Paper 2, Section I, 2F2 F

Topics in Analysis | Part II, 2018

For x∈Rn\mathbf{x} \in \mathbb{R}^{n} we write x=(x1,x2,…,xn)\mathbf{x}=\left(x_{1}, x_{2}, \ldots, x_{n}\right). Define

P:={x∈Rn:xj⩾0 for 1⩽j⩽n}.P:=\left\{\mathbf{x} \in \mathbb{R}^{n}: x_{j} \geqslant 0 \text { for } 1 \leqslant j \leqslant n\right\} .

(a) Suppose that LL is a convex subset of PP, that (1,1,…,1)∈L(1,1, \ldots, 1) \in L and that ∏j=1nxj⩽1\prod_{j=1}^{n} x_{j} \leqslant 1 for all x∈L\mathbf{x} \in L. Show that ∑j=1nxj⩽n\sum_{j=1}^{n} x_{j} \leqslant n for all x∈L\mathbf{x} \in L.

(b) Suppose that KK is a non-empty closed bounded convex subset of PP. Show that there is a u∈K\mathbf{u} \in K such that ∏j=1nxj⩽∏j=1nuj\prod_{j=1}^{n} x_{j} \leqslant \prod_{j=1}^{n} u_{j} for all x∈K\mathbf{x} \in K. If uj≠0u_{j} \neq 0 for each jj with 1⩽j⩽n1 \leqslant j \leqslant n, show that

∑j=1nxjuj⩽n\sum_{j=1}^{n} \frac{x_{j}}{u_{j}} \leqslant n

for all x∈K\mathbf{x} \in K, and that u\mathbf{u} is unique.

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