construction, we can estimate the following expectation w.r.t. both $F$ and
$T$:

$\displaystyle\;\mathbb{E}_{F,T}\operatorname{Tr}\left[\left(\frac{1}{L}\sum_{n\in
F^{-1}(T)}\overline{\rho}_{A,n}\right)^{1+s}\right]$ (115a) $\displaystyle=$
$\displaystyle\;\mathbb{E}_{F}\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}_{A,n}\left(\frac{1}{L}\overline{\rho}_{A,n}+\frac{1}{L}\sum_{n^{\prime}(\neq
n)\in\mathcal{N}\mathrel{\mathop{\mathchar
58\relax}}F(n^{\prime})=F(n)}\overline{\rho}_{A,n^{\prime}}\right)^{s}\right]$
(115b) $\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\;\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}_{A,n}\left(\frac{1}{L}\overline{\rho}_{A,n}+\mathbb{E}_{F}\frac{1}{L}\sum_{n^{\prime}(\neq
n)\in\mathcal{N}\mathrel{\mathop{\mathchar
58\relax}}F(n^{\prime})=F(n)}\overline{\rho}_{A,n^{\prime}}\right)^{s}\right]$
(115c) $\displaystyle=$
$\displaystyle\;\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}_{A,n}\left(\frac{1}{L}\overline{\rho}_{A,n}+\frac{L-1}{L}\overline{\rho}_{A}\right)^{s}\right]$
(115d) $\displaystyle\leq$
$\displaystyle\;\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}_{A,n}\left(\frac{1}{L}\overline{\rho}_{A,n}+\overline{\rho}_{A}\right)^{s}\right]$
(115e) $\displaystyle\stackrel{{\scriptstyle(b)}}{{\leq}}$
$\displaystyle\;\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}_{A,n}\left(\frac{1}{L^{s}}\overline{\rho}^{s}_{A,n}+\overline{\rho}^{s}_{A}\right)\right]$
(115f) $\displaystyle=$
$\displaystyle\;\frac{1}{L^{s}}\operatorname{Tr}\left[\frac{1}{N}\sum_{n\in{\cal
N}}\overline{\rho}^{1+s}_{A,n}\right]+\operatorname{Tr}\left[\overline{\rho}^{1+s}_{A}\right]$
(115g) $\displaystyle\stackrel{{\scriptstyle(c)}}{{=}}$
$\displaystyle\;\frac{1}{L^{s}}e^{-s\overline{\operatorname{H}}_{1+s}(A|F)_{\xi}}+e^{-s\overline{\operatorname{H}}_{1+s}(A)_{\xi}},$
(115h)

where $s\in[0,1]$, $(a)$ follows from the concavity of $x\mapsto x^{s}$ when
$s\in(0,1]$ [85, Table 2.2][86, Appendix A.4] and the Jensen inequality, $(b)$
follows from $(x+y)^{s}\leq x^{s}+y^{s}$ for $x,y>0$, and $(c)$ follows from
(109b) and (109c).

Step 2: We prepare a useful lemma that holds under Assumption 1 (the
multiplicity-free condition) as follows. This lemma will be shown in Step 4.

###### Lemma 16

Assume that the (projective) unitary representation $U$ on $\mathcal{H}_{A}$
is multiplicity-free. Then, the Hilbert space $\mathcal{H}_{A}$ is decomposed
as $\oplus_{k\in\mathcal{K}}\mathcal{H}_{k}$. Let $\\{Q_{A,l}\\}_{l=1}^{L}$ be
a POVM on $\mathcal{H}_{A}$ such that each $Q_{A,l}$ is a projection onto the
invariant subspace $\oplus_{k\in\mathcal{S}_{l}}\mathcal{H}_{k}$, where
$\\{\mathcal{S}_{l}\\}_{l=1}^{L}$ are disjoint subsets of $\mathcal{K}$. That
is, each $Q_{A,l}$ projects into disjoint irreducible subspaces
$\mathcal{H}_{k}$. Define the corresponding conditional projection operator in
$AF$ as

$\displaystyle Q_{AF}\mathrel{\mathop{\mathchar
58\relax}}=\sum_{l=1}^{L}Q_{A,l}\otimes|l\rangle\\!\langle l|_{F},$ (116)

where $\\{\left|l\right\rangle\\}_{l=1}^{L}$ is an orthonormal basis of $F$.
Assume Alice and Fred preshare the pure bipartite state

$\displaystyle\left|\psi_{AF}\right\rangle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{L}}\sum_{l=1}^{L}|\psi_{A,l}\rangle\otimes\left|l\right\rangle_{F},$
(117)

and define $\overline{\sigma}_{A}$ as

$\displaystyle\overline{\sigma}_{A}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{L}\sum_{l=1}^{L}Q_{A,l}\mathcal{G}(\psi_{A,l})Q_{A,l}.$
(118)

We have the following result regarding the one-way LOCC classical
communication capability of $\left|\psi_{AF}\right\rangle$.

Let $M$ be the message size. Let
$E\equiv(\mathtt{g}_{1},\cdots,\mathtt{g}_{M})$ be a random coding such that
each codeword $\mathtt{g}_{m}$ is chosen independently and uniformly from $G$.
We use the typewriter font $\mathtt{g}$ to indicate that it is a random
variable. For a chosen encoder $(g_{1},\cdots,g_{M})$, there exists a one-way
LOCC decoder such that the resulting protocol $\mathcal{C}$’s expected
decoding error is upper bounded as

$\displaystyle\mathbb{E}_{E}e(\mathcal{C}(E))\leq\frac{8M^{s}}{(1-\delta)^{s}}\operatorname{Tr}\overline{\sigma}_{A}^{1+s}+2\delta,$
(119)

where $s\in(0,1)$, $\delta$ is defined as

$\displaystyle\delta\mathrel{\mathop{\mathchar
58\relax}}=1-\left\langle\psi_{AF}\right|Q_{AF}\left|\psi_{AF}\right\rangle=1-\frac{1}{L}\sum_{l=1}^{L}\langle\psi_{A,l}|Q_{A,l}|\psi_{A,l}\rangle=1-\frac{1}{L}\sum_{l=1}^{L}\mathcal{G}(\psi_{A,l}),$
(120)

where the second equation follows from (116) and (117), and the final equation
follows from the invariant property of $Q_{A,l}$ for the group action.

Step 3: We proof Theorem 15 by applying Lemma 16 to the measurement induced by
$2$-universal hash function. That is, we adopt the random coding argument and
show that the expected decoding error for protocols generated by randomly
selecting codewords according to the uniform distribution and measurements
according to $2$-universal hash function is upper bounded.

