Let $f: [0,1]\to \mathbb{R}$ be a lower semi-continuous function, then
$$ \liminf_{x\to a} f(x) \geq f(a), \forall a \in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$\exists x_0 \in [0,1]$ such that $f(x_0) \le f(x)$, $\forall x \in [0,1]$.
This is a problem from a past qualifying exam in Measure Theory. I'm trying to solve it but I do not know how I should begin with this problem. I did not understand where the Inf is taken. What means that limit?
Best Answer
Here is a simple solution.
Let $\{x_n\}\in [0,1]$ be a minimizing sequence, that is, such that
$$\lim_{n\to \infty} f(x_n)= f^*= \inf_{x\in [0,1]}f(x).$$ Such a sequence always exists by the definition of infimum. Since $[0,1]$ is compact, the sequence $\{x_n\}$ has a convergent subsequence. Without loss of generality assume that $x_n\to \bar{x}\in [0,1].$ Then
$$f(\bar{x})\leq \liminf_{n\to \infty}f(x_n)= \lim_{n\to \infty} f(x_n)= f^*.$$ Since $\bar{x} \in [0,1]$ and by the definition of $f^*,$ we can only have
$$f(\bar{x})=f^*,$$ as desired.