MHB Is a Differentiable Function Always Limited on a Closed or Open Interval?

[hopelesss]

A function f is called restricted ( "bounded") on an interval I if there
is a constant K such that | f (x) | ≤ K for all x ∈ I.
(1) Let f be a differentiable function on a closed interval [x1, x2], where x1 and x2
are real numbers such that x1 < x2. Justify that f then is limited.
(2) Let f be a differentiable function in an open interval (x1, x2), where x1 and x2
are real numbers such that x1 < x2. Show that if the derivative f' is
limited in (X1, x2), then f is also limited.

can someone help with this?

 

[caffeinemachine]

A function f is called restricted ( "bounded") on an interval I if there
is a constant K such that | f (x) | ≤ K for all x ∈ I.
(1) Let f be a differentiable function on a closed interval [x1, x2], where x1 and x2
are real numbers such that x1 < x2. Justify that f then is limited.
(2) Let f be a differentiable function in an open interval (x1, x2), where x1 and x2
are real numbers such that x1 < x2. Show that if the derivative f' is
limited in (X1, x2), then f is also limited.

can someone help with this?

Problem (1) can be solved as follows: It is a general fact that the image of a compact set in a topological space under a continuous map is again compact. More precisely, if $f:X\to Y$ is a continuous map, and $X$ is compact, then $f(X)$ is also compact. Now we have a differentiable map $f:[x_1, x_2]\to \mathbf R$. Since diifferentiable maps are in particular continuous, we have $f([x_1, x_2])$ is a compact subspace of $\mathbf R$. Now a compact subspace of $\mathbf R$ is bounded, and thus we have $f$ is bounded.

(In case you do not know the basics of topology, there is a way to do it in a more elementary way. I will post the more elementary solution in case you need it.)

For problem 2), are you acquainted with the mean value theorem for differentiable maps?

 

[hopelesss]

thanks for your reply.
1) I had to turn up topology in the book's index and its chapter 10.
We did not have that chapter yet.

2) yes we learned about the mean value theorem a few weeks ago. i see in my notebook in did many examples using this theorem. But it was more like ''show that sin x <x for all 0''.

 

[caffeinemachine]

thanks for your reply.
1) I had to turn up topology in the book's index and its chapter 10.
We did not have that chapter yet.

2) yes we learned about the mean value theorem a few weeks ago. i see in my notebook in did many examples using this theorem. But it was more like ''show that sin x <x for all 0''.

I will write down an elementary solution for (1) in some time. For (2) you can proceed like this:

Let $M$ be such that $|f'(x)|<M$ for all $x\in (x_1, x_2)$. Such an $M$ can be found because $f'$ is assumed to be bounded in $(x_1 ,x_2)$. Now if $y_1, y_2\in (x_1, x_2)$ with $y_1< y_2$, we have, by the mean value theorem, that there is a $y\in (y_1, y_2)$ such that
$$f'(y)=\frac{f(y_2)-f(y_1)}{y_2-y_1}$$
Thus
$$\left|\frac{f(y_2)-f(y_1)}{y_2-y_1}\right|\leq M$$
giving
$$|f(y_2)-f(y_1)|\leq M(y_2-y_1) \leq M(x_2-x_1)$$
Now can you see why $f$ is bounded in $(x_1, x_2)$?

 

[HallsofIvy]

It is a little strange that you specifically define "bounded" and "restricted" but then ask about a function being "limited" which you did not define.