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In the book (Curso de Análise, volume 1,Elon Lages), there is a suggestion that helps a lot.

First, consider that $$f'(a) =f'(b)=0$$ Then, consider the function $g:[a,b] \rightarrow \mathbb{R}$, where $g(x) = \frac{f(x) - f(a)}{x - a}$ and $g(a) = 0$. Show that $g$ reaches it's maximum or minimum in a point $c \in (a,b)$. For the general case, consider $$g(x) = f(x) - xf'(a)$$

I can see why the first case: if a take the derivative of g, I end up with something like:

$$g'(x) = \frac{1}{x - a} \left( f'(x) - \frac{f(x) - f(a)}{x - a} \right)$$

So, by being continuous (from the differentiable hypothesis) on a compact set, by Weierstrass's Theorem, we have that $g$ must have it's maximum/minimum on $c \in [a,b]$. By being a critical point, we must have $g'(c) = 0$, and assuming $c \neq a$, we have our first conclusion.

But (1) I can't see why it must be a interior point (seriously, I've been on this question for 4 days), and (2) the second suggestion isn't that clear for me.

Any other ideas for solutions will be of great help to me.

Arctic Char
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Igor Souza
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1 Answers1

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There are several steps involved in the first part and here are some hints:

If $g$ does not have a local minimum or a local maximum in $(a,b)$ then $g$ is necessarily strictly monotonic. This implies that $f$ is concave or convex (depending on whether $g$ is increasing or decreasing). Hence $f'$ is monotonic. The fact that $f'(a)=f'(b)=0$ shows that $f'\equiv 0$ and $f$ is a constant. So any $c \in (a,b)$ is good enough.

The second part is straightforward. Just apply the first part to $g(x)=f(x)-xf'(a)$. Since $g'(a)=g'(b)=0$ this is possible). Simplify the equation $g'(c)=\frac {g(c)-g(a)} {c-a}$ to finish the proof.

Kavi Rama Murthy
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