# Supplemental Material: Complete Characterization of Entanglement
Embezzlement

Elia Zanoni<EMAIL_ADDRESS>Thomas Theurer  Gilad Gour  Department of
Mathematics and Statistics, University of Calgary, Calgary, AB T2N 1N4, Canada
Institute for Quantum Science and Technology, University of Calgary, Calgary,
AB T2N 1N4, Canada

(August 28, 2024)

In this Supplemental Material, we provide the proofs omitted in the main text
as well as further details and discussions.

##  I Pure to Mixed State Conversions with LOCC

We begin with the proof of Proposition 1 of the main text, which we restate
for readability.

######  Proposition 1.

For $\psi\in\operatorname{PURE}(AB)$ and $\sigma\in\mathfrak{D}(AB)$,
$\psi\xrightarrow{\operatorname{LOCC}}\sigma$ if and only if there exists a
pure state decomposition $\Set{p_{z},\chi_{z}}$ of $\sigma$ (i.e.,
$\sigma=\sum_{z}p_{z}\chi_{z}$) such that

$\min_{k\in[|A|]}\Bigl{\\{}E_{k}(\psi)-\sum_{z}p_{z}E_{k}(\chi_{z})\Bigr{\\}}\geq
0.$ (SM1)

###### Proof.

We first assume that $\psi$ can be converted into $\sigma$ with
$\operatorname{LOCC}$ operations. Then, according to Refs. [1, 2], there
exists a protocol in which Alice performs a generalized measurement
$\Set{M_{z}}$ and Bob performs a unitary transformation $U_{z}$ conditioned on
the measurement’s outcome such that $\sigma=\sum_{z}(M_{z}\otimes
U_{z})\psi(M_{z}\otimes U_{z})^{\dagger}$. If Alice and Bob record the outcome
of the measurement in a classical system $X$, the output of the protocol is
$\sum_{z}p_{z}{|z\rangle\\!\langle
z|}^{X}\otimes\chi_{z}^{AB}\coloneqq\Set{p_{z},\chi_{z}}$, where

$\Ket{\chi_{z}}=\frac{(M_{z}\otimes U_{z})\Ket{\psi}}{\left\lVert(M_{z}\otimes
U_{z})\Ket{\psi}\right\rVert}\,,\quad p_{z}=\left\lVert(M_{z}\otimes
U_{z})\Ket{\psi}\right\rVert^{2}.$ (SM2)

This shows that Alice and Bob can convert $\psi$ into the ensemble
$\Set{p_{z},\chi_{z}}$ with $\operatorname{LOCC}$ operations, which is
equivalent to the condition [3, 4]

$\min_{k\in[d]}\Bigl{(}E_{k}(\psi)-\sum_{z}p_{z}E_{k}(\chi_{z})\Bigr{)}\geq
0,$ (SM3)

where $E_{k}$ was introduced in the main text and $d=|A|=|B|$ as per our
convention. This proves the necessary condition.

For the reverse, we assume that $\Set{p_{z},\chi_{z}}$ is a pure state
decomposition of $\sigma$ that satisfies

$\min_{k\in[d]}\Bigl{(}E_{k}(\psi)-\sum_{z}p_{z}E_{k}(\chi_{z})\Bigr{)}\geq
0.$ (SM4)

This is equivalent to
$\psi\xrightarrow{\operatorname{LOCC}}\Set{p_{z},\chi_{z}}=\sum_{z}p_{z}{|z\rangle\\!\langle
z|}^{X}\otimes\chi_{z}^{AB}$ [3, 4]. Alice and Bob can trace out the classical
system and they obtain $\sum_{z}p_{z}\chi_{z}=\sigma$, thus $\psi$ can be
converted into $\sigma$ with $\operatorname{LOCC}$ operations. ∎

##  II Purified Conversion Distance

In this section, we present some results concerning the purified conversion
distance [5, 6], which are useful to prove the theorems about the star
conversion distance presented in the main text. The purified distance between
two states $\rho$, $\sigma\in\mathfrak{D}\left(AB\right)$ is defined as

$P\left(\rho,\sigma\right)=\sqrt{1-F^{2}\left(\rho,\sigma\right)},$ (SM5)

where
$F\left(\rho,\sigma\right)=\left\lVert\sqrt{\rho}\sqrt{\sigma}\right\rVert_{1}$
is the fidelity. The purified distance is a metric and according to Uhlmann’s
Theorem [7],

$P\left(\rho,\sigma\right)=\min_{\psi,\varphi}D\left(\psi,\varphi\right),$
(SM6)

where
$D\left(\rho,\sigma\right)=\frac{1}{2}\left\lVert\rho-\sigma\right\rVert_{1}$
is the trace distance and the minimization runs over all purifications $\psi$
and $\varphi$ of $\rho$ and $\sigma$, respectively. Importantly, the purified
distance is topologically equivalent to the trace distance [5],

$D\left(\rho,\sigma\right)\leq
P\left(\rho,\sigma\right)\leq\sqrt{2D\left(\rho,\sigma\right)}.$ (SM7)

We define two conversion distances based on the purified distance, analogously
to what was done in the main text for the trace distance.

######  Definition SM.1.

For $\rho,\sigma\in\mathfrak{D}(AB)$, $\left\lvert A\right\rvert=\left\lvert
B\right\rvert=d$, we define the purified conversion distance as

$d^{P}\left(\rho\to\sigma\right)=\inf_{\mathcal{N}\in\operatorname{LOCC}}P\left(\mathcal{N}\left(\rho\right),\sigma\right),$
(SM8)

and for $\psi,\varphi\in\operatorname{PURE}(AB)$, we define the purified star
conversion distance as

$d^{P}_{\star}\left(\psi\to\varphi\right)=\min_{\mathbf{r}\succ\mathbf{p}}P\left(\mathbf{r},\mathbf{q}\right),$
(SM9)

where $\mathbf{p}$, $\mathbf{q}\in\operatorname{Prob}^{\downarrow}(d)$ are the
Schmidt coefficients of $\psi$ and $\varphi$, respectively, and
$P\left(\mathbf{r},\mathbf{q}\right)=\sqrt{1-F^{2}(\mathbf{r},\mathbf{q})}$ is
the classical version of the purified distance with
$F(\mathbf{r},\mathbf{q})\coloneqq
F(\operatorname{diag}(\mathbf{r}),\operatorname{diag}(\mathbf{q}))=\sum_{x}\sqrt{r_{x}q_{x}}$.

We next prove that the two purified conversion distances introduced in
Definition SM.1 are equal on pure states. To this end, we need the following
Lemma.

######  Lemma SM.2.

Let $\mathbf{q}\in\operatorname{Prob}(d)$. The function
$f:\operatorname{Prob}(d)\to[0,1]$ defined as
$f(\mathbf{v})=\left(\sum_{x=1}^{d}\sqrt{q_{x}v_{x}}\right)^{2}$ is concave.

###### Proof.

For details about concave functions, see Ref. [8]. Here, we are going to use
that a twice differentiable function is concave if and only if its Hessian
matrix is negative semi-definite. The functions
$f_{x,y}(v_{x},v_{y}):=\sqrt{q_{x}v_{x}}\sqrt{q_{y}v_{y}}$ are twice
differentiable in $v_{x},v_{y}$ for $v_{x},v_{y}\in(0,1]$ and their Hessian
matrices are given by

$H\left(v_{x},v_{y}\right)=\frac{1}{4}\begin{bmatrix}-\sqrt{\frac{q_{x}q_{y}v_{y}}{v_{x}^{3}}}&\sqrt{\frac{q_{x}q_{y}}{v_{x}v_{y}}}\\\
\sqrt{\frac{q_{x}q_{y}}{v_{x}v_{y}}}&-\sqrt{\frac{q_{x}q_{y}v_{x}}{v_{y}^{3}}}\end{bmatrix}.$
(SM10)

Since $\det H(v_{x},v_{y})=0$, one of the eigenvalues of $H(v_{x},v_{y})$ is
$0$ and the other is equal to the trace of $H(v_{x},v_{y})$, which is smaller
than or equal to zero. This implies that the functions $f_{x,y}(v_{x},v_{y})$
are concave for $v_{x},v_{y}\in(0,1]$. Moreover, since
$0=f_{x,y}(0,v_{y})=f_{x,y}(v_{x},0)=f_{x,y}(0,0)$, it is easy to see that
$f_{x,y}(v_{x},v_{y})$ is in fact concave for $v_{x},v_{y}\in[0,1]$. From this
follows that $f$ is concave, since

$f\left(\mathbf{v}\right)=\sum_{x}v_{x}q_{x}+\sum_{y}\sum_{x\neq
y}\sqrt{q_{x}v_{x}}\sqrt{q_{y}v_{y}}$ (SM11)

is the sum of concave functions. ∎

We are now ready to prove the promised theorem.

######  Theorem SM.3.

Let $\psi,\varphi\in\operatorname{PURE}(AB)$, then
$d^{P}\left(\psi\to\varphi\right)=d^{P}_{\star}\left(\psi\to\varphi\right)$.

###### Proof.

We assume w.l.o.g. that all pure states are in standard form, i.e. ,
$\Ket{\psi}=\sum_{x}\sqrt{p_{x}}\Ket{xx}_{AB}$, where
$\mathbf{p}\in\operatorname{Prob}^{\downarrow}(d)$ and $\Set{\Ket{x}_{A}}$ and
$\Set{\Ket{x}_{B}}$ are fixed bases for $A$ and $B$, respectively (see main
text for more details). Let
$\mathbf{p},\mathbf{q}\in\operatorname{Prob}^{\downarrow}(d)$ be the Schmidt
coefficients or $\psi$ and $\varphi$, respectively. With
$\mathbf{r}\in\operatorname{Prob}^{\downarrow}(d)$, as a consequence of
Nielsen’s Theorem [9], $\mathbf{r}\succ\mathbf{p}$ if and only if there exists
an $\mathcal{N}\in\operatorname{LOCC}$ such that
$\mathcal{N}(\psi)\in\operatorname{PURE}(AB)$ has Schmidt coefficients
$\mathbf{r}$. We notice that

$P(\mathcal{N}(\psi),\varphi)=\sqrt{1-F^{2}(\mathcal{N}(\psi),\varphi)}=\sqrt{1-\left\lvert\Braket{\mathcal{N}(\psi)}{\varphi}\right\rvert^{2}}=\sqrt{1-\left(\sum_{x}\sqrt{r_{x}q_{x}}\right)^{2}}=\sqrt{1-F^{2}(\mathbf{r},\mathbf{q})}=P(\mathbf{r},\mathbf{q}).$
(SM12)

This implies that

$d^{P}_{\star}\left(\psi\to\varphi\right)=\min_{\mathbf{r}\succ\mathbf{p}}P\left(\mathbf{r},\mathbf{q}\right)=\min_{\begin{subarray}{c}\mathcal{N}\in\operatorname{LOCC}\\\
\mathcal{N}(\psi)\in\operatorname{PURE}(AB)\end{subarray}}P\left(\mathcal{N}\left(\psi\right),\varphi\right)\geq\inf_{\mathcal{N}\in\operatorname{LOCC}}P\left(\mathcal{N}\left(\psi\right),\varphi\right)=d^{P}\left(\psi\to\varphi\right).$
(SM13)

Next, we show that the reverse inequality holds too. Let
$\sigma=\mathcal{M}(\psi)\in\mathfrak{D}(AB)$, where
$\mathcal{M}\in\operatorname{LOCC}$. Also, let $\Set{t_{z},\chi_{z}}$ be a
pure state decomposition of $\sigma$ that satisfies Eq. (SM1) (where the
$\chi_{z}$ are not necessarily in standard form), and $\mathbf{s}^{(z)}$ be
the Schmidt coefficient of $\chi_{z}$ for every $z$. Furthermore, for every
$\chi_{z}$, let $\tilde{\chi}_{z}$ be the pure state in standard form that is
equal to $\chi_{z}$ up to local unitaries, let
$\tilde{\sigma}=\sum_{z}t_{z}\tilde{\chi}_{z}$, and define
$\chi\in\operatorname{PURE}(AB)$ as the pure bipartite state (in standard
form) with Schmidt coefficients $\mathbf{s}=\sum_{z}t_{z}\mathbf{s}^{(z)}$. We
notice that for all $k\in[d]$

$E_{k}(\chi)=1-\sum_{x=1}^{k}s_{x}=1-\sum_{x=1}^{k}\sum_{z}t_{z}s_{x}^{(z)}=1-\sum_{z}t_{z}(1-E_{k}(\chi_{z}))=\sum_{z}t_{z}E_{k}(\chi_{z})\leq
E_{k}(\psi),$ (SM14)

where the last inequality follows from the fact that $\Set{t_{z},\chi_{z}}$
satisfies Eq. (SM1). This implies that $\mathbf{s}\succ\mathbf{p}$.

