Diamond Norm Ancilla Dimension Bound

For fixed $r$, let $M_r$ be the maximum of $\mathrm{\parallel }(\mathrm{\Phi }\otimes {\mathrm{id}}_r)(R_{u,v}){\mathrm{\parallel }}_1$ over unit vectors $u,v\in H\otimes {\mathbb{C}}^r$. Their unit spheres are compact by [F1]. In orthonormal coordinates each entry of $R_{u,v}$ is a finite product of coordinates of $u$ and conjugates of coordinates of $v$; applying the finite linear map $\mathrm{\Phi }\otimes {\mathrm{id}}_r$ remains continuous, and [L4] makes the output trace norm continuous. Hence the maximum exists. By [L1], every $X$ with $\mathrm{\parallel }X{\mathrm{\parallel }}_1\le 1$ has an SVD $X={\sum }_j{s}_j{R}_{u_j,v_j}$ with unit $u_j,v_j$, $s_j\ge 0$, and ${\sum }_j{s}_j\le 1$. Linearity and the trace-norm triangle inequality give $\mathrm{\parallel }(\mathrm{\Phi }\otimes {\mathrm{id}}_r)(X){\mathrm{\parallel }}_1\le {\sum }_j{s}_j{M}_r\le M_r$. For unit $u,v$, the adjoint calculation gives $R_{u,v}^{\ast }{R}_{u,v}=R_{v,v}$, a rank-one orthogonal projection, so its singular values are one $1$ and the remaining zeros; hence $\mathrm{\parallel }{R}_{u,v}{\mathrm{\parallel }}_1=1$. Thus the induced norm equals $M_r$ and is attained.

Now take any finite ancilla $Z$ and unit vectors $x,y\in H\otimes Z$. By [L3], their Schmidt ranks $a,b$ are at most $\min (d,\dim Z)$. Let $m=\max (a,b)$ and pad each Schmidt list to an orthonormal $m$-tuple, obtaining $x^{\mathrm{\prime }},y^{\mathrm{\prime }}\in H\otimes {\mathbb{C}}^m$ by replacing the respective ancilla Schmidt vectors by the standard basis vectors and retaining the Schmidt coefficients and $H$-vectors. Define $V_x,V_y:{\mathbb{C}}^m\to Z$ to send that basis to the respective padded orthonormal ancilla families; thus $x=(I_H\otimes V_x)x^{\mathrm{\prime }}$ and $y=(I_H\otimes V_y)y^{\mathrm{\prime }}$. The tensor-map definition and product pairing show each map $I\otimes V_x$ and $I\otimes V_y$ is an isometry. Factor permutations are unitary because they carry an orthonormal product basis bijectively to another such basis. Thus $(\mathrm{\Phi }\otimes {\mathrm{id}}_Z)(R_{x,y})=(I_K\otimes V_x)(\mathrm{\Phi }\otimes {\mathrm{id}}_m)(R_{x^{\mathrm{\prime }},y^{\mathrm{\prime }}})(I_K\otimes V_y{)}^{\ast }\mathrm{.}$ To verify the equality, expand operators in elementary tensors $A\otimes B$: $\mathrm{\Phi }\otimes \mathrm{id}$ sends this to $\mathrm{\Phi }(A)\otimes B$, and the two isometries send each ancilla matrix unit $R_{e_i,e_j}$ to $R_{z_i,w_j}$. For isometries $U,V$, $(U{A}_0{V}^{\ast }{)}^{\ast }(U{A}_0{V}^{\ast })=V{A}_0^{\ast }{A}_0{V}^{\ast }$. The positive eigenvalues of this operator are those of $A_0^{\ast }{A}_0$: an eigenvector $z$ with positive eigenvalue maps to $Vz$, and any eigenvector with positive eigenvalue must lie in the range of $V$ because the operator vanishes on its orthogonal complement. Hence left/right isometric embeddings preserve the nonzero singular values and trace norm. Embedding ${\mathbb{C}}^m$ isometrically into ${\mathbb{C}}^d$ on both sides is the same case, so the original output has trace norm at most $\mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$.

By step 1.1, the induced norm for each $Z$ is attained by a rank-one unit input, so step 1.2 gives $\mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_Z{\mathrm{\parallel }}_{1\to 1}\le \mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$. Taking the supremum over all finite $Z$ gives $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{\diamond }\le \mathrm{\parallel }\mathrm{\Phi }\otimes {\mathrm{id}}_d{\mathrm{\parallel }}_{1\to 1}$. The reverse inequality holds because $r=d$ is one of the ancilla dimensions in the supremum. Step 1.1 at $r=d$ supplies an attaining rank-one operator.