$\mathcal{B}^{-1}F(t)=\sqrt{2}\int_{\mathbb{C}}F(z)\mathrm{e}^{2\pi
t\overline{z}-\pi t^{2}-\frac{\pi}{2}\overline{z}^{2}-\pi\left\lvert
z\right\rvert^{2}}\,\mathrm{d}z,\qquad t\in\mathbb{R},$

where $F\in\mathcal{F}^{2}(\mathbb{C})$. Moreover, and this should already
seem natural in light of the definitions above, the Gabor and the Bargmann
transform are intimately related via the formula (see Proposition 3.4.1 in
[4])

$\mathcal{G}f(x,-\omega)=\mathrm{e}^{\pi\mathrm{i}x\omega}\mathcal{B}f(z)\mathrm{e}^{-\pi\left\lvert
z\right\rvert^{2}/2},\qquad z=x+\mathrm{i}\omega\in\mathbb{C}.$ (8)

Finally, it is useful to know that any function in the Fock space
$F\in\mathcal{F}^{2}(\mathbb{C})$ satisfies the reproducing kernel Hilbert
space property

$\left\lvert F(z)\right\rvert\leq\left\lVert
F\right\rVert_{\mathcal{F}}\mathrm{e}^{\pi\left\lvert
z\right\rvert^{2}/2},\qquad z\in\mathbb{C}.$

Most importantly, we conclude from this that any $f\in L^{2}(\mathbb{R})$
gives rise to an entire function of order two with upper bound of type $\pi/2$
via

$F(z)=\mathcal{B}f(z)=\mathrm{e}^{-\pi\mathrm{i}x\omega}\mathcal{G}f(x,-\omega)\mathrm{e}^{\pi\left\lvert
z\right\rvert^{2}/2},\qquad z=x+\mathrm{i}\omega\in\mathbb{C},$

and that vice versa any entire function $F$ of order at most two with
sufficiently small type (to be more specific any function in the Fock space
$\mathcal{F}^{2}(\mathbb{C})$) gives rise to an $L^{2}$-function via
$f=\mathcal{B}^{-1}F$. We can now apply our results for entire functions
directly to the Gabor transform.

###  5.1 Intersecting lines

First, we want to consider what happens in the case in which we have two
intersecting lines in the time-frequency plane which meet at an irrational
angle. We want to emphasise that the two following results have first been
proven by Philippe Jaming in 2014 [10].

Remember from Theorem 19 that two finite order entire functions
$F,G\in\mathcal{O}(\mathbb{C})$ satisfying $\left\lvert
F\right\rvert=\left\lvert G\right\rvert$ on
$\mathbb{R}\cup\mathrm{e}^{\mathrm{i}\theta}\mathbb{R}$, for
$\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$, agree up to global phase, i.e.
$F\equiv G$. We can easily transfer this uniqueness to the Gabor transform.

######  Theorem 29 (Cf. Proposition 2).

Let $\theta\in\mathbb{R}\setminus\pi\mathbb{Q}$ and $f,g\in L^{2}(\mathbb{R})$
such that

$\left\lvert\mathcal{G}f(x,0)\right\rvert=\left\lvert\mathcal{G}g(x,0)\right\rvert\mbox{
and
}\left\lvert\mathcal{G}f(x\cos\theta,-x\sin\theta)\right\rvert=\left\lvert\mathcal{G}g(x\cos\theta,-x\sin\theta)\right\rvert,$

for all $x\in\mathbb{R}$. Then, there exists an $\alpha\in\mathbb{R}$ such
that $f=\mathrm{e}^{\mathrm{i}\alpha}g$.

###### Proof.

Let us define the entire functions $F=\mathcal{B}f$ along with
$G=\mathcal{B}g$. Then, both $F$ and $G$ are in the Fock space and thus of
order two. Additionally, by the relation of the Bargmann and the Gabor
transform:

$\left\lvert\mathcal{B}f(x)\right\rvert=\left\lvert\mathcal{G}f(x,0)\right\rvert\mathrm{e}^{\pi
x^{2}/2}=\left\lvert\mathcal{G}g(x,0)\right\rvert\mathrm{e}^{\pi
x^{2}/2}=\left\lvert\mathcal{B}g(x)\right\rvert,\qquad x\in\mathbb{R},$

as well as

$\displaystyle\left\lvert\mathcal{B}f(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert$
$\displaystyle=\left\lvert\mathcal{G}f(x\cos\theta,-x\sin\theta)\right\rvert\mathrm{e}^{\pi
x^{2}/2}=\left\lvert\mathcal{G}g(x\cos\theta,-x\sin\theta)\right\rvert\mathrm{e}^{\pi
x^{2}/2}$
$\displaystyle=\left\lvert\mathcal{B}g(\mathrm{e}^{\mathrm{i}\theta}x)\right\rvert,$

for $x\in\mathbb{R}$. It follows from Theorem 19 that there exists an
$\alpha\in\mathbb{R}$ such that $F=\mathrm{e}^{\mathrm{i}\alpha}G$. Therefore,
we find that

$f=\mathcal{B}^{-1}F=\mathrm{e}^{\mathrm{i}\alpha}\mathcal{B}^{-1}G=\mathrm{e}^{\mathrm{i}\alpha}g.$

∎

Remembering the remark after Theorem 19, we see that two functions $f$ and $g$
whose Gabor transform magnitudes agree on any two lines in the time-frequency
plane intersecting at an irrational angle must agree up to global phase. This
is a generalisation of the well-known uniqueness result for phase retrieval
from Gabor measurements (which guarantees the unique recovery of signals from
measurements taken on the entire time-frequency plane). We can go one step
further, however, if we remember that for the uniqueness result for entire
functions (Theorem 19) it was sufficient to assume that magnitudes agree on
$X\cup\mathrm{e}^{\mathrm{i}\theta}X$, where $X\subset\mathbb{R}$ is any set
with an accumulation point.

######  Theorem 30.

Let $\theta_{1},\theta_{2}\in\mathbb{R}$ such that
$\theta_{1}-\theta_{2}\in\mathbb{R}\setminus\pi\mathbb{Q}$, let
$(x_{k})_{k\geq 1}\in\mathbb{R}$ be a sequence with accumulation point and let
$f,g\in L^{2}(\mathbb{R})$. If

$\left\lvert\mathcal{G}f(x_{k}\cos\theta_{j},-x_{k}\sin\theta_{j})\right\rvert=\left\lvert\mathcal{G}g(x_{k}\cos\theta_{j},-x_{k}\sin\theta_{j})\right\rvert,\qquad
j=1,2,~{}k\geq 1,$

then there exists an $\alpha\in\mathbb{R}$ such that
$f=\mathrm{e}^{\mathrm{i}\alpha}g$.

Next, we will consider lines in the time-frequency plane which intersect at
rational angles, i.e. $\theta\in\pi\mathbb{Q}$. In this case, we will not
obtain a uniqueness results as can be seen from Theorem 21 and can instead
precisely classify the non-uniqueness in the following way.

