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The number of nonzero coordinates of a finite vector is its L0 sparsity count. It vanishes only at zero, but is unchanged under multiplication by a nonzero scalar, so it is not a norm. Minimizing it under linear measurement constraints finds a sparse vector with the smallest possible support of a vector. Sparse injectivity of order $s$ guarantees unique recovery of every sparse vector with at most $s$ nonzero coordinates by this objective.

First the null space property implies sparse injectivity. If a nonzero $v\in \ker A$ had at most $2s$ nonzero coordinates, partition its support of a vector into disjoint $S,T$ with $|S|,|T|\le s$. Applying the null space property to each of these sets yieldswhich is impossible. Hence the null space contains no nonzero sparse vector of order $2s$.

The feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the L0 sparsity count counts its nonzero coordinates. Any different feasible $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ would give a nonzero null space vector $z-x$ with at most $2s$ nonzero coordinates, contrary to sparse injectivity. Consequently every different feasible $z$ has strictly larger L0 sparsity count. The unique sparsest feasible vector is $x$:This proof does not treat the L0 sparsity count as a genuine norm; no triangle inequality for it is needed.

The strict null space property of order $s$ isHere $v_S$ agrees with $v$ on $S$ and is zero elsewhere. Suppose a nonzero null vector had at most $2s$ nonzero entries. Split its support of a vector into disjoint sets $S,T$ of size at most $s$. Applying the null space property to both sets would give $\Vert v_S{\Vert }_1<\Vert v_T{\Vert }_1$ and $\Vert v_T{\Vert }_1<\Vert v_S{\Vert }_1$, a contradiction. Empty parts cause the same contradiction. Thus no such nonzero null vector exists.

For an $s$-sparse $x$, the feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the zero-subscript quantity counts nonzero entries and is not a norm. Any feasible competitor $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ also has at most $s$ nonzero entries. The difference $z-x\in \ker A$ has at most $2s$ nonzero entries, so $z=x$. Competitors with larger support have strictly larger objective. Therefore every s-sparse vector is the unique sparsest feasible vector. This is sparse injectivity; for $s=0$ the only sparse vector is zero and the conclusion is immediate.