Entanglement destroying channels

In a previous post, we were concerned with channels of the form $\Phi\in\Channel(\X,\Y)$ such that  $\bigl(\Phi\otimes \I_{\Lin(\Z)}\bigr)(\rho) \in \Sep(\Y:\Z)$ for every complex Euclidean space $\Z$ and every density operator $\rho\in\Density(\X\otimes\Z)$. Channels of this form have the effect of destroying entanglement that exists between the register they act on and any other registers.

Theorem:
There exist two channels $\Phi_0,\Phi_1\in\Channel(\X,\Y)$, both having the property described above, such that
  \[
  \bigtriplenorm{\Phi_0 - \Phi_1}_1
  > \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1
  \]
  for every $\rho\in\Density(\X)$. (Channels like this have the strange property that they destroy entanglement, and yet evaluating them on an entangled state helps to distinguish them.)


Proof:

For $\lambda\in[0,1]$, consider the two channels $\Phi_0(X),\Phi_0(X)\in\Channel(X)$ defined by
\[\begin{align*}
\Phi_0(X)&=\frac{\lambda}{n+1}(\tr(X)\I_\X+X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X \\
\Phi_1(X)&=\frac{\lambda}{n-1}(\tr(X)\I_\X-X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X
\end{align*}\]
Then for sufficiently small $\lambda\in[0,1]$ both of the Choi representations $J(\Phi_0(X))$ and $J(\Phi_1(X))$ are in a separable neighborhood of the maximally mixed state which implies that they are both separable by some theorem. Therefore, from the results in the previous post, we have that $\Phi_0(X)$ and $\Phi_0(X)$ are entanglement destroying as described in the problem statement.

Now considering that
\[
\Phi_0(\rho) - \Phi_1(\rho)=\frac{-2\lambda}{(n+1)(n-1)}\rho^T,
\]
it follows that
\[
\bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}\bignorm{\rho^T}_1=\frac{2\lambda}{(n+1)(n-1)},
\]
since $\rho\in\Density(\X)$.

Moreover, since
\[\begin{align*}
\bigtriplenorm{\Phi_0 - \Phi_1}_1&=max\{\bignorm{((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)}_1 \ : \ x\in S(\X\otimes\X)\} ,
\end{align*}\]
where $((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)$ gives the partial transpose of $xx^\ast$ (which is at most $n$ since $x\in S(\X\otimes\X)$) multiplied by the scalar quantity $\frac{2\lambda}{(n+1)(n-1)}$. Therefore $\bigtriplenorm{\Phi_0 - \Phi_1}_1=\frac{2\lambda n}{(n+1)(n-1)}$, which implies that
  \[
  \frac{2\lambda n}{(n+1)(n-1)}=\bigtriplenorm{\Phi_0 - \Phi_1}_1
  > \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}.
  \]

Bounding the norm of the Choi representation of a channel in terms of its operator norm

Theorem:

Let $\X$ and $\Y$ be complex Euclidean spaces with $\dim(\X) = n$ and let $\Phi\in\Trans(\X,\Y)$. Let $\norm{\cdot}_1$ denote the usual trace norm of an density operator, and $\triplenorm{\cdot}_1$ the operator norm of a channel.
  \[
  \triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
  \triplenorm{\Phi}_1.  \]

Proof:

Since the Choi representation of $\Phi$ is given by $J(\Phi)=(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)$, then the trace norm is given by and also satisfies
\[\begin{align*}
\norm{J(\Phi)}_1&=\norm{(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)}_1 \\
&\leq \norm{(\Phi\otimes\I_{\Lin(\X)})}_1\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1 \\
&=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1n \\
&=n  \triplenorm{\Phi}_1,
\end{align*}\]
where the last two lines from $\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1=n$ and the definition of the completely bounded trace norm $\triplenorm{\Phi}_1 :=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1$.

Now consider an alternate characterization of the completely bounded trace norm:
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}.
\]
Since this norm satisfies the property
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_0}}_\infty\norm{J(\Phi)}_1\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_1}}_\infty,
\]
and the spectral norm $\norm{A}_\infty$ of an operator $A$ is given by the largest singular value of $A$, then it follows that $\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_a}}_\infty\leq1$ with equality holding in the case where $\rho_a$ is a pure state. Therefore,
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{J(\Phi)}_1,
\]
implying that
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}\leq \norm{J(\Phi)}_1.
\]
Putting these two bounds together then gives
\[
  \triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
  \triplenorm{\Phi}_1.
\]

Separable channels decrease the entaglement of formation

The entanglement of formation of a density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$ is defined as
  \[
  E_{f}(\rho) = \inf\Biggl\{\sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast})
  \,:\, \rho = \sum_{a\in\Sigma} p(a) u_a u_a^{\ast} \Biggr\},
  \]
  where $E(u u^{\ast}) = S(\tr_{\X^{B}}(u u^{\ast}))$ denotes the entanglement entropy of the pure state $u u^{\ast}$ and the infimum is over all expressions of $\rho$ of the given form, where $\Sigma$ is any alphabet, $p\in\P(\Sigma)$ is a probability vector, and $\{u_a\,:\,a\in\Sigma\} \subset \X^{A}\otimes\X^{B}$ is a collection of unit vectors.

