Quantum Source Compression and Classical Communication over Quantum Channels

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 $\sqrt{\mathrm{\Lambda }}\rho \sqrt{\mathrm{\Lambda }}$.

First suppose $P,Q$ are positive definite and $P⪯Q$. By [F2], their eigenvalues are strictly positive: applying the positive quadratic form to an eigenvector gives $\lambda \mathrm{\parallel }v{\mathrm{\parallel }}^2>0$. Define $P^{-1\mathrm{/}2}$ spectrally by replacing each eigenvalue $\lambda$ with ${\lambda }^{-1\mathrm{/}2}$, and set $P^{-1}:=(P^{-1\mathrm{/}2}{)}^2$; multiplication in the eigenbasis shows these are the inverse square root and inverse of $P$. Put $C:=P^{-1\mathrm{/}2}Q{P}^{-1\mathrm{/}2}$. By the adjoint product rule in [F7], $C$ is self-adjoint, and [F7] gives $⟨v,Cv⟩=⟨P^{-1\mathrm{/}2}v,Q{P}^{-1\mathrm{/}2}v⟩>0$ for $v\mathrm{\ne }0$, so $C$ is positive definite. Also $C-I=P^{-1\mathrm{/}2}(Q-P)P^{-1\mathrm{/}2}⪰0$ by the same quadratic-form identity and [F1]. Hence every eigenvalue $\mu$ of $C$ is at least one by [F2]; define $C^{-1}$ spectrally by reciprocating its positive eigenvalues, giving $C^{-1}⪯I$. Since $Q=P^{1\mathrm{/}2}C{P}^{1\mathrm{/}2}$, multiplying this factorization by $P^{-1\mathrm{/}2}{C}^{-1}{P}^{-1\mathrm{/}2}$ on either side gives the identity, so $Q^{-1}=P^{-1\mathrm{/}2}{C}^{-1}{P}^{-1\mathrm{/}2}$. Congruence of $C^{-1}⪯I$ by $P^{-1\mathrm{/}2}$ now yields $Q^{-1}⪯P^{-1}$.

For $t>0$ and nonzero $v$, $⟨v,(A+tI)v⟩\ge t\mathrm{\parallel }v{\mathrm{\parallel }}^2>0$, and likewise for $B$, so $A+tI$ and $B+tI$ are positive definite and ordered. Step 1.1 gives $(A+tI{)}^{-1}⪰(B+tI{)}^{-1}$. Multiplying the order by $t$ and subtracting from $I$ yields $A(A+tI{)}^{-1}=I-t(A+tI{)}^{-1}⪯I-t(B+tI{)}^{-1}=B(B+tI{)}^{-1}\mathrm{.}$

