QSVT action on singular subspaces

Fix $0<{\sigma }_j<1$. In the input basis $(v_j,w_j)$ and output basis $(u_j,{\tilde{w}}_j)$, [F2] says that $U$ is $W_R({\sigma }_j)$, and $U^{\ast }$ has the same matrix because it is real symmetric. The right-projector phase is $Z(\varphi )$ on the input basis since $2{\mathrm{\Pi }}_R-I$ has eigenvalues $+1,-1$ there; the left-projector phase is the same $Z(\varphi )$ on the output basis. Thus the alternating sequence multiplies in the fixed order $Z({\varphi }_0)W_R({\sigma }_j)Z({\varphi }_1)\cdots W_R({\sigma }_j)Z({\varphi }_d)=U_{\mathrm{\Phi }}({\sigma }_j)$, whose upper-left entry is $P_{\mathrm{\Phi }}({\sigma }_j)$ by [F3]–[F4].

If ${\sigma }_j=1$, [F2] gives $U{v}_j=u_j$ and the complementary component is zero; the surviving one-dimensional action is the first-coordinate action of $U_{\mathrm{\Phi }}(1)$, namely $P_{\mathrm{\Phi }}(1)$. For ${\sigma }_j=0$, $U{v}_j$ lies in $\ker {\mathrm{\Pi }}_L$ and $U^{\ast }(U{v}_j)=v_j$. The alternating calls therefore swap this right-kernel vector and its left-complement image exactly as $W_R(0)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ does. If $d$ is odd the final vector lies in $\ker {\mathrm{\Pi }}_L$ and the left compression is zero, equal to $P_{\mathrm{\Phi }}(0)$ because [F4] makes $P_{\mathrm{\Phi }}$ odd. If $d$ is even it returns to $v_j$ with coefficient $P_{\mathrm{\Phi }}(0)$, the prescribed even-kernel term in [F5]. Left-kernel inputs and the residual complement contribute nothing to the stated compression by [F2].

The positive singular planes, the $\sigma =1$ lines, the right zero-kernel lines, and the residual blocks are orthogonal and exhaust the relevant projected spaces by [F2]. Summing the scalar coefficients from steps 1.1–1.2 therefore gives the displayed odd and even compressed-block formulas.

For real $p$ under the stated degree, parity, and interval bounds, [F7] gives an admissible pair $P,Q$ with $\mathrm{Re}P=p$, and [F8] supplies its exact phase list $\mathrm{\Phi }$. At each slot with phase ${\varphi }_j$ about $\mathrm{\Pi }$, use the clean-ancilla toggle $X_{\mathrm{\Pi }}$ from [F11], then apply unconditional $R_z(2{\varphi }_j)$, then a control-one-controlled $R_z(-4{\varphi }_j)$, and then $X_{\mathrm{\Pi }}$ again. In control branch $0$ the middle rotation is $R_z(2{\varphi }_j)$; in branch $1$ it is $R_z(-2{\varphi }_j)$. Since $X_{\mathrm{\Pi }}$ places the ancilla in $\mathrm{\mid }1⟩$ on $\mathrm{ran}\mathrm{\Pi }$ and in $\mathrm{\mid }0⟩$ on $\ker \mathrm{\Pi }$, the induced system phases are respectively $e^{i{\varphi }_j(2\mathrm{\Pi }-I)}$ and $e^{-i{\varphi }_j(2\mathrm{\Pi }-I)}$, and the ancilla returns to $\mathrm{\mid }0⟩$. The same unconditioned signal calls therefore give full branch operators $U_{\mathrm{\Phi }}$ and $U_{-\mathrm{\Phi }}$. Because $W_R(x)$ is real and $Z(-\varphi )=\overline{Z(\varphi )}$, entrywise conjugation gives $U_{-\mathrm{\Phi }}(x)=\overline{U_{\mathrm{\Phi }}(x)}$, and projection of the control from $\mathrm{\mid }+⟩$ to $\mathrm{\mid }+⟩$ has block $\frac{1}{2}(U_{\mathrm{\Phi }}+U_{-\mathrm{\Phi }})$. By step 2.1, its coefficient on each positive singular pair is $\frac{1}{2}(P_{\mathrm{\Phi }}({\sigma }_j)+\overline{P_{\mathrm{\Phi }}({\sigma }_j)})=\mathrm{Re}{P}_{\mathrm{\Phi }}({\sigma }_j)=p({\sigma }_j)$; for an even transform the same identity gives $p(0)$ on $\ker A$. Thus the compressed block is $p^{\mathrm{S}\mathrm{V}}(A)$. If the supplied scalar approximation satisfies ${\sup }_{x\in [-1,1]}\mathrm{\mid }\mathrm{Re}{P}_{\mathrm{\Phi }}(x)-p(x)\mathrm{\mid }\le \eta$, each singular coefficient changes by at most $\eta$, so orthonormality gives operator-norm error at most $\eta$. The proof stays in the reflected convention; [F10]'s scalar is not discarded if a circuit is translated to the SU(2) convention.