######  Lemma 31.

Let $\theta\in\pi\mathbb{Q}\cap(0,\tfrac{\pi}{2}]$, let $n\in\mathbb{N}$ be
the smallest integer such that $\theta n\in\pi\mathbb{N}$, and let $f,g\in
L^{2}(\mathbb{R})$ be non-zero. We can define

$\mathcal{X}=\mathcal{R}_{\ast}(\mathcal{B}f)\cap\mathcal{R}(\mathcal{B}g),\qquad\mathcal{Y}=\left\\{a\in\mathcal{R}_{\ast}(\mathcal{B}f)\,\middle|\,m_{\mathcal{B}f}(a)>M_{\mathcal{B}g}(a)\right\\},$

and

$\mathcal{Y}_{\mathrm{u}}:=\mathcal{Y}\cap\left\\{z\in\mathbb{C}\,\middle|\,\operatorname{\exists}r>0,~{}\phi\in[-\theta,\theta):z=r\mathrm{e}^{\mathrm{i}\phi}\right\\},$

as well as $m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$,

$m_{\mathcal{X}}(a)=\min\\{m_{\mathcal{B}f}(a),m_{\mathcal{B}g}(a)\\},\qquad
a\in\mathcal{X},$

and $m_{\mathcal{Y}}:\mathcal{Y}\to\mathbb{N}$,

$m_{\mathcal{Y}}(a)=m_{\mathcal{B}f}(a)-M_{\mathcal{B}g}(a),\qquad
a\in\mathcal{Y}.$

Then,

$\left\lvert\mathcal{G}f(x,0)\right\rvert=\left\lvert\mathcal{G}g(x,0)\right\rvert\mbox{
and
}\left\lvert\mathcal{G}f(x\cos\theta,-x\sin\theta)\right\rvert=\left\lvert\mathcal{G}g(x\cos\theta,-x\sin\theta)\right\rvert,$

for $x\in\mathbb{R}$, is equivalent to the existence of $r>0$,
$\phi,\psi\in\mathbb{R}$,
$a_{1},a_{2},b_{1},b_{2},b_{2}^{\prime}\in\mathbb{R}$, $m\in\mathbb{N}_{0}$
and $p\in\\{0,1,2\\}$ such that $b_{2}^{\prime}=b_{2}$ if $n\neq 2$ and

$\displaystyle\begin{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2}^{\prime})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2}^{\prime})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

We could alternatively state this result in terms of the Gabor transforms of
$f$ and $g$ via the equation (8) but decided against it to keep the
mathematical formulae as short as possible. In any case, it is clear from the
result above that signals are not uniquely determined from Gabor magnitude
measurements on two lines which intersect at a rational angle. More precisely,
the result above allows for an exact description of all $g\in
L^{2}(\mathbb{R})$ for which the Gabor transform of $g$ agrees with the Gabor
transform of a fixed $f\in L^{2}(\mathbb{R})$ on two such lines. Moreoever, it
simplifies the construction of signals whose Gabor transform magnidues do
agree on two intersecting lines massively. We might namely take any two sets
of isolated points $\mathcal{X}\subset\mathbb{C}_{\ast}$ and
$\mathcal{Y}_{\mathrm{u}}\subset\left\\{z\in\mathbb{C}\,\middle|\,\operatorname{\exists}r>0,~{}\phi\in[-\theta,\theta):z=r\mathrm{e}^{\mathrm{i}\phi}\right\\}$
together with multiplicity functions
$m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$ and
$m_{\mathcal{Y}}:\mathcal{Y}_{\mathrm{u}}\to\mathbb{N}$ such that

$\sum_{a\in\mathcal{X}}\frac{m_{\mathcal{X}}(a)}{\left\lvert
a\right\rvert^{A}}+\sum_{a\in\mathcal{Y}_{\mathrm{u}}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a\right\rvert^{A}}<\infty,$

for some $A\in(0,2)$. It follows from Lemma 11 that the functions

$\displaystyle
F(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{a};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle
G(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k=1}^{n}E\left(\mathrm{e}^{2k\mathrm{i}\theta}\frac{z}{\overline{a}};p\right)^{m_{\mathcal{Y}}(a)},$

for $r>0$, $\phi,\psi\in\mathbb{R}$, $a_{1},b_{1}\in\mathbb{R}$,
$m\in\mathbb{N}_{0}$, and $p>A-1$, are of order at most $A$ and must thus lie
in the Fock space. Therefore, $f=\mathcal{B}^{-1}F$ and $g=\mathcal{B}^{-1}G$
are $L^{2}$-functions and by the prior lemma

$\left\lvert\mathcal{G}f(x,0)\right\rvert=\left\lvert\mathcal{G}g(x,0)\right\rvert\mbox{
and
}\left\lvert\mathcal{G}f(x\cos\theta,-x\sin\theta)\right\rvert=\left\lvert\mathcal{G}g(x\cos\theta,-x\sin\theta)\right\rvert,$

for $x\in\mathbb{R}$. If $\mathcal{Y}_{\mathrm{u}}$ is chosen such that
$\mathcal{Y}=\mathrm{e}^{2\mathrm{i}\theta(\mathbb{Z}/n\mathbb{Z})}\mathcal{Y}_{\mathrm{u}}$
satisfies $\overline{\mathcal{Y}}\neq\mathcal{Y}$, then $f$ and $g$ do not
agree up to global phase. An example of such a choice is given by
$\theta=\tfrac{\pi}{2}$, $\mathcal{X}=\emptyset$,
$\mathcal{Y}_{\mathrm{u}}=\\{\sqrt{2}\mathrm{e}^{\pi\mathrm{i}/4}\\}$ and
$m_{\mathcal{X}}=m_{\mathcal{Y}}=1$. Taking $r=\phi=\psi=a_{1}=b_{1}=m=p=0$
yields

$F(z)=\left(1-\frac{z}{1+\mathrm{i}}\right)\left(1+\frac{z}{1+\mathrm{i}}\right)=1-\frac{z^{2}}{(1+\mathrm{i})^{2}},\qquad
G(z)=1-\frac{z^{2}}{(1-\mathrm{i})^{2}},$

for $z\in\mathbb{C}$, which coincides with the examples we had given at the
beginning of the paragraph “Rational angles” in Section 4. We can now compute
the corresponding $L^{2}$-functions by using the inverse Bargmann transform.
We obtain

$\displaystyle f(t)=\mathcal{B}^{-1}F(t)=\frac{2\pi+(4\pi
t^{2}-1)\mathrm{i}}{\sqrt{2}\pi}\mathrm{e}^{-\pi t^{2}},$ $\displaystyle
g(t)=\mathcal{B}^{-1}G(t)=\frac{2\pi-(4\pi
t^{2}-1)\mathrm{i}}{\sqrt{2}\pi}\mathrm{e}^{-\pi t^{2}},$

for $t\in\mathbb{R}$.

###  5.2 Parallel lines

Let us now turn to the analysis of parallel lines in the time-frequency plane.
The following result follows immediately from Theorem 25 and shows that Gabor
magnitude measurements on two parallel lines are, in general, not sufficient
to recover a signal up to global phase.