Set $f(s)=s^{-1\mathrm{/}2}\mathrm{/}(1+s)$ for $s>0$. It is continuous, hence integrable on every compact interval in its domain by [F10]. For $0<s\le 1$, $f(s)\le s^{-1\mathrm{/}2}$; for $s\ge 1$, $f(s)\le s^{-3\mathrm{/}2}$. For each integer $k\ge 1$, the sequence $a_k={\int }_{1\mathrm{/}k}^{1}f(s)ds$ is nondecreasing by positivity and additivity, and is bounded above by ${\int }_{1\mathrm{/}k}^1{s}^{-1\mathrm{/}2}ds=2(1-k^{-1\mathrm{/}2})\le 2$. For $k>1$ this evaluation follows from the primitive $2\sqrt{s}$, whose derivative is $s^{-1\mathrm{/}2}$ by [F4], and the fundamental theorem in [F4]; for $k=1$ both sides are zero by the equal-endpoint convention in [F5]. The sequence $b_k={\int }_1^{k}f(s)ds$ is likewise nondecreasing and bounded above by ${\int }_1^k{s}^{-3\mathrm{/}2}ds=2(1-k^{-1\mathrm{/}2})\le 2$, using the primitive $-2s^{-1\mathrm{/}2}$ and [F4] when $k>1$, and [F5] when $k=1$. Both sequences therefore converge by [F9]. By additivity [F5], $c_k:={\int }_{1\mathrm{/}k}^{k}f(s)ds=a_k+b_k$ converges to a finite $c$. For every $k\ge 2$, monotonicity [F8] and $f(s)\ge 1\mathrm{/}(3\sqrt{2})$ on $[1,2]$ give $c_k\ge {\int }_1^{2}f(s)ds\ge 1\mathrm{/}(3\sqrt{2})$, so $c>0$. For each fixed $\lambda >0$, define $c_{k,\lambda }:={\int }_{1\mathrm{/}(k\lambda )}^{k\mathrm{/}\lambda }f(s)ds$. If $m\ge 2$ is fixed and $k$ is large enough that $1\mathrm{/}(k\lambda )\le 1\mathrm{/}m$ and $k\mathrm{/}\lambda \ge m$, additivity and monotonicity of the integral with the two displayed tail bounds give $c_m\le c_{k,\lambda }=c_m+{\int }_{1\mathrm{/}(k\lambda )}^{1\mathrm{/}m}f(s)ds+{\int }_m^{k\mathrm{/}\lambda }f(s)ds\le c_m+2(m^{-1\mathrm{/}2}-(k\lambda {)}^{-1\mathrm{/}2})+2(m^{-1\mathrm{/}2}-(\lambda \mathrm{/}k{)}^{1\mathrm{/}2})\le c_m+\frac{4}{\sqrt{m}}\mathrm{.}$ Since $c_m\to c$ and $4\mathrm{/}\sqrt{m}\to 0$, this proves $c_{k,\lambda }\to c$ for each fixed $\lambda >0$. Now [F6] with $t=\lambda s$ in the coefficient integral over the fixed interval $[1\mathrm{/}k,k]$ gives ${\int }_{1\mathrm{/}k}^k{t}^{-1\mathrm{/}2}\frac{\lambda }{\lambda +t}dt=\sqrt{\lambda }{c}_{k,\lambda }⟶c\sqrt{\lambda };$ for $\lambda =0$ the coefficient is identically zero. For a nonnegative $X$, let $P_{\lambda }$ be its spectral projection at eigenvalue $\lambda$ and define $J_k(X):={\sum }_{\lambda \in \mathrm{spec}(X)}\left({\int }_{1\mathrm{/}k}^k{t}^{-1\mathrm{/}2}\frac{\lambda }{\lambda +t}dt\right)P_{\lambda }\mathrm{.}$ The coefficient depends only on $\lambda$, so this definition is independent of choices of eigenvectors within eigenspaces. The coefficient convergence and finite spectral sum imply $⟨v,J_k(X)v⟩\to c⟨v,X^{1\mathrm{/}2}v⟩$ for every vector $v$. For $k\ge 2$, this quadratic form is also the integral of the continuous scalar function $t\mapsto ⟨v,t^{-1\mathrm{/}2}X(X+tI{)}^{-1}v⟩$ on $[1\mathrm{/}k,k]$, as follows from the finite spectral sum. By step 2.1 these functions for $X=A$ and $X=B$ are pointwise ordered; [F8] therefore gives $⟨v,J_k(A)v⟩\le ⟨v,J_k(B)v⟩$. Taking limits and using $c>0$ yields $⟨v,A^{1\mathrm{/}2}v⟩\le ⟨v,B^{1\mathrm{/}2}v⟩$ for every $v$, which is $A^{1\mathrm{/}2}⪯B^{1\mathrm{/}2}$.

If $H=\{0\}$, every endomorphism, including $I_H$, is zero, so the asserted inequality is $0⪯0$. Henceforth assume $H\mathrm{\ne }\{0\}$, as required by [F2]. Set $X=S+T$, $R=\sqrt{X}$, and $U=I-R$. For $c>0$ and endomorphisms $A,B$, expanding $(A-cB{)}^{\ast }(A-cB)⪰0$ and dividing by $c$ gives $A^{\ast }B+B^{\ast }A⪯c^{-1}{A}^{\ast }A+c{B}^{\ast }B$. Apply this with $A=\sqrt{T}R$ and $B=\sqrt{T}U$. Since $R+U=I$, expansion of $(R+U)T(R+U)$ and the preceding inequality give $T⪯(1+c^{-1})RTR+(1+c)UTU\mathrm{.}$ Now $T⪯X=R^2$, so congruence by $U$ gives $UTU⪯U{R}^{2}U=R{U}^{2}R$; the last equality follows because $U=I-R$ commutes with $R$. Since $U^2=I-2R+R^2=I+S+T-2R$, substitution and collection of the two $T$ terms yield $T⪯R((2+c+c^{-1})T+(1+c)(I+S-2R))R\mathrm{.}$ By [F2], $R⪰\sqrt{S}$; [F3] and $0⪯S⪯I$ give $\sqrt{S}⪰S$ by diagonalizing $S$. Therefore $I+S-2R⪯I-S$. Taking $c=1$ in the last display proves $T⪯R(4T+2(I-S))R\mathrm{.}$

