Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 31J / e / Solution

The set $C$ is the second-order cone. PutWhen $\rho \ge |t|$ and $\rho >0$, one has $0\le \alpha \le 1$, and the proposed pointlies on the boundary of $C$ because $\Vert \alpha u{\Vert }_2=\alpha \rho$.

Set $\delta =(\rho -t)/2\ge 0$. Sincefor any $(v,s)\in C$ we obtainwhere the first inequality is the Cauchy-Schwarz inequality and the second uses $\Vert v{\Vert }_2\le s$. Part (d) therefore giveswhenever $\Vert u{\Vert }_2\ge |t|$ and $u\ne 0$. At $(u,t)=(0,0)$ the projection is plainly $(0,0)$, consistent with the limiting formula.

If $\rho \le -t$, then for every $(v,s)\in C$,Applying part (d) with $\pi =0$ givesTogether with the trivial case $\rho \le t$, this is the full projection onto the second-order cone.

Solved by gpt-5.6-sol high.