######  Lemma 32.

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in L^{2}(\mathbb{R})$ be non-zero.
We can define

$\mathcal{X}=\mathcal{R}_{\ast}(\mathcal{B}f)\cap\mathcal{R}(\mathcal{B}g),\qquad\mathcal{Y}=\left\\{a\in\mathcal{R}_{\ast}(\mathcal{B}f)\,\middle|\,m_{\mathcal{B}f}(a)>M_{\mathcal{B}g}(a)\right\\},$

and

$\mathcal{Y}_{\mathrm{u}}:=\mathcal{Y}\cap\left(\mathbb{R}+\mathrm{i}\left[-\tau,\tau\right)\right),$

as well as $m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$,

$m_{\mathcal{X}}(a)=\min\\{m_{\mathcal{B}f}(a),m_{\mathcal{B}g}(a)\\},\qquad
a\in\mathcal{X},$

and $m_{\mathcal{Y}}:\mathcal{Y}\to\mathbb{N}$,

$m_{\mathcal{Y}}(a)=m_{\mathcal{B}f}(a)-M_{\mathcal{B}g}(a),\qquad
a\in\mathcal{Y}.$

Then,

$\left\lvert\mathcal{G}f(x,0)\right\rvert=\left\lvert\mathcal{G}g(x,0)\right\rvert\mbox{
and
}\left\lvert\mathcal{G}f(x,-\tau)\right\rvert=\left\lvert\mathcal{G}g(x,-\tau)\right\rvert,$

for $x\in\mathbb{R}$, is equivalent to the existence of $r>0$,
$\phi,\psi\in\mathbb{R}$,
$a_{1},a_{2},b_{1},b_{1}^{\prime},b_{2},b_{2}^{\prime}\in\mathbb{R}$,
$m\in\mathbb{N}_{0}$, and $p\in\\{0,1,2\\}$ such that

$2x(b_{2}-b_{2}^{\prime})+(b_{1}-b_{1}^{\prime})=\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\Bigg{(}\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert\\\
+\sum_{\ell=1}^{p}\frac{1}{\ell}\operatorname{Re}\left[\left(\frac{x+\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right)^{\ell}-\left(\frac{x+\mathrm{i}\tau}{\overline{a}-2(k-1)\mathrm{i}\tau}\right)^{\ell}\right]\Bigg{)},$

for all $x\in\mathbb{R}$, as well as

$\displaystyle\begin{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1}^{\prime})z+(a_{2}+\mathrm{i}b_{2}^{\prime})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1}^{\prime})z+(a_{2}+\mathrm{i}b_{2}^{\prime})z^{2}}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};p\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};p\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

As before in the case of intersecting lines meeting at a rational angle, this
result allows for a full classification of all $g\in L^{2}(\mathbb{R})$ whose
Gabor transform magnitudes agree with those of any fixed signal $f\in
L^{2}(\mathbb{R})$. In addition, we might use the lemma to find signals whose
Gabor magnitudes agree on parallel lines.

Finally, we can consider what we believe to be the most interesting non-
uniqueness result for the Gabor transform in this section: the result on
infinitely many parallel lines. The following is an immediate corollary of
Theorem 27.

######  Theorem 33.

Let $\tau\in\mathbb{R}_{\ast}$ and let $f,g\in L^{2}(\mathbb{R})$ be non-zero
signals _whose Bargmann transforms have order one_. With the definitions of
$\mathcal{X}$, $\mathcal{Y}$, $\mathcal{Y}_{\mathrm{u}}$, $m_{\mathcal{X}}$
and $m_{\mathcal{Y}}$ from Lemma 32, the following are equivalent:

(a)

For all $x\in\mathbb{R}$ and $n\in\mathbb{Z}$,
$\left\lvert\mathcal{G}f(x,n\tau)\right\rvert=\left\lvert\mathcal{G}g(x,n\tau)\right\rvert$.

(b)

There exist $r>0$, $\phi,\psi\in\mathbb{R}$,
$a^{\prime},b,b^{\prime}\in\mathbb{R}$ as well as $m\in\mathbb{N}_{0}$ such
that

$b=b^{\prime}+\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right)$

as well as

$\displaystyle\begin{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\begin{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},\end{multlined}\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\\\
\cdot\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$.

In particular, the above result allows for the consideration of
$\tau\in\mathbb{R}_{\ast}$ and two sets of isolated points
$\mathcal{X}\subset\mathbb{C}_{\ast}$ and
$\mathcal{Y}_{\mathrm{u}}\subset(\mathbb{R}+\mathrm{i}[-\tau,\tau))$ together
with multiplicity functions $m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$ and
$m_{\mathcal{X}}:\mathcal{Y}_{\mathrm{u}}\to\mathbb{N}$ such that

$\sum_{a\in\mathcal{X}}\frac{m_{\mathcal{X}}(a)}{\left\lvert
a\right\rvert^{1+\epsilon}}+\sum_{a\in\mathcal{Y}_{\mathrm{u}}}\sum_{k\in\mathbb{Z}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a+2k\mathrm{i}\tau\right\rvert^{1+\epsilon}}<\infty,$

for all $\epsilon>0$. If we additionally let $r>0$, $\phi,\psi\in\mathbb{R}$,
$a^{\prime},b,b^{\prime}\in\mathbb{R}$ and $m\in\mathbb{N}_{0}$ be such that

$b=b^{\prime}+\frac{1}{\tau}\sum_{a\in\mathcal{Y}_{\mathrm{u}}}m_{\mathcal{Y}}(a)\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{a+2(k-1)\mathrm{i}\tau}{a+2k\mathrm{i}\tau}\right\rvert-2\tau\operatorname{Im}\frac{1}{a+2k\mathrm{i}\tau}\right),$

and define $f,g\in L^{2}(\mathbb{R})$ via

$\displaystyle\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b)z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{a+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$
$\displaystyle\mathcal{B}g(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a^{\prime}+\mathrm{i}b^{\prime})z}z^{m}\prod_{a\in\mathcal{X}}E\left(\frac{z}{a};1\right)^{m_{\mathcal{X}}(a)}\prod_{a\in\mathcal{Y}_{\mathrm{u}}}\prod_{k\in\mathbb{Z}}E\left(\frac{z}{\overline{a}+2k\mathrm{i}\tau};1\right)^{m_{\mathcal{Y}}(a)},$

for $z\in\mathbb{C}$, then it holds that

$\left\lvert\mathcal{G}f(x,n\tau)\right\rvert=\left\lvert\mathcal{G}g(x,n\tau)\right\rvert,\qquad
x\in\mathbb{R},~{}n\in\mathbb{Z}.$

######  Remark 34.

We note that the assumption

$\sum_{a\in\mathcal{X}}\frac{m_{\mathcal{X}}(a)}{\left\lvert
a\right\rvert^{1+\epsilon}}+\sum_{a\in\mathcal{Y}_{\mathrm{u}}}\sum_{k\in\mathbb{Z}}\frac{m_{\mathcal{Y}}(a)}{\left\lvert
a+2k\mathrm{i}\tau\right\rvert^{1+\epsilon}}<\infty,$

for all $\epsilon>0$, implies that $\mathcal{B}f$ and $\mathcal{B}g$ are
entire functions of order one (according to Lemma 11). It follows that
$\mathcal{B}f$ and $\mathcal{B}g$ are in the Fock space and must thus be well-
defined Bargmann transforms of two signals $f$ and $g$ respectively.

