Let $A$ be a strictly substochastic matrix, let $x$ be a scalar with $x\in[0,1]$, and let $$\Lambda(x)\equiv(I-xA^{T})^{-1}.$$ Let $v$ be a vector satisfying $v_{i}\in[0,1],\forall i$ and $\sum_{i}v_{i}=1$. Consider the matrix $$J(v,x)=\left(\mathrm{diag}\left(\Lambda(x)v\right)\right)^{-1}\Lambda(x)\left[x\mathrm{diag}\left(v\right)+(1-x)\mathrm{diag}\left(\Lambda(x)v\right)\right]\Lambda^{T}(x)\mathrm{diag}\left(I-A\iota\right),$$ which can also be written as $$J(v,x)=S(v,x)\left[xI+(1-x)\left(\mathrm{diag}\left(v\right)\right)^{-1}\mathrm{diag}\left(\Lambda(x)v\right)\right]\Lambda^{T}(x)\mathrm{diag}\left(I-A\iota\right),$$ where $$S(v,x) \equiv \left(\mathrm{diag}\left(\Lambda(x)v\right)\right)^{-1}\Lambda(x)\mathrm{diag}\left(v\right).$$ I want to show that the spectral radius of $J(v,x)$ is weakly lower than one for all $v,x$. Simulations suggest that this is true and that the maximum is achieved for $v$ at a corner (i.e., $v_{i}=1$ for some $i$) and $x=1$, with the spectral radius at that point equal to $1$.
For the case in which $x = 1$ we have $$J(v,1)=S(v,1) (I-A)^{-1}\mathrm{diag}\left(I-A\iota\right).$$ $S(v,\iota)$ is a stochastic matrix, while $$\left(I-A\right)^{-1}\mathrm{diag}\left(I-A\iota\right)\iota=\left(I-A\right)^{-1}\left(I-A\right)\iota=\iota,$$ so matrix $\left(I-A\right)^{-1}\mathrm{diag}\left(I-A\iota\right)$ is also stochastic, so it follows that $J(v,\iota)$ is substochastic and hence has spectral radius weakly lower than one. However, if $v$ is at a corner with $v_{i}=1$ then $S(v,x)$ is a matrix with ones in column $i$, implying that $J(v,1)$ is a matrix with all rows equal to row $i$ of matrix $(I-A)^{-1}\mathrm{diag}\left(I-A\iota\right)$, which adds up to one, and hence has spectral radius equal to one. Thus, the spectral radius of $J(v,1)$ is everywhere weakly lower than one except at corners of $v$ at which it is equal to 1.
If $x = 0$ then $J(v,0) = \mathrm{diag}(I-A\iota).$ This is a diagonal matrix with all elements between zero and one and hence the spectral radius is $1-\min_i \sum_j a_{ij}$, which by assumption is weakly less than one.
In the special case in which $A$ is a diagonal matrix then $S(v,x) = I$ and $$ J(v,x) = \left[xI+(1-x) \Lambda(x)\right]\Lambda^{T}(x)\left(I-A\right).$$ Each element of this diagonal matrix is equal to $$ \left( x + \frac{1-x}{1-x a_{ii}} \right) \frac{1-a_{ii}}{1-x a_{ii}},$$ which is lower than one except at $x = 1$, in which case it is equal to one.
How can I extend this to show that $\rho(J(v,x))\leq1$ for all $v,x$?
One idea is to do this in two steps. In step (1) one would show that for any $v,x$ there is a $k$ such that moving to the corner $v$ with $v_k = 1$ (let's call this corner vector by $v^k$) increases the spectral radius, $\rho(J(v^k,x))\geq \rho(J(v,x))$. In step (2) we would then show that $\rho(J(v^k,1 )) \geq \rho(J(v^k,x))$. Since $ \rho(J(v^k,1))=1$ then these two steps combined would imply that for any $v,x$ we have $\rho(J(v,x)) \leq 1$. Unfortunately, neither of these two steps seems easy, so this could be a false lead.
Something that could be useful is that $$\Lambda^T(x)\mathrm{diag}\left(I-xA\iota\right)\iota=\left(I-xA\right)^{-1}\left(I-xA\right)\iota=\iota,$$ so $\Lambda^T(x)\mathrm{diag}\left(I-xA\iota\right)$ is a stochastic matrix. This implies that $$\Lambda^T(x)\mathrm{diag}\left(I-A\iota\right)\iota \leq \Lambda^T(x)\mathrm{diag}\left(I-xA\iota\right)\iota=\iota,$$ so $\Lambda^T(x)\mathrm{diag}\left(I-A\iota\right)$ is a substochastic matrix. However, we cannot just use the bound $$J(v,x) \leq S(v,x)\left[xI+(1-x)\left(\mathrm{diag}\left(v\right)\right)^{-1}\mathrm{diag}\left(\Lambda(x)v\right)\right]\Lambda^{T}(x)\mathrm{diag}\left(I-xA\iota\right),$$ because for example in the diagonal case above we would no longer have $\rho(J(v,k)) \leq 1$ (the bound is too lose).
Another idea would be to use the fact that $$\rho(J)=\rho(B^{-1}JB)\leq||B^{-1}JB||_{\infty}.$$ The trick here would be to find a matrix $B(v,x)$ such that $$||B^{-1}(v,x)J(v,x)B(v,x)||_{\infty} \leq 1.$$ Since we already have $||J(v^k,1)||_{\infty} = 1$ for any $k$ then presumably one would need that $B(v^k,1) = I$ for any $k$.