Domain of the differentiation operator on $L^2(\mathbb{R})$

Question

Let M and D be the multiplication and differentiation operators on $L^2(\mathbb{R})$:

• $M=x$
• $D=i\frac{\partial}{\partial x}$

I'm trying to understand the maximal domain $\mathcal{D}(D)$ on which $D$ is defined. Letting

• $M_{\epsilon}(f)(x)=e^{i\epsilon x}$f(x)
• $D_{\epsilon}(f)(x)=f(x+\epsilon)$

then by Stone's theorem about one-parameter unitary groups, the maximal domain for infinitesimal generators is given by

• $\mathcal{D}(M)=\{f\in L^2(\mathbb{R}):\lim_{\epsilon\rightarrow 0}\left(\frac{M_\epsilon(f)-1}{i\epsilon}f\right)\text{ converges in }L^2\}$
• $\mathcal{D}(D)=\{f\in L^2(\mathbb{R}):\lim_{\epsilon\rightarrow 0}\left(\frac{D_\epsilon(f)-1}{i\epsilon}f\right)\text{ converges in }L^2\}$

Now, the one-parameter family $\frac{e^{i\epsilon x}-1}{i\epsilon}$ of functions on $\mathbb{R}$ converges to $x$ as $\epsilon\rightarrow 0$ uniformly on compact subsets of $\mathbb{R}$ (due to $\exp$ being analytic on all of $\mathbb{R}$). From this it's straightforward to show that $\mathcal{D}(M)$ can also be characterized as $$\mathcal{D}(M)=\{f\in L^2(\mathbb{R}):\int_{\mathbb{R}}x^2|f(x)|^2 dx < +\infty\}$$ (although I haven't checked the detail, but I'm pretty sure it's correct.)
My question is does $f\in\mathcal{D}(D)$ have a more direct characterization in terms of pointwise differentiation of $f$. For example (just a suggestion)

• $f$ is equal a.e. to a function $g\in L^2(\mathbb{R})$ such that
• $g$ is differentiable a.e. and
• Letting $g^\prime$ be the derivative of $g$ (extended by zero where g is not differentiable), then $g^\prime$ is Lebesgue-measurable and in $L^2(\mathbb{R})$.

Now, I know there are other characterizations of $\mathcal{D}(D)$. For example, since the Fourier Transform maps $M$ to $D$ and vice-versa (up to some constant multiple), then $\mathcal{D}(D)$ can be characterized as the inverse Fourier transform of $\mathcal{D}(M)$. Another approach is the Sobolev approach which is to view derivatives of functions in the sense of distributions, so that $f\in L^2(\mathbb{R})$ has a derivative (in $L^2$) if some other function $g\in L^2(\mathbb{R})$ satisfies $$\int_{\mathbb{R}}g(x)\phi(x) dx = -\int_{\mathbb{R}} f(x)\frac{\partial \phi(x)}{\partial x} dx$$ for every function $\phi$ on $\mathbb{R}$ smooth with compact support. I believe this definition gives the same domain $\mathcal{D}(D)$, although I haven't checked. But both these alternative views of $\mathcal{D}(D)$ are just as opaque as the one coming from Stone's theorem. I'm looking for a characterization of $\mathcal{D}(D)$ in terms of pointwise differentiation of the functions in it.

Thanks.

Answer 1

The weak derivative formulation that you give in your post leads to what you want. It's a question of using the right test functions in order to conclude what you want. For example, suppose that, for some $f,g\in L^2(\mathbb{R})$, the following holds for all compactly supported $\varphi\in C^{\infty}(\mathbb{R})$: $$ \int_{\mathbb{R}} f \frac{d\varphi}{dx}dx = -\int_{\mathbb{R}}g\varphi dx $$ You can take limits of such functions $\varphi$ in order to replace $\varphi$ with $$ \varphi(x)=\int_{-\infty}^{x}\frac{1}{h}\chi_{[a-h,a]}(y)-\frac{1}{k}\chi_{[b,b+k]}(y)dy,\;\;\; a \le b. $$ The conclusion is that $$ \frac{1}{h}\int_{a-h}^{a}fdx-\frac{1}{k}\int_{b}^{b+k}fdx=-\int_{a-h}^{b+k} g\varphi dx. $$ Therefore, for almost every $a,b$, one has $$ f(b)-f(a)=-\int_a^b gdx. $$ So $f$ is equal almost everywhere to an absolutely continuous function with a derivatve equal a.e. to $-g$, which is in $L^2$. And, any such $f$ does satisfy the first equation in this post. So the domain is completely characterized as $H^1(\mathbb{R})$. From there you can get the other properties.