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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 $0<p<q$:

Use the Lq null space property established above. Fix $0\ne v\in \ker A$ and order its magnitudes $a_1\ge \cdots \ge a_N\ge 0$. For a fixed exponent, the sum over the largest $s$ entries is the greatest sum over any support of a vector of size at most $s$, so it suffices to verify the property for this ordered support.

If $s\ge 1$, put $t=a_s$. We have $t>0$: otherwise $v$ would have fewer than $s$ nonzero entries, and applying the Lq null space property to its support would assert a positive number is less than zero. For $0<p<q$, the exponent $p-q$ is negative. ThusThe first inequality uses $a_i\ge t$ on the top part; the last uses $0<a_i\le t$ in the tail. Zero tail entries contribute zero without invoking a negative power of zero. Every other set of size at most $s$ has no larger top sum and no smaller complementary sum, so it too satisfies the strict $p$ inequality. The preceding equivalence provesThis is monotonicity of uniform sparse recovery in the exponent. If $\ker A=\{0\}$ the feasible vector is unique regardless of the objective; if $s=0$, only the zero vector is relevant. These cases need no threshold argument.