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Consider a function $f \colon[0, 1] \rightarrow [0, 1]$. Suppose that $f$ is continuous, strictly increasing, strictly concave on $[0, c]$ where $c \in (0, 1)$, strictly convex on $[c, 1]$, and has three fixed points: $x = 0$, $x = \hat{x} \in (0, 1)$ and $x = 1$. I want to prove that $f(x) > x$ for $x \in (0, \hat{x})$ (i.e. $f$ is always above the 45 degree line between zero and the interior fixed point). However, my proof seems a bit convoluted. Is there any easier way to see this?

Here is my proof:

First case: suppose that $c \geq \hat{x}$. Then $f$ is concave over the full interval $[0, \hat{x}]$. One may then confirm that $f(x) > x$ for all $[0, \hat{x}]$ as claimed (I will skip details).

Second case: suppose that $c < \hat{x}$. Then it remains true that $f(x) > x$ for all $x \in [0, c]$. In particular, $f(c) > c$. Furthermore, if there existed some $x' \in (c, \hat{x})$ such that $f(x') \leq x'$, then either $f(x') = x'$ (impossible since $\hat{x}$ is the only interior fixed point) or else $f(x') < x'$. But if $f(x') < x'$, there would need to exist some $x'' \in (z, x')$ such that $f(x'') = x''$ (by the intermediate value theorem); which is again impossible since $\hat{x}$ is the only interior fixed point. Either way, then, we conclude that $f(x) > x$ for all $x \in (0, \hat{x})$.

As I say, it's a bit convoluted! Is there an easier way? The intuitive idea is that $f$ starts out above the 45 degree line. Also, it is a continuous function and has exactly one fixed point. So it must be above the fixed point to the left of it.

afreelunch
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  • Please give a better title. "This basic fact about functions" does not attract people who know the answer – FShrike May 03 '22 at 19:18
  • OK sure, what would be a better title? – afreelunch May 03 '22 at 19:19
  • You need to be careful with the precise statement of the problem. As written, you know $f$ has the fixed points $0,\hat{x}$ and $1$, not that it does not have any other fixed points. If for example, $f$ can have additional fixed points in $(c,\hat{x})$, then the conclusion is easily seen to be false. A cleaner way to argue would be to define $g(x)=f(x)-x$ on $[c,\hat{x}]$ and get directly a point $x'$ inside for which $g(x')=0$ by intermediate value theorem, assuming $g(x)>0$ is violated on $[c,\hat{x})$. – AnCar May 03 '22 at 19:39
  • Good point, have edited it! – afreelunch May 03 '22 at 19:43
  • This is not true as written. A strictly concave function could start with $y=x$ as a tangent, which will make $f(x)<x$ for $x\in(0,c)$, contrary to what you asserted in the second case. The function is then free to increase up to its other two fixed points. – Theo Bendit May 03 '22 at 19:56
  • @TheoBendit thanks for spotting that -- I forgot to state that the function is convex after the concave region! In light of your comment, the proof clearly needs modification. – afreelunch May 04 '22 at 10:42
  • Three fixed points is the same that $f$ cuts three times the diagonal $y=x$. How to be concave this way? – Piquito May 04 '22 at 10:50
  • @Piquito The function $f$ cannot be globally concave. However, it can be concave and then convex. – afreelunch May 04 '22 at 11:35
  • @afreelunch: Right! So you have to edit your problem (I see you already did). – Piquito May 05 '22 at 14:12

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