In the first stage, Fred performs the projective measurement
$\\{\Pi^{f,t}\\}$, diving system $F$ into $T$ subspaces. When the outcome is
$t$, the post-measurement state on $AF$ is

$\displaystyle|\Psi^{f,t}_{AF}\rangle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{L}}\sum_{n\in
f^{-1}(t)}\left|\psi_{A,n}\right\rangle\left|n\right\rangle_{F}.$ (121)

The outcome is communicated to Bob via a classical noiseless channel. Set
$\bm{Q}_{AF}^{f,t}\mathrel{\mathop{\mathchar 58\relax}}=\sum_{n\in
f^{-1}(t)}Q_{A,n}^{f,t}\otimes|n\rangle\\!\langle n|_{F}$, which is the
conditional version of $\bm{Q}^{f,t}$. One can check that

$\displaystyle\langle\Psi^{f,t}_{AF}|\bm{Q}_{AF}^{f,t}|\Psi^{f,t}_{AF}\rangle=\frac{1}{L}\sum_{n\in
f^{-1}(t)}\langle\psi_{A,n}|Q_{A,n}^{f,t}|\psi_{A,n}\rangle=1-\delta^{f,t}.$
(122)

In the second stage, we apply Lemma 16 to the post-measurement state
$|\Psi^{f,t}_{AF}\rangle$ with corresponding measurement $\bm{Q}^{f,t}$ to
implement classical communication from Alice to Bob. We can do so because the
measurement $\bm{Q}^{f,t}$ satisfies the prerequisite given in Lemma 16.
Consequently, there exists a communication protocol with one-way LOCC decoder
$\mathcal{C}(e,f,t)$ depending on both the encoder $E$ and the $2$-universal
hash function $F$ and $T$ that satisfies the following decoding error
condition

$\displaystyle\mathbb{E}_{E}\epsilon(\mathcal{C}(E,f,t))\leq\frac{8M^{s}}{(1-\delta^{f,t})^{s}}\operatorname{Tr}\left[\left(\frac{1}{L}\sum_{n\in
f^{-1}(t)}Q_{A,n}^{f,t}\overline{\rho}_{A,n}Q_{A,n}^{f,t}\right)^{1+s}\right]+2\delta^{f,t}.$
(123)

Averaging over all possible randomly generated codewords $E$ according to the
uniform distribution and randomly generated measurements according to
$2$-universal hash functions $(F,T)$, we can upper bound the expected value of
the decoding error as follows:

$\displaystyle\;\mathbb{E}_{E,F,T}\epsilon(\mathcal{C}(E,F,T))$ (124a)
$\displaystyle=$
$\displaystyle\;\Pr\left(\delta^{F,T}\geq\frac{1}{2}\right)\mathbb{E}_{E,F,T|\delta^{F,T}\geq
1/2}\epsilon(\mathcal{C}(E,F,T))+\Pr\left(\delta^{F,T}<\frac{1}{2}\right)\mathbb{E}_{E,F,T|\delta^{F,T}<1/2}\epsilon(\mathcal{C}(E,F,T))$
(124b) $\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\;\Pr\left(\delta^{F,T}\geq\frac{1}{2}\right)+\Pr\left(\delta^{F,T}<\frac{1}{2}\right)\mathbb{E}_{E,F,T|\delta^{F,T}<1/2}\epsilon(\mathcal{C}(E,F,T))$
(124c) $\displaystyle\stackrel{{\scriptstyle(b)}}{{\leq}}$
$\displaystyle\;2\mathbb{E}_{F,T}\delta^{F,T}+\Pr\left(\delta^{F,T}<\frac{1}{2}\right)\mathbb{E}_{E,F,T|\delta^{F,T}<1/2}\epsilon(\mathcal{C}(E,F,T))$
(124d) $\displaystyle\stackrel{{\scriptstyle(c)}}{{\leq}}$
$\displaystyle\;4\mathbb{E}_{F,T}\delta^{F,T}$ (124e)
$\displaystyle\quad+\Pr\left(\delta^{F,T}<\frac{1}{2}\right)\mathbb{E}_{F,T|\delta^{F,T}<1/2}\left\\{\frac{8M^{s}}{(1-\delta^{F,T})^{s}}\operatorname{Tr}\left[\left(\frac{1}{L}\sum_{n\in
F^{-1}(T)}Q_{A,n}^{F,T}\overline{\rho}_{A,n}Q_{A,n}^{F,T}\right)^{1+s}\right]\right\\}$
(124f) $\displaystyle\stackrel{{\scriptstyle(d)}}{{\leq}}$
$\displaystyle\;4\mathbb{E}_{F,T}\delta^{F,T}+2^{s+3}M^{s}\mathbb{E}_{F,T}\operatorname{Tr}\left[\left(\frac{1}{L}\sum_{n\in
F^{-1}(T)}Q_{A,n}^{F,T}\overline{\rho}_{A,n}Q_{A,n}^{F,T}\right)^{1+s}\right]$
(124g) $\displaystyle\stackrel{{\scriptstyle(e)}}{{\leq}}$
$\displaystyle\;4\mathbb{E}_{F,T}\delta^{F,T}+2^{s+3}M^{s}\mathbb{E}_{F,T}\operatorname{Tr}\left[\left(\frac{1}{L}\sum_{n\in
F^{-1}(T)}\overline{\rho}_{A,n}\right)^{1+s}\right]$ (124h)
$\displaystyle\stackrel{{\scriptstyle(f)}}{{\leq}}$
$\displaystyle\;4T^{-s^{\prime}}e^{s^{\prime}\overline{\operatorname{H}}_{1-s^{\prime}}(F|A)_{\xi}}+2^{s+3}\frac{M^{s}}{L^{s}}e^{-s\overline{\operatorname{H}}_{1+s}(A|F)_{\xi}}+2^{s+3}M^{s}e^{-s\overline{\operatorname{H}}_{1+s}(A)_{\xi}}$
(124i) $\displaystyle=$ $\displaystyle\;4e^{-s^{\prime}\left[\log
T-\overline{\operatorname{H}}_{1-s^{\prime}}(F|A)_{\xi}\right]}+2^{s+3}e^{-s\left[\overline{\operatorname{H}}_{1+s}(A|F)_{\xi}+\log
N-\log T-\log
M\right]}+2^{s+3}e^{-s\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]}$ (124j) $\displaystyle\stackrel{{\scriptstyle(g)}}{{=}}$
$\displaystyle\;4e^{-s^{\prime}\left[\log
T-\overline{\operatorname{H}}_{1-s^{\prime}}(F|A)_{\xi}\right]}+2^{s+3}e^{-s\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\log
T-\log
M\right]}+2^{s+3}e^{-s\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]},$ (124k)

where

  * •

$\mathbb{E}_{X|B}$ expresses the conditional expectation with respect to the
variable $X$ conditioned on $B$,

  * •

$(a)$ follows from the fact that the decoding error is less than $1$,

  * •

$(b)$ follows from the Markov inequality [140, Eq. (3.31)] that
$\Pr\left(\delta^{F,T}\geq 1/2\right)\leq 2\mathbb{E}_{F,T}\delta^{F,T}$,

  * •

$(c)$ follows from (123),

  * •

$(d)$ follows from the relation that $\delta^{F,T}<1/2$ implies
$(1-\delta^{F,T})^{-s}<2^{s}$. Notice that this relation is the essential
reason why we divide the expectation into two regions – $\delta^{F,T}\geq 1/2$
and $\delta^{F,T}<1/2$ – in (124b), since otherwise we cannot bound the term
$(1-\delta^{F,T})^{-s}$,

  * •

$(e)$ follows from the fact that the measurement element satisfies $0\leq
Q_{A,n}^{f,t}\leq\mathbbm{1}$ and thus the mutual commutativity property
guarantees that
$Q_{A,n}^{f,t}\overline{\rho}_{A,y}Q_{A,n}^{f,t}\leq\overline{\rho}_{A,n}$,
and

  * •

$(f)$ follows from the expectation estimations in (114) and (115) w.r.t. the
$2$-universal hash function.

  * •

$(g)$ follows from the fact that $\xi_{F}$ is the completely mixed state.

