Bourbaki

Exercise 7, § 6

Topological Vector Spaces, II, p. 82

a) Let F, G be two vector spaces in separating duality and C a weakly closed convex cone in F. Let M be a finite dimensional vector subspace of G. Show that, either there exists y0∈Cy_0 \in C such that y0∈M∘y_0 \in M^\circ and y0≠0y_0 \neq 0, or there exists z0∈Mz_0 \in M such that z0∈C∘z_0 \in C^\circ and z0≠0z_0 \neq 0 (argue by induction on the dimension of M). If C does not contain any line and if the two preceding properties are simultaneously satisfied, show that z0z_0 cannot be an internal point of C∘C^\circ. b) Let the two matrices (aij),(bij)(a_{ij}), (b_{ij}) with real entries in nn rows and mm columns, be such that aij>0a_{ij} > 0 for every pair (i,j)(i, j). Show that there is a unique value of λ∈R\lambda \in \mathbf{R} such that there are two vectors x=(xj)∈Rm,y=(yi)∈Rnx = (x_j) \in \mathbf{R}^m, y = (y_i) \in \mathbf{R}^n satisfying the relations x≠0, y≠0, xj≥0, yi≥0x \neq 0,\ y \neq 0,\ x_j \geq 0,\ y_i \geq 0 for all i,ji, j and finally such that

(1)λ∑j=1maijxj≥∑j=1mbijxjfor 1≤i≤n(2)λ∑i=1naijyi≤∑i=1nbijyifor 1≤j≤m. \begin{align*} (1)\quad & \lambda \sum_{j=1}^m a_{ij} x_j \geq \sum_{j=1}^m b_{ij} x_j \quad \text{for}\ 1 \leq i \leq n \\ (2)\quad & \lambda \sum_{i=1}^n a_{ij} y_i \leq \sum_{i=1}^n b_{ij} y_i \quad \text{for}\ 1 \leq j \leq m . \end{align*}

(Putting cij=λaij−bijc_{ij} = \lambda a_{ij} - b_{ij} for 1≤i≤n, 1≤j≤m1 \leq i \leq n,\ 1 \leq j \leq m, and cn+i,j=δijc_{n+i,j} = \delta_{ij} (Kronecker’s index) for 1≤i≤m1 \leq i \leq m show that the problem reduces to finding a vector x∈Rmx \in \mathbf{R}^m and a vector z=(zi)∈Rn+mz = (z_i) \in \mathbf{R}^{n+m} which are non null and satisfy the relations

(3)

∑j=1mcijxj≥0for1≤i≤n+m \sum_{j=1}^m c_{ij} x_j \geq 0 \quad \text{for} \quad 1 \leq i \leq n + m
(4)
∑i=1n+mcijzi=0for1≤j≤m \sum_{i=1}^{n+m} c_{ij} z_i = 0 \quad \text{for} \quad 1 \leq j \leq m
and zi≥0z_i \geq 0 for 1≤i≤n+m1 \leq i \leq n + m. Remark that, if (3) has a solution for one value λ0\lambda_0 of λ\lambda, then it also has a solution for λ≥λ0\lambda \geq \lambda_0, and that if (4) has a solution for λ0\lambda_0, then it also has a solution for λ≤λ0\lambda \leq \lambda_0. Finally use a).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.