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?

  • 1
    $\begingroup$ Hint: try to use the fact that $f$ is lower semicontinuous if and only if $f^{-1}(-\infty,a)$ is open in $[0,1]$ for any real number $a$. Then try to contradict compactness of $[0,1]$ if $f$ has no lower bound. $\endgroup$ – Shalop Aug 31 '17 at 22:22
  • $\begingroup$ Can you explain the if and only if? I do not like to use facts without understanding them first or actually prove the facts first. Thank you anyway, with this hints I think I can proceed but I would like to understand the equivalence. $\endgroup$ – Richard Clare Aug 31 '17 at 22:27
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    $\begingroup$ First of all, there is a mistake on my first comment: $f$ is lower semicontinuous iff $f^{-1}(a, \infty)$ is open, $\forall a \in \Bbb R$. Secondly, $\liminf_{x \to a} f(x)$ is defined to be $ \sup_{\delta>0} \inf_{x \in [a-\delta, a+\delta]} f(x)$. Try to visualize what this means, and then try to prove the equivalence of the two definitions. $\endgroup$ – Shalop Aug 31 '17 at 23:41
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    $\begingroup$ You also seem to have issues understanding the $\liminf$ thing. Well that equation is a (complicated) definition of semi-continuity. Use the simpler version. $f$ is said to be lower semi-continuity at $c$ if for every $\epsilon>0$ there is a neighborhood of $c$ in which $f(x) \geq f(c) - \epsilon$. $\endgroup$ – Paramanand Singh Sep 2 '17 at 8:30

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.

  • $\begingroup$ Can you explain "by the definition of the infimum" and the "without loss of generality assumption? I guess that the fact that $[0,1]$ is compact let you take the $\overline{x} \in [0,1]$ but I'm not sure. Also why the sequence always exists just be the definition of the inf Can you explain how we can take this sequence? If you could do that I will mark this as the answer because it simpler than the other one and easier to understand. $\endgroup$ – Richard Clare Sep 2 '17 at 3:00
  • $\begingroup$ @RichardClare Recall that if $X\subseteq \Bbb R,$ then $$\inf X= a \textrm{ iff }a\leq x \;\forall\;x\in X \textrm{ and if } b\leq x \;\forall\; x\in X, \textrm{ then } b\leq a.$$ With this in mind, in our problem, for any $n\in \Bbb N,$ it is possible to choose $x_n$ such that $f^*\leq f(x_n)\leq f^*+\frac{1}{n},$ which proves the existence of the minimizing sequence. Concerning the 'whithout loss of generality,' note that if $\{x_n\}$ is not convergent, then at the beginning we could have just take the convergent subsequence, which is still a minimizing sequence $\endgroup$ – John D Sep 2 '17 at 7:40
  • $\begingroup$ No further questions. Thank you! $\endgroup$ – Richard Clare Sep 2 '17 at 14:34

The usual proof for continuous functions can be adapted easily to this case. Assume on the contrary that $f$ is unbounded below on $[0,1]$. Then it must be unbounded below on either the left or the right half of the interval $[0,1]$. Select one of the halves where $f$ is unbounded below. And continue this process to get a chain of nested intervals where $f$ is unbounded below. By nested interval principle there is a unique point $c$ which lies in all these nested intervals. Since $f$ is lower semi-continuous at $c$ there is a neighborhood of $c$ in which $f(x) \geq f(c) - 1$. And one of the nested intervals is contained in this neighborhood so that $f$ is bounded below in this nested interval. The contradiction proves that $f$ is bounded below in $[0,1]$.

Next we apply the same nested interval principle to prove that $f$ attains its minimum value on $[0,1]$. Let $m$ be the greatest lower bound (glb) of $f$ on $[0,1]$. Then $m$ is the glb of $f$ on at least one of the two halves of $[0,1]$. Select the half for which $m$ is the glb and repeat the process to generated a sequence of nested intervals such that $m$ is the glb of $f$ on each interval in this sequence. Like before there is a unique point $c$ which lies in all the intervals of this sequence. We prove that $f(c) =m$. On the contrary let's assume that $f(c) >m$ and choose $\epsilon=(f(c) - m) /2>0$. Then by semi-continuous of $f$ at $c$ we have a neighborhood of $c$ in which $$f(x) \geq f(c) - \epsilon=\frac{f(c) +m} {2}>m$$ One of the intervals of the sequence is contained in this neighborhood and by the above inequality the glb of $f$ on this interval is greater than or equal to $(f(c) +m) /2$ and hence greater than $m$. This is a contradiction (because the glb is $m$ on each interval of the sequence). Thus $f(c) =m$ and we are done.

Both the proofs above are based on nested interval principle. It is a fun exercise to prove these properties using various other equivalent formulations of completeness of real numbers (like Dedekind's theorem, Bolzano Weierstrass, Heine Borel Theorem, least upper bound principle). Proving the properties of continuous (semi-continuous) functions on closed intervals via completeness of reals is a key to proper understanding of completeness of real numbers.


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