# Derivative proof involving MVT and Rolle's Theorem

Suppose $$f$$ is a continuous and differentiable function on $$[0,1]$$ and $$f(0) =f(1)$$. Let $$α∈(0,1)$$.

And

$$∀x,y∈(0,1) ,f′(x)\neq 0\wedge f′(y)\neq 0\rightarrow f′(x) \neqαf′(y)$$

Show that there is exactly one x such that $$f(x) =f(αx)$$ and $$f′(x)\neq 0$$

What i tried so far:

Proof.

Let $$a\in (0,1)$$

Assume $$f$$ is continous and differentiable on $$[0,1]$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \$$and $$f(0)=f(1)$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \$$and $$\forall x,y \in (0,1),f'(x)\neq0\wedge f'(y)\neq 0\rightarrow f'(x)\neq af'(y)$$

Show $$\exists !x_0 \in (0,1)s.t. f(x_0)=f(ax_0)\wedge f'(x_0)\neq0$$

Definition:

$$\exists!x_0 \in S,P(x_0)$$

$$\Leftrightarrow\exists x_0 \in S, P(x_0)\wedge(\forall x_1,x_2 \in S, P(x_1)\wedge P(x_2)\rightarrow x_1=x_2)$$

Rewrite what we want to show use definition:

$$\exists x_0 \in (0,1), f(x_0)= f(ax_0)\wedge f'(x_0)\neq0$$

$$\wedge(\forall x_1,x_2 \in (0,1), f(x_1)=f(ax_1)\wedge f'(x_1)\neq0$$

$$\wedge f(x_2)=f(ax_2)\wedge f'(x_2)\neq0$$

$$\rightarrow x_1=x_2)$$

Try to prove this by contradiction.

Assume the negation is true.

Definition:

$$\forall!x_0 \in S,\neg P(x_0)$$

$$\Leftrightarrow\forall x_0 \in S, \neg P(x_0)\vee(\exists x_1,x_2 \in S, P(x_1)\wedge P(x_2)\wedge x_1\neq x_2)$$

Since negation connected with $$\vee$$, one of them is true the whole statement is true, assume the second part is true first.

(Maybe find contradiction from first part later)

Rewrite it as the negation use definition:

Assume:

$$(\exists x_1,x_2 \in (0,1), f(x_1)= f(ax_1)\wedge f'(x_1)\neq0$$

$$\wedge f(x_2)=f(ax_2)\wedge f'(x_2)\neq0$$

$$\wedge x_1\neq x_2)$$

Now try to find some contradictions:

Since either $$x_1 < x_2$$ or $$x_1 > x_2$$

but this don't make much difference for the proof

let's say $$x_1 < x_2$$

Condsider two cases that $$x_1 \leq ax_2$$ or $$x_1 > ax_2$$

Case1: $$x_1 > ax_2$$

Have $$0

Let $$f(x_1)=f(ax_1)=A_1$$

$$f(x_2)=f(ax_2)=A_2$$

By MVT we have :

$$\exists n_1 \in (x_1,x_2),f'(n_1)=\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{A_2-A_1}{x_2-x_1}$$

$$\exists n_2 \in (ax_1,ax_2),f'(n_2)=\frac{f(ax_2)-f(ax_1)}{ax_2-ax_1}=\frac{1}{a}\frac{A_2-A_1}{x_2-x_1}$$

Therefore $$n_1\neq n_2$$

That $$af'(n_2)=f'(n_1)$$

If I can prove $$f'(n_2)\neq 0 \wedge f'(n_1)\neq 0$$ it would be a contradiction.

This is for Case1, I'm kind of stucked here, and for case2, I didn't find the contradiction yet.

I was using Rolle's Thm, but couldn't make $$f'(n_2)\neq 0 \wedge f'(n_1)\neq 0$$.

Any help or hint or suggestion would be appreciated.

• Is, for example, "$0\wedge f'(y)$" in "$f'(x)\neq 0\wedge f'(y)\neq 0$" the meet of $0$ and $f'(y)$ as elements of the ordered set $\mathbb{R}$ (i.e., $\min(0,f'(y))$? Or is it supposed to be read as "$f'(x)\neq 0$ and $f'(y)\neq 0$"? – user10354138 Jun 10 at 8:50
• Did you mean $f′(x)\neq 0$ and $f′(y)\neq 0\rightarrow f′(x) \neqαf′(y)$? – Manx Jun 10 at 8:54
• For $\wedge$ he probably means "and". – lEm Jun 10 at 8:54

If $$f$$ is constant, then it clear.
If $$f$$ is non-constant, then there exist $$x_1\in [0,1]$$, with positive derivative, and also $$f'(x_0)=0$$, for some $$x_0\in (0,1)$$, (Rolle's theorem). Hence, by virtue of Darboux's theorem, there for every $$c\in \big(0,f'(x_1)\big)$$, there exist $$x_c\in (x_1,x_2)$$, such that $$f'(x_c)=c$$.
Therefore, for every $$a\in (0,1)$$, there exist $$y\in (x_0,x_1)$$, such that $$f'(y)=af'(x_1).$$ Hence the assumption in the OP
$$∀x,y∈(0,1) ,f′(x)\neq 0\wedge f′(y)\neq 0\rightarrow f′(x) \neqαf′(y)$$