Let $P$ be the orthogonal projection onto $\mathrm{supp}X$. If $v\in \ker X$, then $0=⟨v,Xv⟩=⟨v,Sv⟩+⟨v,Tv⟩$; both terms are nonnegative, so each is zero. By [F3], these terms are $\mathrm{\parallel }\sqrt{S}v{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }\sqrt{T}v{\mathrm{\parallel }}^2$, hence $Sv=Tv=0$. Since $S,T$ are self-adjoint, they preserve the orthogonal complement $PH$ of $\ker X$ as well. Thus both sides of the desired inequality are block diagonal for $H=PH\oplus \ker X$. On $PH$, congruence of step 1.1 by the inverse of the restriction of $R$ gives $X^{-1\mathrm{/}2}T{X}^{-1\mathrm{/}2}⪯4T+2(I-S)$. On this block, $X^{-1\mathrm{/}2}(S+T)X^{-1\mathrm{/}2}=I_{PH}$, so its left side is $I_{PH}-X^{-1\mathrm{/}2}S{X}^{-1\mathrm{/}2}$. On $\ker X$, the desired inequality is $I⪯2I$, since $S=T=X^{-1\mathrm{/}2}=0$ there. Combining these two block inequalities proves the Statement, including when either block is zero-dimensional.

Define $h(t)=-t\ln t$ for $0<t\le 1$ and $h(0)=0$. It is continuous for positive $t$ by [F3]. Since $\ln$ is increasing and onto, $\ln t\to -\mathrm{\infty }$ as $t\downarrow 0$. Put $u=-\ln t$; then $t=\exp (-u)=1\mathrm{/}\exp u$ by the inverse and reciprocal identities in [F3]. Hence $h(t)=u\mathrm{/}\exp u\to 0$ by [F4] with $m=a=1$, so $h$ is continuous on $[0,1]$.

Let ${\lambda }_1(\rho )\ge \cdots \ge {\lambda }_d(\rho )$ and similarly for ${\rho }_k$. By [F5], density eigenvalues lie in $[0,1]$; [F2] gives $\mathrm{\mid }{\lambda }_j({\rho }_k)-{\lambda }_j(\rho )\mathrm{\mid }\le \mathrm{\parallel }{\rho }_k-\rho \mathrm{\parallel }$ for every $j$. For any $\epsilon >0$, continuity of $h$ at the finitely many ${\lambda }_j(\rho )$ gives one positive radius $\delta$ such that $\mathrm{\parallel }{\rho }_k-\rho \mathrm{\parallel }<\delta$ makes $\mathrm{\mid }h({\lambda }_j({\rho }_k))-h({\lambda }_j(\rho ))\mathrm{\mid }<\epsilon \mathrm{/}d$ for all $j$.

Sum the $d$ inequalities in step 2.1 and use [F1]. This gives $\mathrm{\mid }S({\rho }_k)-S(\rho )\mathrm{\mid }<\epsilon$ whenever $\mathrm{\parallel }{\rho }_k-\rho \mathrm{\parallel }<\delta$, proving continuity. The zero-eigenvalue boundary is covered by step 1.1, so no full-rank assumption is used.

For a realized codebook, set ${\mathrm{\Gamma }}_m=\mathrm{\Pi }{\mathrm{\Pi }}_{X_m}\mathrm{\Pi }$ and $S={\sum }_m{\mathrm{\Gamma }}_m$. Since each factor is a projection, $0⪯{\mathrm{\Gamma }}_m⪯I$. Define ${\mathrm{\Lambda }}_m=S^{-1\mathrm{/}2}{\mathrm{\Gamma }}_m{S}^{-1\mathrm{/}2}$ on $\mathrm{supp}S$ and zero on its kernel; add ${\mathrm{\Lambda }}_0=I-{\sum }_m{\mathrm{\Lambda }}_m$. By [F4], all ${\mathrm{\Lambda }}_m$ are nonnegative and ${\sum }_m{\mathrm{\Lambda }}_m$ is the support projection of $S$, so $({\mathrm{\Lambda }}_0,{\mathrm{\Lambda }}_1,\dots ,{\mathrm{\Lambda }}_M)$ is a POVM.

