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 $r$.
