Showing an operator is (or not) closed on $L^2(\mathbb{R})$

Question

I am linearizing nonlinear waves and get operators of the form below. Everything is considered in $L^2(\mathbb{R})$.

Consider the operator $L_1=\frac{d}{dx}$. The domain is $H^1(\mathbb{R})$ and it is closed.

Consider the operator $L_f:v \mapsto f(x) v(0)$ where $f\in L^2(\mathbb{R})$ and $v(0)$ is the value of $v$ at $x=0$. Correct me if I am wrong but I think the “natural” domain is $H^1$ and it is not even closable (not pre-closed). Just take $\{v_n\}\in H^1$ going to zero in $L^2$, with $v_n(0)=1$ for all $n$, then $L_f(v_n)\to f \neq 0$.

My question: is $L=L_1+L_f$ with domain $H^1$ closable (or closed)? The sequence of $v_n$ described above might not work because $L_1(v_n)$ does not converge.

Answer 1

$L_1+L_f$ is closable. It is perhaps most convenient to take Fourier transforms and consider the operator $$ (Tg)(\xi) = \xi g(\xi) + \left( \int g(\xi)\, d\xi \right) h(\xi) $$ on $L^2(\mathbb R)$. $T$ is now defined on the natural domain of the multiplication operator $D\equiv D(M_{\xi})\subseteq L^1(\mathbb R)$.

Suppose that $g_n\in D$, $g_n\to 0$, $Tg_n\to f$. If $\int g_n$ stays bounded, then we may assume that $\int g_n\to A$, by passing to a subsequence. So $Mg_n\to f-Ah=0$, since $M$ is closed, but then also $\|g_n\|_1 \lesssim \|g_n\|_2 + \|Mg_n\|_2 \to 0$, so $Tg_n\to 0$, as required.

If $A_n=|\int g_n|\to\infty$, then $k_n=g_n/A_n$ satisfies $k_n\to 0$, $Mk_n\to -h$, so $h=0$ and $T=M$ is closed.

Answer 2

The operator $L = L_1 + L_f$ is closed.

Proof. The operator $L_1$ generates a $C_0$-semigroup on $L^2(\mathbb{R})$ (namely, the left shift semigroup) and $L_f$ is a relatively compact perturbation of $L_1$ (since it has finite rank and is continuous with respect to the graph norm of $L_1$). On reflexive Banach spaces, a relatively compact perturbation of a $C_0$-semigroup generator $L_1$ has $L_1$-bound $0$, i.e. for every number $a > 0$ there exists a number $b \ge 0$ such that $$ \lVert L_f v \rVert \le a \lVert L_1 v \rVert + b \lVert v \rVert $$ for all $v \in H^1(\mathbb{R})$, where all norms denote the $L^2$-norm; see for instance [1, Lemma III.2.16(i)] for a proof.
Since $L_f$ has $L_1$-bound $0$, it follows that $L_1 + L_f$ is closed, see e.g. [1, Lemma III.2.4]. $\square$

Reference:

[1] Klaus-Jochen Engel and Rainer Nagel: One-Parameter Semigroups for Linear Evolutions Equations, Springer, 2000 (link to zbMATH)