Block-target approximation bounds logical error and leakage

If $V=0$, both error maps have norm zero by [F2], so both bounds hold. Otherwise, since $T$ preserves both summands, $PTP=UP$ on $V$ and $(I-P)TP=0$. Thus $PGP-UP=P(G-T)P$. For any $y\in V$, the orthogonal decomposition $y=Py+(I-P)y$ gives $\mathrm{\parallel }Py\mathrm{\parallel }\le \mathrm{\parallel }y\mathrm{\parallel }$ and $\mathrm{\parallel }(I-P)y\mathrm{\parallel }\le \mathrm{\parallel }y\mathrm{\parallel }$ by the Pythagorean identity. Hence for every unit vector $x$, $\mathrm{\parallel }P(G-T)Px\mathrm{\parallel }\le \mathrm{\parallel }(G-T)Px\mathrm{\parallel }\le \mathrm{\parallel }G-T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }Px\mathrm{\parallel }\le \delta$. Taking the maximum in [F2] proves the logical block error bound.

In the remaining case $V\mathrm{\ne }0$, $(I-P)GP=(I-P)(G-T)P$ because $(I-P)TP=0$. By the contractions in step 1.1, for every unit $x$, $\mathrm{\parallel }(I-P)(G-T)Px\mathrm{\parallel }\le \mathrm{\parallel }(G-T)Px\mathrm{\parallel }\le \mathrm{\parallel }G-T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }Px\mathrm{\parallel }\le \delta$. Taking the maximum gives the leakage bound.