If $f(a^+)$ and $f(a^-)$ exist, then $f$ is bounded.

Question

Let $f:[0,1] \to \mathbb{R}$ be a function such that for every $a \in [0,1)$ and $b \in (0,1]$ the one-sided limits $$f(a^+)=\lim _{x\to a^+}f(x) \in \mathbb{R}$$ $$f(b^-)=\lim _{x \to b^-} f(x) \in \mathbb {R}$$ exist.

A) Show that $f$ is bounded.

B) Does $f$ necessarily achieve its maximum at some $x \in [0,1]$?

C) Suppose further that $f$ is continuous at $0$ and $1$, and that $f(0) f(1)<0$. Prove that there exists some point $p \in (0,1)$ such that $f(p^-)f(p^+) \leq 0$.

Intuitively, I can see why part A is true, but I am not sure how to prove this formally. For part B, I think the answer is no, but I haven't yet come up with a counterexample. My initial thoughts on part C are to somehow apply the intermediate value theorem, but I am not sure if this is the correct approach or not.

Answer 1

Such an $f$ is called a regulated function. The space of step functions is dense in the space of regulated functions for $\|\cdot\|_{\infty}$, so $\exists \phi$ a step function $[0,1] \to \Bbb R$ such that $\|\phi - f\|_{\infty} \le 1$. Thus $\|f\|_{\infty} \le 1 + \|\phi\|_{\infty} < \infty$.

For $B$, consider the function:

$$f(x) = \begin{cases} x, 0\le x < 1/2 \\ 0, 1/2 \le x \le 1 \end{cases}$$

For $C$, WLOG say $f(0) < 0$ and $f(1) > 0$. As $f$ is continuous at $0$, $\exists 0<\delta < 1$ such that $f(x) < 0$ for all $0\le x \le \delta$. Similarly, there is $\epsilon \in (0,1)$ such that $f(x) > 0$ for all $x \in [1-\epsilon, 1]$.

Let: $$S =\{s \in [0,1]: \text{for all but finitely many } t\in [0,s], f(t) < 0\}$$

Let $p = \sup S$. We have $0<p<1$ since $\delta \in S$ and $S \subset [0,1-\epsilon]$. Clearly $f(p-) < 0$. Now suppose that $f(p+)<0$. We have $f(x) < \frac{f(p+)}{2} < 0$ for all $x \in (p, p+\eta)$ for some small $\eta > 0$. It follows, thus, that $p+\eta/2 \in S$, so $p+\eta/2 \le p$, a contradiction. Therefore $f(p+) \ge 0$, and we get $f(p-)f(p+) \le 0$.

Answer 2

A) Let $x_n$ be such that $|f(x_n)|\to \sup |f|$. By compactness, there exists a convergent subsequence, and furthermore, it can be chosen to be monotone. By assumption, the limit along the monotone subsequence exists. Thus $\sup |f|$ is finite.

B) Answered before.

C) WLOG $f(0)<0$. Then by continuity at $0$ and $1$, there exists $\epsilon\le \frac 12$ such that $f(p-)<0$ for $p$ in $[0,\epsilon)$ and $f(p-)>0$ for $p\in (1-\epsilon,1]$. Let $\bar p =\sup\{p:f(p-)<0\}$. Then ${\bar p} \in [\epsilon,1-\epsilon]$. By assumption, for every $n$, there exists $p_n- \in (\bar p-\frac 1n,\bar p)$ such that $f(p_n-)< 0$ and $p_n+\in (\bar p , \bar p + \frac 1n)$ such that $f(p_n)\ge 0$. Therefore $f(p-)= \lim f(p_n-)\le 0$, while $f(p+)=\lim f(p_n+)\ge 0$.

Answer 3

A) Assume $f$ is not bounded from above. Then there exists a sequence $(x_n)_{n\in\Bbb N}$ such that $f(x_n)\to+\infty$. This sequence has a limit point $x\in[0,1]$. Then a subsequence of $(x_n)$ converges to $x$, either from the right or from the left. But then $f(x_n)\to \infty$ contradicts the existence of $f(x+)$ of $f(x-)$.

B) Consider $f(x)=\begin{cases}x&x<1\\0&x=1\end{cases}$. The masximum is not achieved.

Answer 4

For (C).