For $j\mathrm{\ne }m$, independence of $X_j$ and $X_m$ gives $\mathbb{E}\mathrm{tr}({\mathrm{\Gamma }}_j{\sigma }_{X_m})=\mathrm{tr}(\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi }\bar{\sigma })=\mathrm{tr}((\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi })(\mathrm{\Pi }\bar{\sigma }\mathrm{\Pi }))\le D^{-1}\mathrm{tr}(\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi })\le d\mathrm{/}D\mathrm{.}$ The first equality factors the finite independent expectation into the product of its two means and distributes the finite trace sum using [F8]; the second uses [F7] and the support of the first factor. For the first inequality, $D^{-1}\mathrm{\Pi }-\mathrm{\Pi }\bar{\sigma }\mathrm{\Pi }$ is nonnegative, so [F6] applied to its product with the nonnegative first factor gives the trace bound. For the last inequality, $I-\mathrm{\Pi }$ is nonnegative and [F6]–[F7] give $\mathrm{tr}(\mathrm{\Pi }{\mathrm{\Pi }}_x\mathrm{\Pi })=\mathrm{tr}({\mathrm{\Pi }}_x\mathrm{\Pi })\le \mathrm{tr}({\mathrm{\Pi }}_x)\le d$.

For the transmitted codeword $X_m$, [F3] applied with $S_0={\mathrm{\Gamma }}_m$ and $T_0={\sum }_{j\mathrm{\ne }m}{\mathrm{\Gamma }}_j$ gives the operator inequality $I-{\mathrm{\Lambda }}_m⪯2(I-{\mathrm{\Gamma }}_m)+4{\sum }_{j\mathrm{\ne }m}{\mathrm{\Gamma }}_j$. Since ${\sigma }_{X_m}$ is nonnegative by [F1], [F6] applied to its product with the nonnegative difference of the right and left sides preserves the order after taking trace, and trace linearity [F8] expands the resulting right side to $\mathrm{tr}((I-{\mathrm{\Lambda }}_m){\sigma }_{X_m})\le 2\mathrm{tr}((I-{\mathrm{\Gamma }}_m){\sigma }_{X_m})+4{\sum }_{j\mathrm{\ne }m}\mathrm{tr}({\mathrm{\Gamma }}_j{\sigma }_{X_m})\mathrm{.}$ The gentle bound [F2] for the global projection gives $\mathrm{\parallel }\mathrm{\Pi }{\sigma }_{X_m}\mathrm{\Pi }-{\sigma }_{X_m}{\mathrm{\parallel }}_1\le 2\sqrt{u}$. Since $\mathrm{\parallel }{\mathrm{\Pi }}_{X_m}\mathrm{\parallel }\le 1$, trace-norm duality [F5], cyclicity [F7], and ${\mathrm{\Gamma }}_m=\mathrm{\Pi }{\mathrm{\Pi }}_{X_m}\mathrm{\Pi }$ then give $\mathrm{tr}({\mathrm{\Gamma }}_m{\sigma }_{X_m})=\mathrm{tr}({\mathrm{\Pi }}_{X_m}\mathrm{\Pi }{\sigma }_{X_m}\mathrm{\Pi })\ge \mathrm{tr}({\mathrm{\Pi }}_{X_m}{\sigma }_{X_m})-2\sqrt{u}\ge 1-u-2\sqrt{u}\mathrm{.}$

Average step 2.1 over all messages and codebooks. Its first term is at most $2(u+2\sqrt{u})$ by the acceptance estimate there, and each of the $M-1$ competing terms is at most $4d\mathrm{/}D$ by step 1.2. Thus the expected average error has the claimed bound. Because the codebook sample space is finite, at least one realization has error no greater than its expectation.

Choose Kraus families $(E_a)$ and $(D_b)$ for the encoder and decoder from [F2]. The composite channel has Kraus operators $K_{ba}=D_b{E}_a$, each of rank at most $M$ because it factors through the $M$-dimensional code space. By [F3], ${\sum }_{a,b}{K}_{ba}^{\ast }{K}_{ba}={\sum }_a{E}_a^{\ast }\left({\sum }_b{D}_b^{\ast }{D}_b\right)E_a={\sum }_a{E}_a^{\ast }{E}_a=I\mathrm{.}$