######  Example 35 (Phase retrieval from sampled Gabor magnitudes:
counterexamples).

Let us consider $\tau=1$, $\mathcal{X}=\emptyset$,
$\mathcal{Y}_{\mathrm{u}}=\\{\mathrm{i}/2\\}$ and
$m_{\mathcal{X}}=m_{\mathcal{Y}}=1$. With these choices, we note that

$\sum_{k\in\mathbb{Z}}\frac{1}{\left\lvert\mathrm{i}/2+2k\mathrm{i}\tau\right\rvert^{1+\epsilon}}<\infty,$

for all $\epsilon>0$. We may additionally consider $r=1$,
$\phi=\psi=a^{\prime}=m=0$ as well as $b=\pi/2$ and

$\displaystyle b^{\prime}$
$\displaystyle=\frac{\pi}{2}-\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{\mathrm{i}/2+2(k-1)\mathrm{i}}{\mathrm{i}/2+2k\mathrm{i}}\right\rvert-2\operatorname{Im}\frac{1}{\mathrm{i}/2+2k\mathrm{i}}\right)$
$\displaystyle=\frac{\pi}{2}-\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{4k-3}{4k+1}\right\rvert+\frac{4}{4k+1}\right)=-\frac{\pi}{2}$

to obtain the examples considered in Section 4.2,

$\displaystyle F(z)=\cosh\left(\frac{\pi
z}{2}\right)+\mathrm{i}\sinh\left(\frac{\pi z}{2}\right),$ $\displaystyle
G(z)=\cosh\left(\frac{\pi z}{2}\right)-\mathrm{i}\sinh\left(\frac{\pi
z}{2}\right),$

for $z\in\mathbb{C}$. These entire functions correspond (after taking the
inverse Bargmann transform) to the signals

$\displaystyle
f(t)=\sqrt{2}\mathrm{e}^{-\pi\left(\frac{1}{8}+t^{2}\right)}\left(\cosh\pi
t+\mathrm{i}\sinh\pi t\right),$ $\displaystyle
g(t)=\sqrt{2}\mathrm{e}^{-\pi\left(\frac{1}{8}+t^{2}\right)}\left(\cosh\pi
t-\mathrm{i}\sinh\pi t\right),$

for $t\in\mathbb{R}$, which originally inspired our work in [2]. We note that
the signals $f$ and $g$ do not agree up to global phase but that their Gabor
transform magnitudes agree on $\mathbb{R}\times\mathbb{Z}$.

###  5.3 On the number of roots in disks around the origin of functions in the
Fock space

In this final subsection, we want to discuss an important detail in the
construction of functions in the Fock space through the Hadamard factorisation
theorem. To make this discussion as palpable as possible, we want to start by
presenting an interesting example. Suppose we are given $a,b>0$ and are asked
to construct a non-zero function $f\in L^{2}(\mathbb{R})$ such that
$\mathcal{G}f=0$ on $a\mathbb{Z}\times b\mathbb{Z}$. A natural way of
approaching this problem is to consider that

$\displaystyle\sum_{j,k\geq 2}\frac{1}{(a^{2}j^{2}+b^{2}k^{2})^{1+\epsilon}}$
$\displaystyle=\sum_{j,k\geq
2}\int_{j-1}^{j}\int_{k-1}^{k}\frac{1}{(a^{2}j^{2}+b^{2}k^{2})^{1+\epsilon}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle\leq\sum_{j,k\geq
2}\int_{j-1}^{j}\int_{k-1}^{k}\frac{1}{(a^{2}x^{2}+b^{2}y^{2})^{1+\epsilon}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\int_{1}^{\infty}\int_{1}^{\infty}\frac{1}{(a^{2}x^{2}+b^{2}y^{2})^{1+\epsilon}}\,\mathrm{d}x\,\mathrm{d}y<\infty,$

for $\epsilon>0$. Therefore, it follows that

$\sum_{(j,k)\in\mathbb{Z}^{2}_{\ast}}\frac{1}{(a^{2}j^{2}+b^{2}k^{2})^{1+\epsilon}}<\infty,$

for $\epsilon>0$. According to Lemma 11, we have that the entire function

$F(z)=z\prod_{(j,k)\in\mathbb{Z}^{2}_{\ast}}E\left(\frac{z}{aj+\mathrm{i}bk};2\right),\qquad
z\in\mathbb{C},$

is of order two. It follows that if we can give an upper bound on the type of
$F$, i.e. if we could say that

$\sigma(F):=\limsup_{r\to\infty}\frac{\log\sup_{\left\lvert z\right\rvert\leq
r}\left\lvert F(z)\right\rvert}{r^{2}}<\frac{\pi}{2},$

then $F\in\mathcal{F}^{2}(\mathbb{C})$ and thus we could write
$f=\mathcal{B}^{-1}F\in L^{2}(\mathbb{R})$. So, the important question is: can
we upper bound the type of $F$?

Should we not be able to do this, a natural follow up question would be,
whether we can upper bound the number of roots a general function in the Fock
space can have. If we consider disks centered around the origin, we may use
the well-known Jensen formula from complex analysis to investigate this
question.

######  Lemma 36.

Let $f\in L^{2}(\mathbb{R})$ be non-zero, let $n:(0,\infty)\to\mathbb{N}_{0}$
be such that for all $r>0$, $n(r)$ denotes the number of roots of
$\mathcal{B}f$ (counted with multiplicity) in the disc of radius $r$ centered
at the origin, let $k\in\mathbb{N}_{0}$ denote the multiplicity of the root at
zero of $\mathcal{B}f$ (with the convention that $k=0$ should $\mathcal{B}f$
not have a root at the origin), and let $c\in\mathbb{R}$ be such that

$c=\begin{cases}\log\left(\frac{1}{\sqrt{2}}\left\lVert
f\right\rVert_{2}^{2}\right)-2\log\left\lvert\mathcal{B}f(0)\right\rvert&\mbox{if
}k=0,\\\ \log\left(\frac{1}{\sqrt{2}\mathrm{e}^{k}}\left\lVert
f\right\rVert_{2}^{2}\right)-2\log\left\lvert\lim_{z\to
0}\frac{\mathcal{B}f(z)}{z^{k}}\right\rvert+k&\mbox{if }k>0.\end{cases}$

Then,

$n(r)\leq\pi\mathrm{e}r^{2}-2k\log r+c,\qquad r>0.$

###### Proof.

Let $R\geq r>0$ be such that $R^{2}=\mathrm{e}r^{2}$, let

$F(z):=\frac{\mathcal{B}f(z)}{z^{k}},\qquad z\in\mathbb{C},$

and denote by $n_{F}(t)$ the number of roots of $F$ (counted with
multiplicity) in the disc of radius $t>0$ centered at the origin. Note that
$F$ is an entire function with removable singularity at $z=0$. Therefore, we
have

$\int_{0}^{R}\frac{n_{F}(t)}{t}\,\mathrm{d}t=\frac{1}{2\pi}\int_{0}^{2\pi}\log\left\lvert
F(R\mathrm{e}^{\mathrm{i}\theta})\right\rvert\,\mathrm{d}\theta-\log\left\lvert\lim_{z\to
0}F(z)\right\rvert,$

according to Jensen’s formula. We can use the reproducing property of the Fock
space along with the fact that $2^{1/4}\mathcal{B}$ is an isometry from
$L^{2}$ into the Fock space to see that

$\left\lvert
F(R\mathrm{e}^{\mathrm{i}\theta})\right\rvert=\frac{\left\lvert\mathcal{B}f(R\mathrm{e}^{\mathrm{i}\theta})\right\rvert}{R^{k}}\leq\frac{\left\lVert\mathcal{B}f\right\rVert_{\mathcal{F}}}{R^{k}}\mathrm{e}^{\frac{\pi}{2}R^{2}}=2^{-1/4}\mathrm{e}^{-k/2}\frac{\left\lVert
f\right\rVert_{2}}{r^{k}}\mathrm{e}^{\frac{\pi\mathrm{e}}{2}r^{2}}.$

