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Given that f is twice differentiable on $[a,b]$, prove that there exists $\xi \in (a,b)$ such that $$f(a)-2f\left(\frac{a+b}{2}\right)+f(b)=\frac{1}{4}(b-a)^2f''(\xi) .$$

This problem was given in a book and the hint was to consider the following auxiliary function, and apply Rolle's Theorem:

$$F(t) = f(t) +f(a) -2f\left(\frac{t+a}{2}\right) - \lambda\cdot\frac{(t-a)^2}{4}$$ Where $$\lambda = \frac{f(a)+f(b)-2f\left(\displaystyle\frac{a+b}{2}\right)}{\displaystyle\frac{(b-a)^2}{4}}.$$

I tried and got stuck.

My attempt:

Notice that $F(a) = F(b) = 0$, by Rolle's Theorem, there exists $\xi_1 \in (a,b)$ such that $F'(\xi_1)=0$.

After differentiation, we obtain $$F'(t) = f'(t)-f'\left(\displaystyle\frac{t+a}{2}\right)-\lambda\cdot\displaystyle\frac{t-a}{2}$$

We find that $F'(a) = 0$, thus applying Rolle's Theorem again, there exists $\xi_2 \in (a,\xi_1)$ such that $F''(\xi_2) = 0$

Therefore $$0=f''(\xi_2)-\frac{1}{2}f''\left(\frac{\xi_2+a}{2}\right) - \frac{\lambda}{2}$$

and

$$\lambda = 2f''(\xi_2) - f''\left(\frac{\xi_2+a}{2}\right).$$

I got stuck here and was unable to proceed. I did consider using Darboux's Theorem but I don't think that it is applicable here.

Lee Wei Xuan
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    It's probably easier to prove this using the formula $$f''(x) = \lim_{h \to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$ Indeed, letting $a = x-h$ and $b = x+h$ your claim reduces to showing $f(x-h)-2f(x)+f(x+h) = h^2 f''(\xi)$ for some $\xi \in [x-h,x+h]$, which is basically the same thing. (A proof of the limit formula for the second "symmetric" derivative is easy; if you need a hint, see here) – Brevan Ellefsen Oct 09 '22 at 18:44
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    https://math.stackexchange.com/q/788992/42969 – Martin R Oct 09 '22 at 18:51

2 Answers2

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Your proof is almost complete at the stage when $F'(\xi_1)=0$ ie $$f'(\xi_1)-f'((\xi_1+a)/2)-\lambda\cdot\frac{\xi_1-a}{2}=0$$ Just use the mean value theorem to write $$f'(\xi_1)-f'((\xi_1+a)/2))=\frac{\xi_1-a}{2}f''(\xi)$$ for some $\xi\in(a, b) $ and then you get $\lambda=f'' (\xi) $.

There is no need to differentiate again and deal with $F''(t) $.

Paramanand Singh
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I can only suggest a different approach. Consider $g(\xi) = f(\xi) - \frac{\lambda \xi^2}{2}$, where $\lambda$ is the same as in your attempt. Then $f''(\xi) = \lambda$ is equivalent to $g''(\xi) = 0$. If we assume the converse then $g''$ is positive or is negative on the whole interval $[a,b]$ because $g''$ is continuous. This means that $g''$ is a convex or a concave function. On the other hand, we have $$ g(a) - 2g\left(\frac{a + b}{2}\right) + g(b) = f(a) - 2f\left(\frac{a + b}{2}\right) + f(b) - \frac{\lambda (a^2 - \frac{(a + b)^2}{2}+ b^2)}{2} = f(a) - 2f\left(\frac{a + b}{2}\right) + f(b) - \frac{\lambda (a - b)^2}{4} = 0. $$ Which contradicts the geometric interpretation of convexity/concavity of $g$.

Pavel Gubkin
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