problem: f has the second differential function on $[a,b]$, prove:$\exists \alpha∈(a,b)$ such that : $f(a)-2f(\frac{a+b}{2})+f(b)=\frac{1}{4} (b-a)^2 f^{''}(\alpha)$
My attempt:we choose $λ$ satisfied:
$f(a)-2f(\frac{a+b}{2})+f(b)=\frac{1}{4} (b-a)^2 λ$
And consider Auxiliary function $F(t)=f(t)-2f(\frac{t+a}{2})+f(a)-\frac{1}{4} (t-a)^2 λ$,it is also has the second differential function. We note $F(a)=F(b)=0$,$F^{'}(a)=0$, by mean value theorem, $\exists \beta ∈(a,b)$ such that $F^{''}(\beta)=0$
Then $λ =\frac{f^{''}(\beta)+f^{"}(\frac{\beta +a}{2})}{2}$.
But I want to prove $\exists \omega∈(a,b)$ such that $λ=f^{"}(\omega)$. I think this way is ture,but How to find the $\omega$? I want to use Darboux theorem of differential function to prove$f^{''}(\omega)=\frac{f^{''}(\beta)+f^{''}(\frac{\beta +a}{2})}{2}$. But I didn't succeed.
I hate denominators, so i will introduce $A,B$ with $a=2A$, $b=2B$, and $A<B$. We are searching an intermediary point $c\in[2A,2B]$ with $$ f(2A) -2f(A+B)+f(2B) = (B-A)^2f''(c)\ . $$ In case of a function $f\in \mathcal C^2$, we can write: $$ \begin{aligned} &f(2B) -2f(A+B)+f(2A) \\ &\qquad= \int_A^B(f'(B+u)-f'(A+u))\; du \\ &\qquad= \int_A^B\int_A^Bf''(t+u)\; dt\; du \\ &\qquad= \int_A^B(A-B)f''(\tau+u)\; du \\ &\qquad= (A-B)(A-B)f''(\tau+\nu) \end{aligned} $$ for some intermediate points $\tau,\nu$. The argument can be rewritten to work for a general $f$, using a help function of two variables, $h(t,u)=f(t,u)-f(t,A)-f(A,u)+f(A,A)=(f(t,u)-f(t,A))-(f(A,u)-f(A,A))$.