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Suppose I have a classical-classical-quantum channel $W : \mathcal{X}\times\mathcal{Y} \rightarrow \mathcal{D}(\mathcal{H})$, where $\mathcal{X},\mathcal{Y}$ are finite sets and $\mathcal{D}(\mathcal{H})$ is the set of density matrices on finite dimensional, complex Hilbert space $\mathcal{H}$.

Suppose $p_x$ is the uniform distribution on $\mathcal{X}$ and $p_y$ is the uniform distribution on $\mathcal{Y}$. Further, define for distributions $p_1$ on $\mathcal{X}$ and $p_2$ on $\mathcal{Y}$, the Holevo information $$\chi(p_1, p_2, W) := H\left(\sum_{x,y}p_1(x)p_2(y)W(x,y)\right) - \sum_{x,y}p_1(x)p_2(y)H(W(x,y))$$

where $H$ is the von Neumann entropy.

I would like to show, for $$ p_1 := \sup_{p}\left\{ \chi(p, p_y, W)\right\}, p_2 := \sup_{p}\left\{ \chi(p_x, p, W)\right\}$$ that, $$\chi(p_1, p_2, W) \geq \chi(p_1, p_y, W) \text{ and } \chi(p_1, p_2, W)\geq \chi(p_x, p_2, W).$$

So far, I'm not yet convinced that the statement is true in the first place. I haven't made much progress in proving this, but it seems like some sort of triangle inequality could verify the claim.

Thanks for any suggestions regarding if the statement should hold and tips on how to prove it.

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  • $\begingroup$ As the answer suggests, I did intent to use the argmax and not supremum. $\endgroup$ – Stephen Diadamo May 27 '18 at 15:56
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It appears that the statement is not true in general. Suppose $X = Y = \{0,1\}$, $\mathcal{H}$ is the Hilbert space corresponding to a single qubit, and $W$ is defined as \begin{align} W(0,0) & = | 0 \rangle \langle 0 |,\\ W(0,1) & = | 1 \rangle \langle 1 |,\\ W(1,0) & = | 1 \rangle \langle 1 |,\\ W(1,1) & = \frac{1}{2} | 0 \rangle \langle 0 | + \frac{1}{2} | 1 \rangle \langle 1 |. \end{align} If $p_y$ is the uniform distribution, the optimal choice for $p_1$ is $p_1(0) = 1$ and $p_1(1) = 0$, which gives $\chi(p_1,p_y,W) = 1$, which is the maximum possible value. (I assume you mean to define $p_1$ and $p_2$ as the argmax of those expressions, not the supremum.) Likewise, if $p_x$ is uniform, $p_2(0) = 1$ and $p_2(1) = 0$ is optimal, and the value is the same. However, $\chi(p_1,p_2,W) = 0$, so the inequality does not hold.

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