I stumbled upon a strange question about eulerian graphs.
Let $G$ be a directed Eulerian graph.
Prove that it's normalized adjacency matrix $A$ has $\| A\|_2=1$.
It looked strange to me and I think I have found a counterexample
$$G=(\{1,2,3,4\},\{(1,2),(2,1),(1,3),(3,4),(4,1)\})$$
$$ A_G=\begin{pmatrix} 0 & 1/2 & 1/2 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{pmatrix} $$
Clearly, $G$ is eulerian since $in-deg(1)=out-deg(1)=2$ and all the other degrees are $1$.
We also have $\|A_G\|=\sqrt{2}>1$.
Maybe we need $G$ to be $d$-regular? And if so how does it help?