We set $s^{\prime}=s$ in (124k) and solve the equation w.r.t. the variable $T$

$\displaystyle\log
T-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}=\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\log
T-\log M,$ (125)

yielding

$\displaystyle\log
T=\frac{1}{2}\left(\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}+\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\log
M\right).$ (126)

Based on these choices, we can conclude from (124) that there exists a
concrete communication protocol one-way LOCC decoder $\mathcal{C}(e,f,t)$ for
carefully chosen encoding $e$ and the $2$-universal hash function $f$ such
that its decoding error is upper bounded for $s\in[0,1]$ as follows

$\displaystyle\epsilon\left(\mathcal{C}(e,f,t)\right)\leq$
$\displaystyle(4+2^{s+3})e^{-\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}-\log
M\right]}+2^{s+3}e^{-s\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]}$ $\displaystyle\leq$
$\displaystyle(4+16)e^{-\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}-\log
M\right]}+16e^{-s\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]}.$ (127)

Since $\epsilon\left(\mathcal{C}(e,f,t)\right)\leq 1$, we have

$\displaystyle\epsilon\left(\mathcal{C}(e,f,t)\right)\leq$ $\displaystyle
36\min(1,\max(e^{-\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}-\log
M\right]},e^{-s\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]})).$ (128) $\displaystyle\leq$ $\displaystyle
36\min(1,\max(e^{-\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}-\log
M\right]},e^{-\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right]})).$ (129) $\displaystyle=$ $\displaystyle
36\min(1,e^{-\min(\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}-\log
M\right],\frac{s}{2}\left[\overline{\operatorname{H}}_{1+s}(A)_{\xi}-\log
M\right])}).$ (130) $\displaystyle=$ $\displaystyle
36\min(1,e^{-\frac{1}{2}{\cal L}_{\xi}(-\log M)}).$ (131)

That is,

$\displaystyle-2\log\frac{\epsilon\left(\mathcal{C}(e,f,t)\right)}{36}\geq{\cal
L}_{\xi}(-\log M).$ (132)

Since ${\cal L}$ is monotonically increasing, we have

$\displaystyle-{\cal
L}_{\xi}^{-1}(-2\log\frac{\epsilon\left(\mathcal{C}(e,f,t)\right)}{36})\leq\log
M.$ (133)

This concludes the proof of Theorem 15.

Step 4: Now we show Lemma 16. Define the $L$-th root of unity
$\zeta\mathrel{\mathop{\mathchar 58\relax}}=\exp(2\pi i/L)$. From
$\\{\left|l\right\rangle_{F}\\}_{l=1}^{L}$ we construct the induced Fourier
basis measurement $\\{|\bm{b}_{F}^{l^{\prime}}\rangle\\}_{l^{\prime}=1}^{L}$
via

$\displaystyle|\bm{b}_{F}^{l^{\prime}}\rangle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{L}}\sum_{l=1}^{L}\zeta^{ll^{\prime}}\left|l\right\rangle_{F},\;l^{\prime}=1,\cdots,L.$
(134)

For each $l^{\prime}=1,\cdots,L$, defined the following subnormalized pure
quantum state

$\displaystyle|\phi_{A,l^{\prime}}\rangle\mathrel{\mathop{\mathchar
58\relax}}=\langle\bm{b}_{F}^{l^{\prime}}|Q_{AF}\left|\psi_{AF}\right\rangle=\frac{1}{L}\sum_{l=1}^{L}\zeta^{-ll^{\prime}}Q_{A,l}|\psi_{A,l}\rangle,$
(135)

whose norm can be calculated as

$\displaystyle\langle\phi_{A,l}|\phi_{A,l}\rangle=\frac{1}{L^{2}}\sum_{l=1}^{L}\langle\psi_{A,l}|Q_{A,l}|\psi_{A,l}\rangle=\frac{1-\delta}{L},$
(136)

where the last equality follows from the definition of $\delta$ (120).

Fred now perform this Fourier basis measurement on
$\left|\psi_{AF}\right\rangle$. After measurement, Fred holds the classical
outcome $l$ and Alice holds the post-measurement pure state. This leads to the
classical-quantum state

$\displaystyle\sigma_{AF}\mathrel{\mathop{\mathchar
58\relax}}=\sum_{l=1}^{L}\langle\bm{b}_{F}^{l}|\psi\rangle_{AF}\langle\psi|\bm{b}_{F}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}.$ (137)

Applying the pinching lemma [86, Lemma 3.10] to the quantum state
$|\psi_{AF}\rangle$ and the binary projective measurement
$\\{Q_{AF},\mathbbm{1}_{AF}-Q_{AF}\\}$ gives

$\displaystyle|\psi\rangle\\!\langle\psi|_{AF}\leq
2Q_{AF}|\psi\rangle\\!\langle\psi|_{AF}Q_{AF}+2(\mathbbm{1}_{AF}-Q_{AF})|\psi\rangle\\!\langle\psi|_{AF}(\mathbbm{1}_{AF}-Q_{AF}).$
(138)

Substituting (138) into (137) yields the following inequality regarding
$\sigma_{AF}$:

$\displaystyle\sigma_{AF}$
$\displaystyle\leq\sum_{l=1}^{L}\langle\bm{b}_{F}^{l}|\left(2Q|\psi\rangle\\!\langle\psi|Q+2(\mathbbm{1}-Q)|\psi\rangle\\!\langle\psi|(\mathbbm{1}-Q)\right)|\bm{b}_{F}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}$ (139)
$\displaystyle=2\sum_{l=1}^{L}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|\otimes|l\rangle\\!\langle
l|_{F}+2\sum_{l=1}^{L}\langle\bm{b}_{F}^{l}|(\mathbbm{1}-Q_{AF})|\psi\rangle\\!\langle\psi|(\mathbbm{1}-Q_{AF})|\bm{b}_{F}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}.$ (140)

Define the following classical-quantum channel

$\displaystyle g\mapsto W^{g}_{BF}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{1-\delta}\sum_{l=1}^{L}U_{g}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|U_{g}^{\dagger}\otimes|l\rangle\\!\langle
l|_{F},$ (141)

with uniform probability distribution on $G$. One can verify that such defined
$W^{g}_{BF}$ is indeed a quantum state using (136). This classical-quantum
channel has the following averaged state

$\displaystyle\int_{G}\Big{(}\sum_{l}\left(\frac{1}{1-\delta}U_{g}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|U_{g}^{\dagger}\right)\otimes|l\rangle\\!\langle
l|_{F}\Big{)}\nu(dg)$ $\displaystyle=$
$\displaystyle\;\int_{G}W^{g}_{BF}\nu(dg)$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}$
$\displaystyle\;\frac{1}{(1-\delta)L^{2}}\int_{G}\sum_{l=1}^{L}\sum_{l^{\prime}=1}^{L}\sum_{l^{\prime\prime}=1}^{L}\zeta^{l(l^{\prime}-l^{\prime\prime})}U_{g}Q^{l^{\prime\prime}}|\psi_{A,l^{\prime\prime}}\rangle\langle\psi_{A,l^{\prime}}|Q^{l^{\prime}}U_{g}^{\dagger}\otimes|l\rangle\\!\langle
l|_{F}\nu(dg)$ $\displaystyle\stackrel{{\scriptstyle(b)}}{{=}}$
$\displaystyle\;\frac{1}{(1-\delta)L^{2}}\int_{G}\sum_{l=1}^{L}\sum_{l^{\prime}=1}^{L}U_{g}Q^{l^{\prime}}|\psi_{A,l^{\prime}}\rangle\\!\langle\psi_{A,l^{\prime}}|Q^{l^{\prime}}U_{g}^{\dagger}\otimes|l\rangle\\!\langle
l|_{F}\nu(dg)$ $\displaystyle\stackrel{{\scriptstyle(c)}}{{=}}$
$\displaystyle\;\frac{1}{(1-\delta)L^{2}}\sum_{l=1}^{L}\sum_{l^{\prime}=1}^{L}Q^{l^{\prime}}\mathcal{G}(|\psi_{A,l^{\prime}}\rangle\\!\langle\psi_{A,l^{\prime}}|)Q^{l^{\prime}}\otimes|l\rangle\\!\langle
l|_{F}$ $\displaystyle\stackrel{{\scriptstyle(d)}}{{=}}$
$\displaystyle\;\frac{1}{1-\delta}\overline{\sigma}_{A}\otimes\pi_{F},$ (142)

