Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 36 / 3

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 $S$.

Write $\sigma =\mathrm{sgn}(x_S)$ and $G=A_S^{\ast }{A}_S$. The injectivity of $A_S$ makes this Gram matrix a positive-definite matrix, since $u^{\ast }Gu=\Vert A_{S}u{\Vert }_2^2>0$ for $u\ne 0$. Thus its matrix inverse exists. Construct the least-norm dual certificateOn the active coordinates, $A_S^{\ast }h=G{G}^{-1}\sigma =\sigma$. For $l\notin S$, symmetry of the real Gram matrix and its matrix inverse givesCondition (iii) makes the absolute value of this coordinate strictly less than one. Consequently $h$ is a strict dual certificate for basis pursuit, and part (c) applies. If $S$ is empty, take $h=0$; the zero vector uniquely minimizes the L1 norm on its feasible set. Condition (iii) supplies an explicit certificate and therefore unique recovery. There is no claim that this particular least-norm dual certificate is necessary: other valid certificates may exist when this one fails.