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Compressed sensing reconstructs a sparse vector, or an approximately sparse one, from fewer linear measurements than its ambient dimension. Basis pursuit replaces counting nonzero coordinates by convex ${\ell }^1$ minimization. A null space property characterizes exact uniform recovery, while a robust null space property controls errors from noise and nonsparse tails.

The printed bound is false, even for $s=2$ and zero noise. Let $P=I_7-11^T/7$ and choose $A$ with six orthonormal rows spanning $1^{\perp }$, so $A^{T}A=P$. For a vector supported on at most $s$ entries,Take $x_0=(0,0,1,1,1,1,1)$, $s=2$, and $\eta =0$. Every feasible vector is $x_0+t1$, whose ${\ell }^1$ objective is $2|t|+5|1+t|$. Its unique minimizer is $\hat{x}=(-1,-1,0,0,0,0,0)$. YetNo choice of the noise constant repairs this example. Also $s=1$ must be excluded: $A=(1,1)$ has ${\delta }_1=0$, but its two unit-coordinate vectors have identical measurements and both minimize basis pursuit.

The corrected conclusion, for $s\ge 2$, iswith both constants depending on the restricted isometry constant. The order-$s$ threshold itself is valid for $s\ge 2$; it need not be replaced by an order-$2s$ threshold. A precise sharp robust recovery theorem gives, for $\delta ={\delta }_s<1/3$, the admissible constantsThis is the standard stable recovery result, invoked here explicitly; its order restriction and constants are given in arxiv.org/pdf/1302.1236, Theorem 3.3.

For completeness, the mechanism behind a robust null space property proof is as follows. Put $h=\hat{x}-x_0$, let $S$ index the largest $s$ entries of $x_0$, and write $\sigma =\Vert x_{0,S^c}{\Vert }_1$. Feasibility and minimality give the tube and cone inequalitiesIf $A$ has the robust null space property with $0\le \rho <1$ and $\tau >0$, namely $\Vert h_T{\Vert }_2\le \rho \Vert h_{T^c}{\Vert }_1/\sqrt{s}+\tau \Vert Ah{\Vert }_2$ for every $|T|\le s$, thenApply that property again to the largest $s$ entries $T$ of $h$. Sorted coefficients satisfy $\Vert h_{T^c}{\Vert }_2\le \Vert h{\Vert }_1/\sqrt{s}$, because each tail entry is at most $\Vert h{\Vert }_1/s$. ThereforeThis proves the usual two-constant conclusion under the explicitly stated robust null space property and explains why a fixed coefficient one is not supplied by that argument. The sharp order-$s$ theorem above supplies the conclusion for the actual printed threshold, after the necessary corrections.