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I am wondering about the following question: A strictly convex (concave) differentiable function $f:\mathcal{R}\to\mathcal{R}$ has the geometrical property that its graph lies completely above (below) the tangential line at any point (except at the point of contact). Now, this is a special property of such functions but is a relaxed version of this also valid for all (not necessarily continuously) differentiable functions? More specifically, let's call $x_0$ a convex point if there is a neighborhood of $x_0$ such that $$f(x) > f(x_0) + f'(x_0)(x- x_0)$$ for all $x$ in that neighborhood. Similarly, let's call $x_0$ concave if the reverse inequality holds for some neighborhood: $$f(x) < f(x_0) + f'(x_0)(x- x_0)$$ Then, is it true that any non-linear differentiable function has at least one concave or convex point?

Full disclosure: this question was first stated as true on a different forum without any additional information or explanations. It is possible this is well-known and I have never come across this result, or that this is completely wrong. Regardless, I am hoping to get some additional information as I think this is quite a neat result either way.

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Let $f$ be non linear in the interval $[-1,1]$. We can suppose $f(-1)=f(1)=0$ and $f(x)>0$ for some $x\in(-1,1)$. Now let $N$ be so big that the closed ball of center $(0,-N)$ and with radius $\sqrt{N^2+1}$ doesn't contain the whole graph of $f$. Let $(x_0,f(x_0))$ be one of the points of the graph furthest from $(0,-N)$. Then $f$ is concave at $x_0$.

Indeed, the whole graph of $f$ is contained in the ball of center $(0,-N)$ and boundary passing through $(x_0,f(x_0))$. This implies that the derivative of $f$ at $x_0$ has to be the slope of the tangent to that ball at $(x_0,f(x_0))$, and that tangent only intersects the graph of $f$ at $(x_0,f(x_0))$ (in fact it only intersects the ball at that one point).

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  • $\begingroup$ Thank you for your answer. I am not sure this argument holds if $x_0 = 1$. Could you please elaborate more. Thank you. $\endgroup$
    – Ivan
    Commented Mar 14, 2022 at 0:57
  • $\begingroup$ $x_0$ cannot be $1$ due to the definition of $N$: the point $(0,1)$ is at distance exactly $\sqrt{N^2+1}$ of $(0,-N)$ and there are points of the graph of $f$ further from $(0,-N)$. $\endgroup$
    – Saúl RM
    Commented Mar 14, 2022 at 1:07
  • $\begingroup$ Oops typo, I mean $(1,0)$ instead of $(0,1)$ in the comment above $\endgroup$
    – Saúl RM
    Commented Mar 14, 2022 at 1:22
  • $\begingroup$ I understand now, thank you. However, I don't believe your argument holds because of the requirements of the question of strict convexity and the fact that $x_0$ might not be unique. For example, consider a function's graph that is increasing up to $(-\epsilon, 1)$, constant to $(\epsilon, 1)$, and then decreasing. Then, $x_0=\pm\epsilon$ but the function is not strictly convex there. $\endgroup$
    – Ivan
    Commented Mar 14, 2022 at 1:50
  • $\begingroup$ Your function doesn't work because if $f$ is differentiable and constant from $-\varepsilon$ to $\varepsilon$, the furthest points cannot have $x$-coordinate between $-\varepsilon$ and $\varepsilon$. I'll add more details $\endgroup$
    – Saúl RM
    Commented Mar 14, 2022 at 8:29

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