Is there some $f:\mathbb{R}\to\mathbb{R}$ differentiable at every point such that $\forall x$, the tangent to $f$ at $(x,f(x))$ intersects the graph of $f$ at $2$ points (counting (x,f(x)))?

This problem is delicate, as shown by the example $f(x)=x^3$, where the condition only fails at $x=0$.

Remark: if $f$ is $C^1$, then the function $g:\mathbb{R}\to\mathbb{R}$ which sends $x$ to the only point $y$ such that $(y,f(y))$ is in the tangent line to $x$ seems to be continuous at almost every point. I could not use this effectively though, in the non continuous case maybe you could get some weak version of this using the intermediate value property of the derivative?

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    $\begingroup$ Thinking out loud - don't your conditions imply there's exactly one inflection point and that inflection point will always be an issue. $\endgroup$ Apr 2, 2022 at 4:24
  • $\begingroup$ @MikeO'Connor That happens in the $C^1$ case. Maybe it's true in general $\endgroup$
    – Saúl RM
    Apr 2, 2022 at 7:12
  • $\begingroup$ I would think it's true in general. At the single inflection point, the function would seem to be always curving away from the tangent - on one side curving one way, on the other curving the other way. Maybe first establish that there needs to be exactly one inflection point and see if that point always creates an issue. I don’t think you need the continuity of the derivative for that - but that in turn might imply the continuity of the derivative. $\endgroup$ Apr 2, 2022 at 15:53
  • $\begingroup$ What's the difference between it being true for functions with continuous derivatives and true for functions with derivatives? How are you defining the tangent to the function at a non-differentiable point? $\endgroup$ Apr 8, 2022 at 17:33
  • $\begingroup$ There are differentiable functions with non continuous derivatives, see this question. $\endgroup$
    – Saúl RM
    Apr 8, 2022 at 17:39

2 Answers 2


No $f$ as in the question exists.

Suppose $f$ is as in the statement of the question.

We can suppose that $f(0)=f'(0)=f(1)=0$ and $f>0$ in $(0,1)$ composing $f$ with affine functions or adding affine functions to it. $(*)$

We can then take $a\in(0,1)$ such that $f(a)$ is maximal and let $b\neq a$ with $f(b)=f(a)$. Then $b$ cannot be inside the interval $(0,1)$: in that case, the tangent at the point $c$ of the interval between $a$ and $b$ such that $f(c)$ is minimal cuts the graph of $f$ in at least $3$ points. There are two options for $b$.

If $b>1$, we can find a contradiction: in this case $f(1)=0$ is a local minimum of $f$. Now using the fact that for any $\varepsilon$ there is some $x\in(0,\varepsilon)$ with $f'(x)\in(0,\varepsilon)$ (you can prove it using that $f(0)=f'(0)=0$ and $f(x)>0$ in $(0,1)$), we can find a small $x>0$ such that its tangent line intersects the graph of $f$ at two points near $(1,0)$.

So $b<0$ and $f$ does not have a local minimum at $1$. The graph of $f$ would look something like this:

enter image description here

We have two cases:

Case 1: $f'$ is continuous.

Consider the point $c\in[0,a]$ such that $f'(c)$ is the highest possible, you can check that the tangent to $f$ at $c$, $l_c$, only intersects the graph at $(c,f(c))$. Indeed, if $(d,f(d))$ was another point in $l_c$, we can consider two cases.

  • $d\in(0,a)$. Then there has to be some number $e$ between $c$ and $d$ with $f'(e)>f'(c)$, a contradiction.

  • $d>a$ cannot happen either because the tangent line to $c$ passes above $(a,f(a))$ and $f(x)<f(a)\forall x>a$. A similar argument works for $d<0$.

Case 2: $f'$ is not continuous.

We can suppose that $f'$ is not continuous at $0$ in this case, using a point with discontinuous derivative in $(*)$.

As $(1,0)$ is not a local minimum, there exists some $\varepsilon_1>0$ such that $(-\varepsilon_1,\varepsilon_1)\subseteq f([a,1.1])$ $(**)$.

As $f'$ is not continuous at $0$ and using the intermediate value property of the derivative, for small enough $\varepsilon>0$ there is a sequence $x_n\to 0$ with $\varepsilon<|f'(x_n)|<2\varepsilon$. Let $l_n$ be the tangent line at $x_n$. Then, if we pick $\varepsilon$ small enough, for big enough $n$ the lines $l_n$:

  • Pass below the points $(b,f(b))$ and $(a,f(a))$.

  • Intersect the graph of $f$ at some point of the rectangle $[a,1.1]\times(-\varepsilon_1,\varepsilon_1)$, due to $(**)$

We can consider two cases. If $f'(x_n)<0$ for infinitely many $n$, then take some $x\in (-0.1,0)$ with $|\frac{f(x)}{x}|<\varepsilon$. Then for big enough $n$, $l_n$ intersects the graph of $f$ at some point between $b$ and $x$, which is a contradiction. If $f'(x_n)>0$ for infinitely many $n$, take $x\in (0,a)$ with $\frac{f(x)}{x}<\varepsilon$ and for big enough $n$, $l_n$ intersects the graph at some point between $x$ and $a$, so we have a contradiction again.


While I was working the problem (and it was a pretty interesting problem), a complete answer appeared, but I think I'm still going to post all the results that I got so far. Not all of them are necessary for the final solution, but still interesting nonetheless.

Assume, there exists a differentiable function $f$ as described. Let $g\colon\mathbb{R}\rightarrow\mathbb{R}$ be the fixed point free function as described, that assigns a value $x\in\mathbb{R}$ the abscissa of the unique other point, where the tangent to $f$ at $(x,f(x))$ intersects the graph of $f$. Therefore $g(x)$ is the unique value, so that: \begin{equation} f(x)+f'(x)(g(x)-x)=f(g(x)) \Leftrightarrow f'(x)=\frac{f(g(x))-f(x)}{g(x)-x}. \end{equation}

Conclusion: If $g$ is continuous in $x$, then $f'$ is continuous in $x$.

Proposition: Changing $f(x)$ to $f(x)+\lambda x+\mu$ (with the same sought-after property) doesn't change $g(x)$.

Corollary: The conditions $g(g(x))=x$ and $f'(g(x))=f'(x)$ are equivalent.

Proof: $\Rightarrow$: Take the upper equation for $x$ and $g(x)$, combine them through $f(g(x))$ and use that $g$ doesn't have a fixed point to shorten $g(x)-x$. $\Leftarrow$: Directly use the upper equation and expand the fraction by $-1$. $\square$

Lemma: If $g$ is not continuous in $x$, then either the equivalent conditions of the upper corollary hold or $f'$ is not continuous in $x$.

Notice, that when $f$ is continuously differentiable, the first case always has to hold and that the backwards direction from the second case is the contraposition of the upper conclusion.

Proof: We prove the contraposition. Assume $f'(g(x))\neq f'(x)$ and $f'$ is continuous in $x$. Define $\widetilde{f}(y):=f(y)-f'(x)(y-x)-f(x)$ with $\widetilde{f}(x)=0$ and $\widetilde{f}'(x)=0$ as well as $\widetilde{f}(g(x))=0$ using the upper equation and proposition. (Particularly $\widetilde{f}$ doesn't have any other roots than $x$ and $g(x)$.) $\widetilde{f}'$ is continuous in $x$ and the assumption translates to $\widetilde{f}'(g(x))\neq 0$, so for every neighborhood $U$ of $g(x)$, $\widetilde{f}(U)$ is a neighborhood of $\widetilde{f}(g(x))$. Take a sequence $x_n\xrightarrow{n\rightarrow\infty}x$, then $\widetilde{f}(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}(x)=0$ and $\widetilde{f}'(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}'(x)=0$. The tangents $t_n(y)$ through $(x_n,\widetilde{f}(x_n))$ are given by $t_n(y)=\widetilde{f}(x_n)+\widetilde{f}'(x_n)(y-x_n)$. We have $t_n(g(x))\xrightarrow{n\rightarrow\infty}0$, so for all $n$ above a sufficient high boundary, $t_n(g(x))\in\widetilde{f}(U)$ and therefore $g(x_n)\in U$ (Notice, that a higher boundary may be necessary.), which implies $g(x_n)\xrightarrow{n\rightarrow\infty}g(x)$. $\square$

