definition of dec norm

Question

The dec norm is defined as in the sceenshot. I met with some troubles when I check it is indeed a norm.

It is obvious that$\|u\|_{dec}\geq 0$, but how to check it satifies the triangle inequality and $\|u\|_{dec}=0\Rightarrow u=0$.

Answer 1

Suppose $\|u\|_{dec}=0$. Fix $\varepsilon>0$; then there exist $S_1,S_2$ with $\|S_1\|\leq\varepsilon$, $\|S_2\|\leq\varepsilon$ and $$ Vx=\begin{bmatrix} S_1(x) & u(x)\\ u(x^*)^*& S_2(x)\end{bmatrix} $$ completely positive. Taking $x\geq0$, we get $$ \begin{bmatrix} S_1(x) & u(x)\\ u(x^*)^*& S_2(x)\end{bmatrix} \geq0 $$ in $M_2(B)$. It is well-known that such a $2\times 2$ block matrix being positive implies that there exists a contraction $y(x)$ with $$ u(x)=(S_1(x)+\varepsilon I)^{1/2}y(x)(S_2(x)+\varepsilon I)^{1/2}. $$ Then $$ \|u(x)\|\leq\|(S_1(x)+\varepsilon I)^{1/2}\|\,\|(S_2(x)+\varepsilon I)^{1/2}\| \leq2\varepsilon. $$ As this can be done for any $\varepsilon>0$, we obtain that $u(x)=0$.

For the triangle inequality, if $$ Vx=\begin{bmatrix} S_1(x) & u(x)\\ u(x^*)^*& S_2(x)\end{bmatrix} ,\ \ \ \ Wx=\begin{bmatrix} T_1(x) & v(x)\\ v(x^*)^*& T_2(x)\end{bmatrix} $$ are completely positive, so is $$ (V+W)x=\begin{bmatrix} S_1(x)+T_1(x) & u(x)+v(x) \\ u(x^*)^*+v(x^*)^*& S_2(x)+T_2(x)\end{bmatrix}. $$ Thus \begin{align} \|u+v\|_{dec}&\leq \max\{\|S_1+T_1\|,\|S_2+T_2\|\} \leq\max\{\|S_1\|+\|T_1\|,\|S_2\|+\|T_2\|\}\\[0.3cm] &\leq\max\{\|S_1\|,\|S_2\|\}+\max\{\|T_1\|,\|T_2\|\}. \end{align} Then $$ \|u+v\|_{dec}-\max\{\|S_1\|,\|S_2\|\}\leq\max\{\|T_1\|,\|T_2\|\} $$ for any choice of $S_1,S_2$ and $T_1,T_2$ that appear in the decompositions of $u$ and $v$ respectively. So the left hand side is a lower bound for the right-hand-side, and we get $$ \|u+v\|_{dec}-\max\{\|S_1\|,\|S_2\|\}\leq\|v\|_{dec}. $$ Now rewrite this as $$ \|u+v\|_{dec}-\|v\|_{dec}\leq\max\{\|S_1\|,\|S_2\|\}, $$ to get $\|u+v\|_{dec}-\|v\|_{dec}\leq\|u\|_{dec}$.

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Edit: Existence of the contraction.

If $$\begin{bmatrix}A&B\\ B^*& C\end{bmatrix}\geq0,$$ then$$\begin{bmatrix}A+\varepsilon I&B\\ B^*& C+\varepsilon I\end{bmatrix}\geq0,$$so multiplying left and right with $\begin{bmatrix}(A+\varepsilon I)^{-1/2}&0\\0&(C+\varepsilon I)^{-1/2}\end{bmatrix}$ we have $$\begin{bmatrix}I&(A+\varepsilon I)^{-1/2}B(C+\varepsilon I)^{-1/2}\\(C+\varepsilon I)^{-1/2}B^*(A+\varepsilon I)^{-1/2} & I\end{bmatrix}\geq0,$$ which is equivalent to $\|Y\|\leq1$, where $Y=(A+\varepsilon I)^{-1/2}B(C+\varepsilon I)^{-1/2}$. So $$ B=(A+\varepsilon I)^{1/2}Y(C+\varepsilon I)^{1/2}, $$ with $\|Y\|\leq1$.