Does Every Continuous Function Have an Antiderivative? I was thinking about integrals over continuous function and wanted to know if every continuous function has an antiderivative. Generally, the only way I can think of proving this is through a linear algebra approach.
I wanted to first prove or disprove

Let $D_{x}$ be the differential operator, $C^{1}$ be the set of differentiable functions thats derivative is one time continuous, and $C$ be set of continous functions.
$$C\subseteq D_{x}(C^{1})$$

and then if this were true then wanted to prove or disprove

$D_{x}$ is surjection from $C^{1}$ to $C$

The problem is first I don't know if truth of the propositions would imply that every continuous function has an antiderivative. Next, Im not really comfortable with infinite subspaces. I know that if this did happen to be finite we could use rank nullity theorem and since constant function derivative would be 0 thus in nullspace thus not injective thus not surjective, but I'm pretty sure this type of logic doesn't apply to infinite subspaces.
Also If I didn't state these propositions correctly feel free to edit them
No, e.g., the Cantor function: http://en.wikipedia.org/wiki/Cantor_function is a.e., constant, so that its derivative is a.e. $0$, although it is not constant, i.e., you cannot recover the function from its derivative. It is a canonical example of a function thatis integrable, but without an antiderivative.  I think what you are looking for is for the function to be absolutely-continuous, which will guarantee the existence of an antiderivative.
No, e.g., the Cantor function: http://en.wikipedia.org/wiki/Cantor_function is a.e., constant, so that its derivative is a.e. $0$, although it is not constant, i.e., you cannot recover the function from its derivative. It is a canonical example of a function thatis integrable, but without an antiderivative.  I think what you are looking for is for the function to be absolutely-continuous, which will guarantee the existence of an antiderivative.