Theorem:

For every choice of complex Euclidean spaces $\X^{A}$, $\X^{B}$, $\Y^{A}$, and $\Y^{B}$, every density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$, and every separable channel $\Phi\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$, it holds that
  \[
  E_{f}(\Phi(\rho)) \leq E_{f}(\rho).
  \]

Proof:

Assuming that $\Phi\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$ allows $\Phi$ to be expressed as
\[
\Phi(X)=\sum_{b\in\Gamma}(A_b\otimes B_b)X(A_b^\ast\otimes B_b^\ast),
\]
where $\Gamma$ is some alphabet and $\{A_b : b\in \Gamma\}\subset \Pos(\X^A)$ and $\{B_b : b\in \Gamma\}\subset \Pos(\X^B)$. For $\rho = \sum_{a\in\Sigma} p(a) u_a u_a^{\ast}$, the action of $\Phi$ on $\rho$ is specified by the action of $\Phi$ on each $u_au_a^\ast$ as
\[
\Phi(\rho)=\sum_{a\in\Sigma} p(a) \Phi(u_a u_a^{\ast})=\sum_{a\in\Sigma} p(a)\sum_{b\in\Gamma}(A_b\otimes B_b)u_au_a^*(A_b^\ast\otimes B_b^\ast).
\]
Therefore, represent $\Phi(u_au_a*)$ as
\[
\Phi(u_au_a*)=\sum_{b\in\Gamma}(A_b\otimes B_b)u_au_a^*(A_b^\ast\otimes B_b^\ast)=\sum_{b\in\Gamma}q_a(b) v_{ab} v_{ab}^{\ast},
\]
where $(A_b\otimes B_b)u_a=\sqrt{q_a(b)}v_{ab}$. Now let
\[
C_b=\frac{1}{\sqrt{q_a(b)}}(A_b\otimes B_b),
\]
so that $C_bu_au_a^\ast C_b^\ast=v_{ab}v_{ab}$.

Consider the channel $\Psi_{ab}\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$ defined by
\[
\Psi_{ab}(X)=C_bXC_b^\ast+(\tr(X)-\tr(C_bXC_b^\ast))\sigma,
\]
for some arbitrary $\sigma\in\Density(\Y^A\otimes\Y^B)$. Then $\Psi_{ab}$ is indeed a channel since it is completely positive because it is defined in terms of $C_b$ and $\Phi$ is assumed to be completely positive. Likewise, $\Psi_{ab}$ is separable since $\Phi$ is separable. Moreover, $\Psi_{ab}$ is trace preserving since
\[\begin{align*}
\tr(\Psi_{ab}(X))&=\tr(C_bXC_b^\ast)+(\tr(X)-\tr(C_bXC_b^\ast))\tr(\sigma) \\
&=\tr(C_bXC_b^\ast)+\tr(X)-\tr(C_bXC_b^\ast) \\
&=\tr(X).
\end{align*}\]
By construction $\Psi_{ab}(u_au_a^\ast)=v_{ab}v_{ab}^\ast$.
 Therefore, by a corollary (6.36) to Nielsen's theorem it follows that for every $a\in\Sigma$ with $\rho_a^A=\tr_{\X^B}(u_au_a^\ast)$ and $\sigma_a^A=\tr_{\X^B}(v_{ab}v_{ab}^\ast)$ and $r=min\{rank(\rho_a^A),rank(\sigma^A_a)\}$ it holds that
\[
\lambda_1(\rho_a^A)+\dots+\lambda_1(\rho_m^A)\leq \lambda_1(\sigma_a^A)+\dots+\lambda_1(\sigma_m^A)
\]
for every $m\in\{1,\dots, r\}$.

