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](http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem#Perron.E2.80.93Frobenius_theorem_for_irreducible_matrices), 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? 



Is this function of a matrix convex?
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 $\|v(C)\|\_1-0.5(\|v(A)\|\_1+\|v(B)\|\_1) = -0.0150285...$.


