I'm trying to show whether the function $f(x)=sin(\frac{\pi x_2} {2})$ is convex for $x\in \mathbb{R}^2$, where $0<x_1\le2$ and $0<x_2\le1$
When I compute the Hessian, I find $$H=\nabla^2f=\begin{matrix} 0 & 0\\ 0 &-(\frac{\pi}{2})^2 sin(\frac{\pi x_2}{2})\end{matrix}$$
This matrix has $Trace(H) = 0 $and $det(H) = 0$, which suggests to me that this matrix is positive semidefinite. However, this matrix has a negative eigenvalue and does not satisfy $v^THv\ge0$, which implies that the matrix is not PSD.
How can I rectify this? Is this matrix PSD or no? Thank you!