I am solving an exercise (not Homework)..

Let $E_1$ and $E_2$ be non empty disjoint convex subsets of $X$, with $E_1$ compact and $E_2$ closed in $X$. Then there are $f\in X'$ and $t_1,t_2$ in $\mathbb{R}$ such with $$Re~f(x_1)\leq t_1<t_2\leq Re~f(x_2)$$ for all $x_1\in E_1$ and $x_2\in E_2$.

Existence of a linear functional seperating two convex subsets goes back to at least Hahn Banach seperation theorem... There we need one of the convex subsets to be open.

Recall : $E_1$ is compact, $E_2$ is closed in $X$ with $E_1\cap E_2=\emptyset$ then there exits $r>0$ such that $(E_1+U(0,r))\cap E_2=\emptyset$.

As $U(0,r)$ is open, so is the set $E_3=E_1+U(0,r)$.. Then we have two disjoint convex subsets $E_3$ and $E_2$ of $X$ such that $E_3$ is open.. So, there exists $f\in X'$ and $t\in \mathbb{R}$ such that $$Re~ f(x_3)<t\leq Re~f(x_2)$$ for all $x_3\in E_3$ and $x_2\in E_2$. As $0\in U(0,r)$ we see that $$Re~ f(x_1)<t\leq Re~f(x_2)$$ for all $x_1\in E_1$ and $x_2\in E_2$. So, i am almost done but not sure how to get another $t_0\in \mathbb{R}$ as required..

Please give some hints...

  • $\begingroup$ Sorry, this comment was useless. $\endgroup$ – gerw Nov 11 '15 at 10:33

Look at your second-to-last displayed equation. We have $$ \Re f(x_3) < t \le \Re f(x_2), \qquad x_2 \in E_2, x_3 \in E_1 + U(0,r). $$ That is, $$ \Re f(x_1) + \Re f(u) < t \le \Re f(x_2) \qquad x_2 \in E_2, x_1 \in E_1, u \in U(0,r) $$ Now let $m := \sup_{u \in U(0,r)} \Re f(u)$. As $f \ne 0$, we have $m > 0$. Then $$ \Re f(x_1) + m \le t \le \Re f(x_2), \qquad x_1 \in E_1, x_2 \in E_2 $$ or $$ \Re f(x_1) \le t- m < t \le \Re f(x_2), \qquad x_1 \in E_1, x_2 \in E_2 $$

  • $\begingroup$ Ah... I do not know how i missed it... So, here we are not using compactness of $E_1$.. thanks a lot... $\endgroup$ – user87543 Nov 11 '15 at 12:25

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