In addition, we can use that $n_{F}(t)$ is increasing in $t$ to see that

$\int_{r}^{R}\frac{n_{F}(t)}{t}\,\mathrm{d}t\geq
n_{F}(r)\int_{r}^{R}\frac{1}{t}\,\mathrm{d}t=\frac{n_{F}(r)}{2}$

and thus

$n_{F}(r)\leq 2\int_{r}^{R}\frac{n_{F}(t)}{t}\,\mathrm{d}t\leq
2\int_{0}^{R}\frac{n_{F}(t)}{t}\,\mathrm{d}t.$

We conclude that

$n_{F}(r)\leq\pi\mathrm{e}r^{2}-2k\log
r+\log\left(\frac{1}{\sqrt{2}\mathrm{e}^{k}}\left\lVert
f\right\rVert_{2}^{2}\right)-2\log\left\lvert\lim_{z\to 0}F(z)\right\rvert.$

∎

Let us try to apply this knowledge to our example. We can estimate

$n(r)=\left\lvert(a\mathbb{Z}+\mathrm{i}b\mathbb{Z})\cap\left\\{z\in\mathbb{C}\,\middle|\,\left\lvert
z\right\rvert\leq r\right\\}\right\rvert=\frac{\pi
r^{2}}{ab}+\mathcal{O}(r^{2/3}),\qquad r\to\infty,$

where we used a result of Levitan [11]. Comparing this calculation to the
result in our lemma, we find that $ab\geq\mathrm{e}^{-1}$ is a necessary
condition for $F$ being in the Fock space. In words, if we try to design a
signal $f\in L^{2}(\mathbb{R})$ such that the Gabor transform vanishes on a
grid that is sufficiently fine, then we will fail.

######  Corollary 37 (Sampling for the Gabor transform).

Let $a,b>0$ such that $ab<\mathrm{e}^{-1}$ and let $f,g\in L^{2}(\mathbb{R})$.
If

$\mathcal{G}f=\mathcal{G}g\qquad\mbox{on }a\mathbb{Z}\times b\mathbb{Z},$

then $f=g$.

###### Proof.

Suppose that $\mathcal{G}f=\mathcal{G}g$ on $a\mathbb{Z}\times b\mathbb{Z}$.
Then, $h=f-g\in L^{2}(\mathbb{R})$ satisfies that $\mathcal{G}h=0$ on
$a\mathbb{Z}\times b\mathbb{Z}$ and thus
$H=\mathcal{B}h\in\mathcal{F}^{2}(\mathbb{C})$ satisfies that $H=0$ on
$a\mathbb{Z}\times b\mathbb{Z}$. If assume by contradiction that $h$ is non-
zero, then it follows from the prior lemma and the considerations before this
corollary that $ab\geq\mathrm{e}^{-1}$. The latter contradicts our assumptions
and therefore, the corollary is proven. ∎

We should remark that the assumption $ab<\mathrm{e}^{-1}$ is far from optimal.
In fact, the corollary continues to hold if we assume $ab\leq 1$ according to
the celebrated results by Lyubarskiĭ [12] and Seip [15] from 1992.

We can also directly estimate the type of the $F$ constructed in the beginning
of this subsection. Indeed, $F$ is a special case of the Weierstrass
$\sigma$-function and one may use its quasi-periodicity to show that its type
satisfies [6, 9]

$\sigma(F)\leq\frac{\pi}{2ab}$

and therefore $F\in\mathcal{F}^{2}(\mathbb{C})$ if $ab>1$. It does thus follow
that we can construct a non-zero function $f\in L^{2}(\mathbb{R})$ such that
$\mathcal{G}f=0$ on $a\mathbb{Z}\times b\mathbb{Z}$ if $ab>1$.

Finally, we still believe that Lemma 36 is interesting because it can be used
when constructing signals $f\in L^{2}(\mathbb{R})$ through their Bargmann
transform. In particular, Lemma 36 can be used to get a rough idea about how
dense our choice of the roots of $f$ may be.

##  6 Universal counterexamples in Gabor phase retrieval

In the prior sections (cf. also [2]), we have shown how, given infinitely many
equidistant parallel lines $\Omega$ in the time-frequency plane, one may
construct functions $f,g\in L^{2}(\mathbb{R})$ (depending on $\Omega$) such
that $f$ and $g$ do not agree up to global phase but

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g\right\rvert\mbox{
on }\Omega.$

This raises the question whether $f$ (and $g$) can be chosen independently on
$\Omega$ as well. To partially answer this question, we want to consider the
simplified setup in which $\Omega=\mathbb{R}\times a\mathbb{Z}$, for some
$a>0$.

Let us introduce the equivalence relation $\sim$ on $L^{2}(\mathbb{R})$ via

$f\sim
g:\Leftrightarrow\operatorname{\exists}\alpha\in\mathbb{R}:f=\mathrm{e}^{\mathrm{i}\alpha}g,$

for $f,g\in L^{2}(\mathbb{R})$. We are interested in the following question.

######  Question 38 (Existence of universal counterexamples — I).

Does there exist a function $f\in L^{2}(\mathbb{R})$ such that for all $a>0$,
there exists a $g=g_{a}\in L^{2}(\mathbb{R})$ such that $f\not\sim g$ yet

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g\right\rvert\qquad\mbox{on
}\mathbb{R}\times a\mathbb{Z}?$

In the following, we will show that the answer to this question is negative
because the condition $a>0$ allows for uncountably many different choices of
$a$.

######  Lemma 39.

For all $f\in L^{2}(\mathbb{R})$, there exists an $a>0$ such that for all
$g\in L^{2}(\mathbb{R})$, it holds that

$\left(\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g\right\rvert\mbox{
on }\mathbb{R}\times a\mathbb{Z}\right)\implies f\sim g.$

###### Proof.

Suppose, by contradiction, that there exists an $f\in L^{2}(\mathbb{R})$ such
that for all $a>0$, there exists a $g_{a}\in L^{2}(\mathbb{R})$ satisfying
$f\not\sim g_{a}$ yet

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{a}\right\rvert\quad\mbox{on
}\mathbb{R}\times a\mathbb{Z}.$

Let $a>0$ be arbitrary but fixed for now and consider that our assumption
implies

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{a}\right\rvert\quad\mbox{on
}\mathbb{R}\times\\{0,-a\\},$

and thereby (remember equation (8))

$\left\lvert\mathcal{B}f\right\rvert=\left\lvert\mathcal{B}g_{a}\right\rvert\qquad\mbox{on
}\mathbb{R}+\\{0,\mathrm{i}a\\}.$

At the same time, $f\not\sim g_{a}$ implies that
$\mathcal{B}f\not\sim\mathcal{B}g_{a}$ which together with the equation above
tells us that $\mathcal{B}f$ is non-trivial.

