Suppose that $V$ is a finite dimensional complex Hilbert space. Let $L(V)$ denote the collection of all linear mappings from $V$ to $V$. Let $A\_1,\dots,A\_r:V\rightarrow V$ be linear operators. Then define a completely positive superoperator $\Phi(A\_1,\dots,A\_r):L(V)\rightarrow L(V)$ by letting $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots+A\_rXA\_r^\*$, and define the $L\_2$-spectral radius of $A\_1,\dots,A\_r$ by letting
$\rho\_2(A\_1,\dots,A\_r)=\rho(\Phi(A\_1,\dots,A\_r))^{1/2}$.


The Cauchy-Schwarz inequality holds for the $L\_2$-spectral radius. If $V,W$ are finite dimensional complete Hilbert spaces, and $A\_i:V\rightarrow V,B\_i:W\rightarrow W$ for $1\leq i\leq r$, then $$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq\rho(\Phi(A\_1,\dots,A\_r))^{1/2}\rho(\Phi(B\_1,\dots,B\_r))^{1/2}.$$


Now, define the $L\_{2,d}$-spectral radius of $A\_1,\dots,A\_r$ by letting


$$\rho\_{2,d}(A\_1,\dots,A\_r)$$
$$=\sup\{\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\mid \rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_d(\mathbb{C})\}.$$


From the Cauchy-Schwarz inequality, we observe that $$\rho\_{2,d}(A\_1,\dots,A\_r)\leq\rho\_{2,g}(A\_1,\dots,A\_r)\leq\rho\_{2,\dim(V)}(A\_1,\dots,A\_r)=\rho\_2(A\_1,\dots,A\_r)$$
whenever $1\leq d\leq g$. One should therefore think of $\rho\_{2,d}$ as an approximation to $\rho\_{2}$.


Suppose that $1\leq d<\dim(V)$. Then is there some $r$ along with linear operators $A\_1,\dots,A\_r:V\rightarrow V$ such that $\rho\_{2,d}(A\_1,\dots,A\_r)<\rho\_{2,d+1}(A\_1,\dots,A\_r)?$


Example 0: $\rho\_{2,d}(A)=\rho\_{d}(A)=\rho(A).$ Therefore, $\rho\_{2,d},\rho\_{2}$ should be considered to be a sort of spectral radius but for multiple linear operators.


Example 1: $$\rho\_{2,1}(A\_1,\dots,A\_r)=\sup\{\rho(\alpha A\_1+\dots+\alpha\_r A\_r):|\alpha\_1|^2+\dots+|\alpha\_r|^2=1\}.$$


Example 2: Observe that if $d\geq\dim(V)$, then $$\rho\_{2,d}(A\_1,\dots,A\_r)=\rho(\Phi(A\_1,\dots,A\_r))^{1/2}.$$


Example 3: $$\rho\_{2,1}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix})=1/\sqrt{2}<1=\rho\_{2,2}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}).$$


Example 4: Let $C\_1,\dots,C\_r$ be the $r\times r$-matrices, where $(C\_i)\_{i,i+1\mod r}=1$ and where all the other entries in $C\_i$ are zero. Then
$\rho(C\_1\otimes X\_1+\dots+C\_r\otimes X\_r)=\rho(X\_1\dots X\_r)^{1/r}$. Therefore,
$$\rho\_{2,d}(C\_1,\dots,C\_r)=\sup\{\frac{\rho(X\_1\dots X\_r)^{1/r}}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}:\rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_d(\mathbb{C})\}.$$


Now, suppose that $j\_1,\dots,j\_r$ are natural numbers with $1\leq j\_1\leq\dots\leq j\_r=d$ and where $j\_{i+1}-j\_{i}\in\{0,1\}$ for $1\leq i\leq r$. For $1\leq i\leq r$, let $X\_i$ be the $d\times d$ matrix where $(X\_i)\_{j\_{i},j\_{i+1}}=1$ (here $i$ is taken modulo $r$) and where all the other entries in $X\_i$ are zero. Then $\rho(X\_1\dots X\_r)=1$. Let $n\_{1},\dots,n\_{d}$ be the natural numbers defined by letting $n\_{j}=\{i\mid 1\leq i\leq r,j\_{i}=j\}$. Then $r=n\_1+\dots+n\_d$. Suppose now that $n\_j=r/d$ for all $j$. Then $\rho(\Phi(X\_1,\dots,X\_r))=\frac{r}{d}.$ Therefore, $$\rho\_{2,d}(C\_1,\dots,C\_r)\geq\sqrt{\frac{d}{r}}$$ whenever $d$ is a factor of $r$.



Are these $L_2$-spectral radii approximations strictly increasing?
**Yes.** If $C\_1,\dots,C\_r$ are the matrices in Example 4, then by [my answer to my other question](https://mathoverflow.net/a/424572/22277), $$\rho\_{2,d}(C\_1,\dots,C\_r)=\sqrt{d/r}$$
whenever $1\leq d\leq r$.


