ is faithfully
re-played by $\boldsymbol{\forall}$.

  3. 3.

A ‘copy’ of $\mathcal{N}$ to ensure that any transcript of a run that is
encoded indeed is a real run on the one-counter automaton $\mathcal{N}$. For
each state $q$ of $\mathcal{N}$, we add states in the set $Q_{\mathtt{win}}$,
in this copy. These states are called so, because from here,
$\boldsymbol{\exists}$ can win the letter game, and there is no non-
determinism to resolve.

  4. 4.

A few extra states, to preserve the winner in the succinct reachability game.
These ensure that if a play of the letter game in $\mathcal{M}$ corresponds to
a winning transcript of $\land$ player, then $\boldsymbol{\exists}$ can go to
a state with no non-determinism. However, there are also states
$q_{\heartsuit}$ and $q_{\clubsuit}$ which are the states that
$\boldsymbol{\forall}$ would be able to make $\boldsymbol{\exists}$ reach if
there is a winning play for the $\lor$ player, from where
$\boldsymbol{\exists}$ loses the letter game.

Describing the alphabet of the automaton: The alphabet contains an input
$a_{\delta}$ for each transition from a $\lor$ state. This is to make sure
$\boldsymbol{\forall}$, who picks the letter in the letter game is in charge
of picking the next transition for the player $\lor$ in the game. There is
also a single letter $a$ which creates non-determinism such that in the letter
game, $\boldsymbol{\exists}$ can resolve the non-determinism by picking the
next transition. Later on, to ensure a fair play, $\boldsymbol{\forall}$ is
forced to confirm this non-determinism by reading the letter $a_{\delta}$
corresponding to the transition that $\boldsymbol{\exists}$ had chosen. If
$\boldsymbol{\forall}$ picks a different transition, then the play moves to
the component which we call $Q_{\mathtt{win}}$ from which all transcriptions
of sequences that are ‘valid’ in the original automaton are accepting, and
there is no non-determinism, making it winning for $\boldsymbol{\exists}$ in
the letter game.

There are also some special symbols used in the following way:

  * •

$\$$ is used by $\boldsymbol{\forall}$ to indicate he is at a state $q_{F}\in
F$ with $0$ in the counter. If he reads it anywhere else,
$\boldsymbol{\exists}$ wins the letter game trivially.

  * •

$\heartsuit$ and $\clubsuit$ are to be read immediately after $\$$, but the
states $q_{\heartsuit}$ can only read $\heartsuit$ and $q_{\clubsuit}$ can
only read $\clubsuit$.

Language accepted by the automaton: The language accepted by the automaton
would be any prefix of the words of the form
$a_{\delta_{0}}aa_{\delta_{1}}a_{\delta_{2}}aa_{\delta_{3}}a_{\delta_{4}}\dots
aa_{\delta_{k}}a_{\delta_{k+1}}$ followed by a $\$$ then one of $\heartsuit$
or $\clubsuit$ such that
$\delta_{0}\delta_{1}\delta_{2}\dots\delta_{k}\delta_{k+1}$ is a valid
sequence of transitions in $\mathcal{N}$. Note that the above words
essentially is a sequence of letters of the form $a_{\delta}$ transitions, but
with $a$s read before reading a ‘transition’ from a $\land$ state.

Transitions of the automaton: Now we intuitively describe the transitions of
the automaton that accept such a language above. They are constructed so that
reading each $a_{\delta}$ increments or decrements the counter by the same
amount prescribed by $\delta$.

From the ‘main’ copy of the state belonging to $\lor$, the transitions are
such that if there was a transition from this state in $\mathcal{N}$, a copy
of the transition also is added. Moreover, there are specific letters that one
can read to go to the next state prescribed by the transition. For $\land$
however, these transitions are labelled by $a$, and perform no increment or
decrement. They instead take the run to a temporary copy of of the state, from
which $\boldsymbol{\forall}$ can read the $a_{\delta}$ corresponding to the
delta that $\boldsymbol{\exists}$ had chosen. If the letter does not
correspond to the same transition that $\boldsymbol{\exists}$ had picked, then
the run moves to the states $Q_{\mathtt{win}}$, from where
$\boldsymbol{\exists}$ wins the letter game.

We now define $\mathcal{M}$ formally, given a game on $\mathcal{N}$. We first
describe the $Q_{\mathtt{win}}$, a sub-net, which will form a part of the main
automaton $\mathcal{M}$. This is done so as to make sure the main definition
has less clutter.

##### The $Q_{\mathtt{win}}$ gadget:

We describe the set of states $Q_{\mathtt{win}}$ and the transitions
associated with it in more detail. Recall that this part of the net is to
mainly ensure that only valid transcripts of a run on $\mathcal{N}$ are the
ones that are accepted. For this, we essentially take one copy of $Q$ and one
more copy of $Q_{\land}$. We add one more state to recognise that the
transcript has ended if a $\$$ has been read.

We therefore have, $Q_{\mathtt{win}}=Q\uplus Q_{\land}\uplus\\{q_{\$}\\}$. We
call states $q_{\mathtt{win}}$ with the subscript if $q$ is from the copy of
$Q$ and $q_{\mathtt{twin}}$ for a copy of $q\in Q_{\land}$. There are the
following transitions:

  * •

$(q_{\mathtt{win}},a_{\delta},d,p_{\mathtt{win}})$ for all $q\in Q_{\lor}$ and
$\delta=(q,d,p)$, a transition in $\mathcal{N}$;

  * •

$(p_{\mathtt{win}},a,0,p_{\mathtt{twin}})$ for all $p\in Q_{\land}$

  * •

$(p_{\mathtt{twin}},a_{\delta},d,q_{\mathtt{win}})$ for all $p\in Q_{\land}$
and $\delta=(p,d,q)$, a transition in $\mathcal{N}$;

  * •

$(q_{\mathtt{win}},\$,0,q_{\$})$ for all $q\in Q$.

We describe the automaton
$\mathcal{M}=(Q^{\prime},\Sigma^{\prime},\Delta^{\prime},q_{0},F^{\prime})$
where,

  * •

the set of states
$Q^{\prime}=Q\cup\\{q_{\delta}\mid\delta\in\Delta\\}\cup\\{q_{\heartsuit},q_{\clubsuit},q_{\mathtt{last}}\\}$,
and the states $Q_{\mathtt{win}}$

  * •

the alphabet set is
$\Sigma=\\{\$,\heartsuit,\clubsuit\\}\cup\\{a_{\delta}\mid\delta\in\Delta\\}\cup\\{a\\}$

  * •

the start state is $q_{0}$, which is copy of the start state at $Q$, and

  * •

all states of $Q^{\prime}$ are final states, including $Q_{\mathtt{win}}$

The set of transitions are the union of the sets of transitions below, along
with those of $Q_{\mathtt{win}}$ and some defined from the states to
$Q_{\mathtt{win}}$ and one transitions back from it to $q_{\mathtt{last}}$ to
end the word.

  1. 1.

$\\{(q_{\lor},a_{q},d,q)\mid(q_{\lor},d,q)\in\Delta\text{ and }q_{\lor}\in
Q_{\lor}\\}$

  2. 2.

$\\{(p,a,0,p_{\delta})\mid\delta=(p,d,q)\in\Delta\text{ and }p\in
Q_{\land}\\}$

  3. 3.

$\\{(p_{\delta},a_{\delta},d,q)\mid\delta=(p,d,q)\in\Delta\text{ and }p\in
Q_{\land}\\}$

  4. 4.

$\\{(p_{\delta},a_{\delta^{\prime}},d^{\prime},{q^{\prime}}_{\mathtt{win}})\mid\delta=(p,d,q)\text{
and }\delta^{\prime}=(p,d^{\prime},q^{\prime}),\delta\neq\delta^{\prime}\text{
and }p\in Q_{\land}\\}$

  5. 5.

$\\{(q_{F},\$,-1,q_{\$}),(q_{F},\$,0,q_{\clubsuit}),(q_{F},\$,0,q_{\heartsuit})\mid
q_{F}\in F\\}$

  6. 6.

$\\{(q,\$,0,q_{\$})\mid q\in Q\setminus F\\}$

  7. 7.

$\\{(q_{\heartsuit},\heartsuit,0,q_{\mathtt{last}}),(q_{\clubsuit},\clubsuit,0,q_{\mathtt{last}})\\}$

  8. 8.

From $Q_{\mathtt{win}}$ we have
$(q_{\$},\heartsuit,0,q_{\mathtt{last}}),(q_{\$},\clubsuit,0,q_{\mathtt{last}})\\}$

Figure 5: A snapshot of the $Q_{\mathtt{win}}$ gadget with the states in
$Q_{\mathtt{win}}$ in red, with the ‘main’ vertices which are not in
$Q_{\mathtt{win}}$ in black. If the non-determinism in the run of the succinct
OCN game’s imitation here is incorrectly resolved anywhere then
$\boldsymbol{\exists}$ can take transitions to the red states from which there
is no non-determinism.

Observe that transitions described in items 4, 5, 6 involve transitions to
$Q_{\mathtt{win}}$. In item 4., note that both the transitions $\delta$ and
$\delta^{\prime}$ should be from the same state for this transition to exist.

We now proceed to showing that the above construction is such that

  * $\Rightarrow$

If the reachability game on succinct net $\mathcal{N}$ is won by $\lor$, then
$\mathcal{M}$ constructed is not history-deterministic and
$\boldsymbol{\forall}$ can win the letter game

  * $\Leftarrow$

If the reachability game on $\mathcal{N}$ is won by $\land$, then
$\boldsymbol{\exists}$ has a strategy to win the letter game, and
$\mathcal{M}$ is history-deterministic

##### Winning for $\lor$ reachability game implies winning for
$\boldsymbol{\forall}$ in letter game

Here we prescribe $\boldsymbol{\forall}$’s strategy which is essentially to
follow the reachability strategy of $\lor$. When the game is at a state

  * •

$q\in Q_{\lor}$ with counter value $k$, then he picks $a_{\delta}$ such that
the transition $\delta$ ensures $(q,k)\xrightarrow{d}(q^{\prime},k^{\prime})$
is a winning transition prescribed by a fixed strategy in the game. This leads
to no non-determinism.

  * •

$q\in Q_{\land}$ with counter value $k$, then he picks $a$, but no matter
which transition in the game $\boldsymbol{\exists}$ picks, she reaches a
configuration that is still winning for $\lor$ in the succinct game, this is
because any $(q,k)\xrightarrow{d}(q^{\prime},k^{\prime})$ was a winning
transition in the game.

This strategy maintains an invariant that if the play of a letter game was at
a configuration such that the corresponding configuration in $\mathcal{N}$ was
winning for $\lor$, then $\boldsymbol{\forall}$ can ensure that in the letter
game, any transition that $\boldsymbol{\exists}$ picks also leads to a
configuration where this is true.

Since the game is winning from $(q_{0},0)\in\mathcal{C}(\mathcal{N})$, and as
$\boldsymbol{\forall}$ is mimicking a winning strategy in the letter game, we
know that eventually $\boldsymbol{\forall}$ would reach a state $q_{F}$ that
is in $F$ with a counter value $0$. Once he reaches such a configuration, he
reads $\$$. This ensures that only the transitions
$(q_{F},\$,0,q_{\heartsuit})$ or $(q_{F},\$,0,q_{\clubsuit})$ are enabled.
From here, whichever transition $\boldsymbol{\exists}$ pics, he reads the
other letter corresponding to it to win.

##### Winning for $\land$ in reachability game implies $\boldsymbol{\exists}$
wins letter game

The player $\boldsymbol{\exists}$’s strategy in the letter game is to mimic
the strategy of $\land$ in the underlying reachability game. Let us fix such a
winning strategy for $\land$ in the reachability game. We shall show that this
strategy maintains the invariant that if the play of a letter game was at a
configuration $(q,k)$ where $q\in Q$ is in the main copy of $\mathcal{N}$,
then the corresponding configuration is losing for $\lor$ in the reachability
game.

