A graph is said to have bounded degree if there exists $N \in \mathbb{N}$ such that, for every $x \in V$, one has $\sum\limits_{y \in V} A_{x,y} \le N$. Show that, in this case, for any $f \in l^2(V)$, one has $$ ||Af||_2 \le N ||f||_2 $$
MY IDEA:
$$ ||Af||_2=(\sum\limits_{x \in V} |(Af)(x)|^2)^\frac{1}{2}=(\sum\limits_{x \in V} |\sum\limits_{y \in V}A_{xy}f(y)|^2)^\frac{1}{2} $$
But, by Cauchy-Schwarz, we have $$ (\sum\limits_{x \in V} |\sum\limits_{y \in V}A_{xy}f(y)|^2)^\frac{1}{2} \le (\sum\limits_{x \in V} ||A_x||_2^2 ||f||_2^2)^\frac{1}{2}. $$
But since $||A_x||_2^2=\sum\limits_{y \in V} A_{xy}^2 \le (\sum\limits_{y \in V} A_{xy})^2$, we have
$$ (\sum\limits_{x \in V} ||A_x||_2^2 ||f||_2^2)^\frac{1}{2} \le ||f||_2 (\sum\limits_{x \in V} (\sum\limits_{y \in V} A_{xy})^2)^\frac{1}{2} \le ||f||_2 N |V|^\frac{1}{2}. $$
But this is no the result... where is the mistake?