Optimal Cloning Worst Case Upper Bound

Choose a finite Kraus family $(A_a)$ from [L3], and write ${\mathrm{\Pi }}_n={\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{n}}$ and ${\mathrm{\Pi }}_m={\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m},\mathrm{m}}$. For each unit $u$, put $v=u^{\otimes n}$, $w=u^{\otimes m}$, and $f(u)=\mathrm{tr}(\mathrm{\Phi }(\mathrm{\mid }v⟩⟨v\mathrm{\mid })\mathrm{\mid }w⟩⟨w\mathrm{\mid })$. By [L2], this is the squared root fidelity in the Statement. Since $v$ is a unit symmetric tensor, ${\mathrm{\Pi }}_{n}v=v$ by [F1]. The Kraus formula, the rank-one convention, and Cauchy–Schwarz [F3] give $f(u)={\sum }_a\mathrm{\mid }⟨A_{a}v,w⟩{\mathrm{\mid }}^2={\sum }_a\mathrm{\mid }⟨v,{\mathrm{\Pi }}_n{A}_a^{\ast }w⟩{\mathrm{\mid }}^2\le {\sum }_a\mathrm{\parallel }{\mathrm{\Pi }}_n{A}_a^{\ast }w{\mathrm{\parallel }}^2=⟨Cw,w⟩$, where $C={\sum }_a{A}_a{\mathrm{\Pi }}_n{A}_a^{\ast }$ is self-adjoint and nonnegative: each quadratic form is $\mathrm{\parallel }{\mathrm{\Pi }}_n{A}_a^{\ast }w{\mathrm{\parallel }}^2$.

Integrating this pointwise bound and applying the $m$-th Haar moment [L1] yields $\int f(u)d\mu (u)\le \mathrm{tr}(C{\mathrm{\Pi }}_m)\mathrm{/}{N}_m$. Since $C$ and $I-{\mathrm{\Pi }}_m$ are nonnegative by step 1.1 and [F1], [F2] gives $\mathrm{tr}(C(I-{\mathrm{\Pi }}_m))\ge 0$, hence $\mathrm{tr}(C{\mathrm{\Pi }}_m)\le \mathrm{tr}C$. Trace cyclicity [F2], normalization [L3], and [L1] give $\mathrm{tr}C={\sum }_a\mathrm{tr}({\mathrm{\Pi }}_n{A}_a^{\ast }{A}_a)=\mathrm{tr}{\mathrm{\Pi }}_n=N_n$. Therefore $\int f(u)d\mu (u)\le N_n\mathrm{/}{N}_m$. The function $f$ is continuous in the finite coordinates of $u$, hence measurable; because $\mu$ is a probability measure, its infimum is no larger than its average. Thus ${\inf }_{\mathrm{\parallel }u\mathrm{\parallel }=1}f(u)\le N_n\mathrm{/}{N}_m$, including $d=1$ and $n=m$.