When the game is at a configuration $(q,k)$ in the net $\mathcal{M}$ she does
the following:

  * •

for $q\in Q_{\lor}$ with counter value $k$, if $\boldsymbol{\forall}$ picks
$a_{\delta}$ such that $\delta$ ensures
$(q,k)\xrightarrow{d}(q^{\prime},k^{\prime})$ was a transition, then
$\boldsymbol{\exists}$ needs to make no decisions. If not, the game proceeds
to the copy $Q_{\mathtt{win}}$ and we can show that any sequence of runs that
has a valid run is winning for the player $\boldsymbol{\exists}$ anyway. If
the play instead stays in the ‘main’ copy, then the new configuration reached
maintains the invariant.

  * •

for $q\in Q_{\land}$ with counter value $k$, if $\boldsymbol{\forall}$ picks
$a$, then $\boldsymbol{\exists}$ picks a the transitions $\delta$
corresponding to the configuration prescribed by her winning strategy. Note
that later if $\boldsymbol{\forall}$ does not pick $a_{\delta}$, then
$\boldsymbol{\exists}$ wins by going to $Q_{\mathtt{win}}$. Observe that if
the transition prescribed is such that the run goes below $0$, that run turns
out to be not accepting because of the gadget described. This means
$\boldsymbol{\forall}$ loses the letter game again immediately. If the
transition still stays above 0, then the configuration proceeds to a
$(q^{\prime},k^{\prime})$, prescribed by $\land$’s strategy in the
reachability game to avoid visiting $(q_{F},0)$. The position
$(q^{\prime},k^{\prime})$ is such that there is no winning strategy for $\lor$
from it.

Observe that from any state in $Q_{\mathtt{win}}$, $\boldsymbol{\exists}$ wins
the letter game as there is no non-determinism to resolve, and any play that
does not go to $Q_{\mathtt{win}}$ corresponds to a play in the reachability
game where $\boldsymbol{\exists}$’s resolution of non-determinism in the
letter game corresponds to the choices of the $\land$ player in the
reachability game. But note that from the invariant above,
$\boldsymbol{\exists}$ never reaches $(q_{F},0)$ in such a play, and therefore
she never has to resolve the non-determinism that occurs at $q_{F}$ on reading
$\$$ to states $q_{\heartsuit}$ and $q_{\clubsuit}$. Any infinite play is also
won by $\boldsymbol{\exists}$, and therefore she wins the letter game on
$\mathcal{M}$.

###  0.C.2 Proof of Theorem 5.2

###### Proof

Consider the following problem.

> _DOCA Inclusion:_ Given two deterministic one-counter automata $\mathcal{A}$
> and $\mathcal{B}$, is
> $\mathcal{L}(\mathcal{A})\subseteq\mathcal{L}(\mathcal{B})$?

We reduce DOCA inclusion to the problem of deciding whether a given one-
counter automaton is history-deterministic. Valiant, in Section 5.1 of his
thesis [30] shows that the DOCA inclusion problem is undecidable [31]. This
shows that the problem of deciding if a given OCA is history-deterministic is
also undecidable.

Note that this construction is similar to the one in the construction of Lemma
8. We nevertheless re-state it here, for completeness.

Given DOCA $\mathcal{A}=(Q_{A},\Sigma,\Delta_{A},q^{0}_{A},F_{A})$ and
$\mathcal{B}=(Q_{B},\Sigma,\Delta_{B},q^{0}_{B},F_{B})$, consider the
automaton $\mathcal{B}^{\prime}$, for some symbol $\Cat\notin\Sigma$, defined
as the tuple

$\mathcal{B}^{\prime}=(Q^{\prime}_{B},\Sigma\cup\\{\Cat\\},\Delta^{\prime}_{B},q^{0}_{B},F^{\prime}_{B}),$

where for a new element $q_{*}$ not in $Q$,

  * •

the set of states $Q^{\prime}_{B}=Q_{B}\cup\\{q_{*}\\}$,

  * •

the set of transitions
$\Delta^{\prime}_{B}=\Delta_{B}\cup\\{(q_{0},\mathrm{zero},\Cat,0,q_{*}),(q_{0},\neg\mathrm{zero},\Cat,0,q_{*})\\}$,
and

  * •

the final states $F^{\prime}_{B}=F_{B}\cup\\{q_{*}\\}$.

The automaton $\mathcal{B}^{\prime}$ is deterministic, and the language
accepted by $\mathcal{B}^{\prime}$, is
$\mathcal{L}(\mathcal{B}^{\prime})=\mathcal{L}(\mathcal{B})\cup\\{\Cat\\}$.
Note that $\mathcal{L}(\mathcal{A})\subseteq\mathcal{L}(\mathcal{B})$ if and
only if $\mathcal{L}(\mathcal{A})\subsetneq\mathcal{L}(\mathcal{B}^{\prime})$.

We now describe the automaton $\mathcal{H}$, which is history-deterministic if
and only if $L(A)\subset L(B^{\prime})$. Let
$\mathcal{H}=(Q_{H},\Sigma\cup\\{\Cat,\heartsuit\\},\Delta_{H},q^{0}_{H},F_{H})$,
where

  * •

the set of states $Q_{H}=Q_{A}\cup Q_{B}^{\prime}\cup\\{q^{0}_{H}\\}$,

  * •

the set of transitions
$\Delta_{H}=\Delta_{A}\cup\Delta^{\prime}_{B}\cup\\{(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{A}),(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{B})\\}$

  * •

the set of final states $F_{H}=F_{A}\cup F^{\prime}_{B}$.

Suppose $\mathcal{H}$ constructed as above is history-deterministic. Then,
there is resolver on $q_{H}^{0}$, that chooses the transition
$(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{A})$ or
$(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{B})$. Note that $q^{H}_{0}$ is
the only state where non-determinism occurs, on $\heartsuit$. Since the
language of $\mathcal{B}^{\prime}$ contains the word $\Cat$, which is not in
the language of $\mathcal{A}$, the resolver must choose the transition
$(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{B})$, as otherwise
$\boldsymbol{\exists}$ loses the letter game if $\boldsymbol{\forall}$ gave as
input $\Cat$ after $\heartsuit$. Therefore, any resolver must choose the
transition to the copy of $\mathcal{B}^{\prime}$, on $\heartsuit$. This
implies that for any word accepted by $\mathcal{A}$, the resolver has a
strategy henceforth to produce a sequence of transitions in
$\mathcal{B}^{\prime}$, implying that this word must also be accepted by
$\mathcal{B}$. Hence we have
$\mathcal{L}(\mathcal{A})\subseteq\mathcal{L}(\mathcal{B})$.

For the other direction, suppose
$\mathcal{L}(\mathcal{A})\subseteq\mathcal{L}(\mathcal{B})$. The resolver only
needs to resolve non-determinism in the starting state. Choosing the
transition that takes it to the $\mathcal{B}^{\prime}$ part of the automaton
by selecting $(q^{0}_{H},\mathrm{zero},\heartsuit,0,q^{0}_{B})$ at $q^{0}_{H}$
ensures $\boldsymbol{\exists}$ wins the letter game and hence $\mathcal{H}$ is
history-deterministic.


# Bourgain’s counterexample in the sequential convergence problem for the
Schrödinger equation

Chu-Hee Cho  and  Daniel Eceizabarrena  Department of Mathematical Sciences
and RIM, Seoul National University, Seoul 08826, Republic of Korea
<EMAIL_ADDRESS>Department of Mathematics and Statistics, University of
Massachusetts Amherst, USA<EMAIL_ADDRESS>
###### Abstract.

We study the problem of pointwise convegence for the Schrödinger operator on
$\mathbb{R}^{n}$ along time sequences. We show that the sharp counterexample
to the sequential Schrödinger maximal estimate given recently by Li, Wang and
Yan based in the construction by Lucà and Rogers can also be achieved with the
construction of Bourgain, and we extend it to the fractal setting.

##  1\. Introduction

The Schrödinger operator

(1.1) $e^{it\Delta}f(x)=\int_{\mathbb{R}^{n}}\widehat{f}(\xi)\,e^{2\pi
i(x\cdot\xi+t|\xi|^{2})}\,d\xi$

gives the solution to the free Schrödinger equation on $\mathbb{R}^{n}$ with
initial datum $f$. In 1980, Carleson [4] asked for the smallest Sobolev
regularity $s$ such that

(1.2) $\displaystyle\lim_{t\rightarrow 0}e^{it\Delta}f=f\quad\text{
a.e.},\quad\text{ for all }f\in H^{s}(\mathbb{R}^{n}).$

This problem is now solved; the critical Sobolev exponent is $n/(2(n+1))$ [4,
6, 3, 8, 10]. Interesting variations of this problem have been studied since
then. In this note we focus on two of them: the fractal and the sequential
problems.

In the fractal problem, the convergence property is strengthened to
$\mathcal{H}^{\alpha}$-a.e., where $\alpha\leq n$ is given and
$\mathcal{H}^{\alpha}$ is the $\alpha$-Hausdorff measure. That is, we look for

(1.3) $s_{c}(\alpha)=\inf\\{\,s\geq 0\,:\,\lim_{t\rightarrow
0}e^{it\Delta}f=f\quad\mathcal{H}^{\alpha}\text{- a.e.},\,\,\,\forall f\in
H^{s}(\mathbb{R}^{n})\,\\}.$

Žubrinić [17] and Barceló et al. [2] showed $s_{c}(\alpha)=(n-\alpha)/2$ when
$0<\alpha\leq n/2$, but the problem is open when $n/2<\alpha\leq n$. The best
results are the upper bound

(1.4)
$s_{c}(\alpha)\leq\frac{n}{2(n+1)}+\frac{n}{2(n+1)}(n-\alpha),\quad\text{ for
}\quad\frac{n+1}{2}\leq\alpha\leq n$

by Du–Zhang [10], and a sawtooth-like lower bound111We refer to [11, Theorem
1.1] for the explicit expression of this exponent. $s_{\text{E-PV}}$ by
Eceizabarrena–Ponce-Vanegas [11, Theorem 1.1] that recently improved the
previous best lower bound by Lucà–Rogers [15] and Lucà–Ponce-Vanegas [14]:

(1.5) $\frac{n}{2(n+1)}+\frac{n-1}{2(n+1)}(n-\alpha)\leq
s_{\text{E-PV}}(\alpha)\leq s_{c}(\alpha).$

In the sequential problem, given a sequence of times $t_{k}\to 0$ in a class
$T$ that measures its speed of convergence, we look for the smallest $s$ such
that

(1.6) $\displaystyle\lim_{k\to\infty}e^{it_{k}\Delta}f=f\quad\text{ a.e. \quad
for all }(t_{k})\in T,\qquad\text{ for all }f\in H^{s}(\mathbb{R}^{n}).$

In [7], Dimou and Seeger characterized this regularity for $n=1$ in terms of
Lorentz spaces $T=\ell^{r,\infty}$, which for $0<r<\infty$ are defined by

(1.7)
$(t_{k})\in\ell^{r,\infty}(\mathbb{N})\qquad\Longleftrightarrow\qquad\sup_{\delta>0}\delta^{r}\\#\\{k\in\mathbb{N}:|t_{k}|\geq\delta\\}<\infty.$

Denoting the corresponding critical exponent by

(1.8) $s_{c}(\ell^{r,\infty})=\inf\\{\,s\geq
0\,:\,\lim_{k\to\infty}e^{it_{k}\Delta}f=f\quad\text{a.e.},\,\forall(t_{k})\in\ell^{r,\infty},\,\,\forall
f\in H^{s}(\mathbb{R}^{n})\\},$

we now know from [7, 5, 13] that

(1.9)
$s_{c}(\ell^{r,\infty})=\min\Big{(}\frac{n}{2(n+1)},\frac{rn}{r(n+1)+n}\Big{)},$

which shows a non-trivial change of behavior at $r=n/(n+1)$.

