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Assume that the function $f : \left[0, 1\right] \to \mathbb{R}$ is continuous on $\left[0,1\right]$ and is differentiable on $\left(0,1\right)$. Let $c \in \left(0,1\right)$. Prove that there exist $\xi, \eta \in \left(0, 1\right)$ such that \begin{align} 2 \eta f\left(1\right) + \left(c^2 - 1\right) f^\prime\left(\eta\right) = f\left(\xi\right) . \end{align}

I tried to use the Lagrange mean value theorem and the Rolle mean value theorem on $[0,1]$, but failed.

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2 Answers 2

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Let $g(x)= x^2 f(1)+(c^2-1) f(x)$.

Then $g(1)-g(0)=c^2 f(1)+(1-c^2) f(0)$ so by Rolle there exists $\eta \in (0,1)$ such that: $$g'(\eta)=2 \eta f(1)+(c^2-1) f'(\eta)=\frac{g(1)-g(0)}{1-0}=c^2 f(1)+(1-c^2) f(0)$$ Moreover, by the mean value theorem, as $f$ is continuous and $c^2 f(1)+(1-c^2) f(0) \in [f(0),f(1)]$, there exists $\xi \in [0,1]$ such that $$f(\xi)=c^2 f(1)+(1-c^2) f(0)$$ And so: $$2 \eta f(1)+(c^2-1) f'(\eta)=c^2 f(1)+(1-c^2) f(0)=f(\xi)$$

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Hints:

1) First take $c=\frac{\sqrt{3}}{2}$, $f(x)=(x-\frac{1}{2})^2$, and show that the hypothesis $\xi \in (0,1)$ cannot be satisfied. I suppose that you want $\xi \in [0,1]$.

2) a) Put $u = c^2(f(1)-f(0))+f(0)$. Show that we have $f(0)<u<f(1)$ (if $f(1)>f(0)$) or $f(1)<u<f(0)$ (if $f(1)<f(0)$), and deduce in all cases (the case $f(0)=f(1)$ is left to you) there exists an $\xi$ such that $u=f(\xi)$.

b) Put $g(x)=(c^2-1)f(x)+x^2f(1)$, compute $g(1)-g(0)$ and finish the proof.

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