# A problem about the differential mean value theorem $2ηf(1)+(c^2-1)f'(η)=f(ξ)$

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.

## 2 Answers

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: $$2 \eta f(1)+(c^2-1) f'(\eta)=c^2 f(1)+(1-c^2) f(0)=f(\xi)$$

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 $$f(0) (if $$f(1)>f(0)$$) or $$f(1) (if $$f(1)), and deduce in all cases (the case $$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.