Closed linear operator

In functional analysis, a branch of mathematics, a closed linear operator or often a closed operator is a partially defined linear operator whose graph is closed (see closed graph property). It is a basic example of an unbounded operator.

The closed graph theorem says a linear operator ${\displaystyle f:X\to Y}$ between Banach spaces is a closed operator if and only if it is a bounded operator and the domain of the operator is ${\displaystyle X}$. In practice, many operators are unbounded, but it is still desirable to make them have closed graph. Hence, they cannot be defined on all of ${\displaystyle X}$. To stay useful, they are instead defined on a proper but dense subspace, which still allows approximating any vector and keeps key tools (closures, adjoints, spectral theory) available.

Definition

It is common in functional analysis to consider partial functions, which are functions defined on a subset of some space ${\displaystyle X.}$ A partial function ${\displaystyle f}$ is declared with the notation ${\displaystyle f:D\subseteq X\to Y,}$ which indicates that ${\displaystyle f}$ has prototype ${\displaystyle f:D\to Y}$ (that is, its domain is ${\displaystyle D}$ and its codomain is ${\displaystyle Y}$)

Every partial function is, in particular, a function and so all terminology for functions can be applied to them. For instance, the graph of a partial function ${\displaystyle f}$ is the set ${\displaystyle \operatorname {graph} {\!(f)}=\{(x,f(x)):x\in \operatorname {dom} f\}.}$ However, one exception to this is the definition of "closed graph". A partial function ${\displaystyle f:D\subseteq X\to Y}$ is said to have a closed graph if ${\displaystyle \operatorname {graph} f}$ is a closed subset of ${\displaystyle X\times Y}$ in the product topology; importantly, note that the product space is ${\displaystyle X\times Y}$ and not ${\displaystyle D\times Y=\operatorname {dom} f\times Y}$ as it was defined above for ordinary functions. In contrast, when ${\displaystyle f:D\to Y}$ is considered as an ordinary function (rather than as the partial function ${\displaystyle f:D\subseteq X\to Y}$), then "having a closed graph" would instead mean that ${\displaystyle \operatorname {graph} f}$ is a closed subset of ${\displaystyle D\times Y.}$ If ${\displaystyle \operatorname {graph} f}$ is a closed subset of ${\displaystyle X\times Y}$ then it is also a closed subset of ${\displaystyle \operatorname {dom} (f)\times Y}$ although the converse is not guaranteed in general.

Definition: If X and Y are topological vector spaces (TVSs) then we call a linear map f : D(f) ⊆ X → Y a closed linear operator if its graph is closed in X × Y.

Closable maps and closures

A linear operator ${\displaystyle f:D\subseteq X\to Y}$ is closable in ${\displaystyle X\times Y}$ if there exists a vector subspace ${\displaystyle E\subseteq X}$ containing ${\displaystyle D}$ and a function (resp. multifunction) ${\displaystyle F:E\to Y}$ whose graph is equal to the closure of the set ${\displaystyle \operatorname {graph} f}$ in ${\displaystyle X\times Y.}$ Such an ${\displaystyle F}$ is called a closure of ${\displaystyle f}$ in ${\displaystyle X\times Y}$, is denoted by ${\displaystyle {\overline {f}},}$ and necessarily extends ${\displaystyle f.}$

If ${\displaystyle f:D\subseteq X\to Y}$ is a closable linear operator then a core or an essential domain of ${\displaystyle f}$ is a subset ${\displaystyle C\subseteq D}$ such that the closure in ${\displaystyle X\times Y}$ of the graph of the restriction ${\displaystyle f{\big \vert }_{C}:C\to Y}$ of ${\displaystyle f}$ to ${\displaystyle C}$ is equal to the closure of the graph of ${\displaystyle f}$ in ${\displaystyle X\times Y}$ (i.e. the closure of ${\displaystyle \operatorname {graph} f}$ in ${\displaystyle X\times Y}$ is equal to the closure of ${\displaystyle \operatorname {graph} f{\big \vert }_{C}}$ in ${\displaystyle X\times Y}$).

Examples

A bounded operator is a closed operator by the closed graph theorem. More interesting examples of closed operators are unbounded.

If ${\displaystyle (X,\tau )}$ is a Hausdorff TVS and ${\displaystyle \nu }$ is a vector topology on ${\displaystyle X}$ that is strictly finer than ${\displaystyle \tau ,}$ then the identity map ${\displaystyle \operatorname {Id}$ :(X,\tau )\to (X,\nu )} a closed discontinuous linear operator.^[1]

Consider the derivative operator ${\displaystyle f={\frac {d}{dx}}}$ where ${\displaystyle X=Y=C([a,b])}$ is the Banach space (with supremum norm) of all continuous functions on an interval ${\displaystyle [a,b].}$ If one takes its domain ${\displaystyle D(f)}$ to be ${\displaystyle C^{1}([a,b]),}$ then ${\displaystyle f}$ is a closed operator, which is not bounded.^[2] On the other hand, if ${\displaystyle D(f)}$ is the space ${\displaystyle C^{\infty }([a,b])}$ of smooth scalar valued functions then ${\displaystyle f}$ will no longer be closed, but it will be closable, with the closure being its extension defined on ${\displaystyle C^{1}([a,b]).}$ To show that ${\displaystyle f}$ is not closed when restricted to ${\displaystyle C^{\infty }([a,b])\to C^{\infty }([a,b])}$, take a function ${\displaystyle u}$ that is ${\displaystyle C^{1}}$ but not smooth, such as ${\displaystyle u(x)=x^{3/2}}$. Then mollify it to a sequence of smooth functions ${\displaystyle (u_{n})_{n\in \mathbb {N} }}$ such that ${\displaystyle \|u_{n}-u\|_{\infty }\to 0}$, then ${\displaystyle \|f(u_{n})-u'\|_{\infty }\to 0}$, but ${\displaystyle (u,u')}$ is not in the graph of ${\displaystyle f|_{C^{\infty }([a,b])}}$.

Basic properties

The following properties are easily checked for a linear operator ${\displaystyle f:\operatorname {D} (f)\subseteq X\to Y}$ between Banach spaces:

• If ${\displaystyle A}$ is closed then ${\displaystyle A-\lambda \mathrm {Id} _{\operatorname {D} (f)}}$ is closed where ${\displaystyle \lambda }$ is a scalar and ${\displaystyle \mathrm {Id} _{\operatorname {D} (f)}}$ is the identity function;
• If ${\displaystyle f}$ is closed, then its kernel (or nullspace) is a closed vector subspace of ${\displaystyle X}$;
• If ${\displaystyle f}$ is closed and injective then its inverse ${\displaystyle f^{-1}}$ is also closed;
• A linear operator ${\displaystyle f}$ admits a closure if and only if for every ${\displaystyle x\in X}$ and every pair of sequences ${\displaystyle x_{\bullet }=(x_{i})_{i=1}^{\infty }}$ and ${\displaystyle y_{\bullet }=(y_{i})_{i=1}^{\infty }}$ in ${\displaystyle \operatorname {D} (f)}$ both converging to ${\displaystyle x}$ in ${\displaystyle X}$, such that both ${\displaystyle f(x_{\bullet })=(f(x_{i}))_{i=1}^{\infty }}$ and ${\displaystyle f(y_{\bullet })=(f(y_{i}))_{i=1}^{\infty }}$ converge in ${\displaystyle Y}$, one has ${\displaystyle \lim _{i\to \infty }f(x_{i})=\lim _{i\to \infty }f(y_{i})}$.