# prove that $f(c)=\frac{1}{2}(c-a)(c-b)f''(\xi)$ for $\xi \in (a,b)$

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.

• As often is the case for these issues, you could introduce an auxiliary function like in (math.stackexchange.com/q/105209) – Jean Marie Jan 14 '17 at 8:09
• @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. – Arthur Jan 14 '17 at 8:49
• Then what would be the exact auxiliary function ? – user1942348 Jan 14 '17 at 8:58
• @Arthur Why not try to adapt the auxiliary function in the reference I gave ? – Jean Marie Jan 14 '17 at 9:10

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 $g''(\xi)=0$.

• Would you please elaborate – user1942348 Jan 14 '17 at 9:25
• @user1942348 I've added some details. Compute $C$ and you'll get your answer. – Gabriel Romon Jan 14 '17 at 9:41
• 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 $g(c)=0$? – user1942348 Jan 14 '17 at 9:54
• @user1942348 sure – Gabriel Romon Jan 14 '17 at 9:58

Yes indeed Lagrange MVT will work ,

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

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

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

• Why $\frac{b-a}{2}=\delta_2-\delta_1$ is true? – user1942348 Jan 14 '17 at 9:26
• 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. – Aditya Narayan Sharma Jan 14 '17 at 9:29
• It is a very narrow choice, I think. – user1942348 Jan 14 '17 at 9:30
• Isn't the question very much particular as it asks for the existence of such $\xi$ , we can be specific – Aditya Narayan Sharma Jan 14 '17 at 9:36
• 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 $(c, b)$. – Paramanand Singh Jan 14 '17 at 17:03