Inflection point characterization question

Question

Define the mean of a function $f:\mathbb{R}\to\mathbb{R}$ of differentiability class $\mathcal C^2$ over $[a,b]$ by

$$\operatorname{mean}_{[a,b]}(f) \;=\; \frac{1}{b - a} \displaystyle\int\limits_{a}^{b} f(x) \; \mathrm dx.$$

Fix a point $c\in\mathbb{R}$. We wish to address the two assertions:

1. (a) If for every $\varepsilon$ with $0<\varepsilon<\widetilde\varepsilon$ one has

   $$ \operatorname{mean}_{[\,c-\varepsilon,\;c+\varepsilon\,]}(f)\;=\;f(c), $$

   then $c$ is a point of inflection of $f$.

2. (b) If $c$ is a point of inflection of $f$, then there exists $\widetilde\varepsilon>0$ such that for all $0<\varepsilon<\widetilde\varepsilon$,

   $$ \operatorname{mean}_{[\,c-\varepsilon,\;c+\varepsilon\,]}(f)\;=\;f(c). $$

Here a point of inflection is understood in the classical $\mathcal C^2$-sense: $f''(c)=0$ and, on passing through $c$, the concavity of $f$ changes sign (equivalently, $f''$ changes sign at $c$).

I think I have worked out a fairly straightforward answer to (b).

Consider the function

$$f(x) = x^4 + x^3.$$

It has $f''(x) = 12x ²+ 6x$. There are "true" two inflection points, i.e. two non-undulate inflection points. They are $c_1=0$ and $c_2 = -1/2$. The mean around the origin is

$$\displaystyle\frac{1}{2\varepsilon} \left(\displaystyle\int\limits_{-\varepsilon}^{\varepsilon} (x^4 + x^3) \; \mathrm dx \right) = \displaystyle\frac{1}{\varepsilon} \left(\displaystyle\int\limits_{0}^{\varepsilon} x^4 \; \mathrm dx \right) =\displaystyle\frac{1}{\varepsilon} \left(\displaystyle\frac{2\varepsilon^5}{5}\right) = \displaystyle\frac{2 \varepsilon^4}{5}.$$

Since

$$f(c_0) = c_0^4 + c_0^3 = 0 \neq \displaystyle\frac{2 \varepsilon^4}{5}$$

we have a counterexample to (b).

I am wondering about the (a) part. So far, I have only attempted proving it; it might be that there is a counterexample to it. But unlike (b), this feels like it could be true.

We have the condition

$$\displaystyle\frac{1}{2 \varepsilon} \left( \displaystyle\int\limits_{c - \varepsilon}^{c + \varepsilon} f(x) \; \mathrm dx \right) = f(c).$$

If we differentiate in $c$, we get the condition (see edit) *,

$$f(c+\varepsilon) - f(c-\varepsilon) = 2 \varepsilon f'(c).$$

If we differentiate in $\varepsilon$, we instead end up with

$$\displaystyle\frac{1}{2 \varepsilon} \left( f(c-\varepsilon) + f(c+\varepsilon) \right) - \displaystyle\frac{1}{2 \varepsilon^2} \left( \displaystyle\int\limits_{c - \varepsilon}^{c + \varepsilon} f(x) \; \mathrm dx \right) = 0.$$

Neither of these seem particular helpful to me.

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EDIT:

As remarked in a comment, "differentiation in $c$" is an abuse of notation and is only valid is $c$ "can vary". More formally, we must be able to take a limit of the difference quotient while still being "inside the set" $A := \{ x : f''(x) = 0\}$. But if $c$ is an interior point of $A$, then strictly speaking $c$ is not a "true" inflection point. An undulation point is a point $c \in A$ where the second derivative vanishes, but the sign does not change as one moves along the curve.

Thus on an interval small enough $(c - \varepsilon, \, c + \varepsilon)$, we must have that $f'' \equiv 0$. But then $f'$ is an affine function, and $f$ is a quadratic function. We can thus see that investigate the mean of a quadratic function $f(x) = \alpha x^2 + \beta x + \gamma$.

We have the mean at an undulation point $c \in A$ to $f$,

$$\begin{align*} \operatorname{mean}_{[c - \varepsilon, c + \varepsilon]}(\alpha x^2 + \beta x + \gamma) &= \displaystyle\frac{1}{2 \varepsilon} \left( \displaystyle\int\limits_{c - \varepsilon}^{c + \varepsilon} \left( \alpha x^2 + \beta x + \gamma \right)\; \mathrm dx \right) \\ &= \displaystyle \frac{\alpha \varepsilon^2}{3} + \gamma + c \beta + c^2 \alpha.\end{align*}$$

This depends on $\varepsilon$, so in general we shouldn't expect the mean to be equal to $f(c)$ in this case. Collecting alike terms gives

$$\begin{align*} \operatorname{mean}_{[c - \varepsilon, c + \varepsilon]}(\alpha x^2 + \beta x + \gamma) &= \left(\displaystyle \frac{\varepsilon^2}{3} + c^2 \right) \alpha + c \beta + \gamma. \end{align*}$$

The value $f(c) = \alpha c^2 + \beta c + \gamma$, so we get the extra condition that

$$ \displaystyle \frac{\varepsilon^2}{3} + c^2 = c^2.$$

But this forces $\varepsilon = 0$ (as long as $\alpha \neq 0$), and so the the mean around any point $c$ of a quadratic function cannot be equal to the value at $c$.

Please correct me if this is wrong, otherwise we have solved the undulation point case completely.

Answer 1

The last equation boils down to $$f(c)=\frac{f(c+\varepsilon)+f(c-\varepsilon)}2,$$ i.e., if we translate the graph such that $c=f(c)=0$, then $f$ is an odd function - at least in the $\tilde\varepsilon$ neighbourhood of $c$. And indeed, oddness implies the condition (a). Then $f’$ is even, $f’’$ is odd. And yet there need not be change of concavity because there is no suitable interval of constant sign. Think $f’’(x)=x^2\sin(1/x)$ for $x\ne0$ and $f’’(0)=0$, and integrate twice to get $f$.