The next step is to show that $P(\sigma,\varphi)\geq
P(\mathbf{s},\mathbf{q})$. First, we note that due to the von Neumann trace
inequality [10, 11],

$F^{2}\left(\tilde{\sigma},\varphi\right)=\sum_{z}t_{z}\left(\sum_{x}\sqrt{q_{x}s_{x}^{(z)}}\right)^{2}\geq
F^{2}\left(\sigma,\varphi\right)$ (SM15)

and

$F^{2}(\chi,\varphi)=F^{2}\left(\mathbf{s},\mathbf{q}\right)=\left(\sum_{x}\sqrt{q_{x}\sum_{z}t_{z}s_{x}^{(z)}}\right)^{2}.$
(SM16)

Second, we introduce the concave function
$f\left(\mathbf{v}\right)=\left(\sum_{x}\sqrt{q_{x}v_{x}}\right)^{2}$, for
$\mathbf{v}\in\operatorname{Prob}\left(d\right)$ (see Lemma SM.2), and rewrite
Eq. (SM15) and Eq. (SM16) as

$F^{2}\left(\sigma,\varphi\right)\leq
F^{2}\left(\tilde{\sigma},\varphi\right)=\sum_{z}t_{z}f\left(\mathbf{s}^{(z)}\right)\,,\quad
F^{2}\left(\mathbf{s},\mathbf{q}\right)=f\left(\sum_{z}t_{z}\mathbf{s}^{(z)}\right).$
(SM17)

Finally, the concavity of $f$ and the multidimensional Jensen’s inequality
[12] imply

$F^{2}\left(\sigma,\varphi\right)\leq
F^{2}\left(\tilde{\sigma},\varphi\right)=\sum_{z}t_{z}f\left(\mathbf{s}^{(z)}\right)\leq
f\left(\sum_{z}t_{z}\mathbf{s}^{(z)}\right)=F^{2}\left(\mathbf{s},\mathbf{q}\right),$
(SM18)

which is equivalent to $P(\sigma,\varphi)\geq P(\mathbf{s},\mathbf{q})$. So
far, we have shown that for every $\sigma=\mathcal{M}(\psi)$, with
$\mathcal{M}\in\operatorname{LOCC}$, there exists an
$\mathbf{s}^{(\mathcal{M}(\psi))}\in\operatorname{Prob}^{\downarrow}(d)$ such
that $\mathbf{s}^{(\mathcal{M}(\psi))}\succ\mathbf{p}$ and
$P(\sigma,\varphi)\geq P(\mathbf{s}^{(\mathcal{M}(\psi))},\mathbf{q})$, thus

$d^{P}(\psi\to\varphi)=\inf_{\mathcal{M}\in\operatorname{LOCC}}P\left(\mathcal{M}\left(\psi\right),\varphi\right)\geq\inf_{\mathcal{M}\in\operatorname{LOCC}}P\left(\mathbf{s}^{(\mathcal{M}(\psi))},\mathbf{q}\right)\geq\min_{\mathbf{r}\succ\mathbf{p}}P\left(\mathbf{r},\mathbf{q}\right)=d^{P}_{\star}\left(\psi\to\varphi\right).$
(SM19)

Eq. (SM13) and Eq. (SM19) imply that
$d^{P}(\psi\to\varphi)=d^{P}_{\star}\left(\psi\to\varphi\right)$, and this
concludes the proof. ∎

##  III The Star Conversion Distance

In this section, we discuss properties of the star conversion distance and
provide proofs omitted in the main text. We begin by proving that the star
conversion distance is topologically equivalent to the standard conversion
distance defined via the trace distance. In fact, using Eq. (SM7) and Theorem
SM.3, we obtain

$\displaystyle d(\psi\to\varphi)=$
$\displaystyle\inf_{\mathcal{N}\in\operatorname{LOCC}}D(\mathcal{N}(\psi),\varphi)\leq\inf_{\mathcal{N}\in\operatorname{LOCC}}P(\mathcal{N}(\psi),\varphi)=d^{P}(\psi\to\varphi)=d^{P}_{\star}(\psi\to\varphi)=\min_{\mathbf{r}\succ\mathbf{p}}P(\mathbf{r},\mathbf{q})$
$\displaystyle\leq$
$\displaystyle\min_{\mathbf{r}\succ\mathbf{p}}\sqrt{2D(\mathbf{r},\mathbf{q})}=\sqrt{2d_{\star}(\psi\to\varphi)},$
(SM20)

where $D(\mathbf{p},\mathbf{q})\coloneqq
D(\operatorname{diag}(\mathbf{p}),\operatorname{diag}(\mathbf{q}))$.
Analogously,

$\displaystyle d_{\star}(\psi\to\varphi)=$
$\displaystyle\min_{\mathbf{r}\succ\mathbf{p}}D(\mathbf{r},\mathbf{q})\leq\min_{\mathbf{r}\succ\mathbf{p}}P(\mathbf{r},\mathbf{q})=d^{P}_{\star}(\psi\to\varphi)=d^{P}(\psi\to\varphi)=\inf_{\mathcal{N}\in\operatorname{LOCC}}P(\mathcal{N}(\psi),\varphi)$
$\displaystyle\leq$
$\displaystyle\inf_{\mathcal{N}\in\operatorname{LOCC}}\sqrt{2D(\mathcal{N}(\psi),\varphi)}=\sqrt{2d(\psi\to\varphi)}.$
(SM21)

This proves that $d_{\star}\left(\psi\to\varphi\right)$ is topologically
equivalent to $d\left(\psi\to\varphi\right)$.

In the following, we provide the proof of Theorem 2 of the main text, which we
restate below for readability. In other words, we derive a closed formula for
$d_{\star}\left(\psi\to\varphi\right)$ based on the Schmidt coefficients
$\mathbf{p},\mathbf{q}\in\operatorname{Prob}^{\downarrow}(d)$ of $\psi$ and
$\varphi$, respectively.

######  Theorem 2.

Let $\psi$, $\varphi\in\operatorname{PURE}(AB)$ and let $\mathbf{p}$,
$\mathbf{q}\in\operatorname{Prob}^{\downarrow}(|A|)$ be their corresponding
Schmidt coefficients. Then,

$d_{\star}(\psi\to\varphi)=\max_{k\in[\operatorname{SR}(\psi)]}\Set{\left\lVert\mathbf{p}\right\rVert_{(k)}-\left\lVert\mathbf{q}\right\rVert_{(k)}}.$
(SM22)

If $\xi\in\operatorname{PURE}(AB)$ is separable and $\psi$, $\varphi$ are not,
then

$\displaystyle d_{\star}(\psi\to\varphi)<d_{\star}(\xi\to\varphi).$ (SM23)

###### Proof.

Let $\varepsilon\in[0,1]$ and
$\mathfrak{B}^{\varepsilon}_{\mathbf{q}}=\Set{\mathbf{q}^{\prime}:\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{q}^{\prime}\right\rVert_{1}\leq\varepsilon}$.
As shown in Ref. [13], there exist probability vectors
$\bar{\mathbf{q}}^{(\varepsilon)}\in\mathfrak{B}^{\varepsilon}_{\mathbf{q}}$
called steepest $\varepsilon$-approximations of $\mathbf{q}$ such that
$\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{q}^{\prime}$ for all
$\mathbf{q}^{\prime}\in\mathfrak{B}^{\varepsilon}_{\mathbf{q}}$. Moreover,
these steepest $\varepsilon$-approximations can be constructed explicitly: If
$\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1}\leq\varepsilon$,
then $\bar{\mathbf{q}}^{(\varepsilon)}=\mathbf{e}_{1}$, otherwise let
$k_{\varepsilon}\in[d]$ be the index satisfying

$\sum_{x=1}^{k_{\varepsilon}}q_{x}\leq
1-\varepsilon\quad\text{and}\quad\sum_{x=1}^{k_{\varepsilon}+1}q_{x}>1-\varepsilon.$
(SM24)

The components of $\bar{\mathbf{q}}^{(\varepsilon)}$ are then given by

$\bar{q}^{(\varepsilon)}_{x}=\begin{cases}q_{1}+\varepsilon&\text{if }x=1,\\\
q_{x}&\text{if }x\in\Set{2,\dots,k_{\varepsilon}},\\\
1-\varepsilon-\sum_{x=1}^{k_{\varepsilon}}q_{x}&\text{if
}x=k_{\varepsilon}+1,\\\ 0&\text{otherwise}.\end{cases}$ (SM25)

We now show that

$\displaystyle\min_{\mathbf{r}\succ\mathbf{p}}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}}=\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}.$
(SM26)

First, we notice that if $\mathbf{r}_{\star}$ is an optimizer of
$\min_{\mathbf{r}\succ\mathbf{p}}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}}$
and
$\tilde{\varepsilon}=\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}_{\star}\right\rVert_{1}$,
then $\mathbf{r}_{\star}\in\mathfrak{B}^{\tilde{\varepsilon}}_{\mathbf{q}}$
and therefore
$\bar{\mathbf{q}}^{(\tilde{\varepsilon})}\succ\mathbf{r}_{\star}$. By
transitivity, we also have
$\bar{\mathbf{q}}^{(\tilde{\varepsilon})}\succ\mathbf{p}$, which implies that

$\displaystyle\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}\leq\tilde{\varepsilon}=\min_{\mathbf{r}\succ\mathbf{p}}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}}.$
(SM27)

For the reverse inequality, let $\varepsilon_{\star}$ be the optimizer of
$\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}$.
By definition, $\bar{\mathbf{q}}^{(\varepsilon_{\star})}\succ\mathbf{p}$ and
$\frac{1}{2}\left\lVert\mathbf{q}-\bar{\mathbf{q}}^{(\varepsilon_{\star})}\right\rVert_{1}\leq\varepsilon_{\star}$,
thus

$\displaystyle\min_{\mathbf{r}\succ\mathbf{p}}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}}\leq\varepsilon_{\star}=\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}.$
(SM28)

This shows that

$d_{\star}(\psi\to\varphi)=\min_{\mathbf{r}\in\operatorname{Prob}^{\downarrow}(d)}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}:\mathbf{r}\succ\mathbf{p}}=\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}.$
(SM29)

Our next step is to simplify the above optimization problems by considering
three different cases. We begin with the case $\mathbf{p}=\mathbf{e}_{1}$.
Then there is only one possible choice for $\mathbf{r}$, i.e.,
$\mathbf{r}=\mathbf{e}_{1}$, and

$\displaystyle\min_{\mathbf{r}\in\operatorname{Prob}^{\downarrow}(d)}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}:\mathbf{r}\succ\mathbf{p}}=\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1}=1-q_{1}=\max_{n\in[d]}\sum_{x=1}^{n}(p_{x}-q_{x})=\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{q}\right\rVert_{(n)}),$
(SM30)

where we remember our convention that both $\mathbf{p}$ and $\mathbf{q}$ are
ordered non-increasingly. Second, we consider $\mathbf{q}\succ\mathbf{p}$,
which is equivalent to
$\left\lVert\mathbf{q}\right\rVert_{(k)}\geq\left\lVert\mathbf{p}\right\rVert_{(k)}\forall
k\in[d]$, with equality for $k=d$. This implies that

$\displaystyle\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{q}\right\rVert_{(n)})=0.$
(SM31)

In addition, by choosing $\mathbf{r}=\mathbf{q}$, we find

$\displaystyle
0\leq\min_{\mathbf{r}\in\operatorname{Prob}^{\downarrow}(d)}\Set{\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{r}\right\rVert_{1}:\mathbf{r}\succ\mathbf{p}}\leq\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{q}\right\rVert_{1}=0.$
(SM32)

The last case we consider is $\mathbf{q}\nsucc\mathbf{p}\neq\mathbf{e}_{1}$:
Let $\delta,l$ be defined as

$\displaystyle\delta=\max_{n\in[d]}\sum_{x=1}^{n}(p_{x}-q_{x})=\sum_{x=1}^{l}(p_{x}-q_{x}).$
(SM33)

We begin by showing that
$\delta<\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1}$. To
this end, we note that $l<d$ and $q_{l+1}>0$, which we prove by contradiction.
Indeed, as argued in the second case, $l=d$ implies
$\mathbf{q}\succ\mathbf{p}$. Moreover, if $q_{l+1}$ were equal to zero, then

$\displaystyle\delta=\max_{n\in[d]}\sum_{x=1}^{n}(p_{x}-q_{x})\geq\sum_{x=1}^{l+1}(p_{x}-q_{x})=\sum_{x=1}^{l}(p_{x}-q_{x})+p_{l+1}=\delta+p_{l+1}$
(SM34)

and thus $p_{l+1}=0$. This would imply that $p_{n},q_{n}=0$ for all $n>l$, and
therefore

$\displaystyle\max_{n\in[d]}\sum_{x=1}^{n}(p_{x}-q_{x})=\sum_{x=1}^{l}(p_{x}-q_{x})=\sum_{x=1}^{d}(p_{x}-q_{x})=0,$
(SM35)

which would lead again to the contradiction $\mathbf{q}\succ\mathbf{p}$. If
$l=1$, and since $\mathbf{p}\neq\mathbf{e}_{1}$ by assumption,

$\displaystyle\delta=p_{1}-q_{1}<1-q_{1}=\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1}.$
(SM36)

If $l>1$, then

$\displaystyle\delta=\sum_{x=1}^{l}(p_{x}-q_{x})\leq
1-\sum_{x=1}^{l}q_{x}=1-q_{1}-\sum_{x=2}^{l}q_{x}<1-q_{1}=\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1},$
(SM37)

where we used that $q_{l}\geq q_{l+1}>0$.

