# Proving density of $H^2(\mathbb{R})$ in $L^2(\mathbb{R})$ .

I am trying to use the version of the Lumer-Phillips, which is stated as the following

Let $A$ be a linear operator defined on a dense linear subspace $D(A)$ of the reflexive Banach space $X$. Then $A$ generates a contraction semigroup if and only if $A$ is closed and both $A$ and its adjoint operator $A^∗$ are dissipative.

My operator $\mathcal{L}: D(\mathcal{L}) \subseteq X \longrightarrow X$ is defined by \begin{equation} D(\mathcal{L})=H^2(\mathbb{R}) \end{equation} \begin{equation} X = L^2 (\mathbb{R}) \end{equation} \begin{equation} \mathcal{L} = - \nu \partial_x -\frac{\mu_2}{2} x^2 + \gamma \partial_xx . \end{equation}

My question is how do I show that $D(\mathcal{L})$ is dense in $X$? (Is it dense?) I do not know where to start!

Furthermore, I was wondering how one went about defining the spaces if the operator was to act on a complex variable? (So it made up part of the complex Ginzburg Landau equation).

Kindest regards,

Catherine

• Do you know that $C^\infty_0(\mathbb R)$ is dense in $L^2(\mathbb R)$? – Umberto P. Feb 2 '18 at 15:10
• Just to get the meaning of notation right, is these the space of infinitely-differentiable functions with compact support? – Catherine Drysdale Feb 2 '18 at 15:27

Indeed, smooth (i.e. $C^\infty$ functions) with compact support are dense in $L^2(\mathbb R)$, and this is a subspace of $H^2(\mathbb R)$.
• @CatherineDrysdale You will need to consider complex-valued Sobolev spaces, but this is an easy generalization of the real case: viewing $\mathbb C \cong \mathbb R^2,$ any complex valued function $u$ can be viewed as a pair $(u_1,u_2)$ of real-valued functions. Under this identification one can check $H^k(\Omega,\mathbb C) \simeq H^k(\Omega) \oplus H^k(\Omega)$ and all the properties remain (including density). – ktoi Feb 2 '18 at 18:07