In this paper we work with the combination of these two problems and the
corresponding fractal sequential exponent

(1.10) $s_{c}(\alpha,\ell^{r,\infty})=\inf\\{\,s\geq
0\,:\,\lim_{k\to\infty}e^{it_{k}\Delta}f=f\quad\mathcal{H}^{\alpha}\text{-a.e.},\,\forall(t_{k})\in\ell^{r,\infty},\,\,\forall
f\in H^{s}(\mathbb{R}^{n})\\}.$

By standard arguments, upper bounds for (1.10) follow from maximal estimates

(1.11) $\int_{B(0,1)}\sup_{k}|e^{it_{k}\Delta}f(x)|^{2}d\mu(x)\lesssim
C_{\mu}\|f\|_{H^{s}}^{2},\quad\forall(t_{k})\in\ell^{r,\infty},\quad\forall
f\in H^{s},\quad\forall\mu\in\mathcal{M}_{\alpha},$

where $\mathcal{M}_{\alpha}$ is the set of $\alpha$-dimensional measures, that
is, non-negative Borel measures for which there is a constant $C_{\mu}$ such
that

(1.12) $\displaystyle\mu(B(x,r))\leq C_{\mu}r^{\alpha},\quad\forall
x\in\mathbb{R}^{n},\quad r>0.$

Let us denote the critical Sobolev exponent for the fractal sequential maximal
estimate by

(1.13) $s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})=\inf\\{\,s\geq
0\,:\,\eqref{fractal_maximal_estimate}\text{ holds }\\},$

and by $s^{\text{max}}_{c}(\alpha)$ we denote the exponent for the non-
sequential maximal estimate, where the supremum in (1.11) is taken over
$t\in(0,1)$ instead of over a sequence $t_{k}$.

In [5], Cho, Ko, Koh and Lee proved

(1.14)
$s^{\text{max}}_{c}(\alpha)\leq\frac{n-\alpha}{2}+s^{*}\quad\Longrightarrow\quad
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\min\Big{(}s^{*},\frac{2rs^{*}}{r+2s^{*}}\Big{)}.$

With the upper bounds known for $s^{\text{max}}_{c}(\alpha)$, one gets the
following positive result:

######  Theorem 1.1.

Let $n\geq 1$, $0\leq\alpha\leq n$ and $0<r<\infty$. Then,

  * •

If $\alpha\leq n/2$, then

(1.15) $s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})=\frac{n-\alpha}{2}.$

  * •

If $n/2<\alpha\leq(n+1)/2$, then

(1.16)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\min\Big{(}\frac{2\alpha-n}{4},\frac{r(2\alpha-n)}{2r+2\alpha-n}\Big{)}.$

  * •

If $\alpha\geq(n+1)/2$, then

(1.17)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\min\Big{(}\frac{\alpha}{2(n+1)},\frac{r\alpha}{r(n+1)+\alpha}\Big{)}.$

We briefly explain in Section 2 how to prove this combining (1.14) with the
results by Zubrinić [17], Barceló et al [2], and Du and Zhang [10]. Observe
that when $\alpha\geq(n+1)/2$,

(1.18) $r<\frac{\alpha}{n+1}\qquad\Longrightarrow\qquad
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\frac{r\alpha}{(n+1)r+\alpha},$

improving the upper bound by Du-Zhang. However, there is no improvement if

(1.19) $r\geq\frac{\alpha}{n+1}\qquad\Longrightarrow\qquad
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n}{2(n+1)}(n+1-\alpha).$

When $n/2\leq\alpha\leq(n+1)/2$, we improve the known $n/4$ when
$r<(2\alpha-n)/2$.

Regarding lower bounds, when $\alpha=n$ we know the optimal (1.9). In this
note we give the first lower bounds for
$s_{c}^{\text{max}}(\alpha,\ell^{r,\infty})$ when $\alpha<n$.

In $n=1$, the typical single wave-packet argument gives:

######  Theorem 1.2.

Let $n=1$ and $1/2\leq\alpha\leq 1$. Then,

(1.20)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\geq\frac{1-\alpha}{2}+\min\Big{(}\frac{2\alpha-1}{4},\frac{r(2\alpha-1)}{2r+\alpha}\Big{)}.$

######  Remark 1.3.

Together with Theorem 1.1, this theorem implies:

  1. (1)

If $1/2\leq\alpha\leq 2r$, then
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})=1/4$.

  2. (2)

If $2r\leq\alpha\leq r+1/2$, then

(1.21) $\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}\leq
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{1}{4}.$

  3. (3)

If $r+1/2\leq\alpha\leq 1$, then

(1.22) $\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}\leq
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+2\alpha-1}.$

In particular, the critical exponent is 1/4 if $r\geq 1/2$.

$1/2$$1$$1/4$$1/2$$\frac{1-\alpha}{2}$$\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+(2\alpha-1)}$$\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}$$\boxed{n=1,\,\,r=1/6}$$\alpha$s
Figure 1. Upper (blue) and lower (red) bounds when $n=1$.

When $n\geq 2$, to get (1.9) Li, Wang and Yan [13] adapted the ergodic
counterexample of Lucà and Rogers [15]. We show here that the same result can
also be obtained via Bourgain’s argument in [3], and we generalize it to the
fractal setting.

######  Theorem 1.4.

Let $n\geq 2$.

  1. (1)

If $n-1\leq\alpha\leq n$,

(1.23)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\geq\frac{n-\alpha}{2}+\min\Big{(}\frac{2\alpha-n}{2(n+1)},\frac{r(2\alpha-n)}{r(n+1)+2\alpha-n}\Big{)}.$

  2. (2)

If $n/2<\alpha\leq n-1$,

(1.24)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\geq\frac{n-\alpha}{2}+\min\Big{(}\frac{2\alpha-n}{2(n+1)},\frac{r\alpha}{r(n+1)+n}\Big{)}.$

######  Remark 1.5.

  1. (1)

The exponent in (1.23) is larger than the one in (1.24) due to a better
scaling argument in the range $n-1\leq\alpha\leq n$.

  2. (2)

When $r\geq(2\alpha-n)/(n+1)$, in both cases the exponent is the same as the
one in (1.5) by Lucà-Rogers and Lucà-Ponce-Vanegas. The exponents are new when
$r<(2\alpha-n)/(n+1)$.

  3. (3)

The exponent $s_{\text{E-PV}}$ in (1.5) was obtained in [11] by generalizing
Bourgain’s counterexample with an intermediate space trick from [9]. Combining
our strategy in this note with this intermediate space trick should give a
better lower bound, but it will require heavy computations. We prefer to keep
this note short to highlight the ideas of our construction, so we will not
pursue this further here.

$n/2$$n-1$$n$$n/4$$n/2$$\frac{n-\alpha}{2}$$\frac{n-\alpha}{2}+\frac{r\alpha}{r(n+1)+n}$$\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{2r+(2\alpha-n)}$$\alpha$s
Figure 2. Comparison of the two lower bounds in Theorem 1.4.

#### Notation

$A\lesssim B$ denotes $A\leq CB$ for a constant $C>0$. We write $A\simeq B$
when $A\lesssim B$ and $B\lesssim A$. For $\mu\in\mathcal{M}_{\alpha}$, we
denote the infimum of $C_{\mu}$ in (1.12) by $\langle\mu\rangle_{\alpha}$.

##  2\. Brief explanation of Theorem 1.1

Theorem 1.1 directly follows from the transference result (1.14) proved by Cho
et al., which we reproduce here for convenience of the reader:

(2.1)
$s^{\text{max}}_{c}(\alpha)\leq\frac{n-\alpha}{2}+s^{*}\quad\Longrightarrow\quad
s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\min\Big{(}s^{*},\frac{2rs^{*}}{r+2s^{*}}\Big{)}.$

If $\alpha\leq n/2$, by Barceló et al. [2] (2.1) holds with $s^{*}=0$.
Together with the result by Žubrinić [17], this directly implies that

(2.2)
$s_{c}(\alpha,\ell^{r,\infty})=s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})=\frac{n-\alpha}{2},\qquad\forall
r>0.$

When $n/2\leq\alpha\leq(n+1)/2$, the bound by Barceló et al. [2] and the fact
that $s_{c}^{\text{max}}(\alpha)$ is decreasing imply that
$s^{\text{max}}_{c}(\alpha)\leq n/4$. Hence, (2.1) holds with
$s^{*}=(2\alpha-n)/4$, so

(2.3)
$\begin{split}s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})&\leq\frac{n-\alpha}{2}+\min\Big{(}\frac{2\alpha-n}{4},\frac{r(2\alpha-n)}{2r+2\alpha-n}\Big{)}\\\
&=\min\Big{(}\frac{n}{4},\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{2r+2\alpha-n}\Big{)},\end{split}$

which gives an improvement if $r\leq(2\alpha-n)/2$. Finally, by Du-Zhang [10],
when $\alpha\geq(n+1)/2$ we get (2.1) with $s^{*}=\alpha/(2(n+1))$, which
implies

(2.4)
$s^{\text{max}}_{c}(\alpha,\ell^{r,\infty})\leq\frac{n-\alpha}{2}+\min\Big{(}\frac{\alpha}{2(n+1)},\frac{r\alpha}{r(n+1)+\alpha}\Big{)}.$

This improves the Du-Zhang upper bound if $r\leq\alpha/(n+1)$.

##  3\. Proof of theorem 1.4 \- Dimensions $n\geq 2$

Let $R\gg 1$ be the main parameter. We will construct a sequence of functions
$f_{R}$ and measures $\mu_{R}$ to disprove the maximal estimate (1.11). Let
$S,D$ be auxiliary parameters with $1\leq S\leq R$ and $1\leq D\leq R$, both
of which are powers of $R$ that we determine later.

Let $\phi$ be a smooth function222With a small abuse of notation, $\phi$ and
$\psi$ may represent functions both in $\mathbb{R}$ and $\mathbb{R}^{n-1}$.
with $\operatorname{supp}\widehat{\phi}\subset B_{1}(0)$ and $\phi(0)=1$. Let
$\psi$ be a positive smooth function with support in $1/2\leq|x|\leq 1$.
Define $f_{R}$ by

(3.1)
$\widehat{f}_{R}(\xi)=\widehat{g}_{R}(\xi_{1})\,\widehat{h}_{R}(\xi^{\prime}),\qquad\xi=(\xi_{1},\xi^{\prime})\in\mathbb{R}\times\mathbb{R}^{n-1},$

where

(3.2)
$\widehat{g}_{R}(\xi_{1})=\frac{1}{S}\,\widehat{\phi}\Big{(}\frac{\xi_{1}-R}{S}\Big{)},\qquad\qquad\widehat{h}_{R}(\xi^{\prime})=\sum_{m^{\prime}\in\mathbb{Z}^{n-1}}\psi\Big{(}\frac{D}{R}m^{\prime}\Big{)}\,\widehat{\phi}(\xi^{\prime}-Dm^{\prime}).$

It follows that $\|g_{R}\|_{2}\sim S^{-1/2}$ and
$\|h_{R}\|_{2}\sim(R/D)^{\frac{n-1}{2}}$, so

(3.3)
$\|f_{R}\|=\|g_{R}\|_{2}\|h_{R}\|_{2}\sim\frac{1}{S^{1/2}}\,\Big{(}\frac{R}{D}\Big{)}^{\frac{n-1}{2}}.$

To study the evolution of $f_{R}$, we do so for $g_{R}$ and $h_{R}$
separately. For $x_{1}\in[0,1]$,

