What Is a Core for an Unbounded Operator?

A core is a smaller domain on which an unbounded operator is fully determined by limits. This idea is useful because an operator may be difficult to work with on its entire domain, while a simpler subdomain—such as smooth functions—can still contain all the information needed to recover it by closure.

The key is to measure density in the graph norm, which accounts for both a vector and the operator applied to it. Ordinary density in the underlying Hilbert space is not enough.

Definition of a core

Let $T:D(T)\subseteq H\to H$ be a linear operator on a Hilbert space. For $u\in D(T)$, define the graph norm by

$$\|u\|_T=\bigl(\|u\|^2+\|Tu\|^2\bigr)^{1/2}.$$

A linear subspace $D_0\subseteq D(T)$ is a core for $T$ if it is dense in $D(T)$ with respect to this norm. Equivalently, the graph

$$\{(u,Tu):u\in D_0\}$$

is dense in the graph of $T$ in the product space $H\times H$.

In concrete terms, for every $u\in D(T)$ there must be a sequence $u_n\in D_0$ such that both

$$u_n\to u \quad\text{and}\quad Tu_n\to Tu$$

in $H$. Approximating $u$ alone does not establish that $D_0$ is a core; the operator images must converge to the right limit too.

The closure criterion

If $T$ is closable, its closure $\overline T$ is the operator whose graph is the closure of the graph of $T$. The core condition can then be written as

$$\overline{T|_{D_0}}=\overline T.$$

Thus, starting with the restriction of $T$ to $D_0$ and closing that restriction gives the same operator as closing $T$ on its full domain. If $T$ is already closed, this becomes $\overline{T|_{D_0}}=T$.

This characterization also explains the word “core”: the restricted operator retains enough graph information to reconstruct the whole closed operator.

Worked example: multiplication by the variable

Consider the operator $M_x$ on $L^2(\mathbb R)$ defined by

$$(M_x f)(x)=x f(x), \qquad D(M_x)=\{f\in L^2(\mathbb R):xf\in L^2(\mathbb R)\}.$$

The space $C_c^\infty(\mathbb R)$ of smooth, compactly supported functions is a core for $M_x$. Its graph norm is

$$\|f\|_{M_x}^2=\|f\|_2^2+\|xf\|_2^2 =\int_{\mathbb R}(1+x^2)|f(x)|^2\,dx.$$

To see why smooth compactly supported functions are dense in this norm, take any $f\in D(M_x)$ and first truncate it: let $f_R=f\,\mathbf{1}_{[-R,R]}$. Then

$$\|f-f_R\|_{M_x}^2=\int_{|x|>R}(1+x^2)|f(x)|^2\,dx\longrightarrow 0$$

as $R\to\infty$, because the integrand is integrable by the definition of the domain. For fixed $R$, approximate $f_R$ in ordinary $L^2$ by smooth functions supported in a slightly larger bounded interval. On that interval, $1+x^2$ is bounded, so this approximation also converges in the graph norm. Combining the truncation and smooth approximation gives a sequence from $C_c^\infty(\mathbb R)$ converging to $f$ in graph norm.

Therefore $C_c^\infty(\mathbb R)$ is a core. Notice that this conclusion is stronger than saying it is dense in $L^2(\mathbb R)$: it also guarantees convergence after applying $M_x$.

How to check a proposed core

For a candidate subspace $D_0$, the most direct approach is to take an arbitrary $u\in D(T)$ and construct $u_n\in D_0$ with both $u_n\to u$ and $Tu_n\to Tu$. Alternatively, show that $D_0$ is dense in $D(T)$ for the graph norm. For differential operators, this often means proving approximation in a function-space norm that controls both the function and the relevant derivatives.

When $T$ is closed, it is also enough to show that every graph point $(u,Tu)$ is a limit of graph points $(u_n,Tu_n)$ with $u_n\in D_0$. This is exactly the graph-density condition, expressed without introducing a separate norm.

Common mistakes

• Checking only density in $H$. A subspace can be dense in the Hilbert space but fail to approximate operator images. That does not make it a core.
• Ignoring the domain. A candidate core must be a subspace of $D(T)$. It is not enough for its elements to be in the ambient space.
• Confusing a core with the whole domain. A core may be a proper subdomain; its defining property is that restriction to it leaves the closure unchanged.
• Mixing operator cores with form cores. A core for a quadratic form is defined using the form norm, which is generally different from the graph norm of an operator. State which notion is meant, especially in discussions of self-adjoint operators.