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Section 4.5 Closed Operators

When a linear map \(L\) is continuous then, given any converging sequence \(x_n\to x\text{,}\) we have that \(Lx_n\to Lx\text{.}\) When \(L\) is not bounded, this cannot happen. Still, one might ask for a weaker condition, such as that while \(Lx_n\) might not always converge, when it does at least it does converge to the right point, namely \(Lx\text{.}\) This justifies the following definition.

Definition 4.5.1. Closed Operator.
Let \(X,Y\) be Banach spaces and let
\begin{equation*} L:D(L)\subset X\to Y \end{equation*}
be a linear map whose domain \(D(L)\) is dense in \(X\text{.}\)

We say that \(L\) is closed if, given any sequence \(x_n\) in \(D(L)\) converging to some \(x\in X\text{,}\) then either \(Lx_n\) does not converge or \(x\in D(L)\) and \(Lx_n\to Lx\text{.}\)

We denote by \(\Gamma_L\) the graph of \(L\text{,}\) namely
\begin{equation*} \Gamma_L=\{(x,Lx):x\in D(L)\}, \end{equation*}
and we denote by \(\|\cdot\|_L\) the norm
\begin{equation*} \|x\|_L= \|x\|_X+\|Lx\|_Y\text{ (Graph norm)}. \end{equation*}
(\(1\implies2\)) We need to prove that \(D(L)\) is complete under \(\|\cdot\|_L\text{.}\) Let \(x_n\) be a Cauchy sequence in \((D(L),\|\cdot\|_L)\text{.}\) Then \(x_n\) is Cauchy in \(X\) and \(Lx_n\) is Cauchy in \(Y\text{,}\) so there are \in X\(x\) and \(y\in Y\) such that \(x_n\ x\) and \(Lx_n\to y\text{.}\) Hence, since \(L\) is closed, this means that \(x\in D(L)\text{,}\) so \(D(L)\) is Banach.

(\(2\implies3\)) We need to prove that \(\Gamma_L\) is closed. So, assume that \((x_n,Lx_n)\in\Gamma_L\) converges to some \((x,y)\in X\times Y\text{.}\) This means that \(x_n\to x\) in \(X\) and \(Lx_n\to y\) in \(Y\text{.}\) Since \((D(L),\|\cdot\|_L)\) is complete, this means that \(x\in D(L)\) and \(y=Lx\text{,}\) so \((x,y)\in\Gamma_L\text{.}\) Hence, \(\Gamma_L\) is closed.

(\(3\implies1\)) Let \(L\) be a linear map whose graph is closed and assume that \(x_n\to x\) in \(X\) and \(y_n\to y\) in \(Y\text{.}\) Since the graph is closed, then \(x\in D(L)\) and \(y=Lx\text{,}\) namely \(L\) is closed.

