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A function $f:[a,b] \rightarrow \mathbb R$ is continuous on $[a,b]$ and $f''(x)$ exists $\forall x\in (a,b)$. If $a<c<b$ and $f(a)=f(b)=0$, prove that there exists a point $\xi$ in $(a,b)$ such that $f(c)=\frac12(c-a)(c-b)f''(\xi)$

My attempt: using Lagrange's Mean Value Theorem on $f$ in $[a,c]$ and $[c,b]$, $\exists \xi_1 \in (a,c)$ and $\exists \xi_2 \in (c,b)$ such that $f'(\xi_1)=\frac{f(c)-f(a)}{c-a}=\frac{f(c)}{c-a}$ and $f'(\xi_2)=\frac{f(b)-f(c)}{b-c}=\frac{-f(c)}{b-c}$

This is where I'm stuck. I tried applying MVT again, but it didn't lead me anywhere. I tried using Rolle's Theorem or Darboux Theorem but it only gave me a $f''(\xi_n)=0$. Can someone please tell me what to do next?

Diya
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  • Another possibly nice way to approach the problem is via the identity (established by integration by parts) $\displaystyle (c-b)\int_{a}^{c}(t-a)f''(t),dt + (c-a)\int_{c}^{b}(t-b)f''(t),dt = (b-a)f(c)$ .. then denoting $\varphi(t) := \begin{cases} (c-b)(t-a) & \textrm{ for } t \in [a,c] \ (c-a)(t-b) & \textrm{ for } t \in [c,b]\end{cases}$, we rewrite it as $\displaystyle \int_{a}^{b} \varphi(t)f''(t),dt = (b-a)f(c)$ and apply First mean value theorem for integration, to get the desired result. – r9m Nov 06 '14 at 18:58

1 Answers1

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Consider the function: $\displaystyle g(t) = f(t) - \frac{(t-a)(t-b)}{(c-a)(c-b)}f(c)$, we have $g(a) = g(b) = g(c) = 0$.

So, Apply Rolle's Theorem on $g$ in the intervals $[a,c]$ and $[c,b]$, to conclude there are $\xi_1 \in (a,c)$ and $\xi_2 \in (c,b)$, such that $g'(\xi_1) = g'(\xi_2) = 0$.

Since, $\xi_1 \neq \xi_2$, we can again apply Rolle's Theorem on $g'$ in $[\xi_1 ,\xi_2]$, to conclude there is $\xi \in (\xi_1 ,\xi_2)$, such that $\displaystyle g''(\xi) = f''(\xi) - 2\frac{f(c)}{(c-a)(c-b)} = 0$.

r9m
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