Can anybody prove or disprove this conjecture?

Question

[Supposed Continuous Mean Representation of the second derivative]

$$ f''(x_0) = \lim_{h \to 0} 6 \cdot \frac{\dfrac{\displaystyle\int^{x_0 + h}_{x_0 - h } f(x)\,dx}{2h}-f(x_0)}{h^2}$$

Full conjecture: For functions $f$ from the real line onto the real line for which the second derivative exists in the ordinary sense (as the derivative of the derivative) on the points $x_0$ in the real line (and also is integrable in an open interval around $x_0$ obviously) it is equal to the displayed expression.

I checked for some polynomials and the exponential function. I think this has something to do with the weak derivative and the symmetric derivative but i don't know exactly where the connection is.

Edit: I am now much more confident that this formula does indeed hold for at least a lot of functions. I found that this formula holds if you can substitute the integrand by its second degree taylor expansion without changing the limit. It is reasonable to think that this will always work but the justification for this step is missing. Note that you cant just split up the limit since the integral term and the $f(x_0)$ term on their own will diverge.

Answer 1

The conjecture is true.

For simplicity, we will only consider the case $x_0 = 0$.

Let $f$ be any function continuously differentiable to first degree over some interval $(-\eta, \eta)$ and second differentiable at origin (i.e. $f \in \mathcal{C}^1(-\eta,\eta)$ and $f''(0)$ exists).

Let $f_0 = f(0)$, $f_1 = f'(0)$, $f_2 = f''(0)$ and $g(x) = f'(x) - f_1 - f_2 x$.

Over $(-\eta,\eta)$, we can represent $f(x)$ as an integral of $f'(x)$.

$$f(x) = \begin{cases} f_0 + \int_0^x f'(y)dy,& x \in [0,\eta)\\ f_0 - \int_x^0 f'(y)dy,& x \in (-\eta,0] \end{cases}$$ Integrate first relation over $[0,h] \subset [0,\eta)$, we obtain

$$\begin{align}\int_0^h f(x)dx &= f_0 h + \int_0^h \int_0^x f'(y)dy = f_0h + \int_{0 \le y \le x \le h} f'(y)dx dy\\ &= f_0 h + \int_0^h (h - y) f'(y) dy\\ &= f_0 h + \frac{f_1}{2} h^2 + \frac{f_2}{6} h^3 + \int_0^h (h-y)g(y) dy \end{align}$$ Integrate second relation over $[-h,0] \subset (-\eta,0]$, we find

$$\begin{align}\int_{-h}^0 f(x)dx &= f_0 h - \int_{-h}^0 \int_x^0 f'(y)dy = f_0h - \int_{0 \ge y \ge x \ge -h} f'(y)dx dy\\ &= f_0 h - \int_{-h}^0 (h + y) f'(y) dy\\ &= f_0 h - \frac{f_1}{2} h^2 + \frac{f_2}{6} h^3 - \int_{-h}^0 (h+y)g(y) dy \end{align}$$ Taking average of these relations, divide by $h$ and rearrange, we obtain

$$\frac{1}{2h}\int_{-h}^h f(x) dx - f_0 - \frac{f_2}{6} h^2 = \frac{1}{2h}\int_{-h}^h {\rm sign}(y)(h - |y|)g(y)dy$$

Since $f''(0) = f_2$ exists, for any $\epsilon > 0$, we can find a $\delta > 0$ such that $$\left|\frac{f'(y) - f_1}{y} - f_2\right| < \epsilon \implies |g(y)| < \epsilon|y|$$ whenever $0 < |y| < \delta$.

For any $h \in (0,\delta)$, this leads to

$$\begin{align}\left|\frac{6}{h^2}\left(\frac{1}{2h} \int_{-h}^h f(x) dx - f_0\right) - f_2 \right| & < \frac{3}{h^3}\int_{-h}^h (h-|y|)|g(y)|dy\\ & \le \frac{3\epsilon}{h^3}\int_{-h}^h(h - |y|)|y| dy\\ & = \epsilon\end{align}$$ Since $\epsilon$ can be arbitrary small, we can deduce

$$\lim_{h\to 0+}\frac{6}{h^2}\left(\frac{1}{2h}\int_{-h}^h f(x) dx - f(0) \right) = f''(0)$$

Answer 2

WLOG, $x_0=0$, for convenience.

As the second derivative exists, we have

$$f(h)=f(0)+hf'(0)+\frac{h^2}2f''(0)+o(h^2).$$

Then

$$\int_{-h}^h f(h)=2hf(0)+\frac{h^3}3f''(0)+o(h^3).$$

Hence, after simplification,

$$f''(0)=\lim_{h\to0}(f''(0)+o(1)).$$