# Prove without using mean value theorem

Let $a$ and $b$ be the fixed points of a differentiable function $f$, meaning that $f(a)=a$ and $f(b)=b$. If it is given that $f'(x)$ is never equal to 1, then prove that such a function does not exist.

Using the mean value theorem, on $a$ and $b$, we have for some $c$: $$f'(c)=\frac{f(a)-f(b)}{a-b}=\frac{a-b}{a-b}=1$$

Which is clearly a contradiction. It is a simple application of the mean value theorem. However, I want to see where its provable without using the MVT (or the equivalent Rolle's Theorem), more from the definition of derivative. But I am not able to see any way out to.

If you ask for the motivation, I think it is interesting to prove simple applications of the MVT from something more elementary.

• Do you mean that $f'(x)$ is never $1$? – robjohn Jan 24 '14 at 16:46
• Can you clear up some things. I don't follow your first point. Are you saying $f$ is differentiable and $f(a)=a$ and $f(b)=b$? Then do you mean $f'(x) \neq 1$ for all $x$? – mathematics2x2life Jan 24 '14 at 16:47
• @mathematics2x2life Yes precisely. @ robjhon Yes. Maybe you guys can clear up the language. – Sawarnik Jan 24 '14 at 16:49
• I don't think you can use the definition of the derivative here, since you need then some kind of neighborhood argument. If you say something like for every $a \leq c < d \leq b$ there exists points $x_{1},x_{2} \in [c,d]$ such that $f(x_{1}) = x_{1}$ and $f(x_{2}) = x_{2}$ you can prove this, but using continuity gives you that $f(x) = x$ for all $x$. – Vincent Jan 24 '14 at 16:50
• @Sawarnik: Please edit your post to include the corrections. – Cameron Buie Jan 24 '14 at 16:52

## 1 Answer

Assuming you mean that $f'(x)$ is never $1$, the only thing I can think of is essentially to reprove Rolle and the MVT (that is, the MVT is pretty elementary):

Let $g(x)=f(x)-x$. Then $g(a)=0$ and $g(b)=0$. If $g(x)$ is not identically $0$, then it must attain a maximum or a minimum at some point $c$ between $a$ and $b$. Then $f'(c)-1=g'(c)=0$, but we said that $f'(x)$ is never $1$.

Here is another approach using integrals, but I don't think it is more elementary than the MVT (and it requires $f$ to be continuously differentiable):

If $g'(x)\gt0$ for all $x$ between $a$ and $b$, then $g(b)-g(a)=\int_a^bg'(x)\,\mathrm{d}x\gt0$. Likewise for $g'(x)\lt0$ for all $x$ between $a$ and $b$. If $g'(d)\gt0$ and $g'(e)\lt0$ for some $d$ and $e$ between $a$ and $b$, then the Intermediate Value Theorem says that $g'(c)=0$ for some $c$ between $d$ and $e$.

We might be able to lift the requirement that $f$ be continuously differentiable, but I think we need to use the MVT to show what kinds of discontinuities a derivative can have.

• You basically used Rolle's theorem, which wasn't allowed as its equivalent to MVT. So we can not prove that without using it. Right? If so, why? – Sawarnik Jan 24 '14 at 16:58
• @Sawarnik As robjohn said, you're going to reprove Rolle's or MVT no matter what you do (I think). – David Mitra Jan 24 '14 at 17:12
• @Sawarnik: I have added a second approach, but it leaves a lot to be desired. I think the main point is that the MVT is a pretty basic fact about derivatives. – robjohn Jan 24 '14 at 17:18
• @Sawarnik I would wager your statement is equivalent to the Mean Value Theorem. – David Mitra Jan 24 '14 at 17:22
• @Sawarnik Maybe this can be made rigorous: Take $f$ on $[a,b]$ and translate so that $(a,f(a))$ shifts to $(0,0)$. Call this translation $g$ (so, $g(x)=f(x+a)-f(a)$) and define $h(x)={b g(x)\over g(b)}$. Apply your result to $h$. This gives a $c'\in(0,b)$ with $g'(c')={f(b)/b}$. Now translate back to obtain the $c$ as needed by MVT. – David Mitra Jan 24 '14 at 17:58