where $(a)$ follows (135), $(b)$ follows because $Q_{A,l}$ is a projection
onto disjoint $\oplus_{k\in\mathcal{S}_{l}}\mathcal{H}_{k}$ and the
irreducible representations on $\mathcal{H}_{k}$ are not equivalent. The
multiplicity-free condition is used in this step. In addition, we remark that
the classical-quantum channel (141) can be obtained by Fred performs the
Fourier basis measurement and Alice adopts random coding chosen uniformly from
$G$. $(c)$ follows from the commutativity between $U_{g}$ and $Q^{l^{\prime}}$
and the commutativity follows from the invariance property. $(d)$ follows from
(118).

Applying the direct part of the classical-quantum channel coding theorem [86,
Lemma 4.6], we conclude that there exists an encoder $(g_{1},\cdots,g_{M})$
and a decoder $\bm{\Gamma}\equiv\\{\Gamma^{m}\\}_{m=1}^{M}$ in $BF$ as a one-
way LOCC measurement from Fred to Bob such that the decoding error is upper
bounded for arbitrary $s\in[0,1]$:

$\displaystyle\text{decoding error}\mathrel{\mathop{\mathchar 58\relax}}=$
$\displaystyle\;\frac{1}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[W^{g_{m}}\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(143a) $\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\;{4M^{s}}\int_{G}\operatorname{Tr}\left[(W^{g})^{1-s}\overline{W}^{s}\right]\nu(dg)$
(143b) $\displaystyle=$
$\displaystyle\;4M^{s}\int_{G}\operatorname{Tr}\left[\left(\sum_{l=1}^{L}\frac{L}{1-\delta}U_{g}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|U_{g}^{\dagger}\otimes\frac{1}{L}|l\rangle\\!\langle
l|_{F}\right)^{1-s}\left(\frac{1}{1-\delta}\overline{\sigma}_{A}\otimes\pi_{F}\right)^{s}\right]\nu(dg)$
(143c) $\displaystyle=$
$\displaystyle\;4M^{s}\int_{G}\operatorname{Tr}\left[\sum_{l}\left(\frac{L}{1-\delta}U_{g}|\psi_{A,l}\rangle\\!\langle\psi_{A,l}|U_{g}^{\dagger}\right)^{1-s}\left(\frac{1}{1-\delta}\overline{\sigma}_{A}\right)^{s}\otimes\frac{1}{L}|l\rangle\\!\langle
l|_{F}\right]\nu(dg)$ (143d) $\displaystyle\stackrel{{\scriptstyle(b)}}{{=}}$
$\displaystyle\;4M^{s}\int_{G}\operatorname{Tr}\left[\sum_{l}\left(\frac{L}{1-\delta}U_{g}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|U_{g}^{\dagger}\right)\left(\frac{1}{1-\delta}\overline{\sigma}_{A}\right)^{s}\otimes\frac{1}{L}|l\rangle\\!\langle
l|_{F}\right]\nu(dg)$ (143e) $\displaystyle=$
$\displaystyle\;4M^{s}\operatorname{Tr}\left[\int_{G}\Big{(}\sum_{l}\left(\frac{1}{1-\delta}U_{g}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|U_{g}^{\dagger}\right)\otimes|l\rangle\\!\langle
l|_{F}\Big{)}\left(\Big{(}\frac{1}{1-\delta}\overline{\sigma}_{A}\Big{)}^{s}\otimes
I_{F}\right)\nu(dg)\right]$ (143f)
$\displaystyle\stackrel{{\scriptstyle(c)}}{{=}}$
$\displaystyle\;\frac{4M^{s}}{(1-\delta)}\operatorname{Tr}\left[\Big{(}\sum_{l}\overline{\sigma}_{A}\otimes\frac{1}{L}|l\rangle\\!\langle
l|_{F}\Big{)}\left(\Big{(}\frac{1}{1-\delta}\overline{\sigma}_{A}\Big{)}^{s}\otimes
I_{F}\right)\right]$ (143g) $\displaystyle=$
$\displaystyle\;\frac{4M^{s}}{(1-\delta)^{1+s}}\operatorname{Tr}\left[\sum_{l}\overline{\sigma}_{A}^{1+s}\otimes\frac{1}{L}|l\rangle\\!\langle
l|_{F}\right]$ (143h) $\displaystyle=$
$\displaystyle\;\frac{4M^{s}}{(1-\delta)^{1+s}}\operatorname{Tr}\left[\overline{\sigma}_{A}^{1+s}\right],$
(143i)

where $(a)$ follows from [86, Lemma 4.1], $(b)$ follows from (136) implying
that $\frac{L}{1-\delta}U_{g}|\phi\rangle\\!\langle\phi|_{A,l}U_{g}^{\dagger}$
is a normalized pure state, and $(c)$ follows from (142).

Now we are ready to give a concrete protocol $\mathcal{C}$ achieving the
decoding error concluded in (119). In this protocol, Fred adopts the Fourier
basis measurement $\\{|\bm{b}_{F}^{l}\rangle\\}_{l=1}^{L}$, Alice adopts the
encoding $(g_{1},\cdots,g_{M})$, and Bob adopts the decoder $\bm{\Gamma}$
originally designed for the classical-quantum channel $g\mapsto W_{AF}^{g}$.
This is an communication protocol with one-way LOCC decoder for
$\left|\psi_{AF}\right\rangle$ since $\bm{\Gamma}$ is essentially an one-way
LOCC decoder from Fred to Bob. Thanks to the above analysis, we can evaluate
the decoding error of $\mathcal{C}$ as follows:

$\displaystyle e(\mathcal{C})\mathrel{\mathop{\mathchar 58\relax}}=$
$\displaystyle\;\frac{1}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[\left(U_{g_{m}}\sigma_{AF}U^{\dagger}_{g_{m}}\right)\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(144a) $\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\;\frac{1}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[U_{g_{m}}\left(2\sum_{l=1}^{L}|\phi_{A,l}\rangle\\!\langle\phi_{A,l}|\otimes|l\rangle\\!\langle
l|_{F}\right)U_{g_{m}}^{\dagger}\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(144b)
$\displaystyle\quad+\frac{1}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[U_{g_{m}}\left(2\sum_{l=1}^{L}\langle\bm{b}^{l}|(\mathbbm{1}-Q)|\psi\rangle\\!\langle\psi|(\mathbbm{1}-Q)|\bm{b}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}\right)U_{g_{m}}^{\dagger}\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(144c) $\displaystyle\stackrel{{\scriptstyle(b)}}{{=}}$
$\displaystyle\;\frac{2(1-\delta)}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[W^{g_{m}}\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(144d)
$\displaystyle\quad+\frac{1}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[U_{g_{m}}\left(2\sum_{l=1}^{L}\langle\bm{b}^{l}|(\mathbbm{1}-Q)|\psi\rangle\\!\langle\psi|(\mathbbm{1}-Q)|\bm{b}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}\right)U_{g_{m}}^{\dagger}\left(\mathbbm{1}-\Gamma^{m}\right)\right]$
(144e) $\displaystyle\stackrel{{\scriptstyle(c)}}{{\leq}}$
$\displaystyle\;\frac{8M^{s}}{(1-\delta)^{s}}\operatorname{Tr}\left[\overline{\sigma}_{A}^{1+s}\right]+\frac{2}{M}\sum_{m=1}^{M}\operatorname{Tr}\left[\sum_{l=1}^{L}\langle\bm{b}^{l}|(\mathbbm{1}-Q)|\psi\rangle\\!\langle\psi|(\mathbbm{1}-Q)|\bm{b}^{l}\rangle\otimes|l\rangle\\!\langle
l|_{F}\right]$ (144f) $\displaystyle\stackrel{{\scriptstyle(d)}}{{\leq}}$
$\displaystyle\;\frac{8M^{s}}{(1-\delta)^{s}}\operatorname{Tr}\left[\overline{\sigma}_{A}^{1+s}\right]+2\delta,$
(144g)