Thus, the von Neummann entropy satisfies $S(\sigma_a^A)\leq S(\rho_a^A)$, which implies that the entanglement entropy also satisfies $E(v_{ab}v_{ab}^\ast)\leq E(u_au_a)$ for all $a\in\Sigma$ and $b\in\Gamma$. Then by tracing out system $B$, and taking the weighted average that is described the original state $\rho$ and using the joint convexity of the von Neumann entropy it follows that 
\[\begin{align*}
\sum_{a\in\Sigma} p(a)\sum_{b\in\Gamma}q_a(b) E(v_{ab} v_{ab}^{\ast})
\leq \sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast}).
\end{align*}\]

 Therefore, by definition $E_{f}(\Phi(\rho)) \leq E_{f}(\rho)$.

The SWAP operator and separable measurements

Let $\Sigma$ be an alphabet, let $n = \abs{\Sigma}$, and assume $n\geq 2$. Also let $\X^{A} = \mathbb{C}^{\Sigma}$ and $\X^{B} = \mathbb{C}^{\Sigma}$,  and recall that the swap operator $W\in\Lin(\X^{A}\otimes\X^{B})$ may be defined as
\[
    W = \sum_{a,b\in\Sigma} E_{a,b} \otimes E_{b,a}.
\]
Define $\Pi_0,\,\Pi_1\in\Proj(\X^{A}\otimes\X^{B})$ and $\sigma_0,\sigma_1\in\Density(\X^{A}\otimes\X^{B})$ as follows:
  \[
  \Pi_0 = \frac{1}{2} \I\otimes\I + \frac{1}{2} W,\qquad
  \Pi_1 = \frac{1}{2} \I\otimes\I - \frac{1}{2} W,\qquad
  \sigma_0 = \frac{1}{\binom{n+1}{2}}\Pi_0,\qquad
  \sigma_1 = \frac{1}{\binom{n}{2}}\Pi_1.
  \]

Theorem:

If $\mu:\{0,1\}\rightarrow\Pos(\X^{A}\otimes\X^{B})$ is a separable measurement, then
  \[
  \frac{1}{2} \ip{\mu(0)}{\sigma_0}
  + \frac{1}{2} \ip{\mu(1)}{\sigma_1}
  \leq \frac{1}{2} + \frac{1}{n+1}.
  \]


Proof:

Assuming that $\mu$ is a separable measurement allows $\mu(0)$ to be expressed as
\[
\mu(0)=\sum_{a\in\Gamma}P_a\otimes Q_a,
\]
where $\{P_a : a\in \Gamma\}\subset \Pos(\X^A)$ and $\{Q_a : a\in \Gamma\}\subset \Pos(\X^B)$. Moreover, since $\mu$ is a measurement it must satisfy the completeness condition that $\mu(0)+\mu(1)=\I\otimes\I$ implying that $\mu(1)$ can be expressed in terms of $\mu(0)$ as
\[
\mu(1)=\I\otimes\I-\mu(0)=\I\otimes\I-\sum_{a\in\Gamma}P_a\otimes Q_a.
\]
Write $\sigma_0$ and $\sigma_1$ more explicitly as
\[\begin{align*}
 \sigma_0 &= \frac{1}{\binom{n+1}{2}}\Pi_0=\frac{1}{(n+1)n}(\I\otimes\I + W), \\
 \sigma_1 &= \frac{1}{\binom{n+1}{2}}\Pi_0=\frac{1}{(n-1)n}(\I\otimes\I - W).
\end{align*}\]
Then,
\[\begin{align*}
\ip{\mu(0)}{\sigma_0}
+ \ip{\mu(1)}{\sigma_1}
&= \ip{\mu(0)}{\sigma_0}