Having established that in this third case,
$\delta<\frac{1}{2}\left\lVert\mathbf{q}-\mathbf{e}_{1}\right\rVert_{1}$, the
components of $\bar{\mathbf{q}}^{(\delta)}$ are given by Eq. (SM25). For
$m>k_{\delta}$ (as in Eq. (SM24)),
$\sum_{x=1}^{m}\bar{q}^{(\delta)}_{x}=1\geq\sum_{x=1}^{m}p_{x}$. In addition,
for $m\in[k_{\delta}]$,

$\sum_{x=1}^{m}\bar{q}^{(\delta)}_{x}=\delta+\sum_{x=1}^{m}q_{x}\geq\sum_{x=1}^{m}(p_{x}-q_{x})+\sum_{x=1}^{m}q_{x}=\sum_{x=1}^{m}p_{x}.$
(SM38)

These inequalities are again equivalent to
$\bar{\mathbf{q}}^{(\delta)}\succ\mathbf{p}$ and therefore

$\displaystyle\min\Set{\varepsilon\in[0,1]:\bar{\mathbf{q}}^{(\varepsilon)}\succ\mathbf{p}}\leq\delta.$
(SM39)

Next, we show that $\delta$ is not only an upper bound but the actual minimum:
For $\delta^{\prime}<\delta=\sum_{x=1}^{l}(p_{x}-q_{x})$, we note that $l\leq
k_{\delta^{\prime}}$: If this were not the case, we would have that
$\sum_{x=1}^{l}q_{x}\geq\sum_{x=1}^{k_{\delta^{\prime}}+1}q_{x}>1-\delta^{\prime}$,
and therefore
$\delta^{\prime}>1-\sum_{x=1}^{l}q_{x}\geq\sum_{x=1}^{l}p_{x}-\sum_{x=1}^{l}q_{x}=\delta$,
which is a contradiction. We thus find

$\sum_{x=1}^{l}\bar{q}^{(\delta^{\prime})}_{x}=\delta^{\prime}+\sum_{x=1}^{l}q_{x}<\sum_{x=1}^{l}(p_{x}-q_{x})+\sum_{x=1}^{l}q_{x}=\sum_{x=1}^{l}p_{x},$
(SM40)

which implies that $\bar{\mathbf{q}}^{(\delta^{\prime})}$ does not majorize
$\mathbf{p}$. Combining all three cases, we showed that

$d_{\star}(\psi\to\varphi)=\delta=\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{q}\right\rVert_{(n)}).$
(SM41)

To conclude the proof of the first part of the theorem, we observe that for
$n\geq\operatorname{SR}(\psi^{AB})$,
$\left\lVert\mathbf{p}\right\rVert_{(n)}=1$ and
$\left\lVert\mathbf{q}\right\rVert_{(n)}$ is non-decreasing with $n$, thus we
can restrict the maximization to $n\leq\operatorname{SR}(\psi^{AB})$.

Concerning the second part, due to the previous results, we note that for
$\xi$ separable and $\psi,\varphi$ not,

$\displaystyle d_{\star}(\xi\to\varphi)=1-q_{1}>0,$ (SM42)

because $q_{2}>0$ if $\varphi$ is entangled. If $\mathbf{q}\succ\mathbf{p}$,
$d_{\star}(\psi\to\varphi)=0$ (see the second case above), and if not, we are
in the third case, where we showed that
$d_{\star}(\psi\to\varphi)=\delta<1-q_{1}$, which finishes the proof. ∎

Next, we show that the star conversion distance satisfies a triangle
inequality. Let $\psi$, $\varphi$, and $\chi\in\operatorname{PURE}(AB)$ and
let $\mathbf{p}$, $\mathbf{q}$, and $\mathbf{r}$ be their Schmidt
coefficients. With the help of Eq. (SM41), this implies that

$\displaystyle d_{\star}(\psi\to\chi)$
$\displaystyle=\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{r}\right\rVert_{(n)})$
(SM43)
$\displaystyle=\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{q}\right\rVert_{(n)}+\left\lVert\mathbf{q}\right\rVert_{(n)}-\left\lVert\mathbf{r}\right\rVert_{(n)})$
$\displaystyle\leq\max_{n\in[d]}(\left\lVert\mathbf{p}\right\rVert_{(n)}-\left\lVert\mathbf{q}\right\rVert_{(n)})+\max_{n\in[d]}(\left\lVert\mathbf{q}\right\rVert_{(n)}-\left\lVert\mathbf{r}\right\rVert_{(n)})$
$\displaystyle=d_{\star}(\psi\to\varphi)+d_{\star}(\varphi\to\chi).$

_Remark._ The star conversion distance $d_{\star}(\psi\to\varphi)$ was so far
only defined for pure bipartite states belonging to the same Hilbert space.
This restriction is easily lifted by noting that one can always add separable
auxiliary states such that the Hilbert spaces (or dimensions) match. This can
be done in multiple ways: Let $\psi\in\operatorname{PURE}(AB)$,
$\varphi\in\operatorname{PURE}(A^{\prime}B^{\prime})$, and let $d=\left\lvert
A\right\rvert=\left\lvert B\right\rvert$, $d^{\prime}=\left\lvert
A^{\prime}\right\rvert=\left\lvert B^{\prime}\right\rvert$. Then

$\displaystyle d_{\star}(\psi_{AB}\to\varphi_{A^{\prime}B^{\prime}})\equiv
d_{\star}(\psi_{AB}\otimes|11\rangle\\!\langle
11|_{A^{\prime}B^{\prime}}\to|11\rangle\\!\langle
11|_{AB}\otimes\varphi_{A^{\prime}B^{\prime}}).$ (SM44)

Since in fact we are only interested in the dimension of systems and their
spatial separation, a more compact equivalent notation that we will use later
can be defined as follows: Denote with $m$, $m^{\prime}>0$ the smallest
integers such that $md=m^{\prime}d^{\prime}$, and with $A_{m}$, $B_{m}$,
$A_{m^{\prime}}$, $B_{m^{\prime}}$ Hilbert spaces of corresponding dimensions.
The conversion distance between
$d_{\star}(\psi_{AB}\to\varphi_{A^{\prime}B^{\prime}})$ is then

$d_{\star}(\psi_{AB}\to\varphi_{A^{\prime}B^{\prime}})\equiv
d_{\star}(\psi_{AB}\otimes|11\rangle\\!\langle
11|_{A_{m}B_{m}}\to\varphi_{A^{\prime}B^{\prime}}\otimes|11\rangle\\!\langle
11|_{A_{m^{\prime}}B_{m^{\prime}}}).$ (SM45)

##  IV Universal Embezzling Families

In the main text, we provided the following definition of universal embezzling
families, which we repeat here for readability.

######  Definition 3 (cf. Refs. [14, 15]).

A family of pure bipartite states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is called
a _universal embezzling family_ if
$\lim_{n\to\infty}d(\chi_{n}\to\chi_{n}\otimes\sigma)=0$ for every bipartite
finite dimensional state $\sigma$.

Thanks to the triangle inequality for the star conversion distance proven in
Eq. (SM43), a family of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is an
embezzling family if and only if it can embezzle the state $\Phi_{2}$, as
already shown in Ref [15, Lem. 2].

######  Lemma SM.4 (cf. Ref. [15, Lem. 2]).

Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a family of pure bipartite states.
The following three statements are equivalent

  1. 1.

$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family.

  2. 2.

$\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2})=0$.

  3. 3.

$\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=0$ for every
$m\geq 2$.

###### Proof.

Clearly 2. and 3. follow from 1. due to the definition of universal embezzling
families and the topological equivalence of $d$ and $d_{\star}$. Moreover, 3.
follows from 2. because
$\Phi_{2}^{\otimes\lceil\log_{2}m\rceil}\xrightarrow{\operatorname{LOCC}}\Phi_{m}$,
and therefore

$\displaystyle d_{\star}$ $\displaystyle(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
(SM46) $\displaystyle\leq
d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil})$
$\displaystyle\leq
d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil-1})+d_{\star}(\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil-1}\to\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil})$
$\displaystyle\leq
d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2})+d_{\star}(\chi_{n}\otimes\Phi_{2}\to\chi_{n}\otimes\Phi_{2}^{\otimes
2})+\dots+d_{\star}(\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil-1}\to\chi_{n}\otimes\Phi_{2}^{\otimes\lceil\log_{2}m\rceil})$
$\displaystyle\leq\lceil\log_{2}m\rceil
d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2}).$

By taking the limit $n\to\infty$ on both sides we obtain the desired result.
To conclude, we note that 3. implies 1., since for all $\sigma^{AB}$,
$\Phi_{|A|}\xrightarrow{\operatorname{LOCC}}\sigma^{AB}$. ∎

Next, we derive a formula for the conversion distance
$d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$ using Eq. (SM41) and Eq.
(SM45).

######  Lemma SM.5.

Let $\mathbf{p}^{(n)}\in\operatorname{Prob}^{\downarrow}(d_{n})$ be the
Schmidt coefficients of $\chi_{n}$. Then

$d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\},$
(SM47)

where $a_{k}=\lfloor k/m\rfloor$ and $b_{k}=k-a_{k}m$.

###### Proof.

The Schmidt coefficients of the input and target state are

$\displaystyle\mathbf{p}^{(n)}\otimes\mathbf{e}_{1}=(p_{1}^{(n)},\dots,p_{d_{n}}^{(n)},\underbrace{0,\dots,0}_{d_{n}\cdot(m-1)\,\text{times}}),$
(SM48)
$\displaystyle\mathbf{p}^{(n)}\otimes\mathbf{u}^{(m)}=\frac{1}{m}(\underbrace{p_{1}^{(n)},\dots,p_{1}^{(n)}}_{m\,\text{times}},\dots,\underbrace{p_{d_{n}}^{(n)},\dots,p^{(n)}_{d_{n}}}_{m\,\text{times}}),$

where $\mathbf{u}^{(m)}=(1/m,\dots,1/m)$. It is straightforward to see that

$\displaystyle\left\lVert\mathbf{p}^{(n)}\otimes\mathbf{e}_{1}\right\rVert_{(k)}=\begin{cases}\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}&\text{if
$k\in[d_{n}]$},\\\ 1&\text{if $d_{n}<k\leq d_{n}\cdot m$},\end{cases}$ (SM49)

and by writing $k=a_{k}m+b_{k}$, where $a_{k}=\lfloor k/m\rfloor$,

$\displaystyle\left\lVert\mathbf{p}^{(n)}\otimes\mathbf{u}^{(m)}\right\rVert_{(k)}=\sum_{x=1}^{a_{k}}p_{x}^{(n)}+b_{k}\frac{p^{(n)}_{a_{k}+1}}{m}=\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}+b_{k}\frac{p^{(n)}_{a_{k}+1}}{m}.$
(SM50)

Using the closed formula for the star conversion distance given in Theorem 2,
we obtain

$d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}.$
(SM51)

∎

A simplified version of this expression can be used to characterize embezzling
families.

######  Theorem 4.

A family of pure bipartite states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ with
corresponding Schmidt coefficients $\Set{\mathbf{p}^{(n)}}_{n\in\mathbb{N}}$
is a universal embezzling family if and only if

$\lim_{n\to\infty}\max_{l\in[A_{n}]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{{(2l-1)}}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{{(l-1)}}}=0,$
(SM52)

where $A_{n}=\lceil\operatorname{SR}(\chi_{n})/2\rceil$.

