Bourbaki

Exercise 3, § 1

Topological Vector Spaces, V

Bourbaki marks this one as harder than the rest.

Let E be a real finite dimensional vector space with dimension nn, ff a positive and separating symmetric bilinear form on E, and BfB_f the bounded convex set defined by the relation f(x,x)≤1f(x, x) \leq 1. If a=(a1,...,an)a = (a_1, ..., a_n) is a basis of E and Δ\Delta the discriminant of ff with respect to this basis, we call the volume of BfB_f with respect to aa the number va(f)=γn∣Δ∣−1/2v_a(f) = \gamma_n |\Delta|^{-1/2}, where γn=πn/2/Γ(n2+1)\gamma_n = \pi^{n/2}/\Gamma\left(\frac{n}{2} + 1\right). If b=(b1,...,bn)b = (b_1, ..., b_n) is a second basis of E, and if

a1∧a2∧...∧an=δb1∧b2∧...∧bn, a_1 \wedge a_2 \wedge ... \wedge a_n = \delta b_1 \wedge b_2 \wedge ... \wedge b_n ,
we have vb(f)=∣δ∣va(f)v_b(f) = |\delta| v_a(f). a) Show that, if ff and gg are two positive, separating symmetric bilinear forms such that Bf⊂BgB_f \subset B_g (which is equivalent to g≤fg \leq f), then va(f)≤va(g)v_a(f) \leq v_a(g) (consider a basis for E which is orthogonal for both ff and for gg). b) Let A be a symmetric compact convex set in E, with 0 as an interior point. Show that among all positive, separating symmetric bilinear forms ff on E such that A⊂BfA \subset B_f, there exists one and only one for which the volume of BfB_f (with respect to the given basis of E) is the smallest possible. (To show uniqueness observe that if A is contained in BfB_f and BgB_g, it is in B(f+g)/2B_{(f+g)/2} and that we have va((f+g)/2)≤12(va(f)+va(g))v_a((f+g)/2) \leq \frac{1}{2}(v_a(f) + v_a(g)) for every basis aa of E which is orthogonal for both ff and gg.) c) Let A be a symmetric compact convex set in E, with 0 as an interior point, and let ff be the positive, separating, symmetric bilinear form such that A⊂BfA \subset B_f and that BfB_f has the smallest possible volume with respect to a given basis of E. Show that there exist points x1,...,xn,u1,...,unx_1, ..., x_n, u_1, ..., u_n in E with the following properties : α) For every kk, we have xk∈Ax_k \in A and f(xk,xk)=1f(x_k, x_k) = 1. β) The sequence (u1,...,un)(u_1, ..., u_n) is an orthonormal basis of E for ff. γ) If we put xk=∑j=1nakjujx_k = \sum_{j=1}^n a_{kj} u_j for 1≤k≤n1 \leq k \leq n, we have akj=0a_{kj} = 0 for k<jk < j and akk2≥(n−k+1)/na_{kk}^2 \geq (n - k + 1)/n. (Argue by induction on kk. Suppose x1,...,xk,u1,...,ukx_1, ..., x_k, u_1, ..., u_k have been constructed, let PkP_k be the orthoprojector (for ff) from E onto the subspace generated by u1,...,uku_1, ..., u_k; for every ε>0\varepsilon > 0, consider the bilinear form fεf_\varepsilon defined by
fε(x,y)=(1+ε)k−nf(Pkx,Pky)+(1+ε)kf(x−Pkx,y−Pky) f_\varepsilon(x, y) = (1 + \varepsilon)^{k-n} f(P_k x, P_k y) + (1 + \varepsilon)^k f(x - P_k x, y - P_k y)
and prove that A⊄BfεA \not\subset B_{f_\varepsilon}, using b). For every integer p≥1p \geq 1 choose a point ypy_p in A not belonging to Bf1/pB_{f_{1/p}}; take for xk+1x_{k+1} a limit point of the sequence (yp)(y_p) such that
kf(xk+1−Pkxk+1,xk+1−Pkxk+1)≥(n−k)f(Pkxk+1,Pkxk+1); k_f(x_{k+1} - P_k x_{k+1}, x_{k+1} - P_k x_{k+1}) \geq (n - k) f(P_k x_{k+1}, P_k x_{k+1});
next choose uk+1u_{k+1}.) d) Prove the analogues of b) and c) for the positive separating symmetric bilinear forms such that Bf⊂AB_f \subset A and for which the volume of BfB_f (with respect to a given bases of E) is the largest possible.

Solution

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