Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 340 / 3 / b

It is equivalent to minimize $\Vert z{\Vert }_q$ or its $q$th power $F_q(z)={\sum }_i|z_i{|}^q$, because taking the $q$th root is strictly increasing. For $q<1$, this is an Lq quasi-norm, while $q=1$ gives the L1 norm.

Assume the Lq null space property of order $s$. Let $x$ have support of a vector $S$ with $|S|\le s$, and let $z=x+v$ be any distinct feasible vector. Then $0\ne v\in \ker A$. The triangle inequality for complex magnitudes and the subadditivity $(a+b{)}^q\le a^q+b^q$ giveFor $q=1$ the same inequality follows directly from the triangle inequality. Since $x=0$ off $S$,Thus $x$ is the unique minimizer.

Conversely, suppose every $s$-sparse vector is the unique minimizer for its measurements. Fix $0\ne v\in \ker A$ and any $S$ with $|S|\le s$. The vectors $x=-v_S$ and $z=v_{S^c}$ are distinct and satisfy $Az=Ax$, since $Av=0$. Uniqueness for this $x$ therefore implies $F_q(x)<F_q(z)$, exactly the required strict inequality. HenceThe quantifier is uniform over the entire sparse class. Recovery of one particular signed vector alone would not imply this null space property.