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Let $f,g:[0,1] \rightarrow \mathbb{R}$ are continuous and increasing then there $a,b \in[0,1] $ such that $f-g$ is monotonic on (a,b).

My approach: Let $h=f-g$ and assume this is not true . Therefore, for all $(a,b) \subset [0,1]$ there exist $p,q,r \in (a,b) $ such that $p<q<r$ then either "$\ h(p)>h(q) \ \text{and} \ h(q)<h(r)$" or "$h(p)<h(q) \ \text{and} \ h(q)>h(r)$"

I have no idea how to proceed further please help me.

JANGRA
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1 Answers1

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Such interval doesn't necessary exist.

Let $h$ be a continuous bounded variation nowhere monotonic function (see "Everywhere differentiable, nowhere monotone functions" by Katznelson, Stromberg for example of such function). Any continuous bounded variation function is difference of two continuous increasing functions: let $f$ be total variation function of $h$, and $g = f - h$.

mihaild
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  • Can you please explain it in a simple language? What is total variation function? – JANGRA Mar 23 '23 at 16:12
  • Total variation is $\operatorname{Var}$ from this answer https://math.stackexchange.com/a/141346/659499. What we need from it is that continuous bounded variation (or we can even require bounded derivative) function is difference of two continuous monotonic functions. That it is just difference of two monotonic functions is well-known, but standard proof (for example, in linked answer) actually proves that they are continuous if original function is continuous. – mihaild Mar 23 '23 at 19:02