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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 $0<p<q$:

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.