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 $T=\operatorname{supp}u\cup\operatorname{supp}v$ and $B=A_T^*A_T-I$. The <restricted isometry property> of order $s+t$ says $|z^*Bz|\le\delta_{s+t}\|z\|_2^2$ for every $z$ supported on $T$. Since $B$ is a <Hermitian matrix>, the <finite-dimensional spectral theorem> implies $\|B\|_{2\to2}\le\delta_{s+t}$. Disjoint supports give $\langle u,v\rangle=0$, hence
$$
\boxed{|\langle Au,Av\rangle|=|u_T^*Bv_T|\le\delta_{s+t}\|u\|_2\|v\|_2.}
$$
This argument works over $\mathbb C$ and avoids the loss of a factor caused by treating real and imaginary parts separately.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 340 / 4 / c / Solution

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The restricted isometry property controls $\|Ax\|_2^2/\|x\|_2^2$ uniformly over <sparse vectors>. It supplies quantitative near-orthogonality of disjoint sparse coordinate combinations, not merely normalization of individual columns.

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 Incoming links: Restricted isometry property