where $(a)$ follows from (140), $(b)$ follows from the definition of
$W^{g}_{BF}$ (141), $(c)$ follows from (143) and $\Gamma^{m}\geq 0$, and $(d)$
follows from that $\\{|\bm{b}_{F}^{l}\rangle\\}$ forms an orthonormal basis of
$F$ and the definition of $\delta$ (120). We are done. ∎

Remark 6 Actually, our achievability proof (Theorem 15 and Lemma 16) is
inspired by the proof for [115, Theorem 1], which we refer to as SVW. In the
following, we compare in detail the similarity and uniqueness between our
proof and SVW. In general, both proofs are composed of two parts. In the first
part, we apply a surjective linear hash function. This mimics choosing the
typical subspaces in SVW. In the second part, Fred performs an measurement on
the Fourier basis. This is the one-shot correspondence to the Fourier basis
measurement in SVW. However, our task is different from the task that is
considered in SVW. As so, we need to invent different operations for both the
sender and the receiver, and manage a different evaluation method for the
decoding error probability. On the other hand, SVW does not assume the uniform
distribution on the codewords _a priori_. However, we do have this assumption
due to the special structure of the task under consideration. This uniformity
assumption renders a more complicated proof so that it becomes more difficult
to derive an exponential upper bound.

####  C.1.2 Asymptotic direct part

Based on the one-shot direct part in Theorem 15, we are able to show the
following coding theorem, which concludes the second inequality of Eq. (105).

######  Theorem 17 (Direct part)

Let $\left|\Psi\right\rangle_{AF}$ be a bipartite pure quantum state. When the
(projective) unitary representation $U$ on $\mathcal{H}_{A}$ satisfies
Assumption 1 (the multiplicity-free condition), it holds that

$\displaystyle\operatorname{H}\left(\mathcal{G}(\Psi_{A})\right)\leq
C_{\mathfrak{E}_{\rm g},\mathfrak{D}_{\rightarrow}}(\Psi_{AF}).$ (145)

###### Proof.

We focus on the functions $-s\overline{\operatorname{H}}_{1+s}(AF)_{\xi}$,
$s\overline{\operatorname{H}}_{1-s}(F|A)_{\xi}$, and
$-s\overline{\operatorname{H}}_{1+s}(A)_{\xi}$, which are convex functions for
$s$. Hence, when $s$ is close to $0$, they are approximated to
$-sH(AF)_{\xi}+\frac{s^{2}}{2}V(AF)_{\xi}$,
$sH(F|A)_{\xi}+\frac{s^{2}}{2}V(F|A)_{\xi}$, and
$-sH(A)_{\xi}+\frac{s^{2}}{2}V(A)_{\xi}$. Hence,

$\displaystyle\min\left(s\overline{\operatorname{H}}_{1+s}(AF)_{\xi}-s\overline{\operatorname{H}}_{1-s}(F|A)_{\xi},s\overline{\operatorname{H}}_{1+s}(A)_{\xi}\right)$
(146)

is approximated as $sH(A)_{\xi}-\frac{s^{2}}{2}V_{\xi}$, where
$V_{\xi}\mathrel{\mathop{\mathchar
58\relax}}=\max(V(A)_{\xi}+V(AF)_{\xi},V(F|A)_{\xi})$. Thus, we have

$\displaystyle
s(-nH(A)_{\xi}+\sqrt{n}r)+\min\big{(}s\overline{\operatorname{H}}_{1+s}(AF)_{\xi^{\otimes
n}}-s\overline{\operatorname{H}}_{1-s}(F|A)_{\xi^{\otimes
n}},s\overline{\operatorname{H}}_{1+s}(A)_{\xi^{\otimes n}}\big{)}$ (147)
$\displaystyle=$ $\displaystyle
s(-nH(A)_{\xi}+\sqrt{n}r)+snH(A)_{\xi}-\frac{ns^{2}}{2}V_{\xi}+o(ns^{2})$
(148) $\displaystyle=$
$\displaystyle\sqrt{n}sr-\frac{ns^{2}}{2}V_{\xi}+o(ns^{2})$ (149)
$\displaystyle=$
$\displaystyle-\frac{n}{2}V_{\xi}(s-\frac{r}{\sqrt{n}V_{\xi}})^{2}+\frac{r^{2}}{2V_{\xi}}+o(ns^{2}).$
(150)

Since the maximum of the above value for $s$ is realized around
$s=\frac{r}{\sqrt{n}V_{\xi}}$, we have

$\displaystyle\lim_{n\to\infty}{\cal L}_{\xi^{\otimes
n}}(-nH(A)_{\xi}+\sqrt{n}r)=\frac{r^{2}}{2V_{\xi}},$ (151)

which implies that

$\displaystyle-{\cal L}_{\xi^{\otimes
n}}^{-1}(-\log\epsilon)=nH(A)_{\xi}-\sqrt{-2nV_{\xi}\log\epsilon}+o(\sqrt{n}).$
(152)

Combining the above result with Theorem 15 yields (145). ∎

###  C.2 Strong converse part

In the strong converse part, we show that
$\operatorname{H}\left(\mathcal{G}(\Psi_{A})\right)$ is a strong converse
bound for all the quantities mentioned in Proposition 6 regardless of
Assumption 1 (the multiplicity-free condition). This concludes the first
inequality of Eq. (105).

######  Theorem 18 (Strong converse part)

Let $\left|\Psi\right\rangle_{AF}$ be a bipartite pure quantum state. It holds
that

$\displaystyle C_{\mathfrak{E}_{\rm ppt},\mathfrak{D}_{\rm
ppt}}^{\dagger}\left(\Psi_{AF}\right)$
$\displaystyle\leq\operatorname{H}\left(\mathcal{G}(\Psi_{A})\right),$ (153)
$\displaystyle C_{\mathfrak{E}_{\rm p},\mathfrak{D}_{\rm
sep}}^{\dagger}\left(\Psi_{AF}\right)$
$\displaystyle\leq\operatorname{H}\left(\mathcal{G}(\Psi_{A})\right).$ (154)

###### Proof.

The key tool to prove Eqs. (153) and (154) is the following inequality, which
is shown in [86, (8.217)]. We denote the transpose operation on $F$ by
$\tau_{F}$. For a any bipartite positive semidefinite rank-one operator $X$ on
$\mathcal{H}_{AF}$, we have the relation

$\displaystyle|\tau_{F}(X)|=\sqrt{\operatorname{Tr}_{F}X}\otimes\sqrt{\operatorname{Tr}_{A}X}.$
(155)