\((3\iff4)\text{.}\) If \((x,y)\) is in the graph of \((\lambda\mathbb{1}-L)^{-1}\text{,}\) then \(Ly=\lambda y+x\text{,}\) namely \((y,x+\lambda y)\) belongs to \(\Gamma_L\) (and viceversa). Since the map
\begin{equation*} (x,y)\mapsto(y,x+\lambda y) \end{equation*}
is a continuous invertible map from \(X\times X\) to itself, the two graphs are either both closed or both not closed.
This is an immediate consequence of the closed graph theorem: a continuous map between two topological vector spaces is continuous if and only if its graph is closed.
The linear map
\begin{equation*} \frac{d}{dx}:C^2([0,1])\subset C^0([0,1])\to C^0([0,1]), \end{equation*}
is neither continuous nor closed. Indeed, if a sequence of \(C^2\) functions \(f_n\) converges uniformly to \(g\in C^0([0,1])\) and \(f'_n\) converges uniformly to some \(h\in C^0([0,1])\text{,}\) then \(g\) is differentiable and \(g'=h\) (exchange of limit and derivative for sequences converging uniformly), so \(g\) is \(C^1\text{,}\) but \(g\) is not necessarily \(C^2\text{.}\) In fact, it is easy to write explicit examples of sequences \(f_n\) such that \(g\) is not \(C^2\) (e.g. choose \(f_n\) so that \(f'_n\) converges to \(|x-0.5|\)). Hence, \((d/dx,C^2([0,1]))\) is not closed on \(C^0([0,1])\text{.}\)
The linear map
\begin{equation*} \frac{d}{dx}:C^1([0,1])\subset C^0([0,1])\to C^0([0,1]), \end{equation*}
while not continuous, is closed. Indeed, if a sequence of \(C^1\) functions \(f_n\) converges uniformly to \(g\in C^0([0,1])\) and \(f'_n\) converges uniformly to some \(h\in C^0([0,1])\text{,}\) then \(g\) is differentiable and \(g'=h\) (exchange of limit and derivative for sequences converging uniformly).
The linear map
\begin{equation*} \frac{d^2}{dx^2}:C^2([0,1])\subset C^0([0,1])\to C^0([0,1]), \end{equation*}
is not continuous (e.g. because it is the composition of an unbounded operator with itself) but it is closed. Indeed, let \(f_n\in C^2([0,1])\) such that
\begin{equation*} f_n\to f\in C^0([0,1]),\;f''_n\to g\in C^0([0,1]). \end{equation*}
Then also \(f'_n\) converges in \(C^0([0,1])\text{.}\) Indeed, given a \(C^\) function \(f\text{,}\) by Taylor expansion's theorem we have that, for every \(x\) and \(h\text{,}\) there is a \(\xi\in(-|h|,|h|)\) such that
\begin{equation*} f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(\xi). \end{equation*}
Hence, for every \(h>0\text{,}\)
\begin{equation*} \|f'\|_{C^0} \leq 2\frac{\|f\|_{C^0}}{h}+\frac{h}{2}\|f''\|_{C^0}. \end{equation*}
After minimizing for \(h>0\text{,}\) we finally get that
\begin{equation*} \|f'\|_{C^0} \leq 2\sqrt{\|f\|_{C^0}\|f''\|_{C^0}}. \end{equation*}
Thus, if \(f_n\) and \(f''_n\) are Cauchy, so is \(f'_n\text{.}\) Then, the same arguments used in the previous example show that \(f\in C^2([0,1]))\) and that \(f'_n\to f'\) and \(f''_n\to f''\text{.}\) Hence, the operator is closed.
Definition 4.5.7. Weak Derivative.
Given any \(f\in L^2(\Omega)\text{,}\) we say that \(g\) is the weak derivative of \(f\) if, for every \(\phi\in C_c^\infty(\Omega)\text{,}\)
\begin{equation*} \int_\Omega f(x)\phi'(x)d\mu(x) = -\int_\Omega g(x)\phi(x)d\mu(x). \end{equation*}
One can prove that the weak derivative of a \(L^2\) function, when exists, is unique (modulo sum by a function zero almost everywhere) and that the weak derivative of a \(C^1\) function is the usual derivative.
Let \(\Omega\) be a closed interval on the real line, possibly unbounded, and consider the first-derivative operator
\begin{equation*} \frac{d}{dx}:C^1(\Omega)\subset L^2(\Omega)\to L^2(\Omega). \end{equation*}
Let \(f_n\) be a sequence of \(C^1\) functions converging to \(g(x)=|x|\) (if \(0\) is not in the interior of \(\Omega\text{,}\) translate \(g\) so that the corner point lies in the interior of \(\Omega\)). Then \(f'_n\) converges to the weak derivative of \(g\text{,}\) namely the function \(\sgn(x)\text{,}\) so we have that \(f'_n\) does converge (in \(L^2\)) but \(f_n\) does not converge to an element in \(C^1(\Omega)\text{.}\) Hence, \(d/dx\) is not closed in this environment.
The linear map
\begin{equation*} \frac{d^2}{dx^2}:C^2([0,1])\subset L^2([0,1])\to L^2([0,1]), \end{equation*}
is not continuous (e.g. because it is the composition of an unbounded operator with itself) and not closed. The argument is the same highlighted in the previous example.

Subsection 4.5.1 Closed maps between Fréchet spaces.

Definition 4.5.10. Closed Operators between Fréchet spaces.
Let \(X,Y\) be Fréchet spaces and let
\begin{equation*} L:D(L)\subset X\to Y \end{equation*}
be a linear map whose domain \(D(L)\) is dense in \(X\text{.}\)

