Error recurrence for a chain of block-encoded products

Inductively apply [F1] to the first $k-1$ encodings and the $k$th encoding, using disjoint ancillas at every composition. The resulting compression is $B_1\cdots B_k$ in the stated order, and each step multiplies the normalization, giving ${\prod }_i{\alpha }_i$.

Put $C_i:=A_i\mathrm{/}{\alpha }_i$, so $B_i=C_i+E_i$. The hypotheses give $\mathrm{\parallel }{C}_i{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$. For every unit vector $x$, the vector triangle inequality gives $\mathrm{\parallel }{B}_{i}x\mathrm{\parallel }\le \mathrm{\parallel }{C}_{i}x\mathrm{\parallel }+\mathrm{\parallel }{E}_{i}x\mathrm{\parallel }\le 1+{\epsilon }_i$; taking the maximum in [F2] yields $\mathrm{\parallel }{B}_i{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1+{\epsilon }_i$.

The telescoping identity $B_1\cdots B_k-C_1\cdots C_k={\sum }_{i=1}^k(B_1\cdots B_{i-1})E_i(C_{i+1}\cdots C_k)$ follows by adding and subtracting the intermediate products $B_1\cdots B_i{C}_{i+1}\cdots C_k$. By [F3], [F2], and [step 1.2], its norm is at most ${\sum }_i{\epsilon }_i{\prod }_{j<i}(1+{\epsilon }_j)$, since every later $\mathrm{\parallel }{C}_j\mathrm{\parallel }\le 1$. Each omitted factor $(1+{\epsilon }_j)$ is at least one, so this is at most the displayed sum over all $j\mathrm{\ne }i$. Finally $C_1\cdots C_k=A_1\cdots A_k\mathrm{/}({\prod }_i{\alpha }_i)$, proving the result, including $k=1$.