Show that lower semi-continuous function attains it's minimum. (Proof verification) (By contradiction)

Question

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 $f(x_0) \le f(x)$, $\forall x \in [0,1]$.

My proof:

Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,

Since $f$ is lower semicontinuous at $x_0$,

$$\forall \epsilon > 0, \exists \delta > 0 \mbox{ such that } |x - x_0| < \delta, \Rightarrow f(x) > f(x_0) - \epsilon, \forall x \in (x_0 - \delta,x_0 + \delta)$$

Now since $f$ doesn't attain it's minimum on $[0,1]$, then

$$\forall x_0 \in [0,1], \exists x \in [0,1] \mbox{ s.t } f(x_0) > f(x)$$

Let $\epsilon = f(x_0) - f(x) > 0, \exists \delta > 0$, such that $|x-x_0| < \delta$ and $f(x) > f(x_0) - \epsilon \Rightarrow f(x_0) - f(x) > \epsilon = f(x_0) - f(x)$ which is a contradiction.

I am right? Thanks.

Answer 1

Since $f$ is lower-semicontinuous, for each $x\in [0,1],$ there is an open interval $I_x\subseteq [0,1]$ such that $\inf\{f(y):y\in I_x\}\ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.

So, letting $y=\inf\{f(x):x\in [0,1]\}$, we can find a sequence $(x_n)\subseteq [0,1]$ such that $f(x_n)\to y.$ And of course, there is a subsequence $(x_{n_k})\subseteq (x_n)$ such that $x_{n_k}\to x_0\in [0,1]$.

Then, $f(x_0)\le \liminf_{x\to x_0} f(x)\le \liminf_{k\to \infty}f(x_{n_k})=\lim_{k\to \infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$

Answer 2

Another characterization of lower semi-continuous function that might be helpful:

A function $f:[0,1] \to \Bbb R$ is lower semi-continuous if and only if its upper level set $\{x\in [0,1]:f(x)>c\}$ is open in $[0,1]$ for all $c\in\Bbb R$.

(Indeed above characterization can be applied to every topological space $X$ and $f:X\to\Bbb R$. And the proof below can be generalized to any compact topological space.)

Proof. Assume $x_0\in X$ is such that $f(x_0)>c$. If we choose $0<\epsilon< f(x_0)-c$, we can find $\delta>0$ such that $$x\in [0,1], x\in (x_0-\delta,x_0+\delta)\implies f(x)>f(x_0)-\epsilon.$$ Since $f(x_0)-\epsilon>c$, it follows $$ [0,1]\cap (x_0-\delta,x_0+\delta)\subset \{x : f(x)>c\}. $$ This proves $\{x:f(x)>c\}$ is open in $[0,1]$. $\blacksquare$

By the above proposition, $F_c=\{x:f(x)\le c\}$ is closed in $[0,1]$ for every $c\in\Bbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}\supset F_{-2}\supset \cdots\supset F_{-n}\supset\cdots.$$ Since $$ \bigcap_{n\in\Bbb N}F_{-n} = \varnothing, $$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0\in\Bbb N$ such that $F_{-n_0}=\varnothing$. This implies $f(x)\ge -n_0$ for all $x\in [0,1]$, hence $f$ is lower bounded.

Let us define $S=\{c\in\Bbb R: F_c\ne \varnothing\}$. Since $-n_0\notin S$ and $-n_0\le S$, we know $S$ is lower bounded. Let $m=\inf S$. Then by finite intersection property, $$ F_m=\bigcap_{k\in\Bbb N}F_{m+1/k}\ne \varnothing. $$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $c\notin S$, i.e. $F_c=\varnothing.$ This establishes $m$ is the minimum of $f$.