Blueprint:
s-Numbers of Bounded Linear Operators
between Banach Spaces

8.3.3 The Hilbert numbers of \(I\)

\(h_n(I) \ge (n+1)^{-1}\) for all \(n\).

Proof

The identity of \(\ell _2^{\, n+1}\) factors as \(\mathrm{id}_{\ell _2^{\, n+1}} = \mathrm{proj}_\infty \circ I \circ \mathrm{emb}_1\) with \(\| \mathrm{emb}_1\| , \| \mathrm{proj}_\infty \| \le \sqrt{n+1}\). By the ideal property of the Hilbert numbers and their normalisation \(h_n(\mathrm{id}_{\ell _2^{\, n+1}}) \ge 1\),

\[ 1 \le \| \mathrm{proj}_\infty \| \, h_n(I)\, \| \mathrm{emb}_1\| \le (n+1)\, h_n(I) . \]

\(h_n(I) \le (n+1)^{-1}\) for all \(n\).

Proof

Let \(A \in \mathcal{L}(\ell _2, \ell _1)\) and \(B \in \mathcal{L}(\ell _\infty , \ell _2)\) be non-zero; we must bound \(a_n(B I A) \le \| A\| \, \| B\| /(n+1)\). Factorising \(I = J_2 J_1\) with the norm-one inclusions \(J_1 \colon \ell _1 \to \ell _2\) and \(J_2 \colon \ell _2 \to \ell _\infty \), we obtain \(T := B I A = T_2 T_1\) with \(T_1 := J_1 A\) and \(T_2 := B J_2\) on \(\ell _2\). Applying Lemma 8.8 to \(B\) and to the rows of \(A\),

\[ \| T_2\| _{HS}^2 = \sum _j \| B e_j\| ^2 \le \| B\| ^2 \quad \text{and}\quad \| T_1\| _{HS}^2 = \sum _j \| \mathrm{row}_j(A)\| ^2 \le \| A\| ^2 , \]

which by Lemma 8.9 gives the two hypotheses of Theorem 8.10 along the singular vectors of \(T\). That theorem yields \((n+1)\, a_n(T) \le \| A\| \, \| B\| \) provided \(T\) is compact.

For compactness, replace \(A\) by its coordinate truncation \(A_m\), of rank \(\le m\) and with \(\| A_m\| \le \| A\| \); then \(B I A_m\) has finite rank, hence is compact (Theorem 6.3), and the estimate above applies to it. Since \(\sum _j \| \mathrm{row}_j(A)\| ^2 {\lt} \infty \) we have \(\| \mathrm{row}_j(A)\| \to 0\), so \(\| I(A - A_m)\| \to 0\), and subadditivity of the approximation numbers,

\[ a_n(B I A) \le a_n(B I A_m) + \| B\| \, \| I(A - A_m)\| , \]

lets \(m \to \infty \). Dividing by \(\| A\| \, \| B\| \) and taking the supremum over all admissible \(A, B\) gives \(h_n(I) \le (n+1)^{-1}\).

\(h_n(I) = (n+1)^{-1}\) for all \(n\).

Proof

Combine Proposition 8.12 and Theorem 8.13.