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If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable, $f(x) \neq f'(x)$ for all $x$, show that $\{x\in [0,1] \text{ and } f(x) = 0\}$ is finite.

I have shown that there cannot be an interval $[a,b]$ contained in $[0,1]$ such that $f(x) = 0$ since this would imply that $f$ is constant on this interval which implies that $f'(x) = 0$ on this interval and so there is contradiction since for $x\in [a,b], f(x) = 0$ and $f'(x) = 0$ since the interval is constant, but $f(x) \neq f'(x)$ for all $x$.

However I can't prove that for some infinite sequence $\{x_n\}$, say the rational numbers between $0$ and $1$ the statement holds true.

continental
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3 Answers3

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Use that an infinite sequence of zeros would have a converging subsequence, since $[0,1]$ is compact. The limit $x$ of this subsequence fulfills $f(x)=0=f'(x)$.

MooS
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  • How come the derivative must also be 0? – continental Nov 08 '15 at 09:17
  • $f$ is a priori differentiable, so you can compute the derivative at $x$ by approaching $x$ with any sequence you like. For example the converging subsequence... – MooS Nov 08 '15 at 09:19
  • That makes sense, thanks. – continental Nov 08 '15 at 09:20
  • Countable Choice would be needed to go from "is no infinitely sequence of distinct" to "are only finitely many". $;$ –  Nov 08 '15 at 09:30
  • The fact that the natural numbers admit an injection into any infinte set is a direct consequence of the peano axioms, isn't it? No countable choice needed. – MooS Nov 08 '15 at 10:18
  • @Moos : $;;;$ It's definitely not. $:$ The natural argument (for each n, choose an element that wasn't previously chosen) in fact uses dependent choice, although one can get the result with just Countable Choice. $:$ (For each n, choose an ordered (n+1)-tuple of distinct elements, and then map n to the leftmost entry of the (n+1)-tuple which isn't mapped to by any smaller natural number.) $;;;;;;;;$ –  Nov 08 '15 at 12:13
  • Choosing one single element from a set is not using choice. – MooS Nov 08 '15 at 12:46
  • However, choosing an infinite sequence of elements can easily be using choice. $;$ –  Nov 08 '15 at 13:30
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If $X = \{x\in [0,1] \mid f(x)=0\}$ is infinite, then $X$ has a limit point $\bar x \in [0,1]$. By continuity, $f(\bar x) = 0$. By differentiability, $$ f'(\bar x) = \lim_{\substack{ x \to \bar x\\ x\in X\setminus\{\bar x\}}} \frac {f(x)-f(\bar x)} {x - \bar x} = \lim_{\substack{ x \to \bar x\\ x\in X\setminus\{\bar x\}}} \frac 0 {x - \bar x} = 0, $$ so $f(\bar x) = f(\bar x)$, contrary to the assumption.

BrianO
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  • One can avoid Countable Choice by directly using that $\overline{x}$ is a limit point, rather than asserting the existence of a sequence. $;$ –  Nov 08 '15 at 09:27
  • @RickyDemer I took your suggestion. I have no aversion to countable or dependent choice, but you're right, there's no need to conjure a sequence here. – BrianO Nov 08 '15 at 09:38
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If I am not mistaken, we can even show that $f$ has at most one root on $\mathbb{R}$.

To the contrary, assume that we have two zeros, $x_0 < x_2$. Since $f'(x_0) \ne 0$, we may assume (w.l.o.g.) that $f'(x_0) > 0$. Now we set $x_1 = \inf\{z > x_0 : f(z) = 0\}$. We easily get $f(x_1) = 0$ (by continuity) and $x_1 > x_0$ since $f'(x_0) > 0$.

Now we have $f'(x_1) \ne 0$ and by the intermediate value theorem we can conclude $f'(x_1) < 0$. Now, we define the function $g(z) = f(z) - f'(z)$ and have $g(x_0) < 0 < g(x_1)$. This $g$ is the derivative of $\int f \,\mathrm{d}x - f$. By owing to Darboux's theorem, we find some $z \in (x_0,x_1)$ with $g(z) = 0$, i.e., $f(z) = f'(z)$, which is a contradiction to the assumption.

gerw
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