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Let $f:\mathbb{R}\to (0,\infty)$ be a differentiable function satisfying $$f(f(x))=f^\prime(x)$$for each $x$. Show no such function exists.

I got this problem in an exam. I haven't done anything significant with it. I have found that $f^\prime=f\circ f>0$ so $f(f(x))>f(0)$ hence we have $f^\prime(x)>f(0)$. But I have no idea how to use it. I tried to apply the mean value theorem on $$\frac{f(f(x))-f(0)}{f(x)}=f^\prime(c)=\frac{f^\prime(x)-f(0)}{f(x)}$$ but that doesn't lead anywhere. Can someone help me? Thanks a lot.

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2 Answers 2

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As you say, if there is such an $f$, then $f'(x) > f(0) > 0$ for all $x$. We can now use that lower bound on the derivative to show that $f$ must be negative for sufficiently large, negative $x$:

Let $x < 0$. Then by the MVT, there exists a $c \in (x, 0)$ such that $$\frac{f(0) - f(x)}{-x} = f'(c) > f(0)$$

Hence $f(0) - f(x) > f(0)(-x) \ \ \ $ or $ \ \ \ \ f(x) - f(0) < f(0)x$.

Thus for $x < -1$,

$$f(x) < (x+1)f(0) < 0$$ contradicting the hypothesis that $f(x)$ is always positive.

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PLEASE SEE COMMENTS - THIS IS AN INCORRECT (ATTEMPT AT) A SOLN.

integrate from $-x$ to $x $ for a positive $x$: $$ \int_{-x}^x f(f(t))dt = \int_{-x}^x f'(t)dt = f(x) - f(-x) $$ by Fund. Thm. of Calc.

Now divide by $2x$ and let $x$ decrease to $0$.

The left hand side gives
$$\lim \frac{\int_{-x}^x f(f(t))dt}{2x} = f(f(0))$$ (by Lebesgue's differentiation theorem)

The right hand side gives $$ \lim \frac{f(x) - f(-x)}{2x} = (1/2)\lim \left[ \frac{f(x) - f(0)}{x} - \frac{f(-x) - f(0)}{x} \right] = (1/2)(f'(0)-f'(0))=0. $$

Hence we have f(f(0))=0. A contradiction!

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    $\begingroup$ $\lim_{x\to 0^{+}} \frac{f(-x)-f(0)}{x} = \color{red}{-}f'(0)$. $\endgroup$ Nov 13, 2014 at 2:48
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    $\begingroup$ @achillehui is right, your conclusion should be $f(f(0)) = f'(0)$, which is not surprising at all. $\endgroup$ Nov 13, 2014 at 5:00
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    $\begingroup$ I suspect there is no problem with this function locally at zero. Note that the other solution only works when taking $x < -1$. $\endgroup$ Nov 13, 2014 at 5:01
  • $\begingroup$ Agreed, this is not a solution as written, and may not lead to one! Thanks. Indeed, this gives $f(f(0)) = f'(0)$, which is not noteworthy. $\endgroup$ Nov 13, 2014 at 17:53
  • $\begingroup$ I added a big flag that it wasn't a solution for future users. $\endgroup$ Nov 13, 2014 at 17:54

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