By [F4], write $\mathrm{\mid }\varphi ⟩={\sum }_{j=1}^r{s}_j\mathrm{\mid }{r}_j⟩\mathrm{\mid }{a}_j⟩$, ordering the $s_j$ decreasingly. Expanding $\mathrm{\mid }\varphi ⟩⟨\varphi \mathrm{\mid }$ and contracting either orthonormal factor family by [F7] gives ${\rho }_R:={\mathrm{Tr}}_{A^{\otimes n}}(\mathrm{\mid }\varphi ⟩⟨\varphi \mathrm{\mid })={\sum }_{j=1}^r{s}_j^2\mathrm{\mid }{r}_j⟩⟨r_j\mathrm{\mid },{\rho }^{\otimes n}={\sum }_{j=1}^r{s}_j^2\mathrm{\mid }{a}_j⟩⟨a_j\mathrm{\mid }\mathrm{.}$ Indeed each cross term with indices $j,k$ contracts with coefficient $⟨a_k\mathrm{\mid }{a}_j⟩={\delta }_{jk}$, or $⟨r_k\mathrm{\mid }{r}_j⟩={\delta }_{jk}$, respectively. Thus both reduced operators have exactly the positive eigenvalues $s_j^2$, with any remaining eigenvalues zero. Complete $(r_j{)}_{j=1}^r$ by an orthonormal basis of $\ker {\rho }_R$, available from [F5] applied to the zero operator there. On this full reference basis, ${\rho }_R$ has eigenvalues ${\lambda }_j^{(n)}$, with the Statement's zero padding; in particular ${\sum }_j{\lambda }_j^{(n)}={\sum }_{j=1}^r{s}_j^2=1$.

For each $K=K_{ba}$ let $\mathrm{\mid }{\eta }_K⟩=(I_R\otimes K)\mathrm{\mid }\varphi ⟩$, and let $P_K$ be the projection onto its reference-side Schmidt support (zero if ${\eta }_K=0$). Since $\mathrm{\mid }{\eta }_K⟩\in H_R\otimes \mathrm{ran}K$, apply [F4] on these two factors: there are at most $\dim \mathrm{ran}K\le M$ orthonormal second-factor vectors. Thus $\mathrm{rank}{P}_K\le M$ and $(P_K\otimes I)\mathrm{\mid }{\eta }_K⟩=\mathrm{\mid }{\eta }_K⟩$. Cauchy–Schwarz and the trace-pairing identity in [F7] give $\mathrm{\mid }⟨\varphi \mathrm{\mid }{\eta }_K⟩{\mathrm{\mid }}^2\le ⟨\varphi \mathrm{\mid }(P_K\otimes I)\mathrm{\mid }\varphi ⟩\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2=\mathrm{tr}({\rho }_R{P}_K)\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2\mathrm{.}$

Put $q_j=⟨r_j,P_K{r}_j⟩$. Then $0\le q_j\le 1$, and the full reference basis from step 1.2 gives ${\sum }_j{q}_j=\mathrm{tr}(P_K)=\mathrm{rank}{P}_K\le M$. If $M\ge \dim H_R$, then $\mathrm{tr}({\rho }_R{P}_K)\le 1={\sum }_{j\le M}{\lambda }_j^{(n)}$, since the source has at most $\dim H_R$ positive eigenvalues. Otherwise, ordering the eigenvalues gives $\mathrm{tr}({\rho }_R{P}_K)={\sum }_j{\lambda }_j^{(n)}{q}_j\le {\sum }_{j\le M}{\lambda }_j^{(n)}{q}_j+{\lambda }_M^{(n)}(M-{\sum }_{j\le M}{q}_j)\le {\sum }_{j\le M}{\lambda }_j^{(n)}\mathrm{.}$ Thus the overlap in step 2.1 is bounded by $\left({\sum }_{j\le M}{\lambda }_j^{(n)}\right)\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2$.

Expanding the recovered state in the composite Kraus family from step 1.1 gives $F_e^{(n)}={\sum }_{a,b}\mathrm{\mid }⟨\varphi \mathrm{\mid }{\eta }_{K_{ba}}⟩{\mathrm{\mid }}^2$. By step 3.1 this is at most $\left({\sum }_{j\le M}{\lambda }_j^{(n)}\right){\sum }_{a,b}\mathrm{\parallel }{\eta }_{K_{ba}}{\mathrm{\parallel }}^2$. Step 1.1 gives ${\sum }_{a,b}\mathrm{\parallel }{\eta }_{K_{ba}}{\mathrm{\parallel }}^2=⟨\varphi \mathrm{\mid }{I}_R\otimes ({\sum }_{a,b}{K}_{ba}^{\ast }{K}_{ba})\mathrm{\mid }\varphi ⟩=\mathrm{\parallel }\varphi {\mathrm{\parallel }}^2=1$, proving the bound.