Next, let us remember the notations

$\displaystyle\mathcal{X}=\mathcal{X}_{a}=\mathcal{R}_{\ast}(\mathcal{B}f)\cap\mathcal{R}(\mathcal{B}g_{a}),$
$\displaystyle\mathcal{Y}=\mathcal{Y}_{a}=\left\\{w\in\mathcal{R}_{\ast}(\mathcal{B}f)\,\middle|\,m_{\mathcal{B}f}(w)>M_{\mathcal{B}g_{a}}(w)\right\\},$

as well as $m_{\mathcal{X}}:\mathcal{X}\to\mathbb{N}$,

$m_{\mathcal{X}}(w)=\min\\{m_{\mathcal{B}f}(w),m_{\mathcal{B}g_{a}}(w)\\},\qquad
w\in\mathcal{X},$

and $m_{\mathcal{Y}}:\mathcal{Y}\to\mathbb{N}$,

$m_{\mathcal{Y}}(w)=m_{\mathcal{B}f}(w)-M_{\mathcal{B}g_{a}}(w),\qquad
w\in\mathcal{Y}.$

$M_{\mathcal{X}}$ and $M_{\mathcal{Y}}$ will continue to denote the extensions
of $m_{\mathcal{X}}$ and $m_{\mathcal{Y}}$ to $\mathbb{C}$ by zero,
respectively. If we suppose, by contradiction, that
$\mathcal{Y}_{a}=\emptyset$, then it follows from
$\left\lvert\mathcal{B}f\right\rvert=\left\lvert\mathcal{B}g_{a}\right\rvert$
and the Mc Donald decomposition that there exist $r>0$,
$\phi,\psi\in\mathbb{R}$,
$a_{1},b_{1},b_{1}^{\prime},a_{2},b_{2},b_{2}^{\prime}\in\mathbb{R}$ and
$m\in\mathbb{N}_{0}$ such that

$\displaystyle\mathcal{B}f(z)=r\mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1})z+(a_{2}+\mathrm{i}b_{2})z^{2}}z^{m}\prod_{w\in\mathcal{X}}E\left(\frac{z}{w};2\right)^{m_{\mathcal{X}}(w)},$
$\displaystyle\mathcal{B}g_{a}(z)=r\mathrm{e}^{\mathrm{i}\psi}\mathrm{e}^{(a_{1}+\mathrm{i}b_{1}^{\prime})z+(a_{2}+\mathrm{i}b_{2}^{\prime})z^{2}}z^{m}\prod_{w\in\mathcal{X}}E\left(\frac{z}{w};2\right)^{m_{\mathcal{X}}(w)},$

for $z\in\mathbb{C}$. By

$\displaystyle
1=\frac{\left\lvert\mathcal{B}f(x+\mathrm{i}a)\right\rvert}{\left\lvert\mathcal{B}g_{a}(x+\mathrm{i}a)\right\rvert}=\left\lvert\mathrm{e}^{\mathrm{i}(b_{1}-b_{1}^{\prime})(x+\mathrm{i}a)+\mathrm{i}(b_{2}-b_{2}^{\prime})(x+\mathrm{i}a)^{2}}\right\rvert=\mathrm{e}^{(b_{1}^{\prime}-b_{1})a+2(b_{2}^{\prime}-b_{2})xa},$

for $x\in\mathbb{R}$, we conclude that $b_{1}=b_{1}^{\prime}$ and
$b_{2}=b_{2}^{\prime}$ and thus we find that
$\mathcal{B}f\sim\mathcal{B}g_{a}$ which is a contradiction. Hence, it holds
that $\mathcal{Y}_{a}\neq\emptyset$. We can thus find
$y_{a}\in\mathcal{Y}_{a}\subset\mathcal{R}_{\ast}(\mathcal{B}f)\subset\mathbb{C}_{\ast}$.
By the Lemmata 16 and 24, we can show that (see Subsection 4.2)

$M_{\mathcal{Y}}(z+2a\mathrm{i}k)=M_{\mathcal{Y}}(z),\qquad z\in\mathbb{C}.$

Therefore, it follows that
$y_{a}+2a\mathrm{i}\mathbb{Z}\subset\mathcal{Y}_{a}\subset\mathcal{R}_{\ast}(\mathcal{B}f)$.

As $a>0$ was arbitrary, we have shown that for all $a>0$, there exists a
$y_{a}\in\mathbb{C}_{\ast}$ such that
$y_{a}+2a\mathrm{i}\mathbb{Z}\subset\mathcal{R}_{\ast}(\mathcal{B}f)$.
Therefore, $\mathcal{R}_{\ast}(\mathcal{B}f)$ is uncountable and must thus
contain an accumulation point. As non-zero entire functions have the property
that their roots do not have any accumulation point (except infinity), we have
reached a contradiction. ∎

Interestingly, the proof of the prior lemma relies on the uncountability of
$(0,\infty)$ rather than on the fact that there are arbitrarily small numbers
in $(0,\infty)$. In particular, we can also prove the following result.

######  Lemma 40.

Let $T\subset(0,\infty)$ be uncountable. For all $f\in L^{2}(\mathbb{R})$,
there exists an $a\in T$ such that for all $g\in L^{2}(\mathbb{R})$, it holds
that

$\left(\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g\right\rvert\mbox{
on }\mathbb{R}\times a\mathbb{Z}\right)\implies f\sim g.$

This raises the question whether the same remains true if we replace $T$ by a
countable set containing arbitrarily small numbers such as
$T=\mathbb{N}^{-1}$, for instance.

######  Question 41 (Existence of universal counterexamples — II).

Does there exist a function $f\in L^{2}(\mathbb{R})$ such that for all
$n\in\mathbb{N}$, there exists a function $g_{n}\in L^{2}(\mathbb{R})$ such
that $f\not\sim g_{n}$ yet

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{n}\right\rvert\qquad\mbox{on
}\mathbb{R}\times\frac{1}{n}\mathbb{Z}?$

To answer Question 41, we consider a direct construction that is based on the
work which we have presented thus far in this paper. In particular, we
introduce the function

$F(z):=\prod_{n\in\mathbb{N}}\prod_{k\in\mathbb{Z}}\left(1-\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)\exp\left(\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right),$

for $z\in\mathbb{C}$. Through the theory surrounding the growth of entire
functions and the Hadamard factorisation theorem, we will show that $F$ is a
well-defined entire function of first order.

To be precise, we will apply Lemma 11. We will therefore consider the roots of
$F$ which are given by

$a_{n,k}:=\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n},\qquad
n\in\mathbb{N},~{}k\in\mathbb{Z},$

and consider the convergence of sums of the form

$\displaystyle\sum_{n,k}\frac{1}{\left\lvert a_{n,k}\right\rvert^{\lambda}}$
$\displaystyle=\sum_{n,k}\frac{1}{\left(\mathrm{e}^{2n}+\left(\frac{2k}{n}+\frac{1}{2n}\right)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle=2^{\lambda}\sum_{n\in\mathbb{N}}n^{\lambda}\sum_{k\in\mathbb{Z}}\frac{1}{\left(4n^{2}\mathrm{e}^{2n}+(4k+1)^{2}\right)^{\frac{\lambda}{2}}}.$

Let us distinguish between two different cases: first, if
$k\in\mathbb{N}_{0}$, then $(4k+1)^{2}\geq 16k^{2}$ and thus

$\displaystyle\frac{2^{\lambda}n^{\lambda}}{\left(4n^{2}\mathrm{e}^{2n}+(4k+1)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle\leq\frac{2^{\lambda}n^{\lambda}}{\left(4n^{2}\mathrm{e}^{2n}+16k^{2}\right)^{\frac{\lambda}{2}}}=\frac{n^{\lambda}}{\left(n^{2}\mathrm{e}^{2n}+4k^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle=\frac{1}{\left(\mathrm{e}^{2n}+4\frac{k^{2}}{n^{2}}\right)^{\frac{\lambda}{2}}},$

for $n\in\mathbb{N}$. Secondly, if $k\in\mathbb{N}$, then $(4(k-1)+3)^{2}\geq
16(k-1)^{2}$ such that

