Projected gradient on the nonnegative orthant

The function is a continuously differentiable convex quadratic, with $f((1-\alpha )x+\alpha z)=(1-\alpha )f(x)+\alpha f(z)-\frac{\alpha (1-\alpha )}{2}\mathrm{\parallel }x-z{\mathrm{\parallel }}_2^2(0\le \alpha \le 1)\mathrm{.}$ Its gradient is $\mathrm{\nabla }f(x)=(x_1+2,x_2)$, whose Lipschitz constant is $1$, hence $L=2$ is valid. At $x_0$, the unprojected point is $x_0-\mathrm{\nabla }f(x_0)\mathrm{/}2=(-1,0)$; coordinatewise projection onto $C$ gives $x_1=(0,0)$. Repeating the same computation shows the iterate remains there.

For every $z\in C$, $⟨(-2,0),z-(0,0)⟩=-2z_1\le 0$, hence $(-2,0)\in N_C(0)$ by the outward-cone definition. Equivalently, write $C$ as $-x_i\le 0$ with $G=-I,d=0$; at $x_{\ast }=0$, $\lambda =(2,0)\ge 0$ has zero complementarity and $\mathrm{\nabla }f(0)+G^T\lambda =(2,0)-(2,0)=0$. By [F3], this KKT tuple certifies global minimality and the projected-gradient fixed point.