+ \ip{\I\otimes\I-\mu(0)}{\sigma_1} \\
&=\frac{1}{(n+1)n}\ip{\mu(0)}{\I\otimes\I + W}
+ \frac{1}{(n-1)n}\ip{\I\otimes\I-\mu(0)}{\I\otimes\I - W} \\
&=\frac{1}{(n+1)n}(\ip{\mu(0)}{\I\otimes\I }+\ip{\mu(0)}{ W}) \\
& \ \ \ + \frac{1}{(n-1)n}(\ip{\I\otimes\I}{\I\otimes\I} -\ip{\I\otimes\I}{W} ) \\
& \ \ \ + \frac{1}{(n-1)n}(\ip{\mu(0)}{W}-\ip{\mu(0)}{\I\otimes\I} ).
\end{align*}\]
Now observe that
\[
\ip{\mu(0)}{\I\otimes\I }=\tr\left(\mu(0)^\ast\I\otimes\I\right)=\tr(\sum_{a\in\Gamma}(P_a\otimes Q_a)(\I\otimes\I))=\tr(\sum_{a\in\Gamma}P_a\otimes Q_a)=\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a),
\]
\[
\ip{\I\otimes\I}{\I\otimes\I}=\tr((\I\otimes\I)(\I\otimes\I))=\tr(\I\otimes\I)=n^2,
\]
\[\begin{align*}
\ip{\mu(0)}{ W}=\tr(\sum_{a\in\Gamma}(P_a\otimes Q_a)W)&=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast\otimes e_j^\ast)(P_a\otimes Q_a)W(e_i\otimes e_j) \\
&=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast\otimes e_j^\ast)(P_a\otimes Q_a)(e_j\otimes e_i) \\
&=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}(e_i^\ast P_a e_j)\otimes(e_j^\ast Q_a e_i) \\
&=\sum_{a\in\Gamma}\sum_{i,j\in\Sigma}e_i^\ast P_a e_je_j^\ast Q_a e_i \\
&=\sum_{a\in\Gamma}\sum_{i\in\Sigma}e_i^\ast P_a Q_a e_i \\
&=\sum_{a\in\Gamma}\tr(P_a Q_a), \\
\end{align*}\]
and by similar arguments used in the previous calculation $\ip{\I\otimes\I}{W}=\tr(\I\I)=\tr(\I)=n$.
Therefore, the original expression of interest can be simplified as
\[\begin{align*}
\ip{\mu(0)}{\sigma_0}
+ \ip{\mu(1)}{\sigma_1}&=\frac{1}{(n+1)n}(\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)+\sum_{a\in\Gamma}\tr(P_a Q_a)) \\
& \ \ \ + \frac{1}{(n-1)n}(n^2 -n ) \\
& \ \ \ + \frac{1}{(n-1)n}(\sum_{a\in\Gamma}\tr(P_a Q_a)-\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a) ) \\
&=1+\frac{2}{(n+1)(n-1)n}\left(n\sum_{a\in\Gamma}\tr(P_a Q_a)-\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)\right) \\
&\leq 1+\frac{2}{(n+1)(n-1)n}(n^2-n) \\
&=1+\frac{2}{(n+1)}.
\end{align*}\]
Here, the inequality follows from the fact that the quantity $\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)$ is minimized when the projectors $P_a$ and $Q_a$ both have rank $1$ so that $\tr(P_a)\tr(Q_a)=1$ implying that $\sum_{a\in\Gamma}\tr(P_a)\tr(Q_a)=n$.
Furthermore, in this case,  $\sum_{a\in\Gamma}\tr(P_a Q_a)=n$.