(3.4)
$\begin{split}e^{it\Delta}g_{R}(x_{1})&=\int\frac{1}{S}\,\widehat{\phi}\Big{(}\frac{\xi_{1}-R}{S}\Big{)}\,e^{2\pi
i(x_{1}\xi_{1}+t\xi_{1}^{2})}d\xi_{1}\\\ &=e^{2\pi
i(x_{1}R+tR^{2})}\,\int\widehat{\phi}(\eta_{1})\,e^{2\pi
i(S\eta_{1}(x_{1}+2Rt)+tS^{2}\eta_{1}^{2})}\,d\eta_{1},\end{split}$

where we changed variables $\xi_{1}=R+S\eta_{1}$. For $c>0$ be small enough,

(3.5) $S|x_{1}+2Rt|\leq c,\quad S^{2}t\leq
c\quad\Longrightarrow\quad|e^{it\Delta}g_{R}(x_{1})|\simeq 1.$

On the other hand, for $x^{\prime}\in\mathbb{[}0,1]^{n-1}$ we write

(3.6)
$\begin{split}e^{it\Delta}h_{R}(x^{\prime})&=\sum_{m^{\prime}\in\mathbb{Z}^{n-1}}\psi\Big{(}\frac{D}{R}m^{\prime}\Big{)}\,\int\widehat{\phi}(\xi^{\prime}-Dm^{\prime})\,e^{2\pi
i(x^{\prime}\cdot\xi^{\prime}+t|\xi^{\prime}|^{2})}\,d\xi^{\prime}\\\
&=\sum_{m^{\prime}\in\mathbb{Z}^{n-1}}\psi\Big{(}\frac{D}{R}m^{\prime}\Big{)}\,e^{2\pi
i(Dx^{\prime}\cdot
m^{\prime}+tD^{2}|m^{\prime}|^{2})}\,\int\widehat{\phi}(\eta^{\prime})e^{2\pi
i(\eta^{\prime}\cdot(x^{\prime}+2tDm^{\prime})+t|\eta^{\prime}|^{2})}\,d\eta^{\prime}.\end{split}$

Since $|\eta^{\prime}|\leq 1$ and $D|m^{\prime}|\simeq R$, we ask

(3.7) $t\leq\frac{c}{R}$

to make the phase inside the integral small. Simultaneously, we set

(3.8) $D^{2}t=\frac{p_{1}}{q},\qquad Dx_{i}=\frac{p_{i}}{q}+\epsilon,\quad
i=2,\ldots,n$

with $q\in\mathbb{Z}$, $p_{1},\cdots,p_{n}\in\mathbb{Z}$,
$\operatorname{gcd}(p_{1},q)=1$ and $|\epsilon|\leq cD/R$. Conditions (3.7)
and (3.8) allow us to estimate the evolution (3.6) by a Gauss sum, that is,

(3.9) $\displaystyle|e^{it\Delta}h_{R}(x^{\prime})|$
$\displaystyle\simeq\Big{|}\sum_{m^{\prime}\in\mathbb{Z}^{n-1}}\psi\Big{(}\frac{D}{R}m^{\prime}\Big{)}\,e^{2\pi
i\frac{p^{\prime}\cdot m^{\prime}+p_{1}|m^{\prime}|^{2}}{q}}\,e^{2\pi
i\epsilon\cdot m^{\prime}}\Big{|}$ (3.10)
$\displaystyle\simeq\prod_{i=2}^{n}\Big{(}\frac{R}{Dq}\Big{)}\Big{|}\sum_{m_{i}=1}^{q}e^{2\pi
i\frac{p_{i}\,m_{i}+p_{1}m_{i}^{2}}{q}}\Big{|}$

where $p^{\prime}=(p_{2},\cdots,p_{n})$ and, slightly abusing notation again,
$\epsilon=(\epsilon,\ldots,\epsilon)\in\mathbb{R}^{n-1}$. By properly choosing
$p$ and $q$ (for example, choosing $q$ odd or, alternatively, $q$ even and
$p_{i}\equiv q/2\pmod{2}$), each Gauss sum has size $\sqrt{q}$, so

(3.11)
$|e^{it\Delta}h_{R}(x^{\prime})|\simeq\Big{(}\frac{R}{Dq}\,\sqrt{q}\Big{)}^{n-1}=\Big{(}\frac{R}{D\sqrt{q}}\Big{)}^{n-1}.$

Altogether333 For a detailed proof of estimates (3.9), (3.10) and (3.11), we
refer to [12, Section 3.1] or [16]. , combining (3.3), (3.5) and (3.11) we get

(3.12) $\frac{|e^{it\Delta}f_{R}(x)|}{\lVert f_{R}\rVert_{2}}\simeq
S^{1/2}\,\Big{(}\frac{R}{Dq}\Big{)}^{(n-1)/2}.$

Let us now discuss the conditions we imposed in (3.5), (3.7), (3.8). First,
rewrite the second condition in (3.8) and $\epsilon\leq cD/R$ as

(3.13) $x^{\prime}\in
B_{n-1}\Big{(}\frac{1}{D}\frac{p^{\prime}}{q},\frac{c}{R}\Big{)}.$

On the other hand, from the first conditions in (3.5) and (3.8) we get

(3.14) $x_{1}\in
B_{1}\Big{(}\frac{R}{D^{2}}\frac{p_{1}}{q},\frac{c}{S}\Big{)}.$

Hence, for a new parameter $Q$ (also a power of $R$ to be determined later),
let

(3.15) $X_{R}=\bigcup_{q\simeq
Q}\bigcup_{p}B_{1}\Big{(}\frac{R}{D^{2}}\frac{p_{1}}{q},\frac{c}{S}\Big{)}\times
B_{n-1}\Big{(}\frac{1}{D}\frac{p^{\prime}}{q},\frac{c}{R}\Big{)}\subset[0,1]^{n},$

where $q\simeq Q$ means $Q/2\leq q\leq Q$, possibly with additional
restrictions that do not alter substantially the number of $q$
considered444For example, parity conditions, or imposing $q$ to be prime..
Regarding $t$, (3.5) and (3.7) imply

(3.16) $t\leq c\,\min\Big{(}\frac{1}{S^{2}},\frac{1}{R}\Big{)}.$

This condition determines two regimes, depending on $S\leq R^{1/2}$ or not. In
view of the condition $S|x+2Rt|\leq c$ in (3.5), it affects the set $X_{R}$:

  1. (i)

When $S\lesssim R^{1/2}$, then $t\lesssim 1/R$ so all $|x_{1}|\lesssim 1$ are
allowed. This is the situation of the counterexample in Bourgain’s article
[3].

  2. (ii)

When $R^{1/2}\ll S\leq R$, then $t\lesssim 1/S^{2}\ll 1/R$, so

(3.17)
$t\lesssim\frac{1}{S^{2}}\ll\frac{1}{R}\quad\Longrightarrow\quad|x_{1}|\leq\frac{R}{S^{2}}\ll
1\quad\Longrightarrow\quad X_{R}\subset[0,R/S^{2}]\times[0,1]^{n-1}.$

This situation is the analogue to the one in Li–Wang–Yan [13] to give the
optimal counterexample for the sequential convergence.

From now on, we focus on case (ii) above and assume that $R^{1/2}\leq S\leq
R$. In particular, from $D^{2}t=p_{1}/q\lesssim D^{2}/S^{2}$ and
$|x^{\prime}|\leq c$ we get the restrictions

(3.18) $p_{1}\lesssim\frac{qD^{2}}{S^{2}},\qquad p^{\prime}\lesssim Dq.$

Gathering it all, we have shown that

(3.19) $x\in X_{R}\quad\Longrightarrow\quad\exists
t(x)\in\Big{[}\frac{1}{D^{2}Q},\frac{1}{S^{2}}\Big{]}\,\,:\,\,\frac{|e^{it(x)\Delta}f_{R}(x)|}{\lVert
f_{R}\rVert_{2}}\simeq S^{1/2}\,\Big{(}\frac{R}{DQ}\Big{)}^{(n-1)/2}.$

###  3.1. Lebesgue measure of $X_{R}$

Let us choose the parameters $S,D,Q$ in such a way that $X_{R}$ has the
largest possible measure. In view of (3.17), we aim for $|X_{R}|\simeq
R/S^{2}$. We first argue heuristically:

  * Heuristic argument 1.

It is a consequence of Minkowski’s theorem that if $a_{1},\ldots,a_{n}\geq 1$
are such that $a_{1}+\cdots+a_{n}=n+1$, then any $x\in[0,1]^{n}$ can be
approximated as

(3.20) $\big{|}x_{i}-\frac{p_{i}}{q}\big{|}\leq\frac{1}{q^{a_{i}}},\qquad
i=1,\ldots,n.$

If we rescale $X_{R}$ in (3.15) to balls centered at rationals and write the
radii as $q^{-a_{i}}$, we get

(3.21)
$\frac{D^{2}}{RS}\simeq\frac{1}{Q^{a_{1}}},\qquad\frac{D}{R}\simeq\frac{1}{Q^{a_{i}}},\quad
i=2,\ldots,n.$

Thus, asking the Minkowski condition
$n+1=a_{1}+\ldots+a_{n}=a_{1}+(n-1)a_{2}$,

(3.22)
$\frac{1}{Q^{n+1}}=\frac{1}{Q^{a_{1}}\,Q^{(n-1)a_{2}}}=\frac{D^{2}}{RS}\,\Big{(}\frac{D}{R}\Big{)}^{n-1}\qquad\Longleftrightarrow\qquad(DQ)^{n+1}\simeq
S\,R^{n}.$

  * Heuristic argument 2.

A simpler (but more heuristic) argument is to assume that the union (3.15) of
$X_{R}$ is disjoint, so that

(3.23) $|X_{R}|\simeq
Q\cdot\frac{D^{2}Q}{S^{2}}\cdot(DQ)^{n-1}\cdot\frac{1}{S}\frac{1}{R^{n-1}}\sim\frac{(DQ)^{n+1}}{S^{3}R^{n-1}}.$

The condition that we impose is then

(3.24)
$\frac{(DQ)^{n+1}}{S^{3}R^{n-1}}\simeq\frac{R}{S^{2}}\qquad\Longleftrightarrow\qquad(DQ)^{n+1}\simeq
SR^{n}.$

These heuristics can be made rigorous using Lemma A.1 in Appendix A. Define
the dilation $T(x_{1},x^{\prime})=((D^{2}/R)x_{1},Dx^{\prime})$, so that
$|T(X_{R})|=(D^{2}/R)D^{n-1}|X_{R}|$. The set $T(X_{R})$ has a 1-periodic
structure and it is made of $(D^{2}/S^{2})D^{n-1}$ disjoint translations of
the set

(3.25) $\widetilde{X}_{R}=\bigcup_{q\sim
Q}\bigcup_{p=1}^{q}B_{1}\Big{(}\frac{p_{1}}{q},c\frac{D^{2}}{RS}\Big{)}\times
B_{n-1}\Big{(}\frac{p^{\prime}}{q},c\frac{D}{R}\Big{)}.$

This means that
$(D^{2}/S^{2})D^{n-1}|\widetilde{X}_{R}|=|T(X_{R})|=(D^{2}/R)D^{n-1}|X_{R}|$,
so

(3.26) $|X_{R}|=|\widetilde{X}_{R}|\,\frac{R}{S^{2}}.$

Now, calling $h_{1}(Q)=D^{2}/(RS)$ and $h_{2}(Q)=D/R$, Lemma A.1 implies

(3.27) $(QD)^{n+1}\simeq SR^{n}\quad\Longrightarrow\quad
Q^{n+1}h_{1}(Q)h_{2}(Q)^{n-1}\simeq
1\quad\Longrightarrow\quad|\widetilde{X}_{R}|\simeq 1,$

and therefore

(3.28) $|X_{R}|\simeq R/S^{2}.$

######  Remark 3.1.

In Bourgain’s counterexample, in which $S\lesssim R^{1/2}$, one gets
$|X_{R}|\simeq 1$ for all $R$. From it, one can explicitly construct a
counterexample to the convergence problem by working with the set of
dicergence $X=\limsup_{R}X_{R}$ with positive measure $|X|\simeq 1$. In this
case, however, from (3.17) it is clear that
$X=\limsup_{R}X_{R}\subset\\{0\\}\times[0,1]^{n-1}$, so in particular it has
measure $|X|=0$. Even worse, $\operatorname{dim}_{\mathcal{H}}\leq n-1$. So
even if this disproves the maximal estimate, it is more challenging to
explicitly construct a counterexample to the convergence problem, at least
when $n-1<\alpha\leq n$; we will not pursue this issue in this note.

