Paper 1, Section I, 2H

Topics in Analysis | Part II, 2021

Write

P={xRn:xj0 for all 1jn}P=\left\{\mathbf{x} \in \mathbb{R}^{n}: x_{j} \geqslant 0 \text { for all } 1 \leqslant j \leqslant n\right\}

and suppose that KK is a non-empty, closed, convex and bounded subset of Rn\mathbb{R}^{n} with KIntPK \cap \operatorname{Int} P \neq \emptyset. By taking logarithms, or otherwise, show that there is a unique xKP\mathbf{x}^{*} \in K \cap P such that

j=1nxjj=1nxj\prod_{j=1}^{n} x_{j} \leqslant \prod_{j=1}^{n} x_{j}^{*}

for all xKP\mathbf{x} \in K \cap P.

Show that j=1nxjxjn\sum_{j=1}^{n} \frac{x_{j}}{x_{j}^{*}} \leqslant n for all xKP\mathbf{x} \in K \cap P.

Identify the point x\mathbf{x}^{*} in the case that KK has the property

(x1,x2,,xn1,xn)K(x2,x3,,xn,x1)K\left(x_{1}, x_{2}, \ldots, x_{n-1}, x_{n}\right) \in K \Rightarrow\left(x_{2}, x_{3}, \ldots, x_{n}, x_{1}\right) \in K

and justify your answer.

Show that, given any aIntP\mathbf{a} \in \operatorname{Int} P, we can find a set KK, as above, with x=a\mathbf{x}^{*}=\mathbf{a}.

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