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This comes from Walter Rudin's Functional Analysis p. 388, exercise 21(c): Problem 1

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.

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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.

Johannes Agerskov
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  • Exactly why does $f=+c$? (note that neither $f$ nor $g$ is actually differentable... I'd believe $f=g+c$ if we already knew that $f$ is absolutely continuous. I don't see how we can apply the results in that link, since we don't know that $f\in W^{1,p}$ and we don't know that $f'$ is the weak derivative of $f$.) – David C. Ullrich May 25 '21 at 13:27
  • @DavidC.Ullrich I am not sure I understand, I never wrote $f=+c$? – Johannes Agerskov May 25 '21 at 13:29
  • typo. Why is $f=g+c$? – David C. Ullrich May 25 '21 at 13:30
  • This is why I refered to the other post in here. But basically, you can check that in the distributional sense we have $(f-g)'=0$. It is then a standard result in one dimension that $f-g=c$ as a distribution for some constant $c\in\mathbb{C}$. It then follows that $f=g-c$ in $L^2$ since $C_0^\infty(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. – Johannes Agerskov May 25 '21 at 13:37
  • fine. Except that "checking" that $(f-g)'=0$ seems to me to be as hard as the original problem... – David C. Ullrich May 25 '21 at 14:15
  • @DavidC.Ullrich, Sorry, I just now saw that I did not see your entire first question. Why do you say that we don't know that $f'$ is the weak derivative of $f$? To me, it seems that we exactly know that $f$ and $f'$ are in $L^2$, where $f'$ denotes the weak derivative of $f$. – Johannes Agerskov May 25 '21 at 14:19
  • never mind..... – David C. Ullrich May 25 '21 at 14:22