# Maximal domain of unbounded linear differential operator

Let's consider the following (unbounded) linear Operator. (So called Transport-Operator in some context.) $$\mathrm{T}: \mathcal{H} \supset \mathcal{D}(\mathrm{T}) \to \mathcal{H} , f \mapsto \mathrm{T}f:= v \cdot \partial_xf(x,v),$$ where $$\mathcal{H}$$ is a weighted Hilbert-space defined by $$\mathcal{H} := L^2(\mathbb{T}^1 \times \mathbb{R}, M(v)^{-1}\mathrm{d}x\mathrm{d}v),$$ $$M(v) := \frac{1}{\sqrt{2 \pi}}\mathrm{e}^{-\frac{1}{2}v^2},$$ and the dot product on $$\mathcal{H}$$ is defined by $$(f,g)_{\mathcal{H}} := \int_{\mathrm{T}^1 \times \mathrm{R}}\frac{f(x,v)g(x,v)}{M(v)} \, \, \mathrm{d}x\mathrm{d}v,$$ $$\mathbb{T}^1$$ denotes the one-dimensional Torus.

It is easy to verify that $$\mathrm{T}$$ is formally a skew-symmetric Operator w.r.t. $$(\cdot, \cdot)_{\mathcal{H}}$$.

$$\mathcal{D}(\mathrm{T})$$ denotes the maximal domain of $$\mathrm{T}$$ which should be unique, dense and $$\mathrm{T}$$ should be a closed operator on $$\mathcal{D}(\mathrm{T})$$. $$\mathcal{R}(\mathrm{T})$$ denotes the range of $$\mathrm{T}.$$ My question is now how can I define and determine $$\mathcal{D}(\mathrm{T})$$ and how can I easily describe/characterize $$\mathcal{D}(\mathrm{T})$$ explicitly? The same question for $$\mathcal{R}(\mathrm{T})$$. The derivative $$\partial_x$$ is to be understood in the weak-sense, so for all $$v \in \mathbb{R}$$ we will need $$f(x,v) \in H^1(\mathbb{T}^1)$$ (Sobolev-space) for example. My conjecture would be $$\mathcal{D}(\mathrm{T}) = \left \{f \in \mathcal{H} : \int_{\mathbb{T}^1 \times \mathbb{R}} \frac{v^2(\partial_xf(x,v))^2}{M(v)}\, \, \mathrm{d}x\mathrm{d}v < + \infty \right\} = \{f \in \mathcal{H} : \mathrm{T}f \in \mathcal{H} \},$$ so that $$(\mathrm{T}f, g)_{\mathcal{H}}$$ for arbitrary $$g \in \mathcal{H}$$ is at least well-defined. Would be grateful for any help and ideas!

EDIT: My attempt to show closedness of $$\mathrm{T}$$ on $$\mathcal{D}(\mathrm{T}).$$ Set $$\Omega := \mathbb{T}^1 \times \mathbb{R}$$ and $$\mathrm{d}\mu := M(v)^{-1}\mathrm{d}x\mathrm{d}v.$$ Let $$(f_n)_{n \in \mathbb{N}}$$ a sequence in $$\mathcal{D}(\mathrm{T})$$ with $$f_n \rightarrow f$$ in $$\mathcal{H}$$ and $$\mathrm{T}f_n \rightarrow g$$ in $$\mathcal{H}$$. We have to show: 1.) $$f \in \mathcal{D}(\mathrm{T})$$ and 2.) $$\mathrm{T}f = g.$$ We can proceed as follows. Strong convergence in $$\mathcal{H}=L^2(\Omega, \mathrm{d}\mu)$$ implies weak convergence so we have: $$\int_{\Omega} f_n(x,v) \phi(x,v) \, \mathrm{d}\mu \rightarrow \int_{\Omega} f(x,v) \phi(x,v) \, \mathrm{d}\mu$$ for all $$\phi \in C_c^{\infty}(\Omega),$$ and $$-\int_{\Omega} v \cdot f_n(x,v) \partial_x\phi(x,v) \, \mathrm{d}\mu = \int_{\Omega} v \cdot \partial_xf_n(x,v) \phi(x,v) \, \mathrm{d}\mu \rightarrow \int_{\Omega} g(x,v) \phi(x,v) \, \mathrm{d}\mu$$ for all $$\phi \in C_c^{\infty}(\Omega).$$ Due to $$-\int_{\Omega} v \cdot f_n(x,v) \partial_x\phi(x,v) \, \mathrm{d}\mu \rightarrow -\int_{\Omega} v \cdot f(x,v) \partial_x\phi(x,v) \, \mathrm{d}\mu$$ we have $$-\int_{\Omega} v \cdot f(x,v) \partial_x\phi(x,v) \, \mathrm{d}\mu = \int_{\Omega} g(x,v) \phi(x,v) \, \mathrm{d}\mu,$$ so we can conclude $$g = v \cdot \partial_xf = \mathrm{T}f.$$ Would that be correct?

• If $\mathcal D(\mathrm T)$ is closed and dense in $\mathcal H$, then it is equal to $\mathcal H$. You need to prove that $\mathcal D(\mathrm T)$ is dense and that $\mathrm T$ is a closed operator on $\mathcal D(\mathrm T)$. Commented Jan 27, 2023 at 15:09
• Of course you are absolutely right! The operator $\mathrm{T}$ must be closed! I expressed myself a bit unclear. So I have to show that the graph $\Gamma(\mathrm{T})$ of $\mathrm{T}$ is closed in $\mathcal{H}.$ The density follows because $C_c^{\infty}(\mathbb{T}^1 \times \mathbb{R})$ is dense in $\mathcal{H}$ and $C_c^{\infty}(\mathbb{T}^1 \times \mathbb{R}) \subseteq \mathcal{D}(\mathrm{T}).$ Commented Jan 27, 2023 at 15:17
• But is my conjecture correct with the choose of $\mathcal{D}(\mathrm{T})?$ Commented Jan 27, 2023 at 15:27

## 1 Answer

There is a folklore in functional analysis related to your question. If $$A$$ is a differential operator on some Banach space $$E$$, then some people tend to write $$D(A) = \{f \in E : Af \in E\}$$ and this is the maximal domain by design as you consider every element of the space that gets mapped into the space by the operator. So you can't enlarge this space without enlarging $$E$$ simultanously. So yes, your guess is right. I am not sure if there is a clean characterization of the range of the operator and usually the range is not really a concern in my experience.

• I see. Thank you. Is there good reference for this you could recommend? Because I'm wondering how I can show the closedness of the operator and the "maximality" of this domain as simply as possible? I suppose there is a general strategy. Commented Jan 28, 2023 at 11:06
• I suppose if the operator is densely defined and closed von $\mathcal{D}(\mathrm{T})$ it is already the unique and maximal domain in $\mathcal{H}.$ Commented Jan 28, 2023 at 11:08
• And isn't it possible to have $Af$ in the dual of another space for example? Like for the Laplacian: I can define: $\Delta : L^2 \supset H^2 \cap H^1_0 \to L^2$ or $\Delta : L^2 \supset H^1_0 \to H^{-1}.$ And it is $L^2 \subset H^{-1}.$ Or am I misunderstanding something here? Commented Jan 28, 2023 at 11:35