$\displaystyle\frac{2^{\lambda}n^{\lambda}}{\left(4n^{2}\mathrm{e}^{2n}+(-4k+1)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle=\frac{2^{\lambda}n^{\lambda}}{\left(4n^{2}\mathrm{e}^{2n}+(4(k-1)+3)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle\leq\frac{2^{\lambda}n^{\lambda}}{\left(4n^{2}\mathrm{e}^{2n}+16(k-1)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle=\frac{1}{\left(\mathrm{e}^{2n}+4\frac{(k-1)^{2}}{n^{2}}\right)^{\frac{\lambda}{2}}},$

for $n\in\mathbb{N}$. Therefore, we have

$\displaystyle\sum_{n,k}\frac{1}{\left\lvert a_{n,k}\right\rvert^{\lambda}}$
$\displaystyle=2^{\lambda}\sum_{n\in\mathbb{N}}n^{\lambda}\sum_{k\in\mathbb{Z}}\frac{1}{\left(4n^{2}\mathrm{e}^{2n}+(4k+1)^{2}\right)^{\frac{\lambda}{2}}}$
$\displaystyle\leq 2\sum_{n\in\mathbb{N}}\mathrm{e}^{-\lambda
n}+2\sum_{n,k\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+4\frac{k^{2}}{n^{2}}\right)^{\frac{\lambda}{2}}}$
$\displaystyle=\frac{2\mathrm{e}^{-\lambda}}{1-\mathrm{e}^{-\lambda}}+2\sum_{n,k\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+4\frac{k^{2}}{n^{2}}\right)^{\frac{\lambda}{2}}},$

when $\lambda>0$.

It remains to understand when the rightmost sum converges. The easiest way to
do so seems to be to establish the convergence by some integral test. We will
therefore consider the function

$f(t):=\sum_{n\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}},\qquad
t\in[1,\infty),$

which is well-defined since for every fixed $t\in[1,\infty)$, we have

$\sum_{n\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}}\leq\sum_{n\in\mathbb{N}}\mathrm{e}^{-\lambda
n}=\frac{\mathrm{e}^{-\lambda}}{1-\mathrm{e}^{-\lambda}}.$

By the integral test, the convergence of

$\sum_{k\in\mathbb{N}}f(k)$

is guaranteed by the convergence of

$\int_{1}^{\infty}f(t)\,\mathrm{d}t=\int_{1}^{\infty}\sum_{n\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t.$

By Tonelli’s theorem, we can exchange the summation and the integration to
obtain

$\int_{1}^{\infty}f(t)\,\mathrm{d}t=\int_{1}^{\infty}\sum_{n\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t=\sum_{n\in\mathbb{N}}\int_{1}^{\infty}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t.$

We might now rewrite the integral a little bit by using the substitution
$t=\tfrac{n\exp(n)}{2}s$ and find

$\displaystyle\int_{1}^{\infty}f(t)\,\mathrm{d}t$
$\displaystyle=\sum_{n\in\mathbb{N}}\int_{1}^{\infty}\frac{1}{\left(\mathrm{e}^{2n}+\frac{4}{n^{2}}t^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t=\sum_{n\in\mathbb{N}}\mathrm{e}^{-\lambda
n}\int_{1}^{\infty}\frac{1}{\left(1+\frac{4}{n^{2}\mathrm{e}^{2n}}t^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t$
$\displaystyle=\frac{1}{2}\sum_{n\in\mathbb{N}}n\mathrm{e}^{-(\lambda-1)n}\int_{\frac{2}{n\exp(n)}}^{\infty}\frac{1}{\left(1+s^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t$
$\displaystyle\leq\frac{1}{2}\sum_{n\in\mathbb{N}}n\mathrm{e}^{-(\lambda-1)n}\int_{0}^{\infty}\frac{1}{\left(1+s^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t.$

One may show that

$I(\lambda):=\int_{0}^{\infty}\frac{1}{\left(1+s^{2}\right)^{\frac{\lambda}{2}}}\,\mathrm{d}t<\infty,$

when $\lambda>1$. One way of doing this is demonstrated in the appendix where
we prove that

$I(\lambda)=\frac{\sqrt{\pi}\Gamma\left(\frac{\lambda-1}{2}\right)}{2\Gamma\left(\frac{\lambda}{2}\right)},\qquad\lambda>1,$

where

$\Gamma(z):=\int_{0}^{\infty}t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t,\qquad
z\in\mathbb{C},~{}\operatorname{Re}z>0,$

denotes the Euler gamma function. Finally, we may bound

$\int_{1}^{\infty}f(t)\,\mathrm{d}t\leq\frac{I(\lambda)}{2}\sum_{n\in\mathbb{N}}n\mathrm{e}^{-(\lambda-1)n}<\infty$

by using the ratio test when $\lambda>1$.

We may now conclude from the integral test that

$\sum_{n,k\in\mathbb{N}}\frac{1}{\left(\mathrm{e}^{2n}+4\frac{k^{2}}{n^{2}}\right)^{\frac{\lambda}{2}}}=\sum_{k\in\mathbb{N}}f(k)<\infty$

and thus that

$\sum_{n,k}\frac{1}{\left\lvert a_{n,k}\right\rvert^{\lambda}}<\infty,$

as long as $\lambda>1$. Finally, it follows from Lemma 11 that

$F(z)=\prod_{n\in\mathbb{N}}\prod_{k\in\mathbb{Z}}\left(1-\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)\exp\left(\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right),$

is a well-defined entire function of order at most $1$.

Since $F$ is of first order, it is, in particular, in the Fock space
$\mathcal{F}^{2}(\mathbb{C})$. Therefore, we may define
$f:=\mathcal{B}^{-1}F\in L^{2}(\mathbb{R})$. In the following, we show that
$f$ is a universal counterexample.

######  Theorem 42 (Existence of universal counterexamples — version I).

The function

$f=\mathcal{B}^{-1}\prod_{n\in\mathbb{N}}\prod_{k\in\mathbb{Z}}\left(1-\frac{\cdot}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)\exp\left(\frac{\cdot}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)$

is a well-defined function in $L^{2}(\mathbb{R})$. Additionally, the functions

$g_{m}:=\mathcal{B}^{-1}\mathrm{e}^{-\mathrm{i}b_{m}\cdot}\prod_{k\in\mathbb{Z}}\left(1-\frac{\cdot}{\mathrm{e}^{m}-\frac{\mathrm{i}}{2m}+\frac{2\mathrm{i}k}{m}}\right)\exp\left(\frac{\cdot}{\mathrm{e}^{m}-\frac{\mathrm{i}}{2m}+\frac{2\mathrm{i}k}{m}}\right)\\\
\cdot\prod_{n\in\mathbb{N}\setminus\\{m\\}}\prod_{k\in\mathbb{Z}}\left(1-\frac{\cdot}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)\exp\left(\frac{\cdot}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right),$

where

$b_{m}=m\sum_{k\in\mathbb{Z}}\left(\log\left\lvert\frac{\mathrm{e}^{m}+\frac{\mathrm{i}}{2m}+\frac{2(k-1)\mathrm{i}}{m}}{\mathrm{e}^{m}+\frac{\mathrm{i}}{2m}+\frac{2k\mathrm{i}}{m}}\right\rvert-\frac{2}{m}\operatorname{Im}\frac{1}{\mathrm{e}^{m}+\frac{\mathrm{i}}{2m}+\frac{2k\mathrm{i}}{m}}\right)\in\mathbb{R},$

for $m\in\mathbb{N}$, are well-defined functions in $L^{2}(\mathbb{R})$.
Finally, it holds that $f\not\sim g_{m}$ as well as