###### Proof.

Due to Lemma SM.4 and Eq. (SM47), $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a
universal embezzling family if and only if

$0=\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2})=\lim_{n\to\infty}\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{2}p^{(n)}_{a_{k}+1}\right\\},$
(SM53)

where $a_{k}=\lfloor k/2\rfloor$ and $k=2a_{k}+b_{k}$. First, we prove the
necessary condition. Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a universal
embezzling family. We observe that

$\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{2}p^{(n)}_{a_{k}+1}\right\\}\geq\left\lVert\mathbf{p}^{(n)}\right\rVert_{(1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{1})}-\frac{b_{1}}{2}p^{(n)}_{a_{1}+1}=p_{1}^{(n)}-0-\frac{1}{2}p_{1}^{(n)}>0.$
(SM54)

Taking the limit for $n\to\infty$ in the expression above, we obtain
$\lim_{n\to\infty}p_{1}^{(n)}=0$. Since $p_{1}^{(n)}$ is the largest Schmidt
coefficient, $p_{a_{k}+1}^{(n)}$ converges to zero too. Since
$0\leq\frac{b_{k}}{2}\leq\frac{1}{2}$,

$\displaystyle\lim_{n\to\infty}\frac{b_{k}}{2}p^{(n)}_{a_{k}+1}=0$ (SM55)

and thus

$\displaystyle 0$
$\displaystyle=\lim_{n\to\infty}\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{2}p^{(n)}_{a_{k}+1}\right\\}$
(SM56)
$\displaystyle=\lim_{n\to\infty}\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}\right\\}$
$\displaystyle\geq\lim_{n\to\infty}\max_{l\in[\lceil\operatorname{SR}(\chi_{n})/2\rceil]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l-1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l-1)}}\geq
0.$

For the sufficient condition, we observe that

$0=\lim_{n\to\infty}\max_{l\in[A_{n}]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{{(2l-1)}}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{{(l-1)}}}\geq\lim_{n\to\infty}p_{1}^{(n)},$
(SM57)

which implies that $\lim_{n\to\infty}p_{1}^{(n)}=0$. Furthermore,

$\displaystyle\lim_{n\to\infty}$ $\displaystyle
d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{2})=\lim_{n\to\infty}\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{2}p^{(n)}_{a_{k}+1}\right\\}$
(SM58)
$\displaystyle\leq\lim_{n\to\infty}\max_{k\in[\operatorname{SR}(\chi_{n})]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}\right\\}$
$\displaystyle=\lim_{n\to\infty}\max\Set{\max_{l\in[\lfloor\operatorname{SR}(\chi_{n})/2\rfloor]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l)}},\max_{l\in\Set{0,\dots,\lfloor\frac{\operatorname{SR}(\chi_{n})-1}{2}\rfloor}}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l+1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l)}}}$
$\displaystyle=\lim_{n\to\infty}\max\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2\lfloor\operatorname{SR}(\chi_{n})/2\rfloor)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(\lfloor\operatorname{SR}(\chi_{n})/2\rfloor)},\max_{l\in\Set{0,\dots,\lfloor\frac{\operatorname{SR}(\chi_{n})-1}{2}\rfloor}}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l+1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l)}}}$
$\displaystyle=\lim_{n\to\infty}\max_{l\in\Set{0,\dots,\lfloor\frac{\operatorname{SR}(\chi_{n})-1}{2}\rfloor}}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l+1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l)}}$
$\displaystyle=\lim_{n\to\infty}\max_{l\in\lceil\operatorname{SR}(\chi_{n})/2\rceil}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2l-1)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(l-1)}}=0,$

where the equality between line 4 and line 5 in the equation above is
justified by the fact that

$\displaystyle\left\lvert\left(\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2\lfloor\operatorname{SR}(\chi_{n})/2\rfloor)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(\lfloor\operatorname{SR}(\chi_{n})/2\rfloor)}\right)-\left(\left\lVert\mathbf{p}^{(n)}\right\rVert_{(2\lfloor(\operatorname{SR}(\chi_{n})-1)/2\rfloor)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(\lfloor(\operatorname{SR}(\chi_{n})-1)/2\rfloor)}\right)\right\rvert$
(SM59) $\displaystyle\qquad\leq
p^{(n)}_{2\lfloor\operatorname{SR}(\chi_{n})/2\rfloor}+p^{(n)}_{\lfloor(\operatorname{SR}(\chi_{n})-1)/2\rfloor}$
$\displaystyle\qquad\leq 2p^{(n)}_{1},$

which vanishes in the limit $n\to\infty$. Eq. (SM58) shows that
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is an embezzling family and concludes the
proof. ∎

An important and easy to check necessary condition for a universal embezzling
family is given in the following Corollary.

######  Corollary SM.6 (cf. Ref. [15, Lem. 3]).

If a family of pure bipartite states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a
universal embezzling family, then $\lim_{n\to\infty}p_{1}^{(n)}=0$, where
$\mathbf{p}^{(n)}\in\operatorname{Prob}^{\downarrow}(d_{n})$ are the Schmidt
coefficients of $\chi_{n}$.

##  V Regular Embezzling Families of States

Often, families of states are defined in terms of a positive, monotonic, and
continuous function. This motivates the following definition (compare to Ref.
[15]).

######  Definition SM.7.

A family of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a regular family if
there exists a monotonic function $f:\mathbb{N}\to(0,\infty)$ such that

$\ket{\chi_{n}}=\frac{1}{\sqrt{F_{n}}}\sum_{x=1}^{n}\sqrt{f(x)}\Ket{xx}\quad\text{for
all}\quad n\in\mathbb{N},$ (SM60)

where $F_{n}=\sum_{x=1}^{n}f(x)$.

_Remark._ If $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a regular family, then there
exists a sequence $\Set{n_{j}}_{j\in\mathbb{N}}$ such that the family
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is an embezzling family if and only if
$\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=0$.
However, according to our definition, the family
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is then no longer regular, unless
$n_{j}=j$ for all $j\in\mathbb{N}$.

For the following proofs, it is beneficial to extend $f$ to a monotonic
continuous function on $[1,\infty)$. This can often be done trivially by
simply extending the domain of $f$. An example is the van Dam and Hayden
family where $f(x)=x^{-1}$. Otherwise, we can extend $f$ by connecting two
consecutive points with straight lines. If the function $f$ is multiplied by a
constant factor, this does not change the corresponding family of states.
Therefore, from now on we assume for simplicity that $f(1)=1$.

Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a regular family of states and let
$f$ be the function associated to it. We define the non-increasing functions
$f_{n}$, $n\in\mathbb{N}$, by

$f_{n}(x)=\begin{cases}f(x)&\text{if $f$ is non-increasing,}\\\
\frac{f(n+1-x)}{f(n)}&\text{if $f$ is non-decreasing.}\end{cases}$ (SM61)

By our extension of $f$, $f_{n}$ is also naturally extended to a continuous
function $g(x,y)$ such that $g(x,n)=f_{n}(x)$ via the definition

$g(x,y)=\begin{cases}f(x)&\text{if $f$ is non-increasing,}\\\
\frac{f(y+1-x)}{f(y)}&\text{if $f$ is non-decreasing.}\end{cases}$ (SM62)

The (by definition non-increasing) Schmidt coefficients of $\chi_{n}$ are
therefore given by

$p^{(n)}_{x}=\frac{f_{n}(x)}{F_{n}}.$ (SM63)

Since $p_{1}^{(n)}=\frac{1}{F_{n}}$, we can restate Corollary SM.6 for regular
families.

######  Corollary SM.8 (cf. Ref. [15, Lem. 3]).

If $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a regular universal embezzling family,
then $\lim_{n\to\infty}F_{n}=+\infty$, where $F_{n}$ is defined in Definition
SM.7.

In the following proposition, we present bounds on the limit of the conversion
distance $d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$ in terms of the
function $g(x,y)$ introduced in Eq. (SM62), which will be of use later.

######  Proposition SM.9.

Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a regular family of states, $f$ be
the function associated to it, and $g$ be defined as in Eq. (SM62). If
$\lim_{n\to\infty}F_{n}=\infty$, then

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\geq\liminf_{y\to\infty}\frac{\max_{1\leq
a\leq y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x},$ (SM64)
$\displaystyle\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\leq\limsup_{y\to\infty}\frac{\max_{1\leq
a\leq y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM65)

If $\lim_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ exists, this implies that

$\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\lim_{y\to\infty}\frac{\max_{1\leq
a\leq y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM66)

###### Proof.

For better readability, we divide the proof into steps and use the notation
$G(x,y)=\int_{1}^{x}g(t,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!t$.

_Step 1:_ Starting from Eq. (SM47), we find

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=$
$\displaystyle\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=$
$\displaystyle\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k}m)}+\sum_{x=a_{k}m+1}^{k}p_{x}^{(n)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=$
$\displaystyle\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k}m)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}+\frac{\sum_{x=a_{k}m+1}^{a_{k}m+b_{k}}f_{n}(x)-\frac{b_{k}}{m}f_{n}(a_{k}+1)}{F_{n}}-\right\\}$
$\displaystyle=$
$\displaystyle\liminf_{n\to\infty}\max_{a\in\Set{0,\dots,\lfloor
n/m\rfloor}}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(am)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a)}}$
$\displaystyle=$ $\displaystyle\liminf_{n\to\infty}\max_{a\in[\lfloor
n/m\rfloor]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(am)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a)}},$
(SM67)

where in the last two lines we used that
$\left\lVert\mathbf{p}^{(n)}\right\rVert_{(0)}=0$ and that by construction
$f_{n}(x)\leq 1$: Therefore

$\displaystyle
0\leq\frac{\sum_{x=a_{k}m+1}^{a_{k}m+b_{k}}f_{n}(x)}{F_{n}}\leq\frac{m}{F_{n}},$
$\displaystyle
0\leq\frac{\frac{b_{k}}{m}f_{n}(a_{k}+1)}{F_{n}}\leq\frac{1}{F_{n}}$ (SM68)

and by assumption the right-hand sides converge to zero in the limit
$n\to\infty$. This ensures that the limit inferior is additive.

_Step 2:_ Since $f_{n}(x)$ is by construction non-increasing,

$\displaystyle\int_{1}^{n}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\leq\int_{1}^{n}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x+f_{n}(n)\leq\sum_{x=1}^{n}f_{n}(x)\leq
f_{n}(1)+\int_{1}^{n}f_{n}(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x,$
(SM69)

from which follows

$0\leq\sum_{x=1}^{n}f_{n}(x)-\int_{1}^{n}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\leq f_{n}(1)=1.$ (SM70)

Dividing by $F_{n}=\sum_{x=1}^{n}f_{n}(x)$ and taking the limit $n\to\infty$,
we obtain
$\lim_{n\to\infty}1-\frac{\int_{1}^{n}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\sum_{x=1}^{n}f_{n}(x)}=0$. Since $f_{n}(x)=g(x,n)$, the
limit can be rewritten as $\lim_{n\to\infty}\frac{G(n,n)}{F_{n}}=1$.

_Step 3:_ Analogously to the last step, it holds that

$\frac{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{F_{n}}\leq\frac{\sum_{x=a+1}^{am}f_{n}(x)}{F_{n}}\leq\frac{1+\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{F_{n}},$ (SM71)

and after taking the maximum over $a\in[\lfloor n/m\rfloor]$ and the limit
inferior that

$\displaystyle\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}$
$\displaystyle\leq\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\sum_{x=a+1}^{am}f_{n}(x)}}{F_{n}}$ (SM72)
$\displaystyle\leq\liminf_{n\to\infty}\frac{1+\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}$
$\displaystyle=\liminf_{n\to\infty}\left(\frac{1}{F_{n}}+\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}\right)$
$\displaystyle=\liminf_{n\to\infty}\left(\frac{1}{F_{n}}\right)+\liminf_{n\to\infty}\left(\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}\right)$
$\displaystyle=\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}.$

As a consequence, we have

$\displaystyle\liminf_{n\to\infty}\max_{a\in[\lfloor
n/m\rfloor]}\Set{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(am)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a)}}=$
$\displaystyle\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\sum_{x=a+1}^{am}f_{n}(x)}}{F_{n}}$ $\displaystyle=$
$\displaystyle\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{\int_{a+1}^{am}f_{n}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{F_{n}}$ $\displaystyle=$
$\displaystyle\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{G(am,n)-G(a+1,n)}}{G(n,n)},$ (SM73)

where in the last line we replaced $F_{n}$ by $G(n,n)$, which we can again do
according to step 2.

