This comes from Walter Rudin's Functional Analysis p. 388, exercise 21(c):
But the domain I found on Engel's One-Parameter Semigroups for Linear Evolution Equations p. 66 is $$ D(A)=\{f\in L^2(\mathbb{R}):f \text{ is absolutely continuous, } f' \in L^2(\mathbb{R})\} $$
If both of them are right, then we shall also have $D(A)=\{f \in L^2(\mathbb{R}):\int|y\hat{f}(y)|^2dy<\infty\}$, i.e. Rudin's condition $\iff$ Engel's condition. But how do we prove it? I think it makes sense in one direction because we have $iy\hat{f}(y) = \hat{f'}(y)$. So Rudin was saying that $f' \in L^2$, and it is somewhat clear that Engel's version implies Rudin's version. But given Rudin's condition, how can we tell that $f$ is absolutely continuous (or less rigoursly, is $f$ differentiable under this condition)? I guess I could find related theories on some Fourier analysis book but failed.
As you wrote yourself, it is quite easy to see that Rudin's condition implies $f'\in L^2(\mathbb{R})$. Now if $f'\in L^2(\mathbb{R})$, you can define $g(t)=\int_{[0,t]}f'(t)\text{d} t$ and you have $g=f+c$ a.e. for some constant $c\in\mathbb{C}$ (see e.g. Does fundamental theorem of calculus hold for weakly differentiable function?). Now $g$ is clearly continuous by the dominated convergence theorem, so we may represent the class $f$ by a continuous function as well. Furhtermore, you actually have for any pairwise disjoint sequence of sub-intervals $(x_k,y_k)$ by Cauchy-Schwarz $$\sum_k \lvert f(y_k)-f(x_k)\rvert=\sum_k\lvert g(y_k)-g(x_k)\rvert=\sum_k\left\lvert \int_{[x_k,y_k]}f'(s)\text{d} s \right\rvert\leq\sum_k\int_{[x_k,y_k]}\left\lvert f'(s)\right\rvert\text{d} s \leq \lVert f'\rVert_2 \left(\sum_k(y_k-x_k)\right)^{1/2}. $$ This shows that $f$ is absolutely continuous.