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Is there a non-linear differentiable real function $f:\mathbb{R}\to\mathbb{R}$ so that every $x \in \mathbb{R}$ has the following property:

in every neighbourhood of $x$, the tangent line to $f$ at $x$ intersects at least one other (and therefore infinitely many) points of the graph of $f$? Or in other words, there is no tangent line of $f$ that is "undisturbed" by other points of $f$ in a neighbourhood centred at where the tangent line meets the curve.

I was thinking about starting by drawing a sine curve and then drawing a courser sine curve that wraps around the previous sine curve and repeating ad infinitum, but I'm not sure this would work, or if the function would remain everywhere differentiable.

Maybe some function to do with the Pompeiu derivative can satisfy the requirements?

Edit: Also vaguely relevant: Differentiable function for which the tangent at each point has infinitely many common points with the graph

Adam Rubinson
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  • Interesting question. As you say, a natural place to look would be something involving Fourier series (or more general trigonometric sums) whose coefficients are "choice." But as you say, how to ensure everywhere-differentiability? Are there good theorems on that for Fourier series? That's what comes to mind as an approach. – leslie townes Mar 11 '21 at 16:32
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    Maybe you clarify the question first: do you ask exactly one intersection in every neighborhood? Since otherwise, $f(x) = 0$ does what you ask for... As I read your question (please confirm / infirm): $$ \forall x \forall \epsilon>0 \exists! h, |h|<\epsilon: f(x) + h f'(x) = f(x+h) $$ Is that correct? – Eric Mar 11 '21 at 16:37
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    $f(x)=a$ and $f(x)=ax+b$ would work, the tangent line meeting all points of $f(x)$ in any neighbourhood. – Paul Mar 11 '21 at 17:03
  • Would $f(x) = \begin{cases} x^2 \sin(1/x) \text{ if } x \neq 0 \ 0 \text{ if } x = 0 \end{cases}$ work? – User203940 Mar 11 '21 at 17:23
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    $f(x)$ being linear does work, Eric, like you say, and I'll have to edit my question to not allow linear functions as answers. Is there any non-linear function that satisfies the property? – Adam Rubinson Mar 11 '21 at 18:02
  • @User203940 for that function, only $x=0$ has the desired intersection property. I want all values of $x$ to have the intersection property. – Adam Rubinson Mar 11 '21 at 19:04
  • This example might be a little contrived, but how about $$f(x)=\begin{cases}(x-1)^2 \text{ if $x\geq1$,} \ 0 \text{ if $-1 \leq 0 \leq 1$,} \ (x+1)^2 \text{ if $x\leq-1$?}\end{cases}$$It seems like it would work. – Joe Mar 11 '21 at 20:48
  • @Joe $x = 4$ doesn't have the desired intersection property. The tangent line to $f(x)$ at $x=4$ does not intersect another point on the graph of $f$ in the neighbourhood $x \in (3.9,4.1).$ – Adam Rubinson Mar 11 '21 at 20:52
  • @AdamRubinson: Of course! I missed the 'neighbourhood' part of your question. Anyway, good luck finding an answer. It seems like a nice problem. – Joe Mar 11 '21 at 20:55
  • Just to clarify, Eric, I do not require exactly one intersection in every neighbourhood: in fact, this wouldn't make much sense because if every neighbourhood has at least one point of intersection, then every neighbourhood has infinitely many points of intersection (which, by the way, is a consequence of my supposition). The statement you ask me to confirm/infirm- I'm not sure what the exclamation mark in $\exists!$ means. If you get rid of the "!" then it is correct I think. – Adam Rubinson Mar 11 '21 at 21:56
  • Well, I suppose you could take the example of @User203940 and smear it out over all rationals: start with their function $f$ and use an enumeration ${\lambda_n}$ of the rationals. Then set $f_n(x)=f(x-\lambda_n)$, a copy of $f$ but shifted to have a Rubinson point at $\lambda_n$ instead of at zero. Then define $F(x)=\sum_{n:\lambda_n<x}\frac1{2^n}f_n(x)$. This seems pretty clearly to be good at all rationals, but I suppose it must fail the R test at irrationals. – Lubin Mar 11 '21 at 23:53

1 Answers1

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Not a complete answer. I solve this question if we assume $f$ is twice differentiable.

No. It suffices to show that $f''(x) = 0$ whenever $f$ has the described property at $x$. Note (1) $f(x+h) = f(x)+hf'(x)+\frac{1}{2}h^2 f''(x)+o(h^2)$ as $h \to 0$; the property means $f(x+h) = f(x)+hf'(x)$ for some arbitrarily small $h$, so we can divide (1) by $h^2$ and take the limit along such $h$ to see that $f''(x) = 0$.

mathworker21
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