_Step 4:_ Let $c_{n}$ be the value that maximizes $\max_{1\leq a\leq
n/m}\Set{G(am,n)-G(a+1,n)}$ and let $c_{n}^{\prime}$ be the natural number
satisfying $c_{n}^{\prime}\leq c_{n}<c_{n}^{\prime}+1$. This implies

$G(c_{n}m,n)-G(c_{n}+1,n)=\int_{c_{n}+1}^{c_{n}m}g(t,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!t\leq\int_{c_{n}^{\prime}+1}^{c_{n}^{\prime}m}g(t,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!t+\int_{c_{n}^{\prime}m}^{(c_{n}^{\prime}+1)m}g(t,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!t\leq\int_{c_{n}^{\prime}+1}^{c_{n}^{\prime}m}g(t,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!t+m.$ (SM74)

Dividing by $G(n,n)$, taking the limit inferior, and arguing as in the
previous step, we obtain

$\displaystyle\liminf_{n\to\infty}\frac{\max_{1\leq a\leq
n/m}\Set{G(am,n)-G(a+1,n)}}{G(n,n)}$
$\displaystyle=\liminf_{n\to\infty}\frac{G(c_{n}m,n)-G(c_{n}+1,n)}{G(n,n)}$
(SM75)
$\displaystyle\leq\liminf_{n\to\infty}\frac{G(c_{n}^{\prime}m,n)-G(c_{n}^{\prime}+1,n)+m}{G(n,n)}$
$\displaystyle=\liminf_{n\to\infty}\frac{G(c_{n}^{\prime}m,n)-G(c_{n}^{\prime}+1,n)}{G(n,n)}$
$\displaystyle\leq\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{G(am,n)-G(a+1,n)}}{G(n,n)}.$

The reverse inequality follows from the fact that we maximize over a larger
set. This proves that

$\liminf_{n\to\infty}\frac{\max_{a\in[\lfloor
n/m\rfloor]}\Set{G(am,n)-G(a+1,n)}}{G(n,n)}=\liminf_{n\to\infty}\frac{\max_{1\leq
a\leq n/m}\Set{G(am,n)-G(a+1,n)}}{G(n,n)}.$ (SM76)

_Step 5:_ Combining the results of the previous steps, and in particular Eq.
(V), Eq. (V), and Eq. (SM76), we have proven that

$\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\liminf_{n\to\infty}\frac{\max_{1\leq
a\leq n/m}\Set{\int_{a+1}^{am}g(x,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{n}g(x,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM77)

In the above equation, we can replace
$\int_{a+1}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ with
$\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ because the
difference of the two integrals is finite and divided by a diverging term.
Thus

$\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\liminf_{n\to\infty}\frac{\max_{1\leq
a\leq n/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{n}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM78)

The expression inside the $\liminf$ on the right-hand side of Eq. (SM78) is a
function of $n\in\mathbb{N}\setminus\Set{1}$ and we call it $M(n)$,

$M(n)=\frac{\max_{1\leq a\leq
n/m}\Set{\int_{a}^{am}g(x,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{n}g(x,n)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM79)

We extend this function to real numbers $y>1$ as follows:

$M(y)=\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM80)

Since the natural numbers are a subset of the real numbers,

$\liminf_{n\to\infty}M(n)\geq\liminf_{y\to\infty}M(y),$ (SM81)

which proves Eq. (SM64). Furthermore, Steps 1-4 can be repeated with exactly
the same arguments for the $\limsup$, and they imply

$\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\limsup_{n\to\infty}M(n)\leq\limsup_{y\to\infty}M(y).$
(SM82)

This proves Eq. (SM65).

In the following, we assume that $\lim_{y\to\infty}M(y)$ exists. This implies

$\lim_{y\to\infty}M(y)=\liminf_{y\to\infty}M(y)=\limsup_{y\to\infty}M(y).$
(SM83)

From Eq. (SM81), Eq. (SM82), and Eq. (SM83) we obtain

$\lim_{y\to\infty}M(y)\leq\liminf_{n\to\infty}M(n)\leq\limsup_{n\to\infty}M(n)\leq\lim_{y\to\infty}M(y).$
(SM84)

Therefore, $\lim_{n\to\infty}M(n)$ exists and is equal to
$\lim_{y\to\infty}M(y)$. From Eq. (SM78) and Eq. (SM82) we derive

$\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=\lim_{n\to\infty}M(n)=\lim_{y\to\infty}M(y).$
(SM85)

This proves Eq. (SM66) and concludes the proof.

∎

Corollary SM.8 combined with Proposition SM.9 leads to the following
characterization of embezzling families.

######  Corollary SM.10.

Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a regular family of states, $f$ be
the function associated to it, and $g$ be defined as in Eq. (SM62). If

$\displaystyle\lim_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ (SM86)

exists, then the family of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is an
universal embezzling family if and only if $\lim_{n\to\infty}F_{n}=\infty$ and

$\lim_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}=0.$ (SM87)

In addition, thanks to Proposition SM.9, the task of determining if a family
of states is an embezzling family is converted into an optimization problem.
It is enough to find the maximum of the differentiable function
$G_{y}(a)=\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ on
$1\leq a\leq y/m$, which is easily done with the help of its derivative

$mg(am,y)-g(a,y),\quad 1\leq a\leq y/m.$ (SM88)

##  VI Generalization of the van Dam and Hayden Family

The universal embezzling family $\Set{\chi_{n}}_{n\in\mathbb{N}}$ introduced
by van Dam and Hayden consists of the states

$\Ket{\chi_{n}}=\frac{1}{\sqrt{H_{n}}}\sum_{x=1}^{n}\sqrt{x^{-1}}\Ket{xx},$
(SM89)

where $H_{n}=\sum_{x=1}^{n}x^{-1}$ is the $n$-th harmonic number. We
generalize this family of states as follows: For every $\alpha\in\mathbb{R}$
we introduce the family $\Set{\chi^{(\alpha)}_{n}}_{n\in\mathbb{N}}$, where

$\Ket{\chi^{(\alpha)}_{n}}=\frac{1}{\sqrt{H^{(-\alpha)}_{n}}}\sum_{x=1}^{n}\sqrt{x^{\alpha}}\Ket{xx}$
(SM90)

and $H^{(-\alpha)}_{n}=\sum_{x=1}^{n}x^{\alpha}$ is the $n$-th generalized
harmonic number. Note that the families of states
$\Set{\chi_{n}^{(\alpha)}}_{n\in\mathbb{N}}$ are regular families of states,
and the van Dam and Hayden family is recovered for $\alpha=-1$. The
corresponding Schmidt coefficients, as per our convention arranged in non-
increasing order, are

$\mathbf{p}^{(n|\alpha)}=\begin{cases}\frac{1}{H^{(-\alpha)}_{n}}(1,2^{\alpha},\dots,n^{\alpha})&\text{if
$\alpha\leq 0$},\\\
\frac{1}{H^{(-\alpha)}_{n}}(n^{\alpha},\dots,2^{\alpha},1)&\text{if
$\alpha>0$}.\end{cases}$ (SM91)

In the remaining part of this section, we show that the family of state
$\Set{\chi^{(\alpha)}_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family
if and only if $\alpha=-1$ and derive bounds on the star conversion distance.

### Case $\alpha<-1$.

For $\alpha<-1$, $\lim_{n\to\infty}H^{(-\alpha)}_{n}$ is finite, thus the
regular family of states $\Set{\chi^{\alpha}_{n}}_{n\in\mathbb{N}}$ is not a
universal embezzling family (see Corollary SM.8). The largest Schmidt
coefficient provides a lower bound on the star conversion distance: Using Eq.
(SM41), we obtain

$d_{\star}(\chi^{(\alpha)}_{n}\to\chi^{(\alpha)}_{n}\otimes\Phi_{m})=\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n|\alpha)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n|\alpha)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n|\alpha)}_{a_{k}+1}\right\\}\geq
p_{1}^{(n|\alpha)}\left(1-\frac{1}{m}\right)=\frac{1}{H^{(-\alpha)}_{n}}\left(1-\frac{1}{m}\right),$
(SM92)

where the inequality followed from choosing $k=1$. By taking the limit, we
obtain

$\liminf_{n\to\infty}d_{\star}(\chi^{(\alpha)}_{n}\to\chi^{(\alpha)}_{n}\otimes\Phi_{m})\geq\liminf_{n\to\infty}\frac{1}{H^{(-\alpha)}_{n}}\left(1-\frac{1}{m}\right)=\frac{1}{\zeta(-\alpha)}\left(1-\frac{1}{m}\right),$
(SM93)

where $\zeta(-\alpha)=\sum_{x=1}^{\infty}x^{\alpha}$ is the Riemann Zeta
function. To obtain an upper bound, we observe that, since $\alpha<-1$,

$\sum_{x=a+1}^{(a+1)m}x^{\alpha}\leq(a+1)^{\alpha}+\int_{a+1}^{(a+1)m}x^{\alpha}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x=(a+1)^{\alpha}+\frac{(a+1)^{\alpha+1}}{\alpha+1}\left(m^{\alpha+1}-1\right).$
(SM94)

The right-hand side is decreasing in $a$, and thus

$\displaystyle\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n|\alpha)}\right\rVert_{{(k)}}-\left\lVert\mathbf{p}^{(n|\alpha)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p_{a_{k}+1}^{(n|\alpha)}\right\\}\leq\frac{1}{H^{(-\alpha)}_{n}}\max_{k\in[n]}\Set{\sum_{x=a_{k}+1}^{k}x^{\alpha}}$
(SM95) $\displaystyle\leq\frac{1}{H^{(-\alpha)}_{n}}\max_{a\in\\{0,...,\lfloor
n/m\rfloor\\}}\Set{\sum_{x=a+1}^{(a+1)m}x^{\alpha}}\leq\frac{1}{H^{(-\alpha)}_{n}}\max_{a\in\\{0,...,\lfloor
n/m\rfloor\\}}\Set{(a+1)^{\alpha}+\frac{(a+1)^{\alpha+1}}{\alpha+1}\left(m^{\alpha+1}-1\right)}$
$\displaystyle=\frac{1}{H^{(-\alpha)}_{n}}\left(1+\frac{m^{\alpha+1}-1}{\alpha+1}\right).$

Taking the limit $n\to\infty$, and considering that the conversion distance is
by definition smaller than one, we obtain

$\limsup_{n\to\infty}d_{\star}(\chi^{(\alpha)}_{n}\to\chi^{(\alpha)}_{n}\otimes\Phi_{m})\leq\min\Set{1,\frac{1}{\zeta(-\alpha)}\left(1+\frac{m^{\alpha+1}-1}{\alpha+1}\right)}.$
(SM96)

### Case $\alpha=-1$.

This is the van Dam and Hayden family. Analogously to the previous case, we
observe that

$\sum_{x=a+1}^{(a+1)m}x^{-1}\leq(a+1)^{-1}+\int_{a+1}^{(a+1)m}x^{-1}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x=(a+1)^{-1}+\ln m\leq 1+\ln m.$ (SM97)

Since $H_{n}^{(1)}=H_{n}$ diverges, by following the same steps, we have

$\lim_{n\to\infty}d_{\star}(\chi^{(-1)}_{n}\to\chi^{(-1)}_{n}\otimes\Phi_{m})\leq\lim_{n\to\infty}\frac{1+\ln
m}{H_{n}}=0.$ (SM98)

Thus, the limit for the conversion distance is zero as expected for the van
Dam and Hayden family.

### Case $-1<\alpha<0$.

In this scenario $H^{(-\alpha)}_{n}$ diverges when $n\to\infty$, and we can
use the results of Section V. Here $g(x,y)=x^{\alpha}$, and the derivative
with respect to $a$ of the function
$\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ is

$mg(am,y)-g(a,y)=a^{\alpha}(m^{\alpha+1}-1),$ (SM99)

which is positive in the domain $1\leq a\leq y/m$. This implies that the
function $\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ is
non-decreasing and the maximum

$\displaystyle\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x}$
(SM100)

is obtained for $a=y/m$. Due to Proposition SM.9,

$\lim_{n\to\infty}d_{\star}(\chi^{(\alpha)}_{n}\to\chi^{(\alpha)}_{n}\otimes\Phi_{m})=\lim_{y\to\infty}\frac{\int_{y/m}^{y}x^{\alpha}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}x^{\alpha}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}=1-\frac{1}{m^{\alpha+1}}.$ (SM101)

### Case $\alpha=0$.

Here we have

$p_{x}^{(n)}=\frac{1}{n}\,,\quad(\mathbf{p}^{(n)}\otimes\mathbf{u}^{(m)})_{x}=\frac{1}{nm}.$
(SM102)

We can compute the star conversion distance directly using Eq. (SM41) and
obtain

$d_{\star}(\chi^{(0)}_{n}\to\chi^{(0)}_{n}\otimes\Phi_{m})=\max_{k\in[n]}\Set{\frac{k}{n}-\frac{k}{nm}}=1-\frac{1}{m}.$
(SM103)

### Case $\alpha>0$.

Also here $H^{(-\alpha)}_{n}$ diverges for $n\to\infty$ and we can use the
results of Section V. The function $g(x,y)$ is defined as
$g(x,y)=(y+1-x)^{\alpha}/y^{\alpha}$, because $x^{\alpha}$ is increasing. From
this follows

$mg(am,y)-g(a,y)\geq 0\Leftrightarrow
m(y+1-am)^{\alpha}\geq(y+1-a)^{\alpha}\Leftrightarrow
a\leq\frac{(y+1)\left(m^{\frac{1}{\alpha}}-1\right)}{m^{1+\frac{1}{\alpha}}-1}\coloneqq
a_{\max}.$ (SM104)

Since $a_{\max}$ increases linearly with $y$ and $a_{\max}\leq\frac{y}{m}$ is
equivalent to $y\geq\frac{m\left(m^{1/\alpha}-1\right)}{m-1}$, for large
enough $y$, the value $a_{\max}$ belongs to the interval $[1,y/m]$ and is a
global maximum. Plugging $a_{\max}$ into Proposition SM.9, we obtain

$\lim_{n\to\infty}d_{\star}(\chi^{(\alpha)}_{n}\to\chi^{(\alpha)}_{n}\otimes\Phi_{m})=\lim_{y\to\infty}\frac{\int_{a_{\max}}^{a_{\max}m}\frac{(y+1-x)^{\alpha}}{y^{\alpha}}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}\frac{(y+1-x)^{\alpha}}{y^{\alpha}}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}=(m-1)\left(\frac{m-1}{m^{1+\frac{1}{\alpha}}-1}\right)^{\alpha}.$
(SM105)

This completes the study of regular families of states defined by
$f(x)=x^{\alpha}$. The star conversion distance vanishes only for $\alpha=-1$,
thus the only universal embezzling family of this form is the one introduced
by van Dam and Hayden. Furthermore, exact values for the limit of the star
conversion distance for $\alpha\geq-1$ and lower and upper bounds for
$\alpha<-1$ were provided.

