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Let $f\colon [a,b] \to \mathbb{R}$ be a continuous function. Suppose that there is $y$ such that $f^{-1}(y)$ is uncountable. Can it be show that $$\underline{D}f(x)=\liminf_{z\to x}\frac{f(z)-f(x)}{z-x}=0$$ at some $x\in A=\{ f^{-1}(y)\}$ or perhaps at a lot (all?) of $x\in A$? Can we get better results if we assume $f$ is nowhere monotone ?

If so then according to Existence of a continuous function with pre-image of each point uncountable , there would be a continuous and surjective real-valued function such that $\underline{D}f(x)=0$ for all points in the domain!

If instead we assume only that $A$ is infinite, can we still find at least one $x$ such that $\underline{D}f(x)=0$?

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Student
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  • You're using $y$ for two different things. Once as a constant (the function value that is taken infinitely many times) and once as a variable (in the limit value expression). – Arthur Aug 12 '14 at 21:58
  • @Arthur I will edit it to make it clearer. – Student Aug 12 '14 at 22:00

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The answer to the second question is negative. Take $s(x):[0,1]\to[0,1]$ as the continuous function defined as $2x$ over $[0,1/2]$ and $2-2x$ over $[1/2,1]$, then take $f:[0,1)\to[0,1]$ defined as: $$f(x)=s\left(\left\{\frac{1}{1-x}\right\}\right).$$ For any $y\in[0,1]$ we have $|f^{-1}(y)|=\aleph_0$ but $\underline{D}$ is never zero.

Jack D'Aurizio
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  • Thanks, however I'm not entirely convinced about the second part. For example, as you said $f^(y)$ is closed and by assumption uncountable so it contains a perfect set where all points are limit points. From this we can show that $\underline{D}(x)$ is zero in the perfect set so it is zero in at least an uncountable set (perfect sets are uncountable). – Student Aug 13 '14 at 12:27
  • As for the counter-example did you mean $g=s$ or something else? If you meant $g=s$ I don't understand why $f^{-1}(y)$ is infinite for each $y$... – Student Aug 13 '14 at 14:14
  • I retired the second part of my answer since I saw it was flawed. But for the first part, just consider that $f(x)$ is a triangular wave with an increasing frequency as $x$ approaches $1$. – Jack D'Aurizio Aug 13 '14 at 20:57