I was having trouble with the following question. Any help would be highly appreciated.
Let $A$ be the set of K-dimensional vectors with non-negative components. Let $B$ be the set of K-dimensional vectors with positive components.
Let $f:A\rightarrow \mathbb{R}$ be a continuous function. Let $c\in\mathbb{R}$ be any number and let $C=\{a\in\ A | f(a)\geq c\}$.
Let $b_n$ be a sequence in $B$ that converges to another element $b\in B$ w.r.t. the Euclidean norm.
Show that $\lim_{n\rightarrow \infty} \inf_{a\in C} ab_n = \inf_{a\in C} ab$, where multiplication is the dot product of vectors. Also, $C$ is given as non-empty, and clearly by the construction of $A$ and $B$ the infimum will never be $-\infty$.
I can only show one direction.
Here is a proof sketch.
First of all, for every $x\in B$ there exist $y\in C$ such that $x\cdot y = \inf_{c\in C} x\cdot c$. Indeed let $\alpha = \min_{c\in C} x\cdot c$. Let $R= \{z:z_i \leq \alpha / x_i\}$. Then $$\inf_{c\in C} x\cdot c = \inf_{c\in C\cap R} x\cdot c.$$ But $C$ is a closed set, $R$ is a closed and bounded set, therefore, $C\cap R$ is a compact set. Thus there exist $y\in C\cap R$ that minimizes $x\cdot y$.
Let $a^*\in B$ be the value of $a$ that minimizes $a\cdot b$; let $a_n^*\in B$ be the value of $a$ that minimizes $a\cdot b_n$.
We need to prove that $\lim_{n\to \infty} a_n^*\cdot b_n = a^*\cdot b$. Note that $a_n^*\cdot b_n \leq a^* \cdot b_n$ for every $n$ and thus $\lim\sup_{n\to \infty} a_n^*\cdot b_n \leq \lim_{n\to \infty} a^*\cdot b_n =a^*\cdot b$. Now we prove that $\lim\inf_{n\to \infty} a_n^*\cdot b_n \geq a^*\cdot b$. Let $k_n$ be a subsequence s.t. $a_{k_n}^*\cdot b_{k_n} \to \lim\inf_{n\to \infty} a_n^*\cdot b_n$. Note that $a_n$ is a bounded sequence, and hence $a^*_{k_n}$ is also bounded. Let $a'$ be a limit point of $a_{k_n}^*$. Then $$\lim\inf_{n\to \infty} a_n^*\cdot b_n = \lim_{n\to\infty} a_{k_n}^*\cdot b_{k_n} = a'\cdot b \geq a^*\cdot b.$$