Suppose $f(0)>0$ (as the case $f(0)<0$ is handled similarly). Since $f$ is continuous at $0$, there exists $r\in (0,1)$ such that $\forall x\in [0,r)\;(f(x)>f(0)/2).$ So there exists $r\in (0,1)$ such that $\forall x\in [0,r)\; (f(x^-)\geq 0\land f(x+)\geq 0).$

Let $P(r)\iff \forall x\in [0,r)\;(f(x^-)\geq 0\land f(x^+)\geq 0).$ Let $s=\sup \{r\in (0,1): P(r)\}.$

We have $s>0$.

We cannot have $s=1$: Else, for any $t\in (0,1)$ we have $f(t+)\geq 0$, implying that for any $r>0$ there exists $t'\in (t,1)$ with $f(t')\geq -r$. But the continuity of $f$ at $1$ would then imply $f(1)\geq -r$ for any $r>0$, so $f(1)\geq 0,$ contradicting $f(0)f(1)<0.$

So $s<1.$ We have

(i).... $s>0$, and $s'\in (0,s)\implies f(s'^+)\geq 0.$ So for every $r>0$ and every $s'\in (0,s)$ there exists $s''\in (s',s)$ with $f(s'')>-r.$ So $f(s)\geq -r$ for every $r>0.$ Therefore $$f(s^-)\geq 0.$$

(ii)....We now show that $f(s^+)\leq 0.$

Suppose by contradiction that $f(s^+)>0. $. Since $f(s^-)\geq 0$ by (i), the def'n of $s$ requires that $$\forall s'\in (s,1)\exists x\in (s,s')\; (f(x^+)<0\lor f(x^-)<0).$$ But if $x\in (s,s')$ and $f(x^+)<0$ there exists $y\in (x',s')$ with $f(y)<0,$ while if $x\in (s,s')$ and $f(x^-)<0$ there exists $y\in (s,x)$ with $f(y)<0$.

So for every $s'\in (s,1)$ there exists $y\in (s,s')$ with $f(y)<0,$ implying $f(s+)\leq 0,$ contradicting $f(s+)>0.$ Therefore, by contradiction, we have $$f(s+)\leq 0.$$ Since $f(s^-)\geq 0$ and $f(s^+)\leq 0$ we have $$f(s^+)f(s^-)\leq 0.$$

Remark: The case $f(0)<0$ can be done by applying the above argument to the function $f^*(x)=-f(x).$

Answer 5

Let's extend $f$ to whole of $\mathbb{R}$ by letting $f(x) = f(0)$ for $x < 0$ and $f(x) = f(1)$ for $x > 1$. Then $f$ has left and right hand limits at each point of $[0, 1]$. Therefore for each point $x \in [0, 1]$ there is an open interval $I_{x}$ containing $x$ such that $f$ is bounded on $I_{x}$. By Heine Borel Principle a finite number of these intervals $I_{x}$ cover $[0, 1]$ and $f$ is bounded on each of these selected intervals and hence bounded on $[0, 1]$.

The counter example by Hagen Von Eitzen shows that $f$ is not guaranteed to attain a maximum value on $[0, 1]$.

For part (C) let us assume that no such $p$ exists. Then for every $p \in (0, 1)$ we have $f(p-)$ and $f(p+)$ with same signs. Bisect the interval $[0, 1]$ into two equal parts and note that one of the intervals will be such that the limits of $f$ at end points have different signs. Choose that interval and bisect it again and repeat the procedure to obtain a sequence of nested closed intervals such that limits of $f$ at end points of these intervals have different sign. By nested interval principle there is a unique point $p$ which lies in all the intervals. Now the left right limits of $f$ at $p$ are of same sign and hence $f$ maintains a constant sign in some neighborhood of $p$. This neighborhood of $p$ also contains some intervals from the above sequence and hence $f$ must change sign somewhere in this neighborhood. This contradiction shows that there must be a $p$ with desired properties. Note that the continuity of $f$ at end points is not needed. What is needed is that $f(0+)f(1-) < 0$.

---

The question shows how far we can carry the three properties of continuous functions (boundedness, Intermediate Value Theorem, Extreme Value Theorem) on closed intervals if the condition for continuity is relaxed to existence of one sided limit at each point. The linked blog posts show that these properties of continuous functions are proved using completeness of real numbers and the same holds for the parts (A) and (C) of this question. Thus for example (A) can also be proved via Nested Interval Principle and (C) can be proved via Heine Borel Principle (you should give it a try).