###  3.2. Sequence of times in $\ell^{r,\infty}$

The times $t$ we selected in (3.8) belong in

(3.29) $T=\bigcup_{R}T_{R},\quad\text{ where }\quad
T_{R}=\Big{\\{}\,\frac{1}{D^{2}}\frac{p_{1}}{q}\,:\,q\simeq Q,\,\,1\leq
p_{1}\leq\frac{D^{2}q}{S^{2}}\,\Big{\\}}.$

Observe first that $T_{R}\subset[1/(D^{2}Q),1/S^{2}]$. To make all these
intervals disjoint, we select a sequence of dyadic $R_{n}$ such that
$1/S_{n+1}^{2}\ll 1/(D_{n}^{2}Q_{n})$. This way, we can treat the set of times
$T$ as a sequence that tends to zero. By choosing $S,D,Q$ properly, we will
make it belong in $\ell^{r,\infty}$. For that, we ask

(3.30) $\delta^{r}\\#\\{t:\delta<t<2\delta\\}\lesssim 1,\qquad\text{ for every
}\delta>0.$

Suppose $1/S_{n+1}^{2}\leq\delta\leq 1/S_{n}^{2}$. Only the case
$1/(2D_{n}^{2}Q_{n})\leq\delta\leq 1/S_{n}^{2}$ is relevant, since the set in
(3.30) is empty otherwise. Then,

$\delta^{r}\\#\\{n:\delta<t_{n}<2\delta\\}\leq\delta^{r}\,\cdot\delta\,D^{2}Q\,\cdot
Q\leq\frac{(DQ)^{2}}{S^{2(1+r)}}.$

Hence,

(3.31) $DQ\lesssim
S^{1+r}\qquad\Longrightarrow\qquad(t_{n})\in\ell^{r,\infty}.$

###  3.3. Proof of Theorem 1.4 when $\boldsymbol{\alpha=n}$

For the sequence $(t_{n})\in\ell^{r,\infty}$ that we built, by (3.19) and
(3.28) we get

(3.32)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(0,1)}}{\|f_{R}\|_{2}}\gtrsim{S^{1/2}}\Big{(}\frac{R}{DQ}\Big{)}^{(n-1)/2}|X_{R}|^{1/2}\simeq\Big{(}\frac{R^{n}}{S(DQ)^{n-1}}\Big{)}^{1/2}.$

Maximizing this quantity with the restrictions (3.24) and (3.31), we get

(3.33) $S=R^{\frac{n}{r(n+1)+n}},\quad
DQ=S^{1+r}\quad\Longrightarrow\quad\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(0,1)}}{\|f_{R}\|_{2}}\gtrsim
R^{\frac{rn}{r(n+1)+n}},$

which disproves the maximal estimate when $s=rn/(r(n+1)+n)-\epsilon$ for
arbitrary $\epsilon>0$. Observe that the choice $S\geq R^{1/2}$ makes this
argument be valid only for

(3.34) $R^{\frac{n}{r(n+1)+n}}\geq R^{1/2}\qquad\Longleftrightarrow\qquad
r\leq\frac{n}{n+1}.$

###  3.4. Heuristics for the fractal case $\boldsymbol{\alpha<n}$

Let $\mu$ be an $\alpha$-dimensional measure. To disprove the fractal maximal
estimate, we bound like in (3.32), except now

(3.35)
$\displaystyle\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\mu)}}{\|f_{R}\|_{2}}$
$\displaystyle\gtrsim{S^{1/2}}\Big{(}\frac{R}{DQ}\Big{)}^{(n-1)/2}\mu(X_{R})^{1/2}.$

To make $\mu(X_{R})$ the largest possible, a heuristic argument is to think of
$\mu$ as the Hausdorff measure $\mathcal{H}^{\alpha}$. Then, recalling
$X_{R}\subset[0,R/S^{2}]\times[0,1]^{n-1}$, we make $X_{R}$ scale like
$[0,R/S^{2}]\times[0,1]^{n-1}$ when measured with $\mathcal{H}^{\alpha}$.
First, mimicking Heuristic Argument 2 in Section 3.1, assume that the balls in
$X_{R}$ are disjoint so that

(3.36) $\displaystyle\begin{split}\mathcal{H}^{\alpha}(X_{R})&\simeq
Q\cdot\frac{D^{2}Q}{S^{2}}\cdot(DQ)^{n-1}\cdot\mathcal{H}^{\alpha}\Big{(}B_{1}\Big{(}\frac{1}{S}\Big{)}\times
B_{n-1}\Big{(}\frac{1}{R}\Big{)}\Big{)}\\\
&\simeq\frac{(DQ)^{n+1}}{S^{2}}\frac{R}{S}\cdot\mathcal{H}^{\alpha}\Big{(}B_{n}\Big{(}\frac{1}{R}\Big{)}\Big{)}\\\
&\simeq\frac{(DQ)^{n+1}}{S^{2}}\frac{R}{S}\cdot\frac{1}{R^{\alpha}}\cdot\mathcal{H}^{\alpha}(B_{n}(1)),\end{split}$

where we split the rectangle $B_{1}(1/S)\times B_{n-1}(1/R)$ into $R/S$ copies
of $B_{n}(1/R)$. For the rectangle $[0,R/S^{2}]\times[0,1]^{n-1}$ we argue in
two ways:

  * Scaling 1

As in (3.36), split the rectangle in small cubes of size $R/S^{2}$:

(3.37)
$\displaystyle\begin{split}\mathcal{H}^{\alpha}([0,R/S^{2}]\times[0,1]^{n-1})&\simeq\Big{(}\frac{S^{2}}{R}\Big{)}^{n-1}\cdot\mathcal{H}^{\alpha}\Big{(}B_{n}(R/S^{2})\Big{)}\\\
&\simeq\Big{(}\frac{S^{2}}{R}\Big{)}^{n-\alpha-1}\cdot\mathcal{H}^{\alpha}(B_{n}(1)).\end{split}$

  * Scaling 2

Complete the rectangle to $[0,1]^{n}$ to get the same scaling as for the
Lebesgue measure:

(3.38)
$\mathcal{H}^{\alpha}([0,R/S^{2}]\times[0,1]^{n-1})\simeq\frac{R}{S^{2}}\,\mathcal{H}^{\alpha}([0,1]^{n}).$

Scaling 1 will give a better result, but it will only work when $\alpha\geq
n-1$. When $n/2\leq\alpha\leq n-1$ we will use Scaling 2.

######  Remark 3.2.

That Scaling 1 works only for $\alpha\geq n-1$ is related to the fact that
$(S^{2}/R)^{n-\alpha-1}\to\infty$ when $\alpha<n-1$. Ultimately, this is a
consequence of

(3.39)
$X_{R}\subset[0,R/S^{2}]\times[0,1]^{n-1}\to\\{0\\}\times[0,1]^{n-1}\quad\text{
as }R\to\infty,$

as mentioned in Remark 3.1.

We compute both cases by equaling (3.36) to either (3.37) or (3.38).

####  3.4.1. Computations for Scaling 1

We equate (3.36) and (3.37) to get

(3.40)
$\frac{(DQ)^{n+1}}{SR^{\alpha}}\frac{R}{S^{2}}=\Big{(}\frac{S^{2}}{R}\Big{)}^{n-\alpha-1}\qquad\Longleftrightarrow\qquad\frac{(DQ)^{n+1}}{SR^{n}}=\Big{(}\frac{S}{R}\Big{)}^{2(n-\alpha)}.$

This restriction with the unchanged $\ell^{r,\infty}$ condition in (3.31)
determines

(3.41) $S=R^{\frac{2\alpha-n}{r(n+1)+2\alpha-n}},\qquad DQ=S^{1+r}.$

Hence, plugging the heuristic (3.36) into (3.35) and applying (3.41), we get

(3.42)
$\displaystyle\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\mu)}}{\|f_{R}\|_{2}}$
$\displaystyle\gtrsim R^{\frac{n-\alpha}{2}}\,S^{r}\simeq
R^{\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{r(n+1)+2\alpha-n}},$

which is the exponent in Theorem 1.4 (1). Moreover, $S\geq R^{1/2}$ and (3.41)
require

(3.43)
$\frac{2\alpha-n}{r(n+1)+2\alpha-n}\geq\frac{1}{2}\quad\Longleftrightarrow\quad
r\leq\frac{2\alpha-n}{n+1}.$

####  3.4.2. Computations for Scaling 2

In this case we equate (3.36) and (3.38) to get

(3.44)
$\frac{(DQ)^{n+1}}{SR^{\alpha}}\frac{R}{S^{2}}=\frac{R}{S^{2}}\qquad\Longleftrightarrow\qquad(DQ)^{n+1}=SR^{\alpha}.$

Again, the unchanged $\ell^{r,\infty}$ condition in (3.31) with (3.44)
determines

(3.45) $S=R^{\frac{\alpha}{r(n+1)+n}},\qquad DQ=S^{1+r}.$

Plugging the heuristic (3.36) into (3.35), and applying (3.44), we get

(3.46)
$\displaystyle\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\mu)}}{\|f_{R}\|_{2}}$
$\displaystyle\gtrsim R^{\frac{n-\alpha}{2}}\,S^{r}\simeq
R^{\frac{n-\alpha}{2}+\frac{r\,\alpha}{r(n+1)+n}},$

which is the exponent in Theorem 1.4 (2). The condition $S\geq R^{1/2}$ with
(3.45) requires the same condition

(3.47) $\frac{\alpha}{r(n+1)+n}\geq\frac{1}{2}\quad\Longleftrightarrow\quad
r\leq\frac{2\alpha-n}{n+1}.$

###  3.5. Preparing the rigorous proof of Theorem 1.4

To make the heuristics above rigorous, we will define a measure $\mu_{R}$
supported on $X_{R}$. Following the ideas in [12], we first redefine the
parameters $S,D,Q$ by

(3.48) $R^{A}=DQ,\qquad R^{B}=\frac{D^{2}Q}{R},\qquad R^{\sigma}=S,$

which is unmade by

(3.49) $D=R^{1+B-A},\qquad Q=R^{2A-B-1}.$

$1/2$$\frac{\sigma+1}{2}$$\frac{n+\sigma}{n+1}$$1$$1/2$$\sigma$$\alpha=\displaystyle{\frac{3n+1}{4}}$$\alpha=n$$A$$B$
Figure 3. Region for the parameters $A,B$ defined in (3.48) for a fixed
$\sigma\in[1/2,1]$.

