# Existence of $\vee$ or $\wedge$ for non-monotonic functions

This question is inspired by a discussion in chat with wj32. We allow for equality in the definition of increasing and decreasing and call a function monotonic if it is increasing or decreasing. If $f:\mathbb R\to \mathbb R$ is not monotonic, are there three points $x<y<z$ such that $f(y)<f(x),f(z)$ or $f(y)>f(x),f(z)$? For convenience, call these the $\vee$ and $\wedge$ formations respectively.

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It's a purely combinatorial matter. The main task is to reduce the number of cases to be considered as much as possible.

It suffices to consider a surjective function $f:\ \{1,2,3,4\}\to\{1',\ldots,n'\}$ with the property that there exists a pair $x<y$ with $f(x)<f(y)$ and a pair $u<v$ with $f(u)>f(v)$. Here $n\in\{2,3,4\}$.

I shall only deal with the case $n=4$.

If at least one of $2$ and $3$ is mapped onto $1'$ or $4'$ we are done.

If $f(2)=2$ and $f(3)=3$ then $f(1)=4'$, and we have a $\vee$ over $\{1,2,3\}$.

If $f(2)=3'$ and $f(3)=2'$ then $f(1)=1'$, and we have a $\wedge$ over $\{1,2,3\}$.

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If $f$ is not monotonic, it is not decreasing or increasing. Hence there are points $a<b$ with $f(a)<f(b)$ and points $c<d$ with $f(c)>f(d)$.

If $a=c$, use $adb$ to form $\vee$ or $abd$ to form $\wedge$.

If $a=d$, use $cdb$ to form $\vee$.

If $b=c$, use $abd$ to form $\wedge$.

If $b=d$, use $acd$ to form $\wedge$ or $cab$ to form $\vee$.

Now we are left with the case when $a,b,c,d$ are pairwise distinct.

If $a<b<c<d$, use $abd$ to form $\wedge$ or $acd$ to form $\wedge$.

If $a<c<b<d$, use $abd$ to from $\wedge$ or $acd$ to form $\wedge$.

If $a<c<d<b$, use $acd$ to form $\wedge$ or $cdb$ to form $\vee$.

If $c<a<d<b$, use $cdb$ to form $\vee$ or $cab$ to form $\vee$.

If $c<d<a<b$, use $cdb$ to form $\vee$ or $dab$ to form $\vee$.

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for all $a < b < c\in [p,q]$ $$f(a) \leq f(b) \leq f(c)$$ $\implies$$f is monotone in [p,q] and increasing and for all a < b < c\in [p,q] $$f(a) \geq f(b) \geq f(c)$$ \implies$$f$ is monotone in $[p,q]$ and decreasing

Now, since the function is not monotone, both the above condition are false, and hence, for some $a < b < c \in [p,q]$ we have at least one of the following conditions true.

1. $f(a) < f(b) > f(c)$
2. $f(a) > f(b) < f(c)$
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 @JasperLoy I removed the continuous function thing, since it was superfluous. Anyway, I have not fixed $a,b,c$, Since, if the first two options from the list are not satisfied anywhere in $R$, then either of the other two options (the third one, and then the original assumption) must be satisfied, then I pick one of them and use it to show that, it applies all over the domain. Because, next you can select another point $e$ and do the same thing as has been done with $b,c,d$ to get the same relation with $c,d,e$ and so on. – Jayesh Badwaik Sep 27 '12 at 14:39 But, we first show that there is atleast one, and then prove for all from that single one. About strict inequalities, I guess, whenever there is equality, we get a constant function. – Jayesh Badwaik Sep 27 '12 at 14:45