Different approaches to differentiability in $L^2$ We can use different approaches to differentiability of $L^2(\mathbb{R})$ functions, e.g. we can say that $f\in L^2(\mathbb{R})$ is differentiable iff $f$ has a differentiable version (representative). In this case $f$ has a strong derivative. Or we can consider differentiability almost everywhere, though in this case we can lose a lot of information about a function, for example both constant function and Cantor function would have $0$ as a derivative.
I'm interested in three following approaches to define derivative in $L^2$:
we say that $g$ is derivative of $f$ iff


*

*there exist such $c\in\mathbb{R}$ and version $\tilde{f}$ of $f$ that $\tilde{f}(x)=\int\limits_0^xg(t)dt+c$

*for all $\varphi\in C_0^\infty(\mathbb{R})$ we have $\int\limits_{\mathbb{R}} (g\varphi)=-\int\limits_{\mathbb{R}} (f\varphi')$

*$\hat{g}(\xi)=2\pi i\xi\hat{f}(\xi)$
First one is the most intuitive. For any $g\in L^2(\mathbb{R})$  the integral exists since $L^2[c,x]\subset L^1[c,x]$.
Second approach is based on embedding $L^2(\mathbb{R})$ into the space of generalized functions, taking derivative there and getting back to $L^2(\mathbb{R})$ if possible. Here we have to choose the space of base functions, and the smaller this space is, the wider the set of differentiable functions becomes. So the natural question is what particular space of base function should we choose to make this definition equivalent to the first one (or the third).
Third approach comes from the property of Fourier transform for strong derivatives.
So my questions are about equivalency of those three approaches and the problem of choosing the space of base function in the second one.
 A: Let $f,g \in L^2(\mathbb{R})$. Consider the following three (slightly corrected variants of your) definitions:


*

*The function $g$ is a derivative of $f$ if there exists $\tilde{f}$ with $\tilde{f} = f$ a.e and $\tilde{f}(y) - \tilde{f}(x) = \int_x^y g(t) \, dt$ for all $x,y \in \mathbb{R}$.

*The function $g$ is a derivative of $f$ if for all $\varphi \in C^{\infty}_c(\mathbb{R})$ we have $\int_{\mathbb{R}} g \varphi = -\int_{\mathbb{R}} f \varphi'$.

*The function $g$ is a derivative of $f$ if $\hat{g}(\xi) = 2\pi i \xi \hat{f}(\xi)$.


Definition one was modified because your definition doesn't work even if $f$ is $C^1$ unless $f(c) = 0$ and also to handle functions that might a priori not be continuous. Definition two was modified because even if $f$ is $C^1$, integration by parts introduces a sign factor and we've also replaced the space of tests functions $C^{\infty}_0(\mathbb{R})$ (which I interpret as smooth functions vanishing at infinity) by $C^{\infty}_c(\mathbb{R})$ (smooth functions with compact support) because otherwise, the integrals aren't even necessarily defined.
It turns out that all three variants above are equivalent and in this case, we say that $f$ belongs to the Sobolev space $H^1(\mathbb{R}) = W^{1,2}(\mathbb{R})$. You can find the proof of the equivalences above in most books that talk about Sobolev spaces (for example, in "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Haim Brezis and in "Partial Differential Equations" by Lawrence C. Evans).
