I have an optimization problem of the following form:
$$\min c^T \lambda\\ \text{s.t. } f(x)^T \lambda \ge g(x) \text{ for all } x \in E,$$
where $E$ is an arbitrary set, $c \in \mathbb{R}^n, f \colon E \to \mathbb{R}^n,$ and $g \colon E \to \mathbb{R}.$
Let $$v = \inf\{c^T \lambda \colon f(x)^T \lambda \ge g(x) \text{ for all } x \in E\}$$ be the value of the minimization problem and assume that a) the problem is feasible, meaning that there exists at least one $\lambda \in \mathbb{R}^n$ which meets all constraints and b) the problem is bounded, i.e. $v > - \infty$.
Then optimization theory tells me that the infimum must be attained, i.e. there is a $\lambda^*$ which meets all constraints and for which $c^T \lambda^* = v$.
Why is that? I know that there must be a sequence of $\lambda_n$ with $c^T \lambda_n \to v$, and some books refer to Bolzano-Weierstrass (any bounded sequence has a convergent subsequence), but as far as I can see the set of feasible solutions is unbounded in general. Some authors also refer to duality theory, but all proofs I found there started by assuming that the set of feasible solutions is closed; which I cannot prove either.
I would be very grateful for any help!
This is not true.
A simple counterexample is obtained by setting $n = 2$, $E = (0,\infty)$, $f(x) = (1,1/x^2)$, $g(x) = x + 1/x^3$. Then, it is not hard to see that the feasibility of $\lambda$ is equivalent to $\lambda_1 \, \lambda_2 \ge 0$, $\lambda_1,\lambda_2 \ge 0$.
Setting $c = (1,0)$ yields an infimum of $0$, but this is not attained. The minimizing sequence $\lambda^{(n)} = (1/n,n)$ is unbounded (and the same is true for all minimizing sequences).