Therefore, dividing both sides of the inequality by $2$ gives
\[
  \frac{1}{2} \ip{\mu(0)}{\sigma_0}
  + \frac{1}{2} \ip{\mu(1)}{\sigma_1}
  \leq \frac{1}{2} + \frac{1}{n+1}.
\]

Singular values before and after the action of a unital channel

Let $\X$ be a complex Euclidean space having dimension $n$, let $\Phi\in\Channel(\X)$ be a unital channel, let $X\in\Lin(\X)$ be an operator, and let $Y = \Phi(X)$. Following our usual conventions, let $s_1(X) \geq \cdots \geq s_n(X)$ and $s_1(Y) \geq \cdots \geq s_n(Y)$ denote the singular values of $X$ and $Y$, respectively, ordered from largest to smallest, and where we take $s_k(X) = 0$ when $k > \rank(X)$ and $s_k(Y) = 0$ when $k > \rank(Y)$.

Theorem:

  $s_1(X) + \cdots + s_m(X) \geq s_1(Y) + \cdots + s_m(Y)$ for every $m \in \{1,\ldots,n\}$. 

Proof:

Consider the space $\X\oplus\X$ and let
\[
 \overline{\X}:=
 \begin{pmatrix}
      0 & X\\
      X^{\ast} & 0
    \end{pmatrix}.
\]
Then it holds that $\overline{X}=\overline{X}^\ast$ so that $\overline{X}\in\Herm(\X\oplus\X)$. In addition, consider the channel $\overline{\Phi}\in\Channel(\X\oplus\X)$ defined as
\[
\overline{\Phi} \begin{pmatrix}
      A & B\\
      C & D
    \end{pmatrix}=
    \begin{pmatrix}
      \Phi(A) & \Phi(B)\\
      \Phi(C) & \Phi(D)
    \end{pmatrix}.
\]
Then it follows that
\[
\overline{\Phi}(\I_{\X\oplus\X})     \begin{pmatrix}
      \Phi(\I_{\X}) & 0\\
      0 & \Phi(\I_{\X})
    \end{pmatrix}=
        \begin{pmatrix}
      \I_{\X} & 0\\
      0 & \I_{\X}
    \end{pmatrix}=
    \I_{\X\oplus\X},
\]
 which implies that $\overline{\Phi}$ is unital. Moreover, letting $\Phi(X)=Y$ with $\Phi(X^\ast)=Y^\ast$ yields
\[
 \overline{\Phi}(\overline{X})=
 \begin{pmatrix}
      0&\Phi(X)\\
      \Phi(X^\ast) & 0)
    \end{pmatrix}=
        \begin{pmatrix}
      0 & Y\\
      Y^\ast & 0
    \end{pmatrix}=:\overline{Y}.
\]
 It has now been shown that there exists a unital channel $\overline{\Phi}\in \Trans(\X\oplus\X)$ such that $\overline{\Phi}(\overline{X})=\overline{Y}$, where $\overline{X},\overline{Y}\in\Herm(\X\oplus\X)$. Therefore, by Uhlmann's theorem, this is equivalent to the statement that $\lambda(\overline{Y}) \prec \lambda(\overline{X})$, where $\lambda(\overline{Y})$ and  $\lambda(\overline{X})$ are the vector of eigenvalues of $\overline{Y}$ and $\overline{X}$, respectively.

 In order to determine the singular values of  $\overline{Y}$ and $\overline{X}$, consider the singular value decompositions of $Y$ and $X$:
\[
X=\sum_{k=1}^{r_X}s_k(X)x'_kx_k^\ast \ \ \ \text{and} \ \ \ Y=\sum_{k=1}^{r_Y}s_k(Y)y'_ky_k^\ast,
\]
 where $r_X=\rank(X)$, $r_Y=\rank(Y)$, $s(X)=(s_1(X),\dots,s_{r_X}(X))$ and $s(Y)=(s_1(Y),\dots,s_{r_Y}(Y))$ are the vectors of the non-zero singular values of $X$ and $Y$ (assumed to be written in decreasing order as the index increases), and
\[
 \{x_1,\dots,x_{r_X}\}, \{x'_1,\dots,x'_{r_X}\}\subseteq\X  \ \ \ \text{and} \ \ \ \{y_1\dots y_{r_Y}\}, \{y'_1\dots y'_{r_Y}\}\subseteq\Y
\]
 are orthonormal sets of vectors in their respective spaces.

 Then since the block matrix $\overline{X}$ can be diagonalized as
\[
  \overline{\X}:=
U \begin{pmatrix}
      0 & X\\
      X^{\ast} & 0
    \end{pmatrix}U^\dagger=
 \begin{pmatrix}
      X & 0\\
      0 & -X^{\ast},
    \end{pmatrix}
\]
 with the unitary
\[
U=\frac{1}{\sqrt{2}}\begin{pmatrix}
      \I_\X & \I_\X\\
      \I_\X & -\I_\X
    \end{pmatrix},
\]
the eigenvalues of $\overline{X}$ are given by
\[
 \lambda(\X)=\{s_1(X),\dots,s_{r_X}(X),-s_{r_X}(X),\dots,-s_1(X)),
\]
 where here they have been arranged in decreasing order. An equivalent argument shows that the the eigenvalues of $\overline{Y}$ are similarly given by
\[
 \lambda(\Y)=\{s_1(Y),\dots,s_{r_Y}(Y),-s_{r_Y}(Y),\dots,-s_1(Y)).
\]

However the singular values of $\overline{X}$ and $\overline{Y}$ are related to the eigenvalues via the absolute value. Therefore the singular values $\overline{s}(X)$ of $\overline{X}$ and $\overline{s}(Y)$ of $\overline{Y}$ are positive and there are at least two equal values for each $s_k$. That is,
\[\begin{align*}
\overline{s}(X)&=(s_1(X),s_1(X),\dots,s_{r_X}(X),s_{r_X}(X),\dots) \\
\overline{s}(Y)&=(s_1(Y),s_1(Y),\dots,s_{r_Y}(Y),s_{r_Y}(Y),\dots) ,
\end{align*}\]
where all values $s_{j}(X)$ and $s_{k}(Y)$for $j\geq r_X$ and $k
\geq r_Y$ are assumed to be zero by the convention described in the problem statement.

Then it follows that for all $k\in\{1,\dots, n\}$,
\[
s_1(X)+s_1(X)+\dots+s_{k}(X)+s_{k}(X)\geq s_1(Y)+s_1(Y)+\dots+s_{k}(Y)+s_{k}(Y),
\]
or equivalently that
\[
s_1(X)\dots+s_{k}(X)\geq s_1(Y)+\dots+s_{k}(Y).
\]