We say that \(L\) is closed if, given any sequence \(x_n\) in \(D(L)\) converging to some \(x\in X\text{,}\) then either \(Lx_n\) does not converge or \(x\in D(L)\) and \(Lx_n\to Lx\text{.}\)
The linear map
\begin{equation*} \frac{d}{dx}:C^1(\bR)\subset C^0(\bR)\to C^0(\bR), \end{equation*}
is not continuous for the same reason its cousin defined on \(C^0([0,1])\) is not. As its cousin, though, it is closed. Indeed, assume that a sequence of \(C^1\) functions \(f_n\) converges to \(g\in C^0(\bR)\) and \(f'_n\) converges to some \(h\in C^0(\bR)\) uniformly over compact sets. Then, within any interval \([-N,N]\text{,}\) \(g\) is \(C^1\) and \(g'=h\text{.}\) Hence, this is true over the whole \(\bR\) and so
\begin{equation*} \frac{d}{dx} \end{equation*}
is closed.
The same arguments used in Example 4.5.6 and Example 4.5.11 show that the linear map
\begin{equation*} \frac{d^2}{dx^2}:C^2(\bR)\subset C^0(\bR)\to C^0(\bR), \end{equation*}
is closed.
Notice that the definition and the proposition above works, more generally, in every \(F\)-space provided case (2) above is replaced by the fact that the seminorms are suitably replaced by the distance functions. In particular, the theory of closed operators applies also to the spaces \(L^p\) with \(p\in(0,1)\text{.}\)

Subsection 4.5.2 Closable linear maps.

Some of the examples above show that sometimes, even though an operator is not closed, it has a closed extension. This justifies the following.
Definition 4.5.14.
Given two linear maps \(L:D(L)\subset X\to Y\) and \(M:D(M)\subset X\to Y\text{,}\) we say that \(M\) extends \(L\) (or that \(M\) is an extension of \(L\)) if \(D(M)\supseteq D(L)\) and \(M|_{D(L)}=L\text{.}\) We say that \(M\) is a proper extension of \(L\) if \(D(M)\supset D(L)\text{.}\) We say that \(L\) is closable if it has a closed extension.
The first derivative operator
\begin{equation*} \frac{d}{dx}:C^\infty([0,1])\subset C^0([0,1])\to C^0([0,1]) \end{equation*}
is not closed but it is closable, since the operator
\begin{equation*} \frac{d}{dx}:C^1([0,1])\subset C^0([0,1])\to C^0([0,1]) \end{equation*}
is closed (see Example 4.5.5) and it is an extension of \((C^\infty([0,1]),d/dx)\text{.}\) Notice that \(C^1([0,1])\) is the closure of \(C^\infty([0,1])\) in the graph norm \(\|f\|_\Gamma=\|f\|_{C^0}+\|f'\|_{C^0}\text{.}\)
(\(1\implies2\)) If \(L\) is closable, there exists a closed operator \(L'\) such that \(L'\supset L\text{.}\) Hence. \(\Gamma_L\subset\Gamma_{L'}\) and \(\Gamma_{L'}\) is closed, so \(\overline{\Gamma_L}\subset\Gamma_{L'}\text{.}\) Now, \(\overline{\Gamma_L}\) is not be the graph of an operator if and only if there are \(x\in X\) and \(y,z\in Y\) with \(y\neq z\) such that \((x,y)\in\overline{\Gamma_L}\) and \((x,z)\in\overline{\Gamma_L}\text{.}\) Bet then also \((x,y)\in\Gamma_{L'}\) and \((x,z)\in\Gamma_{L'}\text{,}\) against the fact that \(\Gamma_{L'}\) is a graph.

(\(2\implies3\)) If the closure of the graph of \(L\) is a graph, then \((0,y)\) belongs to the graph of \(L\) if and only if \(y=0\text{.}\) If there were a \(x_n\to0\) such that \(Lx_n\to y\neq0\text{,}\) then \((0,y)\) would belong to the graph, which is impossible.

(\(3\implies1\)) Define \(\bar L\) as the map on
\begin{equation*} D(\bar L)=\{x\in X\,|\,\exists x_n\text{ in }D(L)\text{ s.t. }x_n\to y,\;Lx_n\text{ converges}\} \end{equation*}
defined by \(\bar L(x)=\lim Lx_n\text{.}\) By hypothesis, \(\bar L\) is well defined. Moreover, one can verify that \(\bar L\) is closed. Hence, \(L\) is closable. In fact, below we prove that \(\bar L\) is the smallest closed extension of \(L\text{.}\)
In other words, there is a closed extension \(\bar L\) of \(L\) such that, if \(M\) is a closed extension of \(L\text{,}\) then it is also a closed extension of \(\bar L\text{.}\)
The graph of a closed map is closed. Since the graph of \(\bar L\) is the closure of the graph of \(L\text{,}\) there is no map \(M\) whose graph is closed and strictly smaller than the graph of \(\bar L\text{.}\)
Definition 4.5.18.
Let \(L\) be closable. We denote by \(\bar L\) the smallest extension of \(L\text{.}\) We call such map the closure of \(L\text{.}\)
Remark 4.5.19. How to find the domain \(D(\overline{L})\) of the closure of a closable operator \(L:D(L)\subset X\to Y\text{?}\)
This is often a complicated matter. For instance, the domain of the closure of the Laplacian \((\Delta,C^2(\bR^2))\) on \(C^0(\bR^2)\) is a complicated set, not just a \(C^k\) or \(W^{k,p}\) set. Sometimes it is possible to use case (2) of Proposition 4.5.2 to find \(D(\overline{L})\) as follows. It is often the case that \(L\) is a differential operator and that \(C^\infty_c(\Omega)\subset D(L)\) and it is dense in \(X\text{.}\) Consider now the graph norm \(\|\cdot\|_L\text{.}\) If there is some known norm (say the \(C^k\) norm) that is equivalent to \(\|\cdot\|_L\text{,}\) then it means that \(D(\overline{L})\) equals the corresponding space (say \(C^k(\Omega)\)).