In fact, the reference [86, (8.217)] shows (155) by using the transpose on a
specific basis. While the map $\tau_{F}$ depends on the choice of the basis,
$|\tau_{F}(X)|$ does not depend on it as follows. Consider the map $X\mapsto
U^{\dagger}\tau_{F}(UXU^{\dagger})U$ by using a unitary on $\mathcal{H}_{F}$.
Then, we have

$\displaystyle|U^{\dagger}\tau_{F}(UXU^{\dagger})U|^{2}=U^{\dagger}\tau_{F}(UXU^{\dagger})UU^{\dagger}\tau_{F}(UXU^{\dagger})U=U^{\dagger}\tau_{F}(UXU^{\dagger})\tau_{F}(UXU^{\dagger})U$
(156) $\displaystyle=$ $\displaystyle
U^{\dagger}|\tau_{F}(UXU^{\dagger})|^{2}U=U^{\dagger}\sqrt{\operatorname{Tr}_{F}X}\otimes\sqrt{\operatorname{Tr}_{A}UXU^{\dagger}}U$
(157) $\displaystyle=$
$\displaystyle\sqrt{\operatorname{Tr}_{F}X}\otimes(U^{\dagger}\sqrt{U(\operatorname{Tr}_{A}X)U^{\dagger}}U)=\sqrt{\operatorname{Tr}_{F}X}\otimes\sqrt{\operatorname{Tr}_{A}X}.$
(158)

Hence, $|\tau_{F}(X)|$ does not depend on the choice of basis.

We now show Eq. (153). For any $\epsilon>0$, we choose a sufficiently large
integer $N$ such that any $n\geq N$ satisfies the following two conditions (i)
and (ii):

  1. (i)

There exists a projection $P_{A}$ such that

$\displaystyle[P_{A},U_{g}]$ $\displaystyle=0\hbox{ for }g\in G^{n}$ (159)
$\displaystyle[P_{A},\mathcal{G}(\Psi_{A})^{\otimes n}]$ $\displaystyle=0$
(160) $\displaystyle\operatorname{Tr}P_{A}$ $\displaystyle\leq
e^{n(\operatorname{H}(\mathcal{G}(\Psi_{A}))+\epsilon)},$ (161)
$\displaystyle\operatorname{Tr}(I-P_{A})\mathcal{G}(\Psi_{A})^{\otimes n}$
$\displaystyle\leq\epsilon,$ (162) $\displaystyle\mathcal{G}(P_{A})$
$\displaystyle=P_{A}.$ (163)

  2. (ii)

There exists a projection $P_{F}$ such that

$\displaystyle[P_{F},\Psi_{F}^{\otimes n}]$ $\displaystyle=0,$ (164)
$\displaystyle\operatorname{Tr}P_{F}$ $\displaystyle\leq
e^{n(\operatorname{H}(\Psi_{A})+\epsilon)},$ (165)
$\displaystyle\|P_{F}\Psi_{F}^{\otimes n}P_{F}\|$ $\displaystyle\leq
e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)},$ (166)
$\displaystyle\operatorname{Tr}(I-P_{F})\Psi_{F}^{\otimes n}$
$\displaystyle\leq\epsilon.$ (167)

The conditions (159) and (162) imply that

$\displaystyle\operatorname{Tr}(I-P_{A})\Psi_{A}^{\otimes n}\leq\epsilon.$
(168)

Hence, $[P_{A},P_{F}]=0$ and

$\displaystyle\operatorname{Tr}(I-P_{A}\otimes P_{F})\Psi_{AF}^{\otimes n}\leq
2\epsilon.$ (169)

Let $\mathcal{C}=(\\{\mathcal{E}^{m}\\},\\{\Gamma^{m}\\})\in(\mathfrak{E}_{\rm
ppt},\mathfrak{D}_{\rm ppt})$ be a code for the state $\Psi_{AF}^{\otimes n}$.
Since $\Gamma^{m}$ is a PPT operator and $\mathcal{E}^{m}\in\mathfrak{E}_{\rm
ppt}$, $(\mathcal{E}^{m})^{*}(\Gamma^{m})$ is also a PPT operator, i.e.,

$\displaystyle\tau_{F}((\mathcal{E}^{m})^{*}(\Gamma^{m}))\geq 0.$ (170)

By applying (155) to $(I\otimes P_{F})\Psi_{AF}^{\otimes n}(I\otimes P_{F})$,
the evaluation (166) guarantees that

$\displaystyle\|\tau_{F}((P_{A}P_{F})\Psi_{AF}^{\otimes n}(P_{A}P_{F}))\|$
$\displaystyle=\|P_{A}\otimes I\tau_{F}((I\otimes P_{F})\Psi_{AF}^{\otimes
n}(I\otimes P_{F}))P_{A}\otimes I\|$ (171)
$\displaystyle\leq\|\tau_{F}((I\otimes P_{F})\Psi_{AF}^{\otimes n}(I\otimes
P_{F}))\|$ (172) $\displaystyle\leq
e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}.$ (173)

Hence, we have

$\displaystyle|\tau_{F}((P_{A}\otimes P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes
P_{F}))|\leq e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes P_{F}.$
(174)

Then, we have

$\displaystyle\operatorname{Tr}\Gamma^{m}\mathcal{E}^{m}((P_{A}\otimes
P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes P_{F}))$ (175) $\displaystyle=$
$\displaystyle\operatorname{Tr}(\mathcal{E}^{m})^{*}(\Gamma^{m})(P_{A}\otimes
P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes P_{F})$ (176) $\displaystyle=$
$\displaystyle\operatorname{Tr}\tau_{F}((\mathcal{E}^{m})^{*}(\Gamma^{m}))\tau_{F}((P_{A}\otimes
P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes P_{F}))$ (177)
$\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\operatorname{Tr}\tau_{F}((\mathcal{E}^{m})^{*}(\Gamma^{m}))|\tau_{F}((P_{A}\otimes
P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes P_{F}))|$ (178)
$\displaystyle\stackrel{{\scriptstyle(b)}}{{\leq}}$
$\displaystyle\operatorname{Tr}\tau_{F}((\mathcal{E}^{m})^{*}(\Gamma^{m}))e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes
P_{F}$ (179) $\displaystyle\stackrel{{\scriptstyle(c)}}{{=}}$
$\displaystyle\operatorname{Tr}(\mathcal{E}^{m})^{*}(\tau_{F}(\Gamma^{m}))\mathcal{G}(e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes
P_{F})$ (180) $\displaystyle=$
$\displaystyle\operatorname{Tr}\tau_{F}(\Gamma^{m})\mathcal{E}^{m}(\mathcal{G}(e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes
P_{F}))$ (181) $\displaystyle=$
$\displaystyle\operatorname{Tr}\tau_{F}(\Gamma^{m})\mathcal{G}(e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes
P_{F})$ (182) $\displaystyle=$
$\displaystyle\operatorname{Tr}\tau_{F}(\Gamma^{m})e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}P_{A}\otimes
P_{F}$ (183) $\displaystyle=$ $\displaystyle
e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}\operatorname{Tr}\Gamma^{m}P_{A}\otimes
P_{F},$ (184)

where $(a)$, $(b)$ and $(c)$ follow from (170), (174) and (163), respectively.
Hence, we have

