$f_i(z) = f_i(x,y) = y_i^\alpha - x_i$, where $x,y \in R^n$, $z = (x, y)$, $\alpha \in R$ and $\alpha > 1$, $i = 1, \dots, n$.
Are these functions convex functions ?
For $f_i$, the gradient is $ \begin{pmatrix} 0 \\ \vdots \\ -1 \\ \vdots \\ 0 \\ 0 \\ \vdots \\ \alpha y_i^{\alpha-1} \\ \vdots \\ 0 \\ \end{pmatrix}$,
and the Hessian should be $ \begin{pmatrix} 0 & \cdots & 0 & \cdots & 0 & \cdots & 0\\ \vdots & & \vdots & & \vdots & & \vdots\\ 0 & \cdots & 0 & \cdots & 0 & \cdots & 0\\ 0 & \cdots & 0 & \cdots & 0 & \cdots & 0\\ \vdots & & \vdots & & \vdots & & \vdots\\ 0 & \cdots & 0 & \cdots & \alpha(\alpha-1)y_i^{\alpha-2} & \cdots & 0\\ \vdots & & \vdots & & \vdots & & \vdots\\ 0 & \cdots & 0 & \cdots & 0 & \cdots & 0\\ \end{pmatrix} $
So when $y_i^{\alpha-2} \geq 0$, it's a convex function?