Let \(D\) be the dimension of \(X\) as a real vector space, so \(D {\gt} n\), and let \(\mu \) be a Haar measure on \(X\). Scaling a ball by \(\varepsilon \) scales its measure by \(\varepsilon ^D\), so a cover of \(B_X\) by \(2^n\) balls of radius \(\varepsilon \) gives \(\mu (B_X) \le 2^n \varepsilon ^D \mu (B_X)\) and hence \(1 \le 2^n \varepsilon ^D\), the measure of \(B_X\) being positive and finite. Were \(\varepsilon \le 1/2\), then \(2^n \varepsilon ^D \le 2^n 2^{-(n+1)} = 1/2\), a contradiction.

First let \(T\) be an operator between Hilbert spaces and \(\gamma {\lt} a_n(T)\). The scalar factorisation (Theorem 6.8) provides contractions \(A, B\) with \(B \circ T \circ A = \gamma \cdot \mathrm{id}\) on \(\ell _2^{n+1}\); replacing \(B\) by \(\gamma ^{-1} B\) makes the composition the identity, so

by Proposition 11.8 and the ideal property of \(e_n\). Hence \(a_n(T) \le 2 e_n(T)\).

Now \(h_n(S)\) is the supremum of \(a_n(B \circ S \circ A)/(\| B\| \, \| A\| )\) over factorisations through \(\ell _2\), and each numerator is at most \(2 e_n(B \circ S \circ A) \le 2 \| B\| e_n(S) \| A\| \).