The condition $S\geq R^{1/2}$, the elementary $Q\geq 1$, and forcing the
rectangles in each $q\simeq Q$ level of $X_{R}$ to be disjoint, impose the
basic restrictions

(3.50) $\frac{1}{2}<\sigma\leq 1,\qquad 0\leq A\leq 1,\qquad 0\leq
B\leq\sigma,\qquad 2A\geq B+1.$

###  3.6. Proof using Scaling 1

From (3.40) and (3.41) we set

(3.51)
$\frac{(DQ)^{n+1}}{SR^{n}}=\Big{(}\frac{S}{R}\Big{)}^{2(n-\alpha)},\qquad
DQ=S^{1+r},$

This is rewritten in terms of $A$ and $\sigma$ as

(3.52) $A(n+1)=2\alpha-n+2\sigma(n-\alpha+1/2),\qquad A=\sigma(1+r),$

which are independent555Moreover, the heuristic Sobolev exponent in (3.42)
depends on $\sigma$ (hence on $A$), but not on $B$. of $B$. The first
condition in (3.52) can be rewritten as

(3.53) $\displaystyle A(n+1)$
$\displaystyle=\alpha+\sigma+(2\sigma-1)(n-\alpha)$ (3.54)
$\displaystyle=n+\sigma-2(1-\sigma)(n-\alpha).$

Based on (3.37), define the measure

(3.55) $\mu_{R}(A)=\frac{|A\cap
X_{R}|}{|X_{R}|}\,\Big{(}\frac{S^{2}}{R}\Big{)}^{n-1-\alpha}$

so that $\mu(X_{R})=(S^{2}/R)^{n-1-\alpha}$. Since the computations in Section
3.4.1 give the desired Sobolev exponents, it suffices to prove that $\mu_{R}$
is an $\alpha$ dimensional measure. We first simplify the expression for
$\mu_{R}$. From (3.26), $|X_{R}|=|\widetilde{X}_{R}|R/S^{2}$. Now,

(3.56)
$Q^{n+1}h_{1}(Q)h_{2}(Q)^{n-1}=\frac{(DQ)^{n+1}}{SR^{n}}=\left(\frac{S}{R}\right)^{2(n-\alpha)}\ll
1$

because of (3.51), so Lemma A.1 implies

(3.57) $|\widetilde{X}_{R}|\simeq
Q^{n+1}h_{1}(Q)h_{2}(Q)^{n-1}\simeq\left(\frac{S}{R}\right)^{2(n-\alpha)}\quad\Longrightarrow\quad|X_{R}|\simeq\frac{R}{S^{2}}\,\left(\frac{S}{R}\right)^{2(n-\alpha)}.$

Therefore, we rewrite the measure $\mu_{R}$ as

(3.58) $\frac{1}{|X_{R}|}\,\Big{(}\frac{S^{2}}{R}\Big{)}^{n-1-\alpha}\simeq
R^{n-\alpha}\quad\Longrightarrow\quad\mu_{R}(A)\simeq|A\cap
X_{R}|\,R^{n-\alpha}.$

We now prove that $\mu_{R}$ is $\alpha$-dimensional. Let $r>0$ and $B_{r}$ any
ball of radius $r$. We want to prove that $\mu_{R}(B_{r})\lesssim r^{\alpha}$
with a constant independent of $R$. We will exploit the freedom for $B$ by
choosing it as large as possible. Direct computation from (3.54) shows that
there are two different ranges, which we treat separately:

  1. (1)

$A\leq(1+\sigma)/2\Longleftrightarrow\alpha\leq(3n+1)/4$. In this case, we
choose $B=2A-1$, which corresponds to $Q=1$.

  2. (2)

$A\geq(1+\sigma)/2\Longleftrightarrow\alpha\geq(3n+1)/4$. In this case, we
choose $B=\sigma$.

####  3.6.1. When $B=2A-1$ and $Q=1$.

In this case, the set $X_{R}$ becomes particularly easy because there is no
union in $q\simeq Q=1$, so

(3.59)
$X_{R}=\bigcup_{p}B_{1}\Big{(}\frac{R}{D^{2}}p_{1},\frac{c}{S}\Big{)}\times
B_{n-1}\Big{(}\frac{1}{D}p^{\prime},\frac{c}{R}\Big{)}\subset[0,R/S^{2}]\times[0,1]^{n-1}.$

In particular, all rectangles are disjoint. We measure $B_{r}$ depending on
$r$:

  * •

$r\leq 1/R$. Bound $|B_{r}\cap X_{R}|\leq|B_{r}|=r^{n}$ so

(3.60) $\mu_{R}(B_{r})\leq r^{n}\,R^{n-\alpha}=r^{\alpha}(rR)^{n-\alpha}\leq
r^{\alpha}.$

  * •

$1/R<r\leq 1/D$. The ball $B_{r}$ intersects at most one rectangle, so

(3.61)
$\mu_{R}(B_{r})\leq\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{\alpha}\Big{(}\frac{1}{rR}\Big{)}^{\alpha-1}\leq
r^{\alpha},\qquad\text{ if }\alpha\geq 1.$

  * •

$1/D<r\leq 1/S$. (Observe that $S\leq D=DQ=S^{1+r}$). Now $B_{r}$ intersects
at most $(rD)^{n-1}$ rectangles in direction $x^{\prime}$ and one in direction
$x_{1}$, each of which can contribute at most with a measure $r/R^{n-1}$, so

(3.62)
$\mu_{R}(B_{r})\leq(rD)^{n-1}\,\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{n}\,\frac{D^{n+1}}{R^{\alpha}}\frac{R}{D^{2}}=r^{n}R^{A(n+1)-\alpha-B}.$

By bounding $r\leq R^{-\sigma}$ and using the condition (3.53), we get

(3.63) $\mu_{R}(B_{r})\leq r^{\alpha}\,R^{\sigma-B-(1-\sigma)(n-\alpha)}\leq
r^{\alpha},$

because it follows from (3.54), $2A=B+1$ and $B\leq\sigma\leq 1$ that

(3.64)
$\begin{split}\sigma-B-(1-\sigma)(n-\alpha)&=\sigma-B+\frac{B+1}{2}\frac{n+1}{2}-\frac{\sigma}{2}-\frac{n}{2}\\\
&=\frac{\sigma}{2}+B\frac{n-3}{4}-\frac{n-1}{4}\\\
&\leq(\sigma-1)\frac{n-1}{4}\leq 0.\end{split}$

  * •

$1/S<r\leq R/D^{2}=1/R^{B}$. The ball intersects at most $(rD)^{n-1}$
rectangles in direction $x^{\prime}$ and one in direction $x_{1}$, each
contributing with $1/(SR^{n-1})$. Then, by (3.51),

(3.65)
$\mu_{R}(B_{r})\leq\frac{(rD)^{n-1}}{SR^{n-1}}R^{n-\alpha}=r^{n-1}\frac{D^{n+1}}{SR^{\alpha}}\frac{R}{D^{2}}=r^{n-1}\,\Big{(}\frac{S^{2}}{R}\Big{)}^{n-\alpha}\,\frac{1}{R^{B}}.$

If $\alpha\geq n-1$, we bound $r^{n-1}=r^{\alpha}\,r^{n-1-\alpha}\leq
r^{\alpha}\,S^{1-(n-\alpha)}$, so

(3.66) $\mu_{R}(B_{r})\leq
r^{\alpha}\,\frac{S}{R^{B}}\,\Big{(}\frac{S}{R}\Big{)}^{n-\alpha}=r^{\alpha}\,R^{\sigma-B-(1-\sigma)(n-\alpha)}\leq
r^{\alpha},$

where the last inequality follows from (3.64).

  * •

$1/R^{B}<r\leq R/S^{2}$. The ball intersects at most $(rD)^{n-1}$ rectangles
in direction $x^{\prime}$ and $R^{B}r$ in direction $x_{1}$, each with a
contribution of $1/(SR^{n-1})$. Then, by (3.51),

(3.67) $\mu_{R}(B_{r})\leq
r^{n}\frac{D^{n-1}R^{B}}{SR^{n-1}}\,R^{n-\alpha}=r^{n}\frac{D^{n+1}}{SR^{n}}\,R^{n-\alpha}=r^{\alpha}\Big{(}r\frac{S^{2}}{R}\Big{)}^{n-\alpha}\leq
r^{\alpha}.$

  * •

$R/S^{2}<r\leq 1$. The ball intersects at most $(rD)^{n-1}$ rectangles in
direction $x^{\prime}$ and $D^{2}/S^{2}$ in direction $x_{1}$. So if
$\alpha\geq n-1$, from (3.51) we get

(3.68) $\mu_{R}(B_{r})\leq
r^{n-1}\frac{D^{n+1}}{S^{2}}\,\frac{1}{SR^{n-1}}\,R^{n-\alpha}=r^{n-1}\,\frac{R}{S^{2}}\Big{(}\frac{S^{2}}{R}\Big{)}^{n-\alpha}=r^{\alpha}\Big{(}r\frac{S^{2}}{R}\Big{)}^{n-1-\alpha}\leq
r^{\alpha}.$

Hence $\mu_{R}$ is $\alpha$-dimensional, with underlying constant
$\langle\mu_{R}\rangle_{\alpha}\simeq 1$.

####  3.6.2. When $B=\sigma$ and $Q\gg 1$

Now $X_{R}$ is a superposition of grids of rectangles, one for each $q\simeq
Q$. However, by the choice of $B=\sigma$, for each fixed $x^{\prime}$ the
rectangles form a single tube of length $R/S^{2}$ in direction $x_{1}$. To
control intersections in direction $x^{\prime}$, we slightly adapt a lemma
proved by Lucà and Ponce-Vanegas in [14].

######  Lemma 3.3 (Lemma 13 in [14]).

Let $X_{R}=\cup_{q}\cup_{p}B_{1}\times
B_{n-1}\subset[0,R/S^{2}]\times[0,1]^{n-1}$ as defined in (3.15), which is a
union of $(DQ)^{n+1}/S^{2}$ rectangles. Let $\epsilon>0$. Then, for
$R\gg_{\epsilon}1$, at least $(DQ)^{n+1}/(S^{2}Q^{\epsilon})$ such rectangles
have the following property: if $x=(x_{1},x^{\prime})$ and
$y=(y_{1},y^{\prime})$ are centers of two such rectangles,

(3.69) $|x^{\prime}-y^{\prime}|>\frac{1}{DQ^{\frac{n}{n-1}}}.$

Call $Y_{R}=\bigcup_{q}\bigcup_{p}B_{1}\times B_{n-1}\subset X_{R}$ the union
restricted to the subcollection of rectangles given by Lemma 3.3. Instead of
working with $\mu_{R}$ as in (3.55), we define

(3.70) $\nu_{R}(A)=\frac{|A\cap
Y_{R}|}{|Y_{R}|}\,R^{-\epsilon}\,\Big{(}\frac{S^{2}}{R}\Big{)}^{n-1-\alpha}$

for some $\epsilon>0$. Observe that the rectangles in $Y_{R}$ are pairwise
disjoint because

(3.71)
$\frac{1}{R}\leq\frac{1}{DQ^{n/(n-1)}}\quad\Longleftrightarrow\quad(n+1)A\leq
n+B,$

which holds in view of $B=\sigma$ and (3.54). Hence, comparing with $|X_{R}|$
in (3.57),

(3.72)
$|Y_{R}|\simeq\frac{1}{R^{\epsilon}}\,\frac{(DQ)^{n+1}}{S^{2}}\frac{1}{SR^{n-1}}\simeq\frac{1}{R^{\epsilon}}|X_{R}|.$

Together with (3.58), we get $\nu_{R}(A)\simeq|A\cap Y_{R}|\,R^{n-\alpha}$. We
now prove that $\nu_{R}$ is an $\alpha$-dimensional measure. Let $B_{r}$ be a
ball of radius $r$. Again, recall that $B=\sigma$ implies that for each fixed
$x^{\prime}$, the rectangles form a single tube in direction $x_{1}$.