If $g$ is not continuous in $x$ and the first case of this lemma holds, then $g(x)-x$ changes signs between $x$ and $g(x)$, so there has to be another point where $g$ is not continuous between them, otherwise $\widetilde{g}$ would have a root or equivalently $g$ a fixed point due to the intermediate value theorem.

Assume $x<g(x)$ w.l.o.g. Because of the mean value theorem, there exists a $\xi\in(x,g(x))$ (which especially means $\widetilde{f}(\xi)\neq 0$) with $f'(\xi)=f'(x)\Leftrightarrow\widetilde{f}'(\xi)=0$ using the upper equation. Assume $\widetilde{f}$ is only positive in $(x,g(x))$ w.l.o.g. Let $\xi\in(x,g(x))$ be the value of the global maximum in $(x,g(x))$. It is unique since for two different maxima $\xi_1,\xi_2\in(x,g(x))$ with $\xi_1<\xi_2$ w.l.o.g., there would be a $\zeta\in(\xi_1,\xi_2)$ with $\widetilde{f}'(\zeta)=0$ and $\widetilde{f}(\zeta)<\widetilde{f}(\xi_1)=\widetilde{f}(\xi_2)$. Due to the intermediate value theorem, the horizontal tangent through $(\zeta,\widetilde{f}(\zeta))$ would intersect $\widetilde{f}$ also somewhere in $(x,\xi_1)$ and $(\xi_1,g(x))$. This also implies $g(\xi)\notin[x,g(x)]$. We are left with two cases for which we have to derive a contradiction:

Case 1: Assume $g(\xi)\in(-\infty,x)$, then $f>0$ in $(-\infty,0)$ and $f<f(\xi)$ in $(\xi,\infty)$.

Case 2: Assume $g(\xi)\in(g(x),\infty)$, then $f>0$ in $(g(x),\infty)$ and $f<f(\xi)$ in $(-\infty,x)$.

The argumentation for both cases is analogous and done in the other answer.

enter image description here

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    $\begingroup$ Interesting! I also tried thinking about the function $g$ for a bit, but it seems supposing $g$ is discontinuous doesn't give any easy contradictions :/ $\endgroup$
    – Saúl RM
    Apr 10, 2022 at 13:41
  • $\begingroup$ That's true. I thought it would, but it took longer than I expected. I still think the SF has a very elegant and short solution in the book though. $\endgroup$ Apr 10, 2022 at 14:30
  • $\begingroup$ What is the SF? $\endgroup$
    – Saúl RM
    Apr 11, 2022 at 11:13
  • $\begingroup$ "Half a joke" by Paul Erdös explained by himself here: youtube.com/watch?v=1qeWugmiGt4 He often said, that the SF keeps a book with the most elegant proofs to himself and often complimented collegues by saying that their proof was "right from THE BOOK". After his death, a book called "Proof from THE BOOK", which also included his first paper which he wrote when being just 17. It was a short proof for Bertrand's postulate. $\endgroup$ Apr 11, 2022 at 12:13
  • $\begingroup$ Oh, I see. I suspected you were talking about The Book but I hadn't heard of the SF before haha $\endgroup$
    – Saúl RM
    Apr 11, 2022 at 12:23

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