$\displaystyle s(\mathcal{C})=$ $\displaystyle\frac{1}{|{\cal
M}|}\sum_{m}\operatorname{Tr}\Gamma^{m}\mathcal{E}^{m}(\Psi_{AF}^{\otimes n})$
(185) $\displaystyle\stackrel{{\scriptstyle(a)}}{{\leq}}$
$\displaystyle\frac{1}{|{\cal
M}|}\sum_{m}\operatorname{Tr}\Gamma^{m}\mathcal{E}^{m}((P_{A}\otimes
P_{F})\Psi_{AF}^{\otimes n}(P_{A}\otimes P_{F}))+2\epsilon$ (186)
$\displaystyle\stackrel{{\scriptstyle(b)}}{{\leq}}$
$\displaystyle\frac{1}{|{\cal
M}|}e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}\sum_{m}\operatorname{Tr}\Gamma^{m}P_{A}\otimes
P_{F}+2\epsilon$ (187) $\displaystyle=$ $\displaystyle\frac{1}{|{\cal
M}|}e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}\operatorname{Tr}P_{A}\otimes
P_{F}+2\epsilon$ (188) $\displaystyle\stackrel{{\scriptstyle(c)}}{{\leq}}$
$\displaystyle\frac{1}{|{\cal
M}|}e^{-n(\operatorname{H}(\Psi_{A})-\epsilon)}e^{n(\operatorname{H}(\mathcal{G}(\Psi_{A}))+\epsilon)}e^{n(\operatorname{H}(\Psi_{A})+\epsilon)}+2\epsilon$
(189) $\displaystyle\leq$ $\displaystyle\frac{1}{|{\cal
M}|}e^{n(\operatorname{H}(\mathcal{G}(\Psi_{A}))+3\epsilon)}+2\epsilon,$ (190)

where $(a)$, $(b)$, and $(c)$ follow from the pair of (162) and (167), (184)
and the pair of (161) and (165), respectively. Thus,

$\displaystyle|{\cal
M}|\leq\frac{1}{s(\mathcal{C})-2\epsilon}e^{n(\operatorname{H}(\mathcal{G}(\Psi_{A}))+3\epsilon)},$
(191)

which implies (153).

Next, we show (154). Let
$\mathcal{C}=(\\{\mathcal{E}^{m}\\},\\{\Gamma^{m}\\})\in(\mathfrak{E}_{\rm
p},\mathfrak{D}_{\rm sep})$ be a code for the state $\Psi_{AF}^{\otimes n}$.
Since $\Gamma^{m}$ is a separable operator and
$\mathcal{E}^{m}\in\mathfrak{E}_{\rm p}$, $(\mathcal{E}^{m})^{*}(\Gamma^{m})$
is also a separable operator. Hence, $(\mathcal{E}^{m})^{*}(\Gamma^{m})$ is a
PPT operator, i.e., we have (170). Therefore, in the same way, we can show
(191), which implies (154). ∎

##  Appendix D Proof of Theorem 10

###### Proof.

Applying the channel coding theorem to the classical-quantum channel $g\mapsto
U_{g}\Psi_{A}U_{g}^{\dagger}$ [86], we can easily derive the following
capacity formula:

$\displaystyle C_{\mathfrak{E}_{\rm
g},\mathfrak{D}_{\emptyset}}(\Psi_{AF})=\operatorname{D}\left(\Psi_{A}\|\mathcal{G}(\Psi_{A})\right).$
(192)

Hence, to show (49) it is sufficient to show the strong converse part

$\displaystyle C^{\dagger}_{\mathfrak{E}_{\rm
p},\mathfrak{D}_{\emptyset}}(\Psi_{AF})\leq\operatorname{D}\left(\Psi_{A}\|\mathcal{G}(\Psi_{A})\right).$
(193)

In almost the same way as (43c), the strong converse argument (193) can be
shown by invoking the meta-converse technique originally invented in [132] and
further investigated in [133, Chapter 3]. In the following Proposition 19
(which will be proved shortly), we upper bound the success probability of any
one-shot code $\mathcal{C}\in(\mathfrak{E}_{\rm p},\mathfrak{D}_{\emptyset})$
in terms of the sandwiched quantum Rényi entropy, then the strong converse
bound follows by block coding. Notice that $\lim_{\alpha\to
1}\widetilde{\operatorname{D}}_{\alpha}\left(\Psi_{A}\middle\|\mathcal{G}(\Psi_{A})\right)=\operatorname{D}\left(\Psi_{A}\middle\|\mathcal{G}(\Psi_{A})\right)$.
What’s more, $\widetilde{\operatorname{D}}_{\alpha}$ is continuous and
monotonically decreasing in $\alpha$. Applying the standard argument outlined
in [132, 134], we obtain from Proposition 14 that
$\operatorname{D}\left(\Psi_{A}\middle\|\mathcal{G}(\Psi_{A})\right)$ is
actually a strong converse bound. ∎

###### Proposition 19

Any dense coding code $\mathcal{C}\in(\mathfrak{E}_{\rm
p},\mathfrak{D}_{\emptyset})$ obeys the following bound for arbitrary
$\alpha\in(1,\infty)$:

$\displaystyle
s(\mathcal{C})\leq\exp\left\\{\frac{\alpha-1}{\alpha}\left(\widetilde{\operatorname{D}}_{\alpha}\left(\Psi_{A}\middle\|\mathcal{G}(\Psi_{A})\right)-\log|\mathcal{C}|\right)\right\\},$
(194)

where $\widetilde{\operatorname{D}}_{\alpha}$ is the sandwiched quantum Rényi
entropy defined in (6).

###### Proof.

Given a code
$\mathcal{C}=(\\{\mathcal{E}^{m}\\}_{m},\\{\Gamma^{\widehat{m}}\\}_{\widehat{m}})\in(\mathfrak{E}_{\rm
p},\mathfrak{D}_{\emptyset})$, we define the following two quantum states:

$\displaystyle\rho_{MA}$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{|\mathcal{M}|}\sum_{m}|m\rangle\\!\langle
m|_{M}\otimes\mathcal{E}^{m}(\Psi_{A}),$ (195) $\displaystyle\sigma_{MA}$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\pi_{M}\otimes\mathcal{G}(\Psi_{A}),$ (196)

where $\xi_{MXA}$ serves as a _test state_. The positive operator

$\displaystyle T\mathrel{\mathop{\mathchar
58\relax}}=\sum_{m}|m\rangle\\!\langle m|_{M}\otimes\Gamma^{m}$ (197)

satisfies

$\displaystyle\operatorname{Tr}T\rho_{MA}$
$\displaystyle=\frac{1}{|\mathcal{M}|}\sum_{m}p_{\widehat{M}M}(m|m)=s(\mathcal{C}),$
(198) $\displaystyle\operatorname{Tr}T\sigma_{MA}$
$\displaystyle=\frac{1}{|\mathcal{M}|}\operatorname{Tr}\left[\sum_{m}\Gamma^{m}\mathcal{G}(\Psi_{A})\right]\stackrel{{\scriptstyle(a)}}{{=}}\frac{1}{|\mathcal{M}|}\operatorname{Tr}\left[\mathcal{G}(\Psi_{A})\right]=\frac{1}{|\mathcal{M}|},$
(199)

where $(a)$ follows from the fact that $\\{\Gamma^{m}\\}_{m}$ is a quantum
measurement. Applying the information processing inequality to the binary
measurement $\\{T,I-T\\}$, we have

$\displaystyle
s(\mathcal{C})^{\alpha}\cdot\left(\frac{1}{|\mathcal{C}|}\right)^{1-\alpha}+(1-s(\mathcal{C}))^{\alpha}\cdot\left(1-\frac{1}{|\mathcal{C}|}\right)^{1-\alpha}$
$\displaystyle\leq
e^{(\alpha-1)\overline{\operatorname{D}}_{\alpha}\left(\rho_{MA}\middle\|\sigma_{MA}\right)}.$
(200)

Thus, we have the following:

