Does there exist continuously differentiable function $f:\mathbb{R}\longrightarrow\mathbb{R}$?

Question

Does there exist continuously differentiable function $f:\mathbb{R}\longrightarrow\mathbb{R}$ such that for all $x\in \mathbb{R},\,\,f(x)>0$ and, $f'(x)=(f\circ f)(x)$?

I see this question in the book "Putnam and Beyond". I think it's solution is wrong. Let's solve this problem.

Answer given in the book:

The monotonicity and the positivity of $f$ imply that $f (f (x)) > f (0)$ for all $x$. Thus $f (0)$ is a lower bound for $f'$. Integrating the inequality $f (0) < f'(x)$ we obtain $$ f(x)<f(0)+f(0)x=(x+1)f(0) $$

But then for $x ≤ −1$, we would have $f (x) ≤ 0$, contradicting the hypothesis that $f (x) > 0$ for all $x$. We conclude that such a function does not exist.

And why does the statement of a question from a book have such bad grammar? — Kimball, Jan 26 '15 at 07:49
You suddenly changed the inequality sign. you have $f(0) < f'(x)$ and then you apparently swap sides when proceeding to integrate it (you suddenly have $\int f(x) < \int f(0)$) — user159517, Jan 26 '15 at 10:02
@user159517: It is solution given in the book. it isn't mine. As I said it is wrong. — Ramand, Jan 26 '15 at 10:22
possible duplicate of Strictly increasing function with $f'(x) = f(f(x))$ — Gabriel Romon, Jan 26 '15 at 11:26
You may find 3 solutions here : http://kskedlaya.org/putnam-archive/2010s.pdf — Gabriel Romon, Jan 26 '15 at 11:26
@LeGrandDODOM: This question is not a duplicate of the one you reference, since that other question also assumes that $f$ is increasing. Also, this question asks if a particular proof is valid. — Rory Daulton, Jan 26 '15 at 12:28
One of my questions is http://math.stackexchange.com/questions/1092217/solution-of-f-circ-f-f — Elaqqad, Jan 26 '15 at 12:39

String · Accepted Answer · 2015-01-26T14:05:48.423

3

The book is right on this: Note first that $g'(x)<h'(x)$ does NOT imply $g(x)<h(x)$. Just consider $g(x)=\log x$ and $h(x)=2\log(x)$ on the interval $(0,1)$ as an example. And this is actually the point here.

Let me try to put it differently: We know that $f'(x)>f(0)$ so that $f'(x)-f(0)$ must be strictly positive. Thus since $$ H(x)=f(x)-(x+1)f(0) $$ is a function having derivative $H'(x)=f'(x)-f(0)>0$ it must be strictly increasing. Since $H(0)=0$ this implies $H(x)$ to be negative for $x<0$. Thus $H(-1)<0$ but this means $$ H(-1)=f(-1)<0 $$ which is a contradiction.

The book should have stated explicitly that $f(x)<(x+1)f(0)$ is true only for $x<0$, though.

edited Jan 26 '15 at 14:05

answered Jan 26 '15 at 14:00

String

18,838

Thank you String. Fantastic. Very nice consideration. – Ramand Jan 26 '15 at 14:36
@Ramandi: Always great to help sorting such things out! – String Jan 26 '15 at 14:37

Does there exist continuously differentiable function $f:\mathbb{R}\longrightarrow\mathbb{R}$?

1 Answers1

Linked