C vs C1 Continuous: Understanding Derivatives

[Susanne217]

Homework Statement

Let F be a function defined on a open set E where $$E \in V$$. F is said to be continuous at each point $$x \in V$$. Which according to my textbook at hand can be written as $$F \in \mathcal{C}(E)$$

But the expression $$F \in \mathcal{C}^{1}(E)$$ is that equal to saying that F have first order derivatives defined on the set E?

 

[HallsofIvy]

More than that. $$F \in \mathcal{C}^{1}(E)$$ says that F has a continuous first derivative. For example, f(x)= |x| is continuous in any neighborhood of x= 0 but not differentiable at x= 0 so $$f \in \mathcal{C}^{1}(E)$$ but f is not in $$\mathcal{C}^{1}(E)$$. A slightly more complicated example is $$f(x)= x^2sin(1/x)$$ if x is not 0, f(0)= 0. It is fairly easy to show that f is continuous in any neighborhood of x= 0 and, in fact, that it has a derivative for all x, including x= 0, but the derivative function is not continuous at x= 0.

 

[Susanne217]

Thank you Mr. Hall,

I have a follow-up question.

Let say we have autonomous differential equation

$$x' = f(x)$$ and that we have an open set $$E \subset \mathbb{R}$$ and that $$f \in \mathcal{C}^1(E)$$. We then assume that (I,x) is a solution to the diff-eqn also that $$x(t_1) = x(t_2)$$ where $$t_1, t_2 \in I$$ and that $$t_1 < t_2$$

Show for n = 1 that y is a constant solution.

I know from the definition of f that it has first derivatives on E and that E is open. And since from the definition of the solution to a diff.eqn that $$I \subset \mathbb{R}^n$$. Since I is considered to be a subset of $$R$$ and that E is also a subset of $$\mathbb{R}$$ for n = 1.

Then here is my questions:

1) Do I need to show that f also has derivatives on I?

2) I can't well assume that since f has derivatives on a subset of R then if f's domain consistets of both of R and E. Can this domain be seen a as closed set? And I then assume by Rollo's theorem that there exists some value on f's domain, $$t_3$$. Where if x is a solution of f, and then $$x'(t_1) = f(x(t_1)) = x'(t_2) = 0$$ and then $$x'(t_3) = f(x(t_3))= 0$$ and than the solution x is constant?