Gentle measurement bound in finite dimensions

Choose a purification $\mathrm{\mid }\psi ⟩\in H\otimes R$ of $\rho$ using [F2]. Let $v=(\sqrt{\mathrm{\Lambda }}\otimes I_R)\mathrm{\mid }\psi ⟩$, $q=⟨v,v⟩=\mathrm{tr}(\mathrm{\Lambda }\rho )$, and $s=⟨\psi ,v⟩=\mathrm{tr}(\rho \sqrt{\mathrm{\Lambda }})$. By [F3]–[F4], the eigenvalues $\lambda$ of $\mathrm{\Lambda }$ lie in $[0,1]$ and $\sqrt{\lambda }\ge \lambda$, so $\sqrt{\mathrm{\Lambda }}-\mathrm{\Lambda }$ is nonnegative. Diagonalizing $\rho$ by [F3], $s-q=\mathrm{tr}(\rho (\sqrt{\mathrm{\Lambda }}-\mathrm{\Lambda }))$ is a sum of nonnegative eigenvalues of $\rho$ times nonnegative quadratic forms; hence $s\ge q$. Also $0\le q\le 1$ because $\mathrm{\Lambda }⪯I$ and $\mathrm{tr}\rho =1$.

The Hermitian operator $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }-\mathrm{\mid }v⟩⟨v\mathrm{\mid }$ is zero off the span of $\psi ,v$. If $v$ is not a scalar multiple of $\psi$, write $v=\bar{s}\psi +w$ with $w\perp \psi$; in the orthonormal basis $\psi ,w\mathrm{/}\mathrm{\parallel }w\mathrm{\parallel }$, its $2\times 2$ matrix has trace $1-q$ and determinant $-(q-\mathrm{\mid }s{\mathrm{\mid }}^2)$. By [F7], $q-\mathrm{\mid }s{\mathrm{\mid }}^2\ge 0$, so the eigenvalues have opposite signs or one is zero. Its trace norm is the square root of the characteristic discriminant, $\sqrt{(1+q{)}^2-4\mathrm{\mid }s{\mathrm{\mid }}^2}$. If $v$ is a scalar multiple of $\psi$, the same formula reduces to the one-dimensional trace norm $\mathrm{\mid }1-q\mathrm{\mid }$.

Since $s\ge q$ and $0\le q\le 1$, step 2.1 gives ${\parallel \mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }-\mathrm{\mid }v⟩⟨v\mathrm{\mid }\parallel }_1^2\le (1+q{)}^2-4q^2=(1-q)(1+3q)\le 4(1-q)\le 4u\mathrm{.}$ Taking the partial trace over $R$ gives the claimed bound by [F6], because the partial trace of the first rank-one state is $\rho$ and that of the second is $\sqrt{\mathrm{\Lambda }}\rho \sqrt{\mathrm{\Lambda }}$.