##  VII Uniqueness of the van Dam and Hayden Embezzling Family

In this section, we provide further results on the uniqueness of the van Dam
and Hayden embezzling family. Given a regular family of states
$\Set{\chi_{n}}_{n\in\mathbb{N}}$, the asymptotic behavior of the function $f$
associated to it is relevant to determine whether
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family or not
(e.g., see Corollary SM.8). We thus introduce the following notation to
describe the asymptotic behaviour of a function $h:[1,\infty)\to(0,\infty)$,

$\displaystyle h(x)\nearrow\,\Leftrightarrow\,\exists
N\,\text{s.t.}\,h(x)\,\text{is non-decreasing on}\,(N,+\infty)\,,$
$\displaystyle h(x)\searrow\,\Leftrightarrow\,\exists
N\,\text{s.t.}\,h(x)\,\text{is non-increasing on}\,(N,+\infty).$ (SM106)

We say that $h$ is _asymptotically monotonic_ when it satisfies $h\nearrow$ or
$h\searrow$. Furthermore, we use the notations little-$\omega$ and little-$o$
(see, e.g., Ref. [16]) to describe asymptotic relations between two functions
$g,h:[1,\infty)\to(0,\infty)$,

$h\in\omega(g)\Leftrightarrow\lim_{x\to\infty}\frac{h(x)}{g(x)}=+\infty\,,\quad
h\in o(g)\Leftrightarrow\lim_{x\to\infty}\frac{h(x)}{g(x)}=0.$ (SM107)

Before we state the results about the uniqueness of the van Dam and Hayden
embezzling family, we prove the following Lemma, which is a direct consequence
of Theorem 2 in Ref. [17].

######  Lemma SM.11.

Let $f$, $g$ be continuous, positive functions on $[a,b]$ such that
$f(x)/g(x)$ is a non-decreasing function on $[a,b]$. Then, for every $x_{1}$
such that $a<x_{1}<b$,

$\int_{a}^{b}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{x_{1}}g(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{a}^{x_{1}}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{b}g(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x.$
(SM108)

###### Proof.

Let $\tilde{h}(x)=1$ and $h(x)=\chi_{[a,x_{1}]}(x)$, where $\chi_{S}(x)$ is
the characteristic function of the set $S$. Furthermore, let

$h_{k}(x)=\begin{cases}1&\text{if $a\leq x\leq x_{1}$},\\\
e^{-k(x-x_{1})}&\text{if $x_{1}<x\leq b$}.\end{cases}$ (SM109)

The sequence of functions $\Set{h_{k}}_{k\in\mathbb{N}}$ converges to $h(x)$
and satisfies $0\leq h_{k}(x)\leq 1$. Due to the dominated convergence theorem
(see, e.g., Ref. [18]), we have

$\lim_{k\to\infty}\int_{a}^{b}f(x)h_{k}(x)=\int_{a}^{x_{1}}f(x).$ (SM110)

The same result holds for $g$.

Since both $f(x)/g(x)$ and $\tilde{h}(x)/h_{k}(x)$ are non-decreasing, we can
apply Theorem 2 of Ref. [17] to find

$\int_{a}^{b}f(x)\tilde{h}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{b}g(x)h_{k}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{a}^{b}f(x)h_{k}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{b}g(x)\tilde{h}(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x.$ (SM111)

After taking the limit $k\to\infty$ on both sides, we obtain

$\int_{a}^{b}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{x_{1}}g(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{a}^{x_{1}}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{a}^{b}g(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x,$
(SM112)

which finishes the proof. ∎

With this lemma at hand, we are ready to present the promised results
concerning the uniqueness of the van Dam and Hayden embezzling family.

######  Theorem SM.12.

Let $f$ be a positive non-increasing function such that $f(x)/x^{\alpha}$ is
asymptotically monotonic for all $\alpha\in\mathbb{R}$ and let
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ be the regular family of states associated
to it (see Definition SM.7). Then $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a
universal embezzling family if and only if $f\in\omega(x^{-1-\varepsilon})\cap
o(x^{-1+\varepsilon})\ \forall\varepsilon>0$ and
$\sum_{x=1}^{\infty}f(x)=\infty$. Furthermore, if
$f\notin\omega(x^{-1-\varepsilon})\cap o(x^{-1+\varepsilon})$ for at least one
$\varepsilon>0$, then $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$, where
$\Set{n_{j}}_{j\in\mathbb{N}}$ is any sequence of natural numbers, is not a
universal embezzling family.

###### Proof.

Before we start, we notice that if $\lim_{j\to\infty}n_{j}=J<\infty$, then
$\lim_{j\to\infty}F_{n_{j}}=\sum_{x=1}^{J}f(x)<\infty$, which implies that the
family of states $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is not an embezzling
family (see Corollary SM.8). In this proof we will thus assume, w.l.o.g., that
$\lim_{j\to\infty}n_{j}=\infty$.

_Necessary condition —_ Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be a universal
embezzling family with corresponding function $f$ satisfying the assumptions
above. We already proved in Corollary SM.8 that if
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family, then
$\sum_{x=1}^{\infty}f(x)=+\infty$. To show the remainder, let

$R=\Set{\alpha\in\mathbb{R}}{\lim_{x\to\infty}\frac{f(x)}{x^{\alpha}}=0}\,,\quad
L=\Set{\alpha\in\mathbb{R}}{\lim_{x\to\infty}\frac{f(x)}{x^{\alpha}}=+\infty}.$
(SM113)

Since $\lim_{x\to\infty}f(x)/x^{\alpha}=0$ for $\alpha>0$, we have that
$R\neq\emptyset$. Now we prove by contradiction that also $L\neq\emptyset$.
Let us assume that $L$ is empty. This implies that
$\lim_{x\to\infty}f(x)/x^{\alpha}=l_{\alpha}<\infty$ for all $\alpha$. If
there exists an $\alpha$ such that $l_{\alpha}\neq 0$, then
$\lim_{x\to\infty}f(x)/x^{\alpha-1}=l_{\alpha}(\lim_{x\to\infty}x)=+\infty$.
Thus, $\alpha-1\in L$, and $L\neq\emptyset$, leading to the desired
contradiction. If instead $\lim_{x\to\infty}f(x)/x^{\alpha}=0$ for all
$\alpha$, then $\lim_{x\to\infty}f(x)/x^{-2}=0$. This implies that
$f(x)/x^{-2}$ converges monotonically to zero for large $x$, i.e., there
exists an $N$ such that $f(x)<x^{-2}$ for $x>N$. Thus,
$\sum_{x=N}^{\infty}f(x)<\sum_{x=N}^{\infty}x^{-2}<\infty$. According to
Corollary SM.8, this contradicts the hypothesis that the family under
consideration is universally embezzling. We have therefore shown that
$L\neq\emptyset\neq R$.

The next step is to prove that $\inf R=\sup L$. From the definition of $R$ and
$L$ follows that $\inf R\geq\sup L$. Let us assume that $\inf R>\sup L$, i.e.,
that there exists an $\alpha\in\mathbb{R}$ such that $\sup L<\alpha<\inf R$.
Since $\alpha\notin L\cup R$, there exists a positive real number $l$ such
that $\lim_{x\to\infty}f(x)/x^{\alpha}=l$. Now pick $\alpha_{1}$ such that
$\sup L<\alpha_{1}<\alpha<\inf R$. We observe that
$\lim_{x\to\infty}f(x)/x^{\alpha_{1}}=l(\lim_{x\to\infty}x^{\alpha}/x^{\alpha_{1}})=+\infty$,
thus $\alpha_{1}\in L$. This is in contradiction to the choice
$\alpha_{1}>\sup L$. We therefore showed that $\sup L=\inf R$.

So far, we have shown that if $\Set{\chi_{n}}$ is a universal embezzling
family satisfying our assumptions, then there exists a unique $\alpha$ such
that $f(x)\in o(x^{\alpha+\varepsilon})\cap\omega(x^{\alpha-\varepsilon})$ for
every $\varepsilon>0$. What is left to show is that if $\alpha\neq-1$, then
the family of states corresponding to $f$ cannot be embezzling. From the above
discussion, we know that $\Set{\alpha>0}\subseteq R$, thus we can focus on
$\alpha\leq 0$ and split our discussion into two scenarios, $\alpha<-1$ and
$-1<\alpha\leq 0$.

If $\alpha<-1$, then there exists an $\varepsilon>0$ such that
$\alpha+\varepsilon<-1$. Since $f\in o(x^{\alpha+\varepsilon})$ and
$f(x)/x^{\alpha+\varepsilon}$ converges monotonically to zero for large $x$,
there exists an $N$ such that $f(x)<x^{\alpha+\varepsilon}$ for $x>N$. This
implies that
$\sum_{x=N}^{\infty}f(x)<\sum_{x=N}^{\infty}x^{\alpha+\varepsilon}<\infty$ and
therefore $\sum_{x=1}^{\infty}f(x)<\infty$, which, according to Corollary
SM.8, contradicts the hypothesis that $\Set{\chi_{n}}$ is a universal
embezzling family. Let $\Set{n_{j}}_{j\in\mathbb{N}}$ be any sequence of
natural number such that $\lim_{j\to\infty}n_{j}=\infty$. Then
$\lim_{j\to\infty}\sum_{x=1}^{n_{j}}f(x)=\sum_{x=1}^{\infty}f(x)<\infty$. This
implies, again due to Corollary SM.8, that
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is not a universal embezzling family.

If $-1<\alpha\leq 0$, then there exists an $\varepsilon>0$ such that
$\alpha-\varepsilon>-1$. We notice that $f(x)/x^{\alpha-\varepsilon}$ diverges
to infinity for large $x$, because $f\in\omega(x^{\alpha-\varepsilon})$. This
implies that there exists an $N$ such that $f(x)\geq x^{\alpha-\varepsilon}$
for $x>N$ and therefore
$\sum_{x=N}^{\infty}f(x)\geq\sum_{x=N}^{\infty}x^{\alpha-\varepsilon}=\infty$,
which allows us to use the results of Section V. Since
$f(x)/x^{\alpha-\varepsilon}$ is non-decreasing for $x>N$, for all $y$ such
that $y/m>N$, we can use Lemma SM.11 to obtain

$\int_{N}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{N}^{y/m}x^{\alpha-\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{N}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{N}^{y}x^{\alpha-\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x.$ (SM114)

This implies that

$\frac{(y/m)^{\alpha-\varepsilon+1}-N^{\alpha-\varepsilon+1}}{(y)^{\alpha-\varepsilon+1}-N^{\alpha-\varepsilon+1}}\geq\frac{\int_{N}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{N}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM115)

Next we introduce $\delta:=\alpha-\varepsilon+1>0$ and take on both sides the
limit inferior $y\to\infty$, leading to

$\frac{1}{m^{\delta}}=\liminf_{y\to\infty}\frac{(y/m)^{\alpha-\varepsilon+1}-N^{\alpha-\varepsilon+1}}{(y)^{\alpha-\varepsilon+1}-N^{\alpha-\varepsilon+1}}\geq\liminf_{y\to\infty}\frac{\int_{N}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{N}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}=\liminf_{y\to\infty}\frac{\int_{1}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x},$ (SM116)

where in the last equality, we used that
$\int_{1}^{N}f(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ is finite,
whilst $\int_{N}^{\infty}f(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$
diverges. As last step, we observe that according to Proposition SM.9,

$\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\geq\liminf_{y\to\infty}\frac{\max_{1\leq
a\leq y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}\geq\liminf_{y\to\infty}\frac{\int_{y/m}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}\geq 1-\frac{1}{m^{\delta}}>0.$ (SM117)

Using the remark after Definition SM.7, we obtain that for $-1<\alpha\leq 0$,
there are no sequences $\Set{n_{j}}_{j\in\mathbb{N}}$ such that
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is a universal embezzling family,
contradicting the hypothesis. Once we combine the results for $\alpha<-1$ and
$-1<\alpha\leq 0$, we have that if $\Set{\chi_{n}}$ is a universal embezzling
family satisfying our assumptions, $f(x)\in
o(x^{-1+\varepsilon})\cap\omega(x^{-1-\varepsilon})$ for all $\varepsilon>0$.
We have also shown that if $f\notin
o(x^{-1+\varepsilon})\cap\omega(x^{-1-\varepsilon})$ for at least one
$\varepsilon>0$, then there are no sequences $\Set{n_{j}}_{j\in\mathbb{N}}$
such that the family $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is a universal
embezzling family. This concludes the first part of the proof.