$\displaystyle s(\mathcal{C})^{\alpha}\cdot|\mathcal{C}|^{\alpha-1}$
$\displaystyle=s(\mathcal{C})^{\alpha}\cdot\left(\frac{1}{|\mathcal{C}|}\right)^{1-\alpha}\leq
s(\mathcal{C})^{\alpha}\cdot\left(\frac{1}{|\mathcal{C}|}\right)^{1-\alpha}+(1-s(\mathcal{C}))^{\alpha}\cdot\left(1-\frac{1}{|\mathcal{C}|}\right)^{1-\alpha}$
(201) $\displaystyle\leq
e^{(\alpha-1)\overline{\operatorname{D}}_{\alpha}\left(\rho_{MA}\middle\|\sigma_{MA}\right)}$
(202)
$\displaystyle=\frac{1}{|\mathcal{M}|}\sum_{m}e^{(\alpha-1)\widetilde{\operatorname{D}}_{\alpha}\left(\mathcal{E}^{m}(\Psi_{A})\middle\|\mathcal{G}(\Psi_{A})\right)}$
(203)
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}\frac{1}{|\mathcal{M}|}\sum_{m}e^{(\alpha-1)\widetilde{\operatorname{D}}_{\alpha}\left(\mathcal{E}^{m}(\Psi_{A})\middle\|\mathcal{E}^{m}\circ\mathcal{G}(\Psi_{A})\right)}$
(204)
$\displaystyle\stackrel{{\scriptstyle(b)}}{{\leq}}\frac{1}{|\mathcal{M}|}\sum_{m}e^{(\alpha-1)\widetilde{\operatorname{D}}_{\alpha}\left(\Psi_{A}\middle\|\mathcal{G}(\Psi_{A})\right)},$
(205)

where $(a)$ follows from the condition
$\mathcal{E}^{m}\circ\mathcal{G}=\mathcal{G}$ for arbitrary
$\mathcal{E}^{m}\in\mathfrak{E}_{\rm p}$ (cf. the definition in Eq. (37)) and
$(b)$ follows from the information processing inequality of the sandwiched
quantum Rényi entropy for positive maps [131, Theorem 2]. ∎

##  Appendix E Proof of Theorems 12 and 13

Proof of Theorem 12 Notice that Eq. (66) can be shown in the same way as
(43c).

Now we show Eq. (67). For a state $\rho_{AF}$, we have a trace-preserving
positive operation $\mathcal{E}_{F}$ on $\mathcal{H}_{F}$ such that
$\rho_{AF}=\mathcal{E}_{F}(\Psi_{AF})$, where $\Psi_{AF}$ is a purification of
$\rho_{A}$. Let
$\mathcal{C}=(\\{\mathcal{E}^{m}\\},\\{\Gamma^{m}\\})\in(\mathfrak{E}_{\rm
p},\mathfrak{D}_{\rm sep})$ be a code for the state $\rho_{AF}^{\otimes n}$.
Since $\\{\mathcal{E}_{F}^{*}(\Gamma^{m})\\}$ is a separable measurement,
where $\mathcal{E}_{F}^{*}$ is the dual map of $\mathcal{E}_{F}$, we define a
code
$\hat{\mathcal{C}}=(\\{\mathcal{E}^{m}\\},\\{\mathcal{E}_{F}^{*}(\Gamma^{m})\\})$
in the encoder-decoder pair $\in(\mathfrak{E}_{\rm p},\mathfrak{D}_{\rm sep})$
for the state $\Psi_{AF}^{\otimes n}$. Since
$s(\mathcal{C})=s(\hat{\mathcal{C}})$, (67) follows from (154).

Next, we show (68). For a state $\rho_{AF}^{\prime}$, we have a trace-
preserving operation $\mathcal{E}_{F}^{\prime}\in\mathscr{C}(F\to F)_{\rm
ppt}$ such that $\rho_{AF}=\mathcal{E}_{F}^{\prime}(\Psi_{AF})$. Let
$\mathcal{C}=(\\{\mathcal{E}^{m}\\},\\{\Gamma^{m}\\})\in(\mathfrak{E}_{\rm
ppt},\mathfrak{D}_{\rm ppt})$ be a code for the state
${\rho_{AF}^{\prime}}^{\otimes n}$. Since
$\\{{\mathcal{E}_{F}^{\prime}}^{*}(\Gamma^{m})\\}$ is a separable measurement,
we define a code
$\hat{\mathcal{C}}^{\prime}=(\\{\mathcal{E}^{m}\\},\\{{\mathcal{E}_{F}^{\prime}}^{*}(\Gamma^{m})\\})$
in the encoder-decoder pair $\in(\mathfrak{E}_{\rm ppt},\mathfrak{D}_{\rm
ppt})$ for the state $\Psi_{AF}^{\otimes n}$. Since
$s(\mathcal{C})=s(\hat{\mathcal{C}}^{\prime})$, (68) follows from (153).
$\blacksquare$

Proof of Theorem 13 First, we show Eq. (69). Let
$\mathcal{C}=(\\{\mathcal{E}^{m}\\},\\{\Lambda^{x}\\},\\{\Gamma^{\widehat{m}|x}\\})$
be a code in the encoder-decoder pair $(\mathfrak{E}_{\rm
g},\mathfrak{D}_{\rightarrow})$ for the state $\Psi_{AF}^{\otimes n}$. Since
the operation $\mathcal{E}_{F}$ is a trace-preserving positive operation, the
operation ${\mathcal{E}_{F}}^{*}$ is a unit-preserving positive operation.
Since $\\{{\mathcal{E}_{F}}^{*}(\Lambda^{x})\\}$ is a POVM on
$\mathcal{H}_{F}^{\otimes n}$, we define a code
$\hat{\mathcal{C}}=(\\{\mathcal{E}^{m}\\},\\{{\mathcal{E}_{F}}^{*}(\Lambda^{x})\\},\\{\Gamma^{\widehat{m}|x}\\})$
in the encoder-decoder pair $(\mathfrak{E}_{\rm
g},\mathfrak{D}_{\rightarrow})$ for the state $\rho_{AF}^{\otimes n}$. Since
$s(\mathcal{C})=s(\hat{\mathcal{C}})$, (69) follows from (145).

Next, we show Eq. (70). Let
$\mathcal{C}=(\\{{\mathcal{E}^{m}}^{\prime}\\},\\{{\Lambda^{x}}^{\prime}\\},\\{{\Gamma^{\widehat{m}|x}}^{\prime}\\})$
be a code in the encoder-decoder pair $(\mathfrak{E}_{\rm
g},\mathfrak{D}_{\rightarrow})$ for the state ${\Psi_{AF}^{\prime}}^{\otimes
n}$. Since $\\{{{\mathcal{E}_{F}^{\prime}}}^{*}(\Lambda^{x})\\}$ is a POVM on
$\mathcal{H}_{F}^{\otimes n}$, we define a code
$\hat{\mathcal{C}}^{\prime}=(\\{{\mathcal{E}^{m}}^{\prime}\\},\\{{{\mathcal{E}_{F}^{\prime}}}^{*}({\Lambda^{x}}^{\prime})\\},\\{{\Gamma^{\widehat{m}|x}}^{\prime}\\})$
in the encoder-decoder pair $(\mathfrak{E}_{\rm
g},\mathfrak{D}_{\rightarrow})$ for the state ${\rho_{AF}^{\prime}}^{\otimes
n}$. Since $s(\mathcal{C})=s(\hat{\mathcal{C}}^{\prime})$, (70) follows from
(145). $\blacksquare$