If F'(x) is bounded so is F(x)

[dpesios]

Hello everybody,
A few years ago i tried to join a mathematics department and in the relevant exams
i came up against the following problem. I apologise beforehand if the statement of the problem is a little bit ambiguous because i do not remember it exactly. However, I am sure you will get the point.

Homework Statement

if the first derivative of the real function F(x) is continuous and bounded over the interval [a,b] (or (a,b) ?) , prove that F(x) also is bounded on the interval (a,b) (or [a,b] ?) and the vice versa.

Homework Equations

So we can see that m =< F'(x) =< M.
How can we get from this into the boundness of the F(x) without falling into pitfalls ?
What about the vice versa ?
Should we use the definition or something else ?

The Attempt at a Solution

I will not attempt to publish the solution I proposed because many of you may laugh.

 

[Robert1986]

For starters, for the "vice-versa" thing to have a fighting chance, it would have to be on a closed interval. For example, x^1/2 is bounded on (0,1] but its derivative is not.

 

[dpesios]

Thanks for the reply,

Indeed the "vice-versa" does not hold for every function F(x) as it has also been discussed https://www.physicsforums.com/showthread.php?t=515616&highlight=bounded+function+derivative".

How can we argue that F(x) is bounded ? By intuition the statement is obvious ...

Is integrating over (a,t) where t a variable among a and b the inequality m =< F'(x) <= M a good argument ? We will get rid off the derivative by doing so ...

I have an engineering background so make allowances ...

 

[Office_Shredder]

That sounds like an excellent place to start

 

[Robert1986]

Hmmm, you have a function F and you know that is derivative, F' is bounded on [a,b]. But nothing says that F' is continuous, which you would need to prove to be able to use FTC. For example, FTC1 says that if f is continuous on [a,x] and if F(x) is the integral of F(x) from a to x, then F'(X) = f(x). The proof 100% relies on the continuity of f at x. FTC2 (which is the one you are wanting to use) requires the continuity of f, as well. Now of course, you could still use FTC. However, I would suggest this approach: you know that F'(x) is bounded and you know that it exists at every x in [a,b]. This implies that F is continuous, but you are dealing with a compact (closed and bounded) interval and therefore F is uniformly continuous. Now, say that F is unbounded at some point c. Let epsilon = 1 and show that no matter how close you bring x to c, |F(x) - F(c)| > 1. Now, this is pretty clear since every F(x), x not equal c, is going to be finite, but F(c) isn't finite. Now, there is some care you will have to take to ensure that F(x) is, infact finite for x not equal c.