Norm of the operator $Tv(y)=\int_{-1}^{1}h(x,y)v(x)dx$ is equal to $\sup_{\|v\|=1, v \geq 0} \|Tv\|$ if $h(x,y) \geq 0$

Question

Let $h \in L^2([-1,1]\times[-1,1])$ such that $h(x,y) \geq 0$ for all $x,y \in [-1,1]$ and let $T:L^2([-1,1]) \to L^2([-1,1]) $ the operator defined by $$ Tv(y)=\int_{-1}^{1}h(x,y)v(x)dx.$$ I need to show that $$\|T\|=\sup_{\|v\|=1, v \geq 0} \|Tv\|.$$ Can you give me a hint, please?