If $f : [0,\infty) \to [0,\infty)$ is such that,
- $f(f(x)) = f(x)$,
- $f$ is differentiable,
is it true that $f$ is either a constant or $f = x$?
I think so anyway.
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Welcome to Mathematics Stack Exchange. If $f(x)=x$, we don't generally have $f(x)^2=f(x)$ – J. W. Tanner Mar 03 '20 at 15:30
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@J.W.Tanner I think they might mean $f(f(x))$ but extra clarification would be nice. – Ninad Munshi Mar 03 '20 at 15:32
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I think the OP means $f\circ f$ – saulspatz Mar 03 '20 at 15:32
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Oops. I meant (f•f) (x) = f(x)... . – Yuanny Mar 03 '20 at 15:33
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1I can't quite tell what you mean in your most recent comment since I have become so accustomed to seeing things written in MathJax and $\LaTeX$. Can you confirm with words whether you mean for $f^2$ to be interpreted as $f$ composed with $f$? (commonly written with an open circle $\circ$) Or do you mean for $f^2$ to be interpreted as $f$ multiplied by $f$? (commonly written with a small dot $\cdot$) What you wrote it looks more like a bullet point than either of these. – JMoravitz Mar 03 '20 at 15:35
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1See also Find examples of continuous functions $f:[0,1]\to[0,1]$ that satisfy $\forall x\in[0,1], f(f(x))=f(x)$ other than $f(x)=x$. – Dietrich Burde Mar 03 '20 at 15:37
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See also $f(f(x))=f(x)$ question and Describe all the continuous function $f:\mathbb R \to \mathbb R$ such that $f \circ f=f$. Again describe all linear maps with the same property.. – Dietrich Burde Mar 03 '20 at 15:38
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This question requires that $f$ be differentiable though, so I don't see that it's a duplicate. – saulspatz Mar 03 '20 at 15:50
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I think the condition of differentiability is more restrictive than just continuity. – Yuanny Mar 03 '20 at 19:07
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It will in fact force f(x) to be x on the whole of [0,infinity), assuming there are at least two points along the x-axis which do not map to the same point on the y-axis. – Yuanny Mar 03 '20 at 19:09