Explicit finite-dimensional contraction error bounds for energy and gradient

Apply [F2] to each column $c_p$ of $C$: $\mathrm{\parallel }{c}_p{\mathrm{\parallel }}_1\le \sqrt{n}\mathrm{\parallel }{c}_p{\mathrm{\parallel }}_2\le \sqrt{n}B$. Therefore every transformed one-body error satisfies $\mathrm{\mid }\mathrm{\Delta }{H}_{pq}^{\mathrm{\prime }}\mathrm{\mid }\le {\epsilon }_2\mathrm{\parallel }{c}_p{\mathrm{\parallel }}_1\mathrm{\parallel }{c}_q{\mathrm{\parallel }}_1\le n{B}^2{\epsilon }_2$. With $r^2$ entries, $\mathrm{\parallel }\mathrm{\Delta }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_F\le rn{B}^2{\epsilon }_2$.

Each transformed four-index error is bounded by ${\epsilon }_4{\prod }_{j\in \{p,q,r,s\}}\mathrm{\parallel }{c}_j{\mathrm{\parallel }}_1\le n^2{B}^4{\epsilon }_4$. There are $r^4$ entries, so $\mathrm{\parallel }\mathrm{\Delta }{G}^{\mathrm{\prime }}{\mathrm{\parallel }}_F\le r^2{n}^2{B}^4{\epsilon }_4$. The rank-two coefficient tensor $(D_{pq}{D}_{rs})$ has Frobenius norm $\mathrm{\parallel }D{\mathrm{\parallel }}_F^2\le M^2$.

For a fixed entry $C_{a\mu }$, differentiating the two occurrences of $C$ in $H^{\mathrm{\prime }}$ yields two sums. Each row of $D$ or column of $D$ has ${\mathrm{\ell }}^1$ norm at most $\sqrt{r}M$ by [F2], while ${\sum }_b\mathrm{\mid }\mathrm{\Delta }{H}_{ab}\mathrm{\mid }\mathrm{\mid }{C}_{bq}\mathrm{\mid }\le \sqrt{n}B{\epsilon }_2$. Thus each gradient entry from $H$ has magnitude at most $2\sqrt{nr}BM{\epsilon }_2$, and the Frobenius norm over $nr$ entries is at most $2nrBM{\epsilon }_2$.

Differentiating the four factors of $C$ in $G^{\mathrm{\prime }}$ gives four sums. For each fixed slot, the sum over the other three basis indices is at most $n^{3\mathrm{/}2}{B}^3{\epsilon }_4$ by [F2]. The remaining coefficient sum over the three orbital indices is at most $r^{3\mathrm{/}2}{M}^2$: one row or column of $D$ has ${\mathrm{\ell }}^1$ norm at most $\sqrt{r}M$ and all entries of the other copy sum to at most $rM$. Including the factor $1\mathrm{/}2$ and four slots, each gradient entry is bounded by $2n^{3\mathrm{/}2}{r}^{3\mathrm{/}2}{B}^3{M}^2{\epsilon }_4$. Multiplying by $\sqrt{nr}$ for its $nr$ entries gives $2n^2{r}^2{B}^3{M}^2{\epsilon }_4$.

By finite Cauchy–Schwarz, the one-body energy error is at most $\mathrm{\parallel }D{\mathrm{\parallel }}_F\mathrm{\parallel }\mathrm{\Delta }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_F\le Mrn{B}^2{\epsilon }_2$. Pairing $(D_{pq}{D}_{rs})$ with $\mathrm{\Delta }{G}^{\mathrm{\prime }}$ gives the two-body bound $\frac{1}{2}{M}^2{r}^2{n}^2{B}^4{\epsilon }_4$. Adding them proves the energy estimate.

The full gradient error is the sum of the one-body and two-body gradient errors. The triangle inequality and steps 1.3 and 1.4 give the claimed bound.