# Rolle's Theorem

• Feb 19th 2010, 08:10 PM
ABigSmile
Rolle's Theorem
This is not exactly Rolle's Theorem but it's a problem that was given to me that is similar that is bugging me. It states "if $f$ is differentiable on $(a,b)$, and $f(a) = f(b) = 0$, then $f$ is uniformly continuous on $[a,b]$" I understand since $f$ is differentiable on $(a,b)$ that it must also be continuous on $(a,b)$.
But continuity doesn't imply uniform continuity. So is there any way to show uniform continuity? Because if there is, I haven't been able to figure it out so far. Is this statement always false since our interval is open? Would providing a simple counter example suffice if that's the case?
• Feb 19th 2010, 08:17 PM
Drexel28
Quote:

Originally Posted by ABigSmile
This is not exactly Rolle's Theorem but it's a problem that was given to me that is similar that is bugging me. It states "if $f$ is differentiable on $(a,b)$, and $f(a) = f(b) = 0$, then $f$ is uniformly continuous on $[a,b]$" I understand since $f$ is differentiable on $(a,b)$ that it must also be continuous on $(a,b)$.
But continuity doesn't imply uniform continuity. So is there any way to show uniform continuity? Because if there is, I haven't been able to figure it out so far. Is this statement always false since our interval is open? Would providing a simple counter example suffice if that's the case?

I don't understand the question? Does $f$ have to be continuous on $[a,b]$ otherwise this follows directly since any continuous function on a compact space is uniformly continuous.
• Feb 19th 2010, 08:27 PM
ABigSmile
Quote:

Originally Posted by ABigSmile
"if $f$ is differentiable on $(a,b)$, and $f(a) = f(b) = 0$, then $f$ is uniformly continuous on $[a,b]$"

The problem I am trying to solve is this. It doesn't say prove it so I am assuming it could be a true or false statement. Also I am pretty sure $f$ does not have to be continuous on $[a,b]$ .Does that clear anything up? Sorry for the confusion.
• Feb 19th 2010, 08:31 PM
Drexel28
Quote:

Originally Posted by ABigSmile
The problem I am trying to solve is this. It doesn't say prove it so I am assuming it could be a true or false statement. Also I am pretty sure $f$ does not have to be continuous on $[a,b]$ .Does that clear anything up? Sorry for the confusion.

Ok. So why doesn't the function $f(x)=\begin{cases} 0 & \mbox{if} \quad x=0,1\\ 1 & \mbox{if} \quad 0 serve as a counter example?. Clearly $f(a)=f(b)=0$ and $f$ is differentiable on $(0,1)$ but it isn't continuous, let alone uniformly continuous, on $[0,1]$.