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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.

For a real vector $x_0$ with support $S$, uniqueness in basis pursuit is equivalent to $|⟨\mathrm{sgn}(x_{0,S}),h_S⟩|<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. The supporting-line inequality for the absolute value proves sufficiency. Taking small positive and negative multiples of $h$ before any active sign changes proves necessity.

The null space property relative to $S$ is $\Vert h_S{\Vert }_1<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. It is equivalent to exact basis pursuit recovery of every vector supported in $S$. Sufficiency follows from the triangle inequality; for necessity, compare $-h_S$ and $h_{S^c}$, which have the same image under $A$. Recovery of a single fixed signed vector can hold without this uniform property.

Let $S=\mathrm{supp}x$ be the support of a vector $x$, with $|S|\le s$. Suppose the null space property holds. Every other feasible vector is $z=x+v$, where $0\ne v\in \ker A$. Splitting its L1 norm over $S$ and $S^c$ and using the triangle inequality givesThus the sparse vector $x$ is the unique minimizer in basis pursuit. Notice that the argument works for complex coordinates: it uses the absolute value inequality, rather than a real sign function.

Conversely, suppose basis pursuit uniquely recovers every sparse vector of order $s$. Fix $0\ne v\in \ker A$ and any $S$ with $|S|\le s$. Take $x=-v_S$ and $z=v_{S^c}$. These vectors have the same measurements, because $A{v}_S+A{v}_{S^c}=0$, and they are distinct since $z-x=v\ne 0$. The vector $x$ has at most $s$ nonzero coordinates, so uniqueness givesThis is the null space property for every such $S$. Uniform unique recovery by basis pursuit is equivalent to the order-$s$ null space property.

Let $\sigma =\mathrm{sgn}(x_S)$, using the real sign function on the support of a vector $x$. The fixed-sign null space condition in the PDF uses $S^c$ on its right-hand side. This complement is essential. For any real coordinate $x_j\ne 0$, the supporting-line inequality for the absolute value isEvery distinct feasible vector is $x+v$ with $0\ne v\in \ker A$. Summing the coordinate inequalities on $S$ and adding the L1 norm on $S^c$ givesThe strict final inequality is precisely the fixed-sign null space condition. Hence $x$ is the unique basis pursuit minimizer. Unlike the null space property of Question 1, this condition concerns the particular sign function values of $x$, rather than every vector supported in $S$.

Assume $x$ is the unique basis pursuit minimizer. Fix $0\ne v\in \ker A$ and setThere is $\epsilon >0$ such that the sign function of $x_j+t{v}_j$ equals that of $x_j$ at every $j\in S$ whenever $|t|<\epsilon$. To choose it, take less than the minimum of $|x_j|/|v_j|$ over the nonzero $v_j$ in $S$; if that set is empty, any positive $\epsilon$ works. On this interval the L1 norm has the exact expressionFor $0<t<\epsilon$, both $x+tv$ and $x-tv$ are distinct feasible vectors. Uniqueness forces their objective differences to be strictly positive, so $\alpha +\beta >0$ and $-\alpha +\beta >0$. ThereforeThe fixed-sign null space condition is necessary as well as sufficient. This argument also covers an empty support of a vector: then $\alpha =0$ and the nonzero null space vector has $\beta >0$. Strictness is indispensable: equality would make a sufficiently short feasible segment have the same objective as $x$.

Let $u=A^{\ast }h$ be the strict dual certificate for basis pursuit supplied by condition (ii). For $0\ne v\in \ker A$, we haveHere the adjoint operator is the real transpose. The injectivity of $A_S$ ensures $v_{S^c}\ne 0$: otherwise $A_S{v}_S=0$ would force $v=0$. Thus at least one nonzero term lies outside the support of a vector $x$. Since $|u_l|<1$ at every such index,This proves the fixed-sign null space condition, so part (a) gives uniqueness in basis pursuit. If $S^c$ is empty, the injectivity of $A_S=A$ instead means there is no nonzero null space vector, and the feasible set is a singleton. Injective active columns and a strict dual certificate for basis pursuit ensure unique recovery. Both ingredients matter: strictness outside $S$ cannot detect a nonzero null space direction supported entirely inside $S$.

Under the usual definition, the null space property relative to $S$ is $\Vert h_S{\Vert }_1<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. It characterizes recovery by basis pursuit of every vector supported in $S$. Indeed, if it holds, then for any such $x_0$ and any nonzero $h\in \ker A$,Conversely, failure for some $h$ makes $x_0=-h_S$ and $z=h_{S^c}$ feasible with the same measurements and $\Vert z{\Vert }_1\le \Vert x_0{\Vert }_1$. ThusFor the single fixed vector in the printing, necessity is false with that definition. For example,has $\ker A=\mathrm{span}\{(1,1,1)\}$. Along its feasible line, the objective is $|1+t|+|-1+t|+|t|$, uniquely minimized at $t=0$, although the null space property would require $2<1$.

The correct fixed-sign null space condition isSufficiency follows from the supporting-line inequality for the absolute value. Necessity follows by considering $x_0\pm th$ for small positive $t$, when all signs on $S$ remain unchanged: a nonpositive directional increase gives either a decrease or another minimizer. If “relative to $S$” was intended to include these signs, this is the precise condition needed.

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.

The displayed property holds for every $h$ and every $|S|\le s$, with $0\le \rho <1$ and $\tau >0$. It converts the cone inequality from basis pursuit minimality and the tube inequality from noisy feasibility into a two-constant bound $\Vert \hat{x}-x_0{\Vert }_2\le C_1\eta +C_2{\sigma }_s(x_0)/\sqrt{s}$. Both constants are needed in general; a fixed coefficient one on the approximation term does not follow.

For $s\ge 2$ and $\delta ={\delta }_s(A)<1/3$, noisy basis pursuit has error $\Vert \hat{x}-x_0{\Vert }_2\le C_1\eta +C_2{\sigma }_s(x_0)/\sqrt{s}$, with constants depending only on $\delta$. One admissible pair isThese constants follow from Theorem 3.3 of the sharp restricted-isometry recovery analysis, with the actual noise bounded by the tolerance $\eta$. The restriction $s\ge 2$ is necessary: equal unit columns give ${\delta }_1=0$ without unique recovery. The coefficient of the approximation term cannot universally be one.

For a real vector $x$ with support of a vector $S$, this certificate is a vector $h$ whose measurements under the adjoint operator match the active sign function values and lie strictly between minus one and one on the inactive coordinates. If $A_S$ is injective, such a certificate proves that $x$ is the unique basis pursuit minimizer. For a nonzero $v\in \ker A$, injectivity ensures $v_{S^c}\ne 0$, and $0=⟨A^{\ast }h,v⟩$ implies the fixed-sign null space condition. The strict inequality is interpreted coordinatewise if $S^c$ is empty. The condition with the sign function written here is for real variables; complex basis pursuit uses the unit phases of active coordinates instead.