  * •

If $r\leq 1/R$, then $|Y_{R}\cap B_{r}|\leq|B_{r}|=r^{n}$, so

(3.73) $\nu_{R}(B_{r})\leq r^{n}R^{n-\alpha}=r^{\alpha}(rR)^{n-\alpha}\leq
r^{\alpha}.$

  * •

$1/R<r\leq 1/(DQ^{n/(n-1)})$. Then $B_{r}$ intersects at most one tube, for
which

(3.74)
$\nu_{R}(B_{r})\leq\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{\alpha}\Big{(}\frac{1}{rR}\Big{)}^{\alpha-1}\leq
r^{\alpha},\qquad\text{ if }\alpha\geq 1.$

  * •

If $1/(DQ^{n/(n-1)})<r\leq R/S^{2}$, then $B_{r}$ intersects at most
$(rDQ^{n/(n-1)})^{n-1}$ tubes, so

(3.75) $\displaystyle\nu_{R}(B_{r})\leq
r^{n-1}D^{n-1}Q^{n}\,\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{n}\,\frac{D^{n-1}Q^{n}}{R^{\alpha-1}}=r^{n}\,\frac{R^{A(n+1)}}{R^{\alpha}\,R^{B}}.$

Since $B=\sigma$, by (3.53) we get $A(n+1)-\alpha-B=(2\sigma-1)(n-\alpha)$.
Together with $r\leq R/S^{2}=R^{1-2\sigma}$, this implies

(3.76) $\nu_{R}(B_{r})\leq
r^{n}\,R^{(2\sigma-1)(n-\alpha)}=r^{\alpha}(rR^{2\sigma-1})^{n-\alpha}\leq
r^{\alpha}.$

  * •

If $R/S^{2}<r\leq 1$, then $B_{r}$ intersects again at most
$(rDQ^{n/(n-1)})^{n-1}$ tubes, but the measure of each tube is now
$(R/S^{2})(1/R^{n-1})$, so

(3.77) $\displaystyle\nu_{R}(B_{r})\leq
r^{n-1}D^{n-1}Q^{n}\,\frac{R}{S^{2}}\frac{1}{R^{n-1}}\,R^{n-\alpha}=r^{n-1}\frac{R}{S^{2}}\frac{R^{A(n+1)}}{R^{\alpha}\,R^{B}}$

Again, by $B=\sigma$ and (3.53) we get
$A(n+1)-\alpha-B=(2\sigma-1)(n-\alpha)$, so

(3.78) $\nu_{R}(B_{r})\leq
r^{n-1}\,R^{1-2\sigma}\,R^{(2\sigma-1)(n-\alpha)}=r^{\alpha}\,(rR^{2\sigma-1})^{n-\alpha-1}\leq
r^{\alpha},$

if $\alpha\geq n-1$.

Hence, $\nu_{R}$ is $\alpha$-dimensional, with constant
$\langle\nu_{R}\rangle_{\alpha}\simeq 1$.

####  3.6.3. Concluding the proof

Let $n-1\leq\alpha<n$. Let $R\gg 1$. Define the measure $\lambda_{R}=\mu_{R}$
if $\alpha\leq(3n+1)/4$, and $\lambda_{R}=\nu_{R}$ if $\alpha\geq(3n+1)/4$. In
both cases, $\lambda_{R}$ is an $\alpha$-dimensional measure and
$\langle\lambda_{R}\rangle_{\alpha}\simeq 1$. By (3.19), (3.55) and (3.70),

(3.79)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\lambda_{R})}}{\|f_{R}\|_{2}}\gtrsim
S^{\frac{1}{2}}\,\Big{(}\frac{R}{DQ}\Big{)}^{\frac{n-1}{2}}\lambda_{R}(X_{R})^{\frac{1}{2}}\simeq
S^{\frac{1}{2}}\,\Big{(}\frac{R}{DQ}\Big{)}^{\frac{n-1}{2}}\,R^{-\epsilon}\,\Big{(}\frac{S^{2}}{R}\Big{)}^{\frac{n-1-\alpha}{2}}.$

Solving for $DQ$ and $S$ from (3.51), we get
$S=R^{\frac{2\alpha-n}{r(n+1)+2\alpha-n}}$. Since $S\geq R^{1/2}$, this forces
$r\leq(2\alpha-n)/(n+1)$. Hence

(3.80)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\lambda_{R})}}{\|f_{R}\|_{2}}\gtrsim
R^{\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{r(n+1)+2\alpha-n}-\epsilon}.$

If we assume the fractal maximal estimate for $\alpha$-dimensional measures in
$H^{s}$, then

(3.81)
$R^{\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{r(n+1)+2\alpha-n}-\epsilon}\lesssim\langle\lambda_{R}\rangle_{\alpha}\,R^{s}\lesssim
R^{s}\qquad\forall R\gg_{\epsilon}1,$

which gives a contradiction for
$s<\frac{n-\alpha}{2}+\frac{r(2\alpha-n)}{r(n+1)+2\alpha-n}$. Finally, when
$r=\frac{2\alpha-n}{n+1}$, the exponent is
$\frac{n-\alpha}{2}+\frac{2\alpha-n}{2(n+1)}$, which immediately implies
$s_{c}^{\operatorname{max}}(\alpha,\ell^{r,\infty})\geq\frac{n-\alpha}{2}+\frac{2\alpha-n}{2(n+1)}$
for $r\geq\frac{2\alpha-n}{n+1}$. Combining these two gives the statement in
Theorem 1.4.(1). ∎

###  3.7. Proof using Scaling 2

From (3.44) and (3.45), now set

(3.82) $(DQ)^{n+1}=SR^{\alpha},\qquad DQ=S^{1+r},$

which is rewritten in terms of $A$ and $\sigma$ as

(3.83) $A(n+1)=\sigma+\alpha,\qquad A=\sigma(1+r).$

Define the measure

(3.84) $\mu_{R}(A)=\frac{|A\cap X_{R}|}{|X_{R}|}\,\frac{R}{S^{2}}.$

Proceeding like in (3.56) and (3.57), we get

(3.85)
$|X_{R}|\simeq\frac{(DQ)^{n+1}}{SR^{n}}\,\frac{R}{S^{2}}\simeq\frac{1}{R^{n-\alpha}}\,\frac{R}{S^{2}}\quad\Longrightarrow\quad\mu_{R}(A)\simeq|A\cap
X_{R}|\,R^{n-\alpha}.$

We prove now that $\mu_{R}$ is an $\alpha$-dimensional measure. We separate
two cases again:

####  3.7.1. When $B=2A-1$ and $Q=1$.

We repeat the procedure in Section 3.6.1 by working with the simpler, disjoint
structure of $X_{R}$ in (3.59). Let $r>0$.

  * •

If $r\leq 1/R$, bound $|B_{r}\cap X_{R}|\leq|B_{r}|=r^{n}$ so

(3.86) $\mu_{R}(B_{r})\leq r^{n}\,R^{n-\alpha}=r^{\alpha}(rR)^{n-\alpha}\leq
r^{\alpha}.$

  * •

If $1/R<r\leq 1/D$, then $B_{r}$ intersects at most one rectangle, so

(3.87)
$\mu_{R}(B_{r})\leq\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{\alpha}\Big{(}\frac{1}{rR}\Big{)}^{\alpha-1}\leq
r^{\alpha},\quad\text{ if }\alpha-1\geq 0.$

  * •

If $1/D<r\leq 1/S$, $B_{r}$ intersects at most $(rD)^{n-1}$ rectangles in
direction $x^{\prime}$ and one in direction $x_{1}$, each contributing at most
with $r/R^{n-1}$. By (3.82),

(3.88)
$\mu_{R}(B_{r})\leq(rD)^{n-1}\,\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{n}\,\frac{D^{n+1}}{R^{\alpha}}\frac{R}{D^{2}}=r^{n}\,\frac{S}{R^{B}}.$

Then, since $r^{n-\alpha}\leq 1/S^{n-\alpha}$,

(3.89) $\mu_{R}(B_{r})\leq\frac{r^{\alpha}}{R^{B}\,S^{n-\alpha-1}}\leq
r^{\alpha},\qquad\text{ if }\alpha\leq n-1.$

  * •

If $1/S<r\leq 1/R^{B}$, $B_{r}$ intersects at most $(rD)^{n-1}$ rectangles in
direction $x^{\prime}$ and one in direction $x_{1}$, each contribution being
$1/(SR^{n-1})$. By (3.82),

(3.90)
$\mu_{R}(B_{r})\leq\frac{(rD)^{n-1}}{SR^{n-1}}R^{n-\alpha}=r^{n-1}\,\frac{D^{n+1}}{SR^{\alpha}}\,\frac{R}{D^{2}}=r^{\alpha}\,\frac{r^{n-1-\alpha}}{R^{B}}.$

Since $r\leq 1/R^{B}$, we get

(3.91) $\mu_{R}(B_{r})\leq\frac{r^{\alpha}}{R^{B}\
R^{B(n-\alpha-1)}}=\frac{r^{\alpha}}{R^{B(n-\alpha)}}\leq
r^{\alpha},\qquad\text{ if }\alpha\leq n-1.$

  * •

$1/R^{B}<r\leq 1$. The ball intersects at most $(rD)^{n-1}$ rectangles in
direction $x^{\prime}$ and $rR^{B}$ in direction $x_{1}$, each contribution
being $1/(SR^{n-1})$. By (3.82),

(3.92)
$\mu_{R}(B_{r})\leq\frac{(rD)^{n-1}rR^{B}}{SR^{n-1}}R^{n-\alpha}=r^{n}\,\frac{D^{n+1}}{SR^{\alpha}}\,\frac{R}{D^{2}}\,R^{B}=r^{n}\leq
r^{\alpha}.$

Hence, $\mu_{R}$ is an $\alpha$-dimensional measure with constant
$\langle\mu_{R}\rangle\simeq 1$.

####  3.7.2. When $B=\sigma$ and $Q\gg 1$.

As in Section 3.6.2, using Lemma 3.3 we extract from $X_{R}$ a subcollection
$Y_{R}$ of $(DQ)^{n+1}/(S^{2}Q^{\epsilon})$ rectangles. The condition for
disjointness in the $x^{\prime}$ direction in (3.71) is also satisfied because
of (3.82). As before, since $B=\sigma$, for each fixed $x^{\prime}$ the
rectangles form a single tube of length $R/S^{2}$ in direction $x_{1}$. Define

(3.93) $\nu_{R}(A)=\frac{|A\cap
Y_{R}|}{|Y_{R}|}\,R^{-\epsilon}\,\frac{R}{S^{2}}.$

By disjointness, proceeding like in (3.72), we get
$|Y_{R}|\simeq|X_{R}|/R^{\epsilon}$, so from (3.85) we get
$\nu_{R}(A)\simeq|A\cap Y_{R}|R^{n-\alpha}$. Let now $r>0$.

  * •

If $r\leq 1/R$, we bound $|Y_{R}\cap B_{r}|\leq|B_{r}|=r^{n}$, so

(3.94) $\nu_{R}(B_{r})\leq r^{n}R^{n-\alpha}=r^{\alpha}(rR)^{n-\alpha}\leq
r^{\alpha}.$

  * •

If $1/R<r\leq 1/(DQ^{n/(n-1)})$, $B_{r}$ intersects at most one tube, for
which

(3.95)
$\nu_{R}(B_{r})\leq\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{\alpha}\Big{(}\frac{1}{rR}\Big{)}^{\alpha-1}\leq
r^{\alpha},\quad\text{ if }\alpha\geq 1.$

  * •

If $1/(DQ^{n/(n-1)})<r\leq 1$, $B_{r}$ intersects at most
$(rDQ^{n/(n-1)})^{n-1}$ tubes:

(3.96) $\displaystyle\nu_{R}(B_{r})\leq
r^{n-1}D^{n-1}Q^{n}\,\frac{r}{R^{n-1}}\,R^{n-\alpha}=r^{n}\,\frac{R^{A(n+1)}}{R^{\alpha}\,R^{B}}.$

Since $B=\sigma$, from (3.83) we get $A(n+1)=\alpha+B$, so

(3.97) $\nu_{R}(B_{r})\leq r^{n}\leq r^{\alpha}.$

Hence, $\nu_{R}$ is an $\alpha$-dimensional measure with constant
$\langle\nu_{R}\rangle\simeq 1$.

