 Solution

The printed definition of the restricted isometry constant omits the quantifier: the two inequalities must hold for every vector with at most $s$ nonzero entries. The constant is the smallest nonnegative value, or the infimum over positive values; it can be zero.

Let $T=\mathrm{supp}u\cup \mathrm{supp}v$ and $B=A_T^{\ast }{A}_T-I$. The restricted isometry property of order $s+t$ says $|z^{\ast }Bz|\le {\delta }_{s+t}\Vert z{\Vert }_2^2$ for every $z$ supported on $T$. Since $B$ is a Hermitian matrix, the finite-dimensional spectral theorem implies $\Vert B{\Vert }_{2\to 2}\le {\delta }_{s+t}$. Disjoint supports give $⟨u,v⟩=0$, henceThis argument works over $\mathbb{C}$ and avoids the loss of a factor caused by treating real and imaginary parts separately.