4 The smallest and the largest s-numbers
On a Hilbert space, every s-number sequence agrees with the approximation numbers: \(s_n(S) = a_n(S)\) for every continuous linear map \(S\) between Hilbert spaces.
The upper bound is Theorem 4.2. For the lower bound \(a_n(S) \le s_n(S)\): given \(c {\lt} a_n(S)\), the scalar factorisation (Theorem 6.8) gives contractions with \(B \circ S \circ A = c \cdot \mathrm{id}_{\ell _2^{n+1}}\); chaining (S3) with the (S5) normalisation \(s_n(\mathrm{id}_{\ell _2^{n+1}}) = 1\) and homogeneity yields \(c \le s_n(S)\). Letting \(c \to a_n(S)\) gives \(a_n(S) \le s_n(S)\).
For every s-number sequence \(s\) and every operator \(S\),
For the upper bound \(s_n(S) \le a_n(S)\), let \(A\) be any operator with \(\operatorname {rank}(A) \le n\); then axiom (S2) yields \(s_n(S) \le s_n(A) + \| S - A\| = \| S - A\| \) by (S4), and taking the infimum over \(A\) finishes. For the lower bound \(h_n(S) \le s_n(S)\): each ratio \(a_n(B S A)/(\| B\| \| A\| )\) defining \(h_n(S)\) has \(B S A\) an operator between Hilbert spaces, so the coincidence theorem (Theorem 4.1) gives \(a_n(B S A) = s_n(B S A) \le \| B\| \, s_n(S) \, \| A\| \) by (S3); take the supremum.