Let $\mathcal{N}_{n}$ be the set of symmetric nonnegative irreducible matrices. For a matrix $A \in \mathcal{N}_{n}$ let $v^{A}$ be its Perron vector, normalized so that $||v^{A}||_{2}=1$.
Define the function $f:\mathcal{N}_{n} \rightarrow \mathbb{R}$ by $$ f(A)=||v^{A}||_{1}. $$
According to my observations, $f$ is apparently convex.
Is this true/known?
Maybe I am misinterpreting something, because according to my experiments, this function is neither convex nor concave.
The following is a counterexample (EDIT: I changed the example to use symmetric matrices):
\begin{equation*} A=\begin{pmatrix}8 &4\\ 4 & 6\end{pmatrix},\quad B=\begin{pmatrix} 4 & 4\\ 4 & 6\end{pmatrix},\quad C=\frac{A+B}{2}=\begin{pmatrix}6 & 4\\ 4 &6 \end{pmatrix}. \end{equation*} For this choice, we have
\begin{equation*} v(A) = (.7882, .6154),\quad v(B)=(.6154,.7882),\quad v(C)=(.7071, .7071). \end{equation*} But $\|v(C)\|_1 > 0.5\|v(A)\|_1 + 0.5\|v(B)\|_1$ (notice all there vectors have unit 2-norm as required).
A similar counterexample to potential concavity is also easy to find.
EDIT 2: Here is a counterexample to concavity.
\begin{equation*} A = \begin{pmatrix}16&2\\ 2&16\end{pmatrix},\quad B = \begin{pmatrix}14&8\\8 &2 \end{pmatrix},\quad C = (A+B)/2 \end{equation*} Then, we have \begin{equation*} v(A) = \begin{pmatrix}\tfrac{1}{\sqrt{2}}\\\tfrac{1}{\sqrt{2}}\end{pmatrix},\quad v(B) =\begin{pmatrix}\tfrac{2}{\sqrt{5}}\\\tfrac{1}{\sqrt{5}}\end{pmatrix},\quad v(C)= \begin{pmatrix}\tfrac{3+\sqrt{34}}{\sqrt{68+6 \sqrt{34}}}\\\tfrac{5}{\sqrt{68+6 \sqrt{34}}} \end{pmatrix}. \end{equation*} Doing the numerics with this shows that $\|v(C)\|_1-0.5(\|v(A)\|_1+\|v(B)\|_1) = -0.0150285...$.