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{m}\right\rvert\mbox{
on }\mathbb{R}\times\frac{1}{m}\mathbb{Z}.$

###### Proof.

We have already shown that $f\in L^{2}(\mathbb{R})$ is well-defined. Let us
now consider $m\in\mathbb{N}$ arbitrary but fixed. We remark that the
considerations before the statement of the theorem imply that

$\sum_{k\in\mathbb{Z}}\frac{1}{\left\lvert\mathrm{e}^{m}-\frac{\mathrm{i}}{2m}+\frac{2\mathrm{i}k}{m}\right\rvert^{\lambda}}+\sum_{n\in\mathbb{N}\setminus\\{m\\}}\sum_{k\in\mathbb{Z}}\frac{1}{\left\lvert
a_{n,k}\right\rvert^{\lambda}}<\infty.$

Therefore, it follows from Lemma 11 that

$G_{m}(z):=\prod_{k\in\mathbb{Z}}\left(1-\frac{z}{\mathrm{e}^{m}-\frac{\mathrm{i}}{2m}+\frac{2\mathrm{i}k}{m}}\right)\exp\left(\frac{z}{\mathrm{e}^{m}-\frac{\mathrm{i}}{2m}+\frac{2\mathrm{i}k}{m}}\right)\\\
\cdot\prod_{n\in\mathbb{N}\setminus\\{m\\}}\prod_{k\in\mathbb{Z}}\left(1-\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right)\exp\left(\frac{z}{\mathrm{e}^{n}+\frac{\mathrm{i}}{2n}+\frac{2\mathrm{i}k}{n}}\right),$

for $z\in\mathbb{C}$, is a well-defined entire function of first order. At the
same time, the convergence of the sums mentioned above also implies that the
sum defining $b_{m}$ is convergent. It follows that
$G_{m}\mathrm{e}^{-\mathrm{i}b_{m}\cdot}\in\mathcal{F}^{2}(\mathbb{C})$ and
thus $g_{m}=\mathcal{B}^{-1}(G_{m}\mathrm{e}^{-\mathrm{i}b_{m}\cdot})\in
L^{2}(\mathbb{R})$ is well-defined.

Next, we may see that $f\not\sim g_{m}$ by definition. Indeed,
$\mathrm{e}^{m}+\tfrac{\mathrm{i}}{2m}$ is a root of $\mathcal{B}f$ but not a
root of $\mathcal{B}g_{m}$. Finally, we may apply Theorem 27 with
$\tau=\tfrac{1}{m}$, $r=1$, $\phi=\psi=a^{\prime}=b=0$, $b^{\prime}=-b_{m}$,
$m=0$,

$\mathcal{Y}_{\mathrm{u}}=\left\\{\mathrm{e}^{m}+\tfrac{\mathrm{i}}{2m}\right\\},$

$m_{\mathcal{Y}}=1$,

$\mathcal{X}=\left\\{\mathrm{e}^{n}+\tfrac{\mathrm{i}}{2n}+\tfrac{2\mathrm{i}k}{n}\,\middle|\,n\in\mathbb{N}\setminus\\{m\\},~{}k\in\mathbb{Z}\right\\},$

and $m_{\mathcal{X}}=1$. In this way, we see that

$\left\lvert\mathcal{B}f\right\rvert=\left\lvert\mathcal{B}g_{m}\right\rvert\mbox{
on }\mathbb{R}+\mathrm{i}\frac{1}{m}\mathbb{Z}.$

According to the relation of the Gabor transform and the Bargmann transform
(see equation (8)), we conclude

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{m}\right\rvert\mbox{
on }\mathbb{R}\times\frac{1}{m}\mathbb{Z}.$

∎

The following statement about existence of universal counterexamples follows
directly from the prior theorem.

######  Corollary 43 (Existence of universal counterexamples — version II).

There exists an $f\in L^{2}(\mathbb{R})$ such that for all $n\in\mathbb{N}$,
there exists a $g_{n}\in L^{2}(\mathbb{R})$ such that $f\not\sim g_{n}$ and
yet

$\left\lvert\mathcal{G}f\right\rvert=\left\lvert\mathcal{G}g_{n}\right\rvert\mbox{
on }\mathbb{R}\times\frac{1}{n}\mathbb{Z}.$

##  Appendix A Relating the convergence of a certain integral to the Euler
gamma function

The following lemma might be well-known. We found it by comparing it to the
well-known relation of the Euler gamma function and the beta function.

######  Lemma 44.

It holds that

$\int_{0}^{\infty}\frac{1}{(1+t^{2})^{\lambda/2}}\,\mathrm{d}t=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{\lambda-1}{2}\right)}{\Gamma\left(\frac{\lambda}{2}\right)},\qquad\lambda>1.$

###### Proof.

Let us consider

$\Gamma\left(\frac{\lambda}{2}\right)\cdot\int_{0}^{\infty}\frac{1}{(1+t^{2})^{\lambda/2}}\,\mathrm{d}t=\int_{0}^{\infty}\int_{0}^{\infty}\frac{u^{\lambda/2-1}\mathrm{e}^{-u}}{(1+t^{2})^{\lambda/2}}\,\mathrm{d}t\,\mathrm{d}u.$

We may now use the substitution $u=x+y$, $t=\sqrt{x/y}$ with Jacobian

$J=\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial
y}\\\ \frac{\partial t}{\partial x}&\frac{\partial t}{\partial
y}\end{pmatrix}=\begin{pmatrix}1&1\\\
\frac{1}{2\sqrt{xy}}&-\frac{\sqrt{x}}{2y^{3/2}}\end{pmatrix}.$

We compute

$\left\lvert\det
J\right\rvert=\frac{1}{2\sqrt{xy}}+\frac{\sqrt{x}}{2y^{3/2}}=\frac{x+y}{2x^{1/2}y^{3/2}}$

and note that

$x=\frac{ut^{2}}{1+t^{2}},\qquad y=\frac{u}{1+t^{2}},$

which implies that the domain of integration remains unchanged. Therefore,

$\displaystyle\Gamma\left(\frac{\lambda}{2}\right)\cdot\int_{0}^{\infty}\frac{1}{(1+t^{2})^{\lambda/2}}\,\mathrm{d}t$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}\frac{u^{\lambda/2-1}\mathrm{e}^{-u}}{(1+t^{2})^{\lambda/2}}\,\mathrm{d}t\,\mathrm{d}u$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}\left(\frac{u}{1+t^{2}}\right)^{\lambda/2}\mathrm{e}^{-u}u^{-1}\,\mathrm{d}t\,\mathrm{d}u$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}y^{\lambda/2}\mathrm{e}^{-x-y}(x+y)^{-1}\frac{x+y}{2x^{1/2}y^{3/2}}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\frac{1}{2}\cdot\int_{0}^{\infty}\int_{0}^{\infty}y^{\lambda/2-3/2}x^{-1/2}\mathrm{e}^{-x-y}\,\mathrm{d}x\,\mathrm{d}y$
$\displaystyle=\frac{1}{2}\cdot\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{\lambda-1}{2}\right).$

The lemma follows from the well-known facts that $\Gamma(1/2)=\sqrt{\pi}$ and
that the Gamma function has no zeros. ∎

##### Acknowledgements

We want to thank Rima Alaifari and Giovanni S. Alberti for fruitful
discussions. In addition, the author would like to acknowledge funding through
the SNSF Grant 200021_184698.

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