####  3.7.3. Concluding the proof

Let $n/2\leq\alpha<n$. Let $R\gg 1$. Define the measure $\lambda_{R}=\mu_{R}$
or $\lambda_{R}=\nu_{R}$, depending on the value of $\alpha$. In both cases,
$\lambda_{R}$ is an $\alpha$-dimensional measure and
$\langle\lambda_{R}\rangle_{\alpha}\simeq 1$. By (3.19),

(3.98)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\lambda_{R})}}{\|f_{R}\|_{2}}\gtrsim
S^{\frac{1}{2}}\,\Big{(}\frac{R}{DQ}\Big{)}^{\frac{n-1}{2}}\lambda_{R}(X_{R})^{\frac{1}{2}}\simeq
S^{\frac{1}{2}}\,\Big{(}\frac{R}{DQ}\Big{)}^{\frac{n-1}{2}}\,R^{-\epsilon}\,\Big{(}\frac{R}{S^{2}}\Big{)}^{1/2}.$

Solving for $DQ$ and $S$ from (3.82) we get $S=R^{\frac{\alpha}{r(n+1)+n}}$
and $DQ=S^{1+r}$, which forces $r\leq(2\alpha-n)/(n+1)$. Hence,

(3.99)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}f_{R}|\|_{L^{2}(d\lambda_{R})}}{\|f_{R}\|_{2}}\gtrsim
R^{\frac{n-\alpha}{2}+\frac{r\alpha}{r(n+1)+n}-\epsilon}.$

If we assume the fractal maximal estimate for $\alpha$-dimensional measures in
$H^{s}$, then

(3.100)
$R^{\frac{n-\alpha}{2}+\frac{r\alpha}{r(n+1)+n}-\epsilon}\lesssim\langle\lambda_{R}\rangle_{\alpha}\,R^{s}\lesssim
R^{s}\qquad\forall R\gg_{\epsilon}1,$

which gives a contradiction for
$s<\frac{n-\alpha}{2}+\frac{r\alpha}{r(n+1)+n}$. Evaluating this exponent at
$r=\frac{2\alpha-n}{n+1}$, we immediately get
$s_{c}^{\operatorname{max}}(\alpha,\ell^{r,\infty})\geq\frac{n-\alpha}{2}+\frac{2\alpha-n}{2(n+1)}$
for $r\geq\frac{2\alpha-n}{n+1}$. Combining these two gives the statement in
Theorem 1.4.(2). ∎

##  4\. Proof of Theorem 1.2 \- Dimension $n=1$

###  4.1. When $\boldsymbol{\alpha=1}$

We reprove the result by Dimou and Seeger [7]. We work with the function
$g_{R}$ defined in (3.2). In (3.5) we got

(4.1) $S|x_{1}+2Rt|\leq c,\quad S^{2}t\leq
c\quad\Longrightarrow\quad|e^{it\Delta}g_{R}(x_{1})|\simeq 1.$

As we did for $n\geq 2$ in Section 3, we work with $S\geq R^{1/2}$. Define

(4.2) $T_{R}=\Big{\\{}\,\frac{p}{Q}\,\,:\,\,1\leq
p\leq\frac{Q}{S^{2}}\,\Big{\\}},\quad\text{ and }\quad
X_{R}=\bigcup_{p=1}^{Q/S^{2}}B\Big{(}\frac{R}{Q}\,p,\frac{1}{S}\Big{)}.$

We impose the restrictions $S^{2}\leq Q$ and $Q\leq RS$, the first one for the
sets in (4.2) not to be empty, and the second for the intervals in $X_{R}$ to
be disjoint. Actually, since $X_{R}\subset[0,R/S^{2}]$, we get the maximum
measure when

(4.3)
$Q=RS\quad\Longrightarrow\quad|X_{R}|\simeq\frac{Q}{S^{2}}\,\frac{1}{S}=\frac{R}{S^{2}}.$

On the other hand, observe that $T_{R}\subset[1/Q,1/S^{2}]$. Asking $R_{n}$ to
be a dyadic sequence that grows fast enough, all intervals
$[1/Q_{n},1/S_{n}^{2}]$ are disjoint. Moreover, for any $\delta>0$, place it
in $1/S_{n+1}^{2}\leq\delta\leq 1/S_{n}^{2}$ so that

(4.4) $\delta^{r}\,\\#\\{\,t\in
T_{R_{n}}\,:\,\delta<t<2\delta\,\\}\leq\delta^{1+r}Q_{n}\leq\frac{Q_{n}}{S_{n}^{2(1+r)}}.$

Hence, defining our sequence of times as $\\{t_{n}\\}=\bigcup_{n}T_{R_{n}}$,

(4.5) $Q\lesssim
S^{2(1+r)}\qquad\Longrightarrow\qquad\\{t_{n}\\}\in\ell^{r,\infty}.$

Gathering all, we get

(4.6)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}g_{R}|\|_{L^{2}}}{\|g_{R}\|_{2}}\gtrsim
S^{1/2}\,|X_{R}|^{1/2}=\Big{(}\frac{R}{S}\Big{)}^{1/2}.$

Combining (4.3) and taking the smallest $S$ in (4.5), we get $S=R^{1/(1+2r)}$,
hence

(4.7)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}g_{R}|\|_{L^{2}(X)}}{\|g_{R}\|_{2}}\gtrsim
R^{\frac{r}{2r+1}}.$

Hence the maximal estimate cannot work in $H^{s}$ for $s<r/(2r+1)$. Observe
also that the choice $S\geq R^{1/2}$ restricts $r\leq 1/2$.

Since when $r=1/2$ the exponent in (4.7) is 1/4, this immediately implies that
$s_{c}^{\operatorname{max}}(\ell^{r,\infty})\geq 1/4$ for all $r\geq 1/2$. ∎

###  4.2. When $\boldsymbol{\alpha<1}$

With the same function $g_{R}$, the constructions (4.2) and the condition
(4.5) intact, we alter condition (4.3). Heuristically, we want $X_{R}$ and
$[0,R/S^{2}]$ to have the same $\mathcal{H}^{\alpha}$-scaling. Since

(4.8)
$\mathcal{H}^{\alpha}(X_{R})\simeq\frac{Q}{S^{2}}\frac{1}{S^{\alpha}}\,\mathcal{H}^{\alpha}([0,1]),\quad\text{
and
}\quad\mathcal{H}^{\alpha}\Big{(}\Big{[}0,\frac{R}{S^{2}}\Big{]}\Big{)}\simeq\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha}\,\mathcal{H}^{\alpha}([0,1]),$

we choose

(4.9)
$\frac{Q}{S^{2}}\frac{1}{S^{\alpha}}=\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha}\quad\Longleftrightarrow\quad
Q=R^{\alpha}\,S^{2-\alpha}.$

In particular, $Q=RS(S/R)^{1-\alpha}\ll RS$. With that, heuristically we
expect to get

(4.10)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}g_{R}|\|_{L^{2}(\mathcal{H}^{\alpha})}}{\|g_{R}\|_{2}}\gtrsim
S^{1/2}\,\mathcal{H}^{\alpha}(X_{R})^{1/2}\simeq
S^{1/2}\,\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha/2}=\Big{(}\frac{R^{\alpha}}{S^{2\alpha-1}}\Big{)}^{1/2}.$

Taking the smallest $S$ in (4.5), from (4.9) we get
$S=R^{\alpha/(2r+\alpha)}$, hence

(4.11)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}g_{R}|\|_{L^{2}(\mathcal{H}^{\alpha})}}{\|g_{R}\|_{2}}\gtrsim
R^{\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}},$

which due to $S\geq R^{1/2}$ will be subject to the restriction
$r\leq\alpha/2$.

To make this rigorous, as in previous sections, define the measure

(4.12) $\mu_{R}(A)=\frac{|A\cap
X_{R}|}{|X_{R}|}\,\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha}.$

It suffices to prove that it is $\alpha$-dimensional. First observe that $Q\ll
RS$ implies that the balls in $X_{R}$ are disjoint, so
$|X_{R}|=(Q/S^{2})/S=R^{\alpha}/S^{1+\alpha}$, and hence

(4.13) $\mu_{R}(A)=|A\cap X_{R}|\,S^{1-\alpha}.$

Now, for any ball $B_{r}$ of radius $r>0$,

  * •

If $r\leq 1/S$, then $|B_{r}\cap X_{R}|\leq|B_{r}|=r$, so

(4.14) $\mu_{R}(B_{r})\leq rS^{1-\alpha}=r^{\alpha}\,(rS)^{1-\alpha}\leq
r^{\alpha}.$

  * •

If $1/S\leq r\leq R/Q$, then $|B_{r}\cap X_{R}|\leq 1/S$, so

(4.15) $\mu_{R}(B_{r})\leq\frac{1}{S}S^{1-\alpha}=\frac{1}{S^{\alpha}}\leq
r^{\alpha}.$

  * •

If $R/Q\leq r\leq R/S^{2}$, then $B_{r}$ intersects at most $rQ/R$ intervals
of $X_{R}$, so

(4.16) $\mu_{R}(B_{r})\leq
r\frac{Q}{R}\,\frac{1}{S}S^{1-\alpha}=r\,\Big{(}\frac{S^{2}}{R}\Big{)}^{1-\alpha}=r^{\alpha}\,\Big{(}r\frac{S^{2}}{R}\Big{)}^{1-\alpha}\leq
r^{\alpha}.$

  * •

If $R/S^{2}\leq r\leq 1$, then $B_{r}$ may intersect all $X_{R}$, so

(4.17)
$\mu_{R}(B_{r})\leq\frac{Q}{S^{2}}\,\frac{1}{S}\,S^{1-\alpha}=\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha}\leq
r^{\alpha}.$

Hence, $\mu_{R}$ is $\alpha$-dimensional with constant
$\langle\mu_{R}\rangle\simeq 1$, so

(4.18)
$\frac{\|\sup_{n}|e^{it_{n}\Delta}g_{R}|\|_{L^{2}(d\mu_{R})}}{\|g_{R}\|_{2}}\gtrsim
S^{1/2}\,\mu(X_{R})^{1/2}=S^{1/2}\,\Big{(}\frac{R}{S^{2}}\Big{)}^{\alpha/2}=R^{\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}},$

where $S=R^{\alpha/(2r+\alpha)}$ from (4.5) and (4.9). Hence, when
$r\leq\alpha/2$, the $\alpha$-dimensional fractal maximal estimate cannot hold
in $H^{s}$ for $s<\frac{1-\alpha}{2}+\frac{r(2\alpha-1)}{2r+\alpha}$. Finally,
when $r=\alpha/2$ the exponent is 1/4, so
$s_{c}^{\operatorname{max}}(\alpha,\ell^{r,\infty})\geq 1/4$ for all
$r\geq\alpha/2$. ∎

##  Appendix A A lemma to compute the measure of $X_{R}$

The following lemma is an adaptation of Proposition 4.2 in [1] by An, Chu and
Pierce. We reproduce it in a slightly more general form than in [11, Lemma
3.4.], and we refer to [11, Lemma 3.4.] and [12, Proposition 5.2.] for a
detailed proof.

######  Lemma A.1.

Let $Q\gg 1$. Define

(A.1) $X_{Q}=\bigcup_{\begin{subarray}{c}q\simeq Q\\\ q\text{ odd
}\end{subarray}}\bigcup_{\begin{subarray}{c}(p_{1},p^{\prime})\in[0,q)^{n}\\\
\gcd(p_{1},q)=1\end{subarray}}B_{1}\Big{(}\frac{p_{1}}{q},h_{1}(Q)\Big{)}\times
B_{n-1}\Big{(}\frac{p^{\prime}}{q},h_{2}(Q)\Big{)}$

such that $h_{1}(Q)\simeq 1/Q^{\alpha}$ and $h_{2}(Q)\simeq 1/Q^{\beta}$ for
some $\alpha,\beta\geq 1$. Then,

(A.2)
$|X_{Q}|\simeq\frac{Q^{n+1}h_{1}(Q)h_{2}(Q)^{n-1}}{1+Q^{n+1}h_{1}(Q)h_{2}(Q)^{n-1}}.$

## Acknowledgements

C.-H. Cho is supported by NRF-RS202300239774 and NRF-2022R1A4A1018904. D.
Eceizabarrena is funded in part by the Simons Foundation Collaboration Grant
on Wave Turbulence (Nahmod’s award ID 651469) and by the American Mathematical
Society and the Simons Foundation under an AMS-Simons Travel Grant for the
period 2022-2024.

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