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A function $f(x)$ is continuous on $[a,b]$ and $f''(x)$ exists for all $x\in (a,b)$. IF $c\in (a,b)$ $f(a)=f(b)=0$ prove that $f(c)=\frac{1}{2}(c-a)(c-b)f''(\xi)$ for $\xi \in (a,b)$.

I have no idea to solve it. I know that I have to use Lagrange MVT. Please help.

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  • $\begingroup$ As often is the case for these issues, you could introduce an auxiliary function like in (math.stackexchange.com/q/105209) $\endgroup$ – Jean Marie Jan 14 '17 at 8:09
  • $\begingroup$ @JeanMarie The immediate attempt at auxillary function $g(x) = f(x) - (x-a)(x-b)\frac{f(c)}{(c-a)(c-b)}$ (which makes $g(a) = g(b) = g(c) = 0$) did not work out well, I think. The condition on $\xi$ does not become any easier as far as I can see. If you can think of another auxillary function, go ahead. $\endgroup$ – Arthur Jan 14 '17 at 8:49
  • $\begingroup$ Then what would be the exact auxiliary function ? $\endgroup$ – user1942348 Jan 14 '17 at 8:58
  • $\begingroup$ @Arthur Why not try to adapt the auxiliary function in the reference I gave ? $\endgroup$ – Jean Marie Jan 14 '17 at 9:10
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Consider $g:x\mapsto (b-a)f(x)-\frac 12 (a-b)(b-x)(x-a)C$

You can find $C$ such that $g(c)=0$.

Note that $g(a)=g(b)=g(c)=0$, so there's some $\xi$ such that $g''(\xi)=0$.

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  • $\begingroup$ Would you please elaborate $\endgroup$ – user1942348 Jan 14 '17 at 9:25
  • $\begingroup$ @user1942348 I've added some details. Compute $C$ and you'll get your answer. $\endgroup$ – Gabriel Romon Jan 14 '17 at 9:41
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    $\begingroup$ both $b-a$ and $a-b$ in $(b-a)f(x)-\frac 12 (a-b)(b-x)(x-a)C$ are redundant, they will cancelled out. Is it not? Simply one can take $g(x)=f(x)-(b-x)(x-a)C$ with $g(c)=0$? $\endgroup$ – user1942348 Jan 14 '17 at 9:54
  • $\begingroup$ @user1942348 sure $\endgroup$ – Gabriel Romon Jan 14 '17 at 9:58
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Yes indeed Lagrange MVT will work ,

Let $a<\delta_1<c<\delta_2<b$ and $\xi \in (\delta_1,\delta_2)$ .

If we start with the following expression,

$\displaystyle \frac{1}{b-a}\left(\frac{f(c)-f(a)}{c-a}-\frac{f(c)-f(b)}{c-b}\right)$

Now, There must exist $\delta_1\in(a,c)$ & $\delta_2\in(c,b)$ such that $\displaystyle f'(\delta_1)=\frac{f(c)-f(a)}{c-a}$ & $\displaystyle f'(\delta_2)=\frac{f(c)-f(b)}{c-b}$ by Lagrange MVT

Now suppose $\delta_1<\xi<\delta_2$ then since $f'(x)$ exists in $(a,b)$ and continuous on $[\delta_1,\delta_2]$ so again by Lagrange MVT we have ,

$\displaystyle f''(\xi)=\frac{f'(\delta_2)-f'(\delta_1)}{\delta_2-\delta_1}$

So the very first expression turns,

$\displaystyle \frac{1}{b-a}\left(\frac{f(c)-f(a)}{c-a}-\frac{f(c)-f(b)}{c-b}\right) = \frac{f'(\delta_2)-f'(\delta_1)}{b-a}=\frac{f'(\delta_2)-f'(\delta_1)}{2(\delta_2-\delta_1)}=\frac{1}{2}f''(\xi)$

where $\displaystyle \left(\frac{b-a}{2}=\delta_2-\delta_1\right)$

Since $f(a)=f(b)=0$ this is equivalent to $\displaystyle f(c)=\frac{1}{2}(c-a)(c-b)f''(\xi)$

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  • $\begingroup$ Why $\frac{b-a}{2}=\delta_2-\delta_1$ is true? $\endgroup$ – user1942348 Jan 14 '17 at 9:26
  • $\begingroup$ SInce $\delta_1,\delta_2$ are arbitrary and the question is "There exists" , so there exists two such entities whose difference is the average length of the interval. For that exact case we can have the condition fulfilled. $\endgroup$ – Aditya Narayan Sharma Jan 14 '17 at 9:29
  • $\begingroup$ It is a very narrow choice, I think. $\endgroup$ – user1942348 Jan 14 '17 at 9:30
  • $\begingroup$ Isn't the question very much particular as it asks for the existence of such $\xi$ , we can be specific $\endgroup$ – Aditya Narayan Sharma Jan 14 '17 at 9:36
  • $\begingroup$ You can not choose $\delta_{1},\delta_{2}$ to suit your needs. These numbers are not arbitrary and the only information we have on them is that one lies in $(a, c) $ and the other lies in $(c, b) $. $\endgroup$ – Paramanand Singh Jan 14 '17 at 17:03

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