 Solution

Use the stated characterization by the Lq null space property. Raising an Lq quasi-norm comparison to its positive exponent preserves its order, so the hypothesis at $q$ says that, for every nonzero $v\in \ker A$ and every $|S|\le s$,We prove the corresponding $p$ inequality; this is monotonicity of uniform sparse recovery in the exponent.

Fix a nonzero null space vector and arrange its coordinate magnitudes as $a_1\ge \cdots \ge a_N\ge 0$. Assume $s\ge 1$. The Lq null space property rules out a nonzero null space vector supported on at most $s$ coordinates, so $s<N$ and $t=a_s>0$. Since $p-q<0$, the factors $a_j^{p-q}$ are at most $t^{p-q}$ for $j\le s$ and at least $t^{p-q}$ for $j>s$ with $a_j>0$. Terms with $a_j=0$ contribute zero and require no negative power of zero. ThusThe largest $s$ coordinates maximize the $p$-power sum on any set of size at most $s$. Its complement therefore has the smallest complementary $p$-power sum. The displayed strict inequality proves the Lq null space property at exponent $p$ for every allowed support of a vector. The stated recovery characterization now applies to the Lq quasi-norm at $p$.

If $\ker A=\{0\}$, the measurement constraint already singles out $x$, for every objective; if $s=0$, only the zero sparse vector needs recovery. These cases do not need a positive threshold. Uniform recovery at $q$ implies uniform recovery at every $0<p<q$: