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This question is motivated by this posting. There, OP asked to prove that an increasing function $f: [0,1] \to [0,1]$ has a fixed point.

Pondering upon this question, it came to me that it may suffice to assume only an infinitesimal version of monotonicity. So I formulated the following question:

Problem. Suppose that $f : [0, 1] \to [0, 1]$ has both left-limit and right-limit at every point on $[0, 1]$ and satisfies $$\lim_{s \to x^-} f(s) \leq f(x) \leq \lim_{x \to x^+} f(s).$$ Is it necessary that $f$ has a fixed point?

I guess I have a rough idea of the proof, but was not successful to give a full justification. So my question is,

Q. If this claim is true, is there any proof or a reference to this problem? Otherwise, is there any counter-example?

Here is my trial:

The assumption on $f$ implies that $f$ has only countably many discontinuities (see this, for instance). Now define $F : [0, 1] \to 2^{[0,1]}$ by

$$ F(x) = \Big[ \lim_{s \to x^-} f(s), \lim_{x \to x^+} f(s) \Big].$$

That is, values of $F$ are closed intervals in $I$ with endpoints given by left-limit and right-limit.

We can prove that the graph $\Gamma = \{ (x, y) \in I^2 : y \in F(x) \}$ of $F$ is closed and connected in $I^2$. Using this, we can prove that $c \in F(c)$ for some $c \in I$. (Notice that this is also a consequence of the Kakutani fixed point theorem.)

If this $c$ is the point of continuity of $f$, then $F(c)$ is a singleton and we are done. But if $F(c)$ happens to be a non-degenerate interval, then we are out of luck. In the second case, let us write $F(c) = [p, q]$. Then either $p < f(c)$ or $f(c) < q$, and then we may initiate the same argument on either of the square $[0, c]^2$ or $[c,1]^2$ with due modification of $f$.

I guess that we can carefully manage this process so that, even when this procedure does not halt, sort of squeezing argument enters in and prove that we can find a point of continuity if $f$ which is also a fixed point of $F$. But I am not sure if it will actually work.

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Sangchul Lee
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  • What does the right hand limit $\lim_{x \to 0^-}f(s)$ refer to? – coffeemath Jan 25 '17 at 16:12
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    @coffeemath, Thank you, that is a good point. When $x = 0$, we simply use $f(0)$ instead of $\lim_{x \to 0^-} f(x)$ and likewise, we use $f(1)$ instead of $\lim_{x\to1^+} f(x)$. – Sangchul Lee Jan 25 '17 at 16:14

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The answer of the quoted question can be modified to accommodate this case too: Define $A=\{x:f(x)\geq x\}$ and $a=\sup A$.

If $a<f(a)=a+\delta$, then use the fact that there exists a $\delta'$ such that for $a'\in[a,a+\delta']$, one has $f(a')>f(a)-\delta/2$, in particular all $a'\in[a,a+\delta'\wedge(\delta/2)]$ one has $f(a')> a+\delta-\delta/2\geq a'$, so $a'\in A$, which is a contradiction.

Similarly, if $a>f(a)=a-\delta$, then there exists a $\delta'$ such that for $a'\in[a-\delta',a]$, one has $f(a')< f(a)+\delta/2$, in particular all $a'\in[a-\delta'\wedge(\delta/2),a]$ one has $f(a')<a-\delta+\delta/2\leq a'$, so $a'\notin A$, and so $\sup A \leq a-\delta'\wedge(\delta/2)$, which is a contradiction.

So $a=f(a)$.

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