Lagrange: Show that there are two points where $f'$ equals zero

Question

Let $f: \mathbb R \to \Bbb R$ a differentiable function. Let $T>0 \in R$ such that $f(x+T)=f(x) \forall x \in \Bbb R$

Show that the interval $[0,T)$ has two points where the function $f'$ get equals to $0$(means that $f'(x) = 0$

What I've done: If I substitute $0$ I get $f(T) = f(0)$ and by Rolle Theorem - If a real-valued function f is continuous on a proper closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that $f'(c)=0$.

Now I'm trying to find the second point I used Lagrange Theorem(Mean value theorem - https://en.wikipedia.org/wiki/Mean_value_theorem)

I get $f'(c) =$ $f(T) - f(0) \over T-0$ = $f(T) - f(T) \over T$ = $0$

but how can I confirm that it's a different point.

Can someone help me please?

Thanks in advance

Answer 1

Another way to proceed would be to just remember how Rolle's theorem is proved in the first place. Assume $f$ is not constant (in which case $f'(x)=0$ everywhere). Since $[0,T]$ is a closed interval and $f$ is continuous, we know (by a theorem) that $f$ achieves an absolute max and an absolute min on $[0,T]$, say at points $x_\max, x_\min \in [0,T]$. However, by the periodicity, these are actually absolute extrema for the function on the whole of $\mathbb{R}$. Note also that $x_\max \neq x_\min$, since we supposed the function was not constant. Now, from the fact that the derivative vanishes at an extreme point, we get two points where $f'$ is zero.

Answer 2

• What you can do if $f$ is continuously differentiable :

Suppose there is no such point. Then your function is strictly increasing or decreasing, and cannot be periodic. Now suppose there is only one such point, say $x_0$. Then you have that, for $x\in]x_0,x_0+T[$, $f'(x)>0$ (for example). Therefore, $$0<\int_{x_0}^{x_0+T}f'(x)\,dx=f(x_0+T)-f(x_0)=0,$$ which is a contradiction. So you have at least two such points.

• What you can do in all cases :

Apply Rolle's theorem between $0$ and $T$, get a such $x_0\in(0,T)$, then re-apply Rolle's theorem between $x_0$ and $x_0+T$, then get $x_1\in(x_0,x_0+T)$ with $x_0\neq x_1$.You finally just have to translate $x_1$ in $[0,T)$ by adding some $k\cdot T,k\in\mathbb{Z}$, if it is not already inside, and you have your points.

Answer 3

To get an intuition of what is happening, I suggest you draw a picture. You have to be in one of the following (non mutually exclusive) cases : either the function is flat at $0$ (meaning f'(0)=0), or it takes the value $f(0)$ at least three times (allowing you to apply Rolle's theorem twice). Let me give the full proof.

Assume we have $f'(0) = 0$. From Rolle's theorem, because $f(0)=f(T)$, there also exists $c_1 \in ]0,T[$ such that $f'(c_1)= 0$. So we're done because we have found two distinct points (namely $0$ and $c_1$).

Otherwise, we have $f'(0)\ne 0$, say $f'(0)> 0$ (the other case is similar). So there exists $x \in ]0,T[$ such that $f(x)> f(0)$ (because $f$ is increasing in a neighborhood of $0$). And similarly $f'(T)=f'(0)>0$ so there exists $y \in ]0,T[$ such that $f(y)<f(T)$ (because $f$ is increasing in a neighborhood of $T$). We have $f(x) > f(0) > f(y)$, so from the intermediate value theorem, there exists $z$ between $x$ and $y$ (and thus in $]0,T[$) such that $f(0)=f(z)=f(T)$. Now you can apply Rolle's theorem twice and find $0 < c_1 < z < c_2 < T$ such that $f'(c_1)=f'(c_2)=0$.