3.2 Bernstein numbers \(b_n\)
For \(S \in \mathcal{L}(X, Y)\) and \(n \in \mathbb {N}_0\),
Let \(\mathbb {K}\) be a non-trivially normed field and \(X, Y\) normed spaces over \(\mathbb {K}\). Then the Bernstein numbers \(b = (b_n)_n\) form a strict s-number sequence: they satisfy (S1)–(S5) and the strengthening (S5\('\)).
Write \(\gamma (S, M) := \inf _{0 \ne x \in M} \| Sx\| /\| x\| \) for the gain of \(S\) on a subspace \(M\), so that \(b_n(S) = \sup _{\dim M = n+1} \gamma (S, M)\). On a one-dimensional \(M = \operatorname {span}\{ v\} \) the gain is exactly \(\| Sv\| /\| v\| \), and the supremum over \(v\) gives \(b_0(S) = \| S\| \); monotonicity in \(n\) holds because every \((n+2)\)-dimensional subspace contains an \((n+1)\)-dimensional one, and shrinking \(M\) only enlarges the gain. Subadditivity (S2) is the pointwise triangle inequality \(\| (S+T)x\| /\| x\| \le \| Sx\| /\| x\| + \| T\| \), pushed through the infimum and the supremum. For the ideal property (S3) fix an \((n+1)\)-dimensional \(M \subseteq W\): if \(A\) is not injective on \(M\), the gain of \(BSA\) on \(M\) vanishes; if it is, then \(M' := A(M)\) has dimension \(n+1\) and \(\gamma (BSA, M) \le \| B\| \, \gamma (S, M') \, \| A\| \le \| B\| \, b_n(S) \, \| A\| \). For the rank axiom (S4), an operator of rank \(\le n\) has nontrivial kernel on every \((n+1)\)-dimensional subspace by rank–nullity, forcing the gain to zero. Finally (S5\('\)): the identity has gain \(1\) on every nontrivial subspace, and \(\dim X {\gt} n\) guarantees an \((n+1)\)-dimensional subspace exists.
The argument needs only a non-trivially normed scalar field \(\mathbb {K}\) — no completeness, and no Riesz lemma.