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

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.

• It's wrong. You should have said "$\forall\epsilon>0\forall x_0\in [0,1]$... such that $|x-x_0|<\delta \implies f(x)>f(x_0)\color{red}- \epsilon$". – Myeonghyeon Song Jan 21 '19 at 17:04
• It is $-\epsilon$ in the definition, and the $\forall x$ has to be after the “such that”. – Mindlack Jan 21 '19 at 17:04
• Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now. – Richard Clare Jan 21 '19 at 17:06
• No. The $\delta$ is allowed to depend on $\epsilon$ and $x_0$, not $x$. – Mindlack Jan 21 '19 at 17:16
• An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here. – Daniel Jan 21 '19 at 17:34

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 $$f(x_0)=y.$$

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 $$c, it holds $$c\notin S$$, i.e. $$F_c=\varnothing.$$ This establishes $$m$$ is the minimum of $$f$$.

• In this proof, did we use that $F_c$ is compact? Thanks! – Toasted_Brain Dec 5 '20 at 7:25