(Re)Defining Rolle's theorem and its converse

Question

My textbook has stated Rolle's theorem as:

Let $f : [a,b] \to \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a, b)$, such that $f(a) = f(b)$, where $a$ and $b$ are some real numbers. Then there exists some $c$ in $(a,b)$ such that $f'(c) = 0$.

I'm asked to verify whether its converse is true or not. Problem is, I don't know exactly what its converse is. In the given statement, the IF condition is not exactly clear to me. So I made a new equivalent statement and then found its converse.

I want to know whether my new statement of the theorem is equivalent to the given (textbook) statement and did I define its converse correct.

New Statement (Rolle's Theorem).

Given $f : [x,y] \to \mathbb{R}$ is continuous on $[x,y]$ and differentiable on $(x,y)$, if there exists $(a,b) \subseteq (x,y)$ such that $f(a) = f(b)$, then there exists $c \in (a,b)$ such that $f'(c) = 0$.

Converse of the above statement:.

Given that $f : [x,y] \to \mathbb{R}$ is continuous on $[x,y]$ and differentiable on $(x,y)$, if there exists $c \in (x,y)$ such that $f'(c) = 0$, then there exists interval $(a,b) \subseteq (x,y)$ such that $c \in (a,b)$ and $f(a) = f(b)$.

Answer 1

For simplicity, let us agree that we are dealing with the Rolle’s theorem stated for functions defined on closed intervals of $\mathbb{R}$.

Theorem (Rolle). Let $f \colon [a,b] \to \mathbb{R}$ be a function, where $a, b$ are real numbers such that $a < b$. Suppose that $f$ is continuous on $[a,b]$, differentiable on $]a,b[$, and $f(a) = f(b)$. Then, there exists some real number $c \in ]a,b[$ such that $f’(c) = 0$.

Given any function $f \colon [a,b] \to \mathbb{R}$, in order to apply Rolle’s theorem, we must guarantee that $f$ satisfies the following conditions:

1. $f$ is continuous on $[a,b]$;
2. $f$ is differentiable on $]a,b[$;
3. $f(a) = f(b)$.

One could say that these three conditions are the hypotheses of Rolle’s theorem (which is not incorrect). In this case, the converse of Rolle’s theorem would be the following.

Let $f \colon [a,b] \to \mathbb{R}$ be a function, where $a, b$ are real numbers such that $a < b$. If there exists a real number $c \in ]a,b[$ such that $f’(c) = 0$, then $f$ is continuous on $[a,b]$, differentiable on $]a,b[$, and $f(a) = f(b)$.

We could work with this. But that is not what we usually do. Instead, the first two conditions mentioned above are usually included in the “let $f \colon [a,b] \to \mathbb{R}$ be a function” part of the theorem.

In other words, Rolle’s theorem states a result concerning a function $f \colon [a,b] \to \mathbb{R}$ that is continuous on $[a,b]$ and differentiable on $]a,b[$. If you open any book on Real Analysis, you will see various results concerning functions satisfying these conditions (for example, Cauchy’s and Lagrange’s theorems).

It is the third condition stated above that distinguishes Rolle’s theorem from other theorems concerning functions $f$ as discussed above. In this way, we usually say that the hypothesis of Rolle’s theorem is that $f(a) = f(b)$. So, if $f(a) = f(b)$, then there exists some real number $c \in ]a,b[$ such that $f’(c) = 0$. In this case, the converse of Rolle’s theorem would be:

Rolle’s theorem converse. Let $f \colon [a,b] \to \mathbb{R}$ be a function, where $a, b$ are real numbers such that $a < b$. Suppose that $f$ is continuous on $[a,b]$ and differentiable on $]a,b[$. If there exists some real number $c \in ]a,b[$ such that $f’(c) = 0$, then $f(a) = f(b)$.

And (as you wrote) this statement is false. Just consider the map $$ \begin{array}{rrcl} f \colon & [-1,1] & \to & \mathbb{R} \\ & x & \mapsto & x^3. \end{array} $$