_Sufficient condition —_ Let $f$ be a function satysfying our assumptions.
Since $f(x)/x^{-1}$ is asymptotically monotonic by assumption,
$\lim_{x\to\infty}f(x)/x^{-1}$ exists in $[0,+\infty]$. A priori, it can be
either $0$, $0<l\in\mathbb{R}$, or $+\infty$. If
$\lim_{x\to\infty}f(x)/x^{-1}=l>0$, then by definition of the limit, for any
$l>\tilde{\varepsilon}>0$ there exists an $N$ such that
$\left\lvert\frac{f(x)}{x^{-1}}-l\right\rvert<\tilde{\varepsilon}$ for all
$x>N$. This is equivalent to
$(l-\tilde{\varepsilon})x^{-1}<f(x)<(l+\tilde{\varepsilon})x^{-1}$ for all
$x>N$. Since
$\sum_{x=N}^{\infty}f(x)\geq(l-\tilde{\varepsilon})\sum_{x=N}^{\infty}x^{-1}=+\infty$,
we can compute the conversion distance using Proposition SM.9, and find

$\displaystyle\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle=\lim_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ (SM118)
$\displaystyle\leq\lim_{y\to\infty}\frac{\max_{1\leq
a<N}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}+\max_{N\leq a\leq
y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ $\displaystyle=\lim_{y\to\infty}\frac{\max_{N\leq a\leq
y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ $\displaystyle\leq\lim_{y\to\infty}\frac{\max_{N\leq a\leq
y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{N}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle\leq\frac{l+\tilde{\varepsilon}}{l-\tilde{\varepsilon}}\lim_{y\to\infty}\frac{\max_{N\leq
a\leq y/m}\Set{\int_{a}^{am}x^{-1}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{N}^{y}x^{-1}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle=\frac{l+\tilde{\varepsilon}}{l-\tilde{\varepsilon}}\lim_{y\to\infty}\frac{\log
m}{\log y-\log N}$ $\displaystyle=0.$

Thus if $\lim_{x\to\infty}f(x)/x^{-1}=l\neq 0$, then $\Set{\chi_{n}}$ is a
universal embezzling family.

Suppose now $\lim_{x\to\infty}f(x)/x^{-1}=0$. Since $f(x)/x^{-1}$ is non-
increasing for large $x$, there exists an $\tilde{N}$ such that $f(x)<x^{-1}$
for all $x>\tilde{N}$. For any fixed $\varepsilon>0$, by hypothesis,
$\lim_{x\to\infty}f(x)/x^{-1-\varepsilon}=+\infty$. Furthermore, since by
assumption $\int_{1}^{\infty}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x=\infty$, while
$\int_{1}^{\infty}x^{-1-\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x<\infty$, there exists an $\tilde{N}_{\varepsilon}$, such that
$\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{1}^{y}x^{-1-\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x$ for $y>\tilde{N}_{\varepsilon}$. Let
$N_{\varepsilon}=\max\Set{\tilde{N},\tilde{N}_{\varepsilon}}$. Using again
Proposition SM.9 and performing the same steps as in Eq. (SM118), we obtain

$\displaystyle\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle\leq\limsup_{y\to\infty}\frac{\max_{N_{\varepsilon}\leq a\leq
y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ (SM119)
$\displaystyle\leq\limsup_{y\to\infty}\frac{\max_{N_{\varepsilon}\leq a\leq
y/m}\Set{\int_{a}^{am}x^{-1}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}x^{-1-\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ $\displaystyle=\log
m\lim_{y\to\infty}\frac{\varepsilon}{1-y^{-\varepsilon}}$
$\displaystyle=\varepsilon\log m.$

Since this is true for all $\varepsilon>0$,

$\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\leq\lim_{\varepsilon\to
0}\varepsilon\log m=0.$ (SM120)

This implies that
$\lim_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=0$, i.e.,
$\\{\chi_{n}\\}$ is a universal embezzling family.

Let us now consider the case $\lim_{x\to\infty}f(x)/x^{-1}=\infty$. From this
immediately follows that $\sum_{x=1}^{\infty}f(x)=+\infty$, so we can use the
results of Section V. Furthermore, there exists an $N$ such that $f(x)/x^{-1}$
is non-decreasing for $x\geq N$. We write $f(x)=x^{-1}h(x)$ and thus $h(x)$ is
non-decreasing for $x\geq N$. Computing the derivative of
$\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ for $a\geq N$,
we find

$\frac{\mathop{\mathrm{\mathstrut
d}}\nolimits\\!\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\mathop{\mathrm{\mathstrut
d}}\nolimits\\!a}=mf(am)-f(a)=\frac{h(am)-h(a)}{a}\geq 0.$ (SM121)

Thus, $\max_{N\leq a\leq y/m}\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x=\int_{y/m}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x$ and, following the same steps as in Eq. (SM118), we obtain

$\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\leq\limsup_{y\to\infty}\frac{\max_{N\leq
a\leq y/m}\Set{\int_{a}^{am}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}=\limsup_{y\to\infty}\frac{\int_{y/m}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}.$ (SM122)

Let us fix $\varepsilon>0$. Since $f(x)/x^{-1+\varepsilon}$ is non-increasing
for $x>N_{\varepsilon}$ by hypothesis, $x^{-1+\varepsilon}/f(x)$ is non-
decreasing for $x>N_{\varepsilon}$. By applying Lemma SM.11 (for $y$ large
enough), we obtain

$\int_{N_{\varepsilon}}^{y}x^{-1+\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{N_{\varepsilon}}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\geq\int_{N_{\varepsilon}}^{y/m}x^{-1+\varepsilon}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x\int_{N_{\varepsilon}}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x.$ (SM123)

From this follows that

$\frac{\int_{N_{\varepsilon}}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{N_{\varepsilon}}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}\geq\frac{(y/m)^{\varepsilon}-N_{\varepsilon}^{\varepsilon}}{y^{\varepsilon}-N_{\varepsilon}^{\varepsilon}}.$
(SM124)

Taking the $\limsup$ on both sides, and adding the finite contributions
$\int_{1}^{N_{\varepsilon}}f(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$ to
the diverging integrals
$\int_{N_{\varepsilon}}^{y/m}f(x)\mathop{\mathrm{\mathstrut d}}\nolimits\\!x$
and $\int_{N_{\varepsilon}}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x$, we obtain

$\limsup_{y\to\infty}\frac{\int_{1}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}\geq\frac{1}{m^{\varepsilon}}.$ (SM125)

Combining Eq. (SM122) and Eq. (SM125) we get

$\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\leq
1-\limsup_{y\to\infty}\frac{\int_{1}^{y/m}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}\leq 1-\frac{1}{m^{\varepsilon}}.$ (SM126)

This is true for all $\varepsilon>0$, thus taking the limit $\varepsilon\to 0$
we obtain the desired result

$\limsup_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})\leq\lim_{\varepsilon\to
0}1-\frac{1}{m^{\varepsilon}}=0.$ (SM127)

This concludes the proof of the sufficient condition. ∎

The above result on the uniqueness of the van Dam and Hayden family can be
expressed as follows: Any regular family of states
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ satisfying the conditions of Theorem SM.12
is a universal embezzling family if and only if $f$, the function associated
to $\Set{\chi_{n}}_{n\in\mathbb{N}}$, is asymptotically close to $x^{-1}$, the
function associated to the van Dam and Hayden family, where asymptotically
close means $f\in\omega(x^{-1-\varepsilon})\cap o(x^{-1+\varepsilon})$ for all
$\varepsilon>0$.

The assumption that $f(x)/x^{\alpha}$ is asymptotically monotonic for every
$\alpha\in\mathbb{R}$ is crucial for our proof and does not follow from the
monotonicity of $f$. There are functions that are non-increasing, but
oscillate asymptotically when multiplied by powers of $x$. An example is the
function

$f(x)=\frac{1+(1+\sin\ln\ln x)\ln x}{x},$ (SM128)

which is non-increasing, but $xf(x)$ oscillates between $1$ and $\infty$.
Theorem SM.12 does not provide any information about families of states
associated to such functions, and it cannot be used to determine whether such
families are universally embezzling or not.

Next, we prove two related propositions concerning regular families of states
associated to non-decreasing functions.

######  Proposition SM.13.

Let $f$ be a positive non-decreasing function such that
$f(x)/x^{\alpha}\searrow$ for at least one $\alpha>0$. Then the regular family
of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ associated to $f$ (see Definition
SM.7) is not a universal embezzling family. Furthermore, there are no
sequences $\Set{n_{j}}_{j\in\mathbb{N}}$ such that
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is a universal embezzling family.

###### Proof.

Since $f(x)$ is non-decreasing and $\sum_{x=1}^{\infty}f(x)$ diverges, we can
use Proposition SM.9 and Lemma SM.11, to obtain

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle\geq\liminf_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int_{a}^{am}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}g(x,y)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ (SM129)
$\displaystyle=\liminf_{y\to\infty}\frac{\max_{1\leq a\leq
y/m}\Set{\int^{y-a+1}_{y-am+1}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle\geq\liminf_{y\to\infty}\frac{\Set{\int^{y(1-1/m)+1}_{1}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle\geq\liminf_{y\to\infty}\frac{\Set{\int^{y(1-1/m)+1}_{1}x^{\alpha}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}}{\int_{1}^{y}x^{\alpha}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ $\displaystyle=\left(1-\frac{1}{m}\right)^{\alpha+1}.$

The last inequality is based on Lemma SM.11 and is derived as in the previous
cases. Using the remark after Definition SM.7, we obtain that
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is not a universal embezzling family.
Furthermore, there are no sequences $\Set{n_{j}}_{j\in\mathbb{N}}$ such that
$\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$ is a universal embezzling family. ∎

######  Proposition SM.14.

Let $f$ be a positive non-decreasing function such that $f(x)/e^{kx}\nearrow$
for at least one $k>0$. Then the regular family of states
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ associated to $f$ is not a universal
embezzling family. Furthermore, there are no sequences
$\Set{n_{j}}_{j\in\mathbb{N}}$ such that $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$
is a universal embezzling family.

###### Proof.

From Eq. (SM47) follows that

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle\geq\left(1-\frac{1}{m}\right)\liminf_{n\to\infty}p_{1}^{(n)}$
(SM130)
$\displaystyle=\left(1-\frac{1}{m}\right)\liminf_{n\to\infty}\frac{f(n)}{\sum_{x=1}^{n}f(x)}$
$\displaystyle=\left(1-\frac{1}{m}\right)\liminf_{y\to\infty}\frac{f(y)}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle\geq\left(1-\frac{1}{m}\right)\liminf_{y\to\infty}\frac{\int_{y-1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}f(x)\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$
$\displaystyle\geq\left(1-\frac{1}{m}\right)\liminf_{y\to\infty}\frac{\int_{y-1}^{y}e^{kx}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}{\int_{1}^{y}e^{kx}\mathop{\mathrm{\mathstrut
d}}\nolimits\\!x}$ $\displaystyle=\left(1-\frac{1}{m}\right)(1-e^{-k}).$

Also here the last inequality is due to Lemma SM.11 and the family of states
is not a universal embezzling family. Furthermore, there are no sequences
$\Set{n_{j}}_{j\in\mathbb{N}}$ such that $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$
is a universal embezzling family. ∎

##  VIII Asymptotically Regular Families

In the definition of universal embezzling families, Definition 3, and in all
the results about embezzlement, only the asymptotic behaviour of a family of
states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is relevant. This motivates the
following definition, which is a generalization of regular families.

######  Definition SM.15.

A family of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is called asymptotically
regular if there exists an asymptotically monotonic function
$f:\mathbb{N}\to(0,\infty)$ such that

$\ket{\chi_{n}}=\frac{1}{\sqrt{F_{n}}}\sum_{x=1}^{n}\sqrt{f(x)}\Ket{xx}\quad\text{for
all}\quad n\in\mathbb{N},$ (SM131)

where $F_{n}=\sum_{x=1}^{n}f(x)$.

We next show that our results hold for asymptotically regular families too. To
this end, we start with the following theorem.

######  Theorem SM.16.

Let $\Set{\chi_{n}}_{n\in\mathbb{N}}$ be an asymptotically regular family and
let $f$ be the function associated to it (see Definition SM.15). Then one can
construct a function $\tilde{f}$ that satisfies

  1. 1.

$\tilde{f}$ corresponds to a regular family
$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ (see Definition SM.7),

  2. 2.

$\lim_{x\to\infty}\tilde{f}(x)/f(x)=1$,

  3. 3.

$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family if
and only if $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a universal embezzling
family,

  4. 4.

