Optimal Cloner Attains Bound

Put $E=I-{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{n}}\ge 0$ by [L1], so $E^{1\mathrm{/}2}$ exists by [F5]. The first summand of $\mathrm{\Phi }$ is completely positive: by [F4], choose an orthonormal product basis $(g_j)$ of the added registers and set $A_j=\sqrt{N_n\mathrm{/}{N}_m}{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}(I\otimes \mathrm{\mid }{g}_j⟩)$. Then ${\sum }_j{A}_{j}X{A}_j^{\ast }=\frac{N_n}{N_m}{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}(X\otimes I^{\otimes (m-n)}){\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}$. When $m=n$, the added tensor unit has its single basis vector and this map is the identity before compression. By [F3], write $\omega ={\sum }_{\mathrm{\ell }}{s}_{\mathrm{\ell }}\mathrm{\mid }{f}_{\mathrm{\ell }}⟩⟨f_{\mathrm{\ell }}\mathrm{\mid }$ in an orthonormal eigenbasis and choose an input orthonormal basis $(e_j)$ by [F4]. The second summand has Kraus operators $K_{\mathrm{\ell }j}=\sqrt{s_{\mathrm{\ell }}}\mathrm{\mid }{f}_{\mathrm{\ell }}⟩⟨e_j\mathrm{\mid }{E}^{1\mathrm{/}2}$, using [F6] and the outer-product convention, since ${\sum }_{\mathrm{\ell },j}{K}_{\mathrm{\ell }j}X{K}_{\mathrm{\ell }j}^{\ast }=\mathrm{tr}(EX)\omega$. Thus both summands and their sum are completely positive by the finite Kraus theorem [L3].

The Haar moment [L2], followed by the coordinate partial trace over the last $m-n$ factors, gives ${\mathrm{Tr}}_{m-n}({\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}})=(N_m\mathrm{/}{N}_n){\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{n}}$, because each product-basis partial trace sends $\mathrm{\mid }u⟩⟨u{\mathrm{\mid }}^{\otimes m}$ to $\mathrm{\mid }u⟩⟨u{\mathrm{\mid }}^{\otimes n}$. Finite-dimensionality allows the partial trace to pass through the coordinatewise integral. Therefore $\mathrm{tr}\text{\hspace{0.17em}⁣}(\frac{N_n}{N_m}{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}(X\otimes I){\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}})=\mathrm{tr}(X{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{n}}),$ where cyclicity, ${\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}^2={\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}$, and [L3] were used. The second summand has trace $\mathrm{tr}(EX)\mathrm{tr}(\omega )=\mathrm{tr}(EX)$ by [F2]. Their sum is $\mathrm{tr}X$, so [F1] and step 1.1 make $\mathrm{\Phi }$ a channel.

For unit $u$, $\mathrm{\mid }u{⟩}^{\otimes n}$ lies in the symmetric subspace, so the second summand vanishes. Also $\mathrm{\mid }u{⟩}^{\otimes m}$ is fixed by ${\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}$. Its expectation in the first summand is therefore $(N_n\mathrm{/}{N}_m)⟨u^{\otimes m}\mathrm{\mid }(\mathrm{\mid }u⟩⟨u{\mathrm{\mid }}^{\otimes n}\otimes I)\mathrm{\mid }{u}^{\otimes m}⟩=N_n\mathrm{/}{N}_m$. This value is independent of $u$, proving the claim.