Continuous function composed with itself is equal to propagation of a differential equation.This question has been bugging me for a while. It was given as the last question of a first year undergrad analysis exam and so should be solvable with little machinery, yet it seems to point straight at ODEs which have yet to be covered. Here is the question:Let&nbsp;F:R→R&nbsp;be a bounded&nbsp;&nbsp;C1&nbsp;function and&nbsp;(ft:R→R)t∈R&nbsp;a family of continuous functions with&nbsp;f0(x)=x&nbsp;such thatlimh→0ft+h(x)−ft(x)h=F(ft(x))Show that there exists a continuous function&nbsp;g:R→R&nbsp;such that&nbsp;f1=g∘g.I have attempted this with a classmate and we've come to various levels of understanding of the question, but the required conclusion keeps escaping us. Our main issues with the question are that we have very little understanding on continuous functions composed with themselves, and additionally we haven't managed to use the&nbsp;&nbsp;C1&nbsp;and boundedness condition on F.Note that the question also states that we are allowed to assume that the family of functions&nbsp;(ft)t∈R&nbsp;is uniquely determined by the given conditions: this again points to the domain of differential equations, which ideally we should not need to refer to in order to solve this.

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Continuous function composed with itself is equal to propagation of a differential equation.This question has been bugging me for a while. It was given as the last question of a first year undergrad analysis exam and so should be solvable with little machinery, yet it seems to point straight at ODEs which have yet to be covered. Here is the question:Let&amp;nbsp;F:R→R&amp;nbsp;be a bounded&amp;nbsp;&amp;nbsp;C1&amp;nbsp;function and&amp;nbsp;(ft:R→R)t∈R&amp;nbsp;a family of continuous functions with&amp;nbsp;f0(x)=x&amp;nbsp;such thatlimh→0ft+h(x)−ft(x)h=F(ft(x))Show that there exists a continuous function&amp;nbsp;g:R→R&amp;nbsp;such that&amp;nbsp;f1=g∘g.I have attempted this with a classmate and we&#039;ve come to various levels of understanding of the question, but the required conclusion keeps escaping us. Our main issues with the question are that we have very little understanding on continuous functions composed with themselves, and additionally we haven&#039;t managed to use the&amp;nbsp;&amp;nbsp;C1&amp;nbsp;and boundedness condition on F.Note that the question also states that we are allowed to assume that the family of functions&amp;nbsp;(ft)t∈R&amp;nbsp;is uniquely determined by the given conditions: this again points to the domain of differential equations, which ideally we should not need to refer to in order to solve this.

glikozyd3s68 · Accepted Answer

Fix s∈R and define ht(x)=Δft+s(x). Then, it is trivial to conclude that ddtht(x)=F(ht(x)). Further, define wt(x)=ft(fs(x)). Then, again, it is trivial to show that ddtwt(x)=F(wt(x)). From the referred uniqueness, (wt)t∈R=(ht)t∈R. In other words, ft+s(x)=ft(fs(x)) for any x and now choose g=f12 so that g∘g=f12∘f12=f1.

Continuous function composed with itself is equal to

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