$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ contains a universal embezzling
subfamily if and only if $\Set{\chi_{n}}_{n\in\mathbb{N}}$ contains a
universal embezzling subfamily.

###### Proof.

Since $f$ is by assumption asymptotically monotonic, $\lim_{x\to\infty}f(x)$
exists in the extended domain $[0,\infty]$. We first study the case
$\lim_{x\to\infty}f(x)=l$, with $0<l\in\mathbb{R}$. In this case, we choose
$\tilde{f}(x)=l$, which corresponds to a regular family. Moreover, from the
definition of $\tilde{f}$ follows that $\lim_{x\to\infty}f(x)/\tilde{f}(x)=1$.
We also notice that for every $0<\varepsilon<l$, there exists an $N$ such that
$l-\varepsilon<f(x)<l+\varepsilon$ for all $x\geq N$. This implies that
$\sum_{x=1}^{\infty}f(x)\geq\sum_{x=N}^{\infty}f(x)\geq\sum_{x=N}^{\infty}(l-\varepsilon)=\infty$.
Similarly, $\sum_{x=1}^{\infty}\tilde{f}(x)=\sum_{x=1}^{\infty}l=\infty$.
Since
$\lim_{n\to\infty}\frac{1}{F_{n}}=\lim_{n\to\infty}\frac{1}{\tilde{F}_{n}}=0$,
we obtain that $\lim_{n\to\infty}p_{1}^{(n)}=0$ (where $\mathbf{p}^{(n)}$ are
the Schmidt coefficients of $\chi_{n}$), and therefore, according to Eq.
(SM47),

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
(SM132)
$\displaystyle=\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}\right\\},$

where $a_{k}=\lfloor k/m\rfloor$ and $b_{k}=k-ma_{k}$. The Schmidt
coefficients are by definition non-increasing, i.e., obtained by reordering
$\Set{f(x)/F_{n}}_{x\in\mathbb{N}}$. Since there are at most $N$ natural
numbers $x$ that do not satisfy the condition
$l-\varepsilon<f(x)<l+\varepsilon$, there are at most $N$ Schmidt coefficients
that do not satisfy
$\frac{l-\varepsilon}{F_{n}}<p^{(n)}_{x}<\frac{l+\varepsilon}{F_{n}}$. We call
the set of indices corresponding to these Schmidt coefficients $A$ (thus
$\left\lvert A\right\rvert\leq N$) and observe that for any
$a,b\in\Set{N,\dots,n}$ such that $a\leq b$,

$\displaystyle\sum_{x=a}^{b}p^{(n)}_{x}$
$\displaystyle=\sum_{x\in\Set{a,\dots,b}\setminus
A}p^{(n)}_{x}+\sum_{x\in\Set{a,\dots,b}\cap A}p^{(n)}_{x}$ (SM133)
$\displaystyle=\sum_{x\in\Set{a,\dots,b}\setminus
A}p^{(n)}_{x}+\sum_{x\in\Set{a,\dots,b}\cap
A}\frac{l}{F_{n}}+\sum_{x\in\Set{a,\dots,b}\cap
A}\left(p^{(n)}_{x}-\frac{l}{F_{n}}\right).$

This implies that

$\sum_{x=a}^{b}\frac{(l-\varepsilon)}{F_{n}}+\sum_{x\in\Set{a,\dots,b}\cap
A}\left(p^{(n)}_{x}-\frac{l}{F_{n}}\right)<\sum_{x=a}^{b}p^{(n)}_{x}<\sum_{x=a}^{b}\frac{(l+\varepsilon)}{F_{n}}+\sum_{x\in\Set{a,\dots,b}\cap
A}\left(p^{(n)}_{x}-\frac{l}{F_{n}}\right),$ (SM134)

and therefore

$\displaystyle\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}\frac{l-\varepsilon}{F_{n}}\right\\}\leq\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}\right\\}\leq\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}\frac{l+\varepsilon}{F_{n}}\right\\},$
(SM135)

which implies

$(l-\varepsilon)\liminf_{n\to\infty}\frac{n-\lfloor
n/m\rfloor}{F_{n}}\leq\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}\right\\}\leq(l+\varepsilon)\liminf_{n\to\infty}\frac{n-\lfloor
n/m\rfloor}{F_{n}}.$ (SM136)

Observing that
$\sum_{x=1}^{N}f(x)+(l-\varepsilon)(n-N)<F_{n}<\sum_{x=1}^{N}f(x)+(l+\varepsilon)(n-N)$,
we obtain

$\frac{l-\varepsilon}{l+\varepsilon}\left(1-\frac{1}{m}\right)\leq\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}\right\\}\leq\frac{l+\varepsilon}{l-\varepsilon}\left(1-\frac{1}{m}\right).$
(SM137)

Since Eq. (SM137) holds for every $\varepsilon>0$, we conclude that
$\liminf_{n\to\infty}\max_{k\in\Set{N,\dots,n}}\left\\{\sum_{x=a_{k}+1}^{k}p^{(n)}_{x}\right\\}=1-\frac{1}{m}$.
Inserting this result into Eq. (SM132), we have

$\liminf_{x\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})=1-\frac{1}{m}.$
(SM138)

The function $\tilde{f}$ is a rescaling of $x^{0}$, which we already studied
in Section VI. This implies that the family
$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ is equal to the family
$\Set{\chi^{(0)}_{n}}_{n\in\mathbb{N}}$ and

$\liminf_{x\to\infty}d_{\star}(\tilde{\chi}_{n}\to\tilde{\chi}_{n}\otimes\Phi_{m})=\liminf_{x\to\infty}d_{\star}(\chi_{n}^{(0)}\to\chi_{n}^{(0)}\otimes\Phi_{m})=1-\frac{1}{m}=\liminf_{x\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m}).$
(SM139)

This proves that neither $f$ nor $\tilde{f}$ correspond to families with
universally embezzling subfamilies (and are therefore also not universally
embezzling themselves).

We consider now the case $\lim_{x\to\infty}f(x)=0$, thus $f$ is asymptotically
non-increasing. Let $N$ be such that $f$ is non-increasing on $(N,\infty)$ and
let $a=\min_{x\in\Set{1,\dots,N}}f(x)$. Since $f$ is positive, $a>0$. Let
$M>N$ be such that $f(x)<a$ for every $x\geq M$ (such $M$ exists because
$\lim_{x\to\infty}f(x)=0$). In this case, we define $\tilde{f}$ as

$\tilde{f}(x)=\begin{cases}f(M)&\text{if }x\leq M,\\\ f(x)&\text{if
}x>M.\end{cases}$ (SM140)

Clearly, the family $\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ associated to it
is regular and $\lim_{x\to\infty}f(x)/\tilde{f}(x)=1$. Furthermore, the
ordered Schmidt coefficients satisfy for all $x\geq M$

$p^{(n)}_{x}=\frac{f(x)}{F_{n}}=\frac{\tilde{f}(x)}{F_{n}}=\tilde{p}^{(n)}_{x}\frac{\tilde{F}_{n}}{F_{n}}.$
(SM141)

If $F_{n}$ converges, $\tilde{F}_{n}$ converges too and by Corollary SM.6
neither $\Set{\chi_{n}}_{n\in\mathbb{N}}$ nor
$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ are universal embezzling families
(and do not contain universal embezzling subfamilies). If instead $F_{n}$
diverges, so does $\tilde{F}_{n}$ and
$\lim_{n\to\infty}\frac{\tilde{F}_{n}}{F_{n}}=\lim_{n\to\infty}\frac{\sum_{x=1}^{n}\tilde{f}(x)}{\sum_{x=1}^{n}f(x)}=\lim_{n\to\infty}\frac{\sum_{x=M}^{n}\tilde{f}(x)}{\sum_{x=M}^{n}f(x)}=1$.
Using Eq. (SM47) and Eq. (SM141), we obtain

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
(SM142)
$\displaystyle=\liminf_{n\to\infty}\max_{k\in\Set{mM,\dots,n}}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}\frac{\tilde{F}_{n}}{F_{n}}\max_{k\in\Set{mM,\dots,n}}\left\\{\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(k)}-\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}\tilde{p}^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(k)}-\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}\tilde{p}^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}d_{\star}(\tilde{\chi}_{n}\to\tilde{\chi}_{n}\otimes\Phi_{m}),$

where again $a_{k}=\lfloor k/m\rfloor$ and $b_{k}=k-ma_{k}$. Also in this
case, we therefore proved that $\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a
universal embezzling family if and only if
$\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family
(and the same holds for subfamilies).

Lastly, we consider the case $\lim_{x\to\infty}f(x)=\infty$, thus $f$ is
asymptotically non-decreasing. Analogously to the previous case, let $N$ be
such that $f$ is non-decreasing for $x\in(N,\infty)$. Let $a$ be the maximum
of $f(x)$ for $x\in\Set{1,\dots,N}$ and let $M>N$ be such that $f(x)>a$ for
$x\geq M$. Also here, we define $\tilde{f}$ via

$\tilde{f}=\begin{cases}f(M)&\text{if }x\leq M,\\\ f(x)&\text{if
}x>M.\end{cases}$ (SM143)

The family of states $\Set{\tilde{\chi}_{n}}_{n\in\mathbb{N}}$ is regular,
$\lim_{x\to\infty}\tilde{f}(x)/f(x)=1$, and
$\lim_{x\to\infty}\tilde{F}_{n}/F_{n}=1$. The Schmidt coefficients associated
to $\chi_{n}$ and $\tilde{\chi}_{n}$ are related by

$p^{(n)}_{x}=\frac{f(n+1-x)}{F_{n}}=\frac{\tilde{f}(n+1-x)}{F_{n}}=\tilde{p}^{(n)}_{x}\frac{\tilde{F}_{n}}{F_{n}}\quad\forall
x\leq n-M.$ (SM144)

Using Eq. (SM47) again and the relation between Schmidt coefficients derived
in Eq. (SM144) we obtain

$\displaystyle\liminf_{n\to\infty}d_{\star}(\chi_{n}\to\chi_{n}\otimes\Phi_{m})$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
(SM145)
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n-M]}\left\\{\left\lVert\mathbf{p}^{(n)}\right\rVert_{(k)}-\left\lVert\mathbf{p}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}p^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}\frac{\tilde{F}_{n}}{F_{n}}\max_{k\in[n-M]}\left\\{\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(k)}-\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}\tilde{p}^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}\max_{k\in[n]}\left\\{\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(k)}-\left\lVert\tilde{\mathbf{p}}^{(n)}\right\rVert_{(a_{k})}-\frac{b_{k}}{m}\tilde{p}^{(n)}_{a_{k}+1}\right\\}$
$\displaystyle=\liminf_{n\to\infty}d_{\star}(\tilde{\chi}_{n}\to\tilde{\chi}_{n}\otimes\Phi_{m}).$

This proves the theorem. ∎

Thanks to Theorem SM.16, Theorem SM.12 also holds for asymptotically regular
families.

######  Corollary SM.17.

Let $f\searrow$ be a positive function such that $f(x)/x^{\alpha}$ is
asymptotically monotonic for all $\alpha\in\mathbb{R}$ and let
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ be the asymptotically regular family of
states associated to $f$ (see Definition SM.15). Then
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ is a universal embezzling family if and only
if $f\in\omega(x^{-1-\varepsilon})\cap o(x^{-1+\varepsilon})$ for all
$\varepsilon>0$ and $\sum_{x=1}^{\infty}f(x)=\infty$. Furthermore, if
$f\notin\omega(x^{-1-\varepsilon})\cap o(x^{-1+\varepsilon})$ for at least one
$\varepsilon>0$, then $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$, where
$\Set{n_{j}}_{j\in\mathbb{N}}$ is any sequence of natural numbers, is not a
universal embezzling family.

###### Proof.

Combine Theorem SM.16 and Theorem SM.12. ∎

For the same reasons, Proposition SM.13 and Proposition SM.14 also hold for
asymptotically regular families.

######  Corollary SM.18.

Let $f$ be a positive function such that $f\nearrow$ and
$f(x)/x^{\alpha}\searrow$ for at least one $\alpha>0$. Then the asymptotically
regular family of states $\Set{\chi_{n}}_{n\in\mathbb{N}}$ associated to $f$
is not a universal embezzling family. Furthermore, there are no sequences
$\Set{n_{j}}_{j\in\mathbb{N}}$ such that $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$
is a universal embezzling family.

######  Corollary SM.19.

Let $f$ be a positive function such that $f\nearrow$ and $f(x)/e^{kx}\nearrow$
for at least one $k>0$. Then the asymptotically regular family of states
$\Set{\chi_{n}}_{n\in\mathbb{N}}$ associated to $f$ is not a universal
embezzling family. Furthermore, there are no sequences
$\Set{n_{j}}_{j\in\mathbb{N}}$ such that $\Set{\chi_{n_{j}}}_{j\in\mathbb{N}}$
is a universal embezzling family.

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