Unbounded & Closed Operators

Differentiation amplifies high frequencies, so no uniform output bound can hold under a norm that measures only function amplitude. Treating such an operator as “a linear map with an inconvenient constant” loses the key structure. An unbounded operator is specified by both its rule and its domain.

The domain belongs to the operator #

An operator from $X$ to $Y$ is a pair $(T,D(T))$ with a linear subspace $D(T)\subseteq X$ and a linear map $T:D(T)\to Y$. Two operators with the same formula but different domains are different: they may have different kernels, ranges, adjoints, closures, and spectra.

For example, differentiation on $L^2(0,1)$ may use periodic, Dirichlet, or other boundary conditions. Those conditions live in $D(T)$ and change the spectral problem.

The graph is

$$G(T)=\{(x,Tx):x\in D(T)\}\subseteq X\times Y.$$

$T$ is closed when $G(T)$ is closed: whenever $x_n\in D(T)$, $x_n\to x$ in $X$, and $Tx_n\to y$ in $Y$, then $x\in D(T)$ and $Tx=y$. Closedness says that simultaneous convergence of input and output cannot produce an inadmissible limiting pair.

Closed does not mean bounded #

Let $X=Y=C[0,1]$ with the sup norm and

$$D(T)=C^1[0,1],\qquad Tf=f'.$$

The operator is unbounded: for $f_n(x)=\sin(nx)/n$, $\|f_n\|_\infty\le1/n$ while $\|Tf_n\|_\infty=1$. It is nevertheless closed. If $f_n\to f$ and $f_n'\to g$ uniformly, then

$$f_n(x)-f_n(0)=\int_0^x f_n'(t)dt$$

passes to the limit, giving $f(x)-f(0)=\int_0^x g(t)dt$; hence $f\in C^1$ and $f'=g$.

There is no conflict with the closed graph theorem because $T$ is not defined on all of $C[0,1]$. An everywhere-defined closed linear map between Banach spaces is bounded.

Graph norm makes a closed operator complete #

Define

$$\|x\|_T=\|x\|_X+\|Tx\|_Y, \qquad x\in D(T).$$

If $X$ and $Y$ are Banach, then $T$ is closed iff $D(T)$ is Banach under the graph norm. The graph norm records both convergence of the object and of the quantity produced by the operator. Sobolev norms are systematic versions of this idea for differential operators.

Closable operators and closure #

$T$ is closable when the closure of $G(T)$ is itself the graph of an operator $\overline T$. Equivalently,

$$x_n\to0,quad Tx_n\to y\quad\Longrightarrow\quad y=0.$$

If a nonzero $y$ were possible, the closed graph would contain both $(0,0)$ and $(0,y)$ and could not define a function. The closure $\overline T$ is the smallest closed extension of $T$.

A densely defined operator on a Hilbert space is closable iff the domain of its adjoint is dense. Density is what lets inner products against $D(T)$ determine candidate adjoint vectors uniquely.

Resolvents require domain-aware bijectivity #

For a closed operator $T$ on a Banach space, $\lambda$ lies in the resolvent set when

$$T-\lambda I:D(T)\to X$$

is bijective and its inverse is bounded on all of $X$. The inverse maps into $D(T)$, usually with graph-norm control. Ignoring the domain can turn two different boundary-value problems into the same symbol and produce the wrong spectrum.

Diagnostic checklist #

• Write $D(T)$ beside the formula every time.
• State the ambient norm and the graph norm.
• Test closedness with simultaneous input/output convergence.
• Test closability with the zero-input criterion.
• Do not invoke closed graph unless the map is everywhere-defined between Banach spaces.
• Treat boundary conditions as domain data, not annotations.