prove the second part, relation (\[QT3\]). We derive it from the first part, (\[AJ4\]). Denote $$\label{KC4}
e^{{\rm q }}_{n_1}(J,F)={\sigma }$$ and fix any $\delta>0$. Let $A=((A_l)_{l=0}^{k-1},(b_l)_{l=0}^{k-1},\varphi)$ be a quantum algorithm from $F$ to $Y$ with $n_q(A)\le n_1$ and $$\label{QT4}
e(J,A,F)\le {\sigma }+\delta.$$ Let $\zeta$ be a random variable with distribution $A(f)$. Observe that, by definition (see section 2), $\zeta$ takes values in the finite set $$Y_0={\varphi }\left(\prod_{l=0}^{k-1}{{\mathbb Z}}[0,2^{m_l})\right)\subset Y.$$ Define $\mathcal{E}_0\subset Y$ to be the closed, absolutely convex hull of $F\cup Y_0$, and put $$\mathcal{E}=B_Y \cap \frac{2}{{\sigma }+\delta}\mathcal{E}_0.$$ Since $F\cup Y_0$ is precompact in $Y$, so are $\mathcal{E}_0$ and $\mathcal{E}$. Moreover, there is a ${\gamma }>0$ such that $F\cup Y_0\subseteq {\gamma }B_Y$. Hence, $$\label{KC3}
F\cup Y_0\subseteq \max\left({\gamma },\frac{{\sigma }+\delta}{2}\right)\mathcal{E}.$$ For any $f\in F$ we have $$f-\zeta\in 2\mathcal{E}_0$$ and by (\[QT4\]), with probability at least 3/4, $$f-\zeta\in ({\sigma }+\delta) B_Y.$$ Consequently, with probability at least 3/4, $$\label{QT5}
f-\zeta\in ({\sigma }+\delta) \mathcal{E}.$$ Since $\mathcal{E}$ is a closed, absolutely convex, and bounded subset of $Y$, it defines a norm $\|\,.\,\|_{\mathcal{E}}$ on $E={\rm span}(\mathcal{E})$ as follows $$\|\,y\,\|_{\mathcal{E}}=\inf\{\theta>0\,:\, y\in \theta \mathcal{E}\}\quad (y\in E),$$ and $\mathcal{E}$ is the unit ball of $(E,\|\;\|_{\mathcal{E}})$. By (\[KC3\]), $F\subset E$ and $Y_0\subset E$. Define $J_E: F\to E$ and ${\varphi }_E$ 