See Example 4.5.31 for an application of this method.
The first partial derivative operator
\begin{equation*} \frac{\partial}{\partial x}:C^1(\bR^2)\subset C^0(\bR^2)\to C^0(\bR^2) \end{equation*}
is not closed under the Fréchet compact-open topology on \(C^0(\bR^2)\text{.}\) Indeed, let \(f_n\in C^0(\bR^2)\) with \(f_n\to f\in C^0(\bR^2)\) and \(\partial_x f_n\to g\in C^0(\bR^2)\text{.}\) Then one can prove that, over each compact (and so over the whole \(\bR^2\)), \(f\) is \(C^1\) in the \(x\) variable. On the other side, no particular regularity can be claimed with respect to the \(y\) variable. Indeed, take \(h\in C^0(\bR)\) and consider the constant sequence
\begin{equation*} f_n(x,y) = h(y). \end{equation*}
Then \(f_n\to h\) and \(\partial_x f_n\to 0\) but certainly \(h\) is not \(C^1\) in both \(x\) and \(y\text{.}\)

About closability, suppose that \(f_n\to 0\) and \(\partial_x f_n\to g\text{.}\) This means that, for every \(y\text{,}\) \(f_n(x,y)\to 0\) and \(f'_n(x,y)\to g(x)\) uniformly over each interval \([-N,N]\text{.}\) Hence \(g(x,y)=0\) for every \(x,y\in\bR^2\) and so the operator is closable. The domain of its closure is the linear subspace of continuous functions that are \(C^1\) in the variable \(x\text{.}\)
The first partial derivative operator
\begin{equation*} X=2y\frac{\partial}{\partial x}+(1-y^2)\partial_y:C^1(\bR^2)\subset L^1_{loc}(\bR^2)\to L^1_{loc}(\bR^2) \end{equation*}
is closable. Indeed, suppose that \(f_n\to 0\) and \(X f_n\to g\) and denote by \(F^t\) the flow of \(X\text{,}\) seen as a vector field on the plane. Then, for any function \(g\in C^1(\bR^2)\text{,}\)
\begin{equation*} Xg(x,y) = \frac{d}{dt}g(F^t(x,y))\bigg|_{t=0}, \end{equation*}
so that
\begin{equation*} f_n(F^t(x,y)) = f_n(x,y) + \int_0^t (Xf_n)(F^s(x,y))ds. \end{equation*}
By taking the limit for \(n\to\infty\text{,}\) we get that
\begin{equation*} 0 = 0 + \int_0^t g(F^s(x,y))ds. \end{equation*}
for all \(t\geq0\) and all \((x,y)\in\bR^2\text{.}\) Hence, \(g=0\text{.}\)

Now it makes sense to ask whether or not \(X\) is closed on \(L^1_{loc}(\bR^2)\) or not. So, let \(f_n\in C^1(\bR^2)\) with \(f_n\to f\in L^1_{loc}(\bR^2)\) and \(X f_n\to g\in L^1_{loc}(\bR^2)\text{.}\) Then one can prove that, over each compact (and so over the whole \(\bR^2\)), \(f\) is \(W^{1,1}_{loc}\) in the direction \(X\) (thinking of \(X\) as a vector field) -- this is a direct consequence of the fact that \(\left|\int_0^1\ln s\,ds\right|<\infty\) and that \(Xf=2\text{.}\) On the other side, no particular regularity can be claimed with respect to any other direction. Indeed, take \(h\in C^0(\bR)\) and consider the sequence
\begin{equation*} f_n(x,y) = \log\left(|1+y|+\frac{1}{n}\right) - \log\left(|1-y|+\frac{1}{n}\right). \end{equation*}
and the function
\begin{equation*} f(x,y) = \log|1+y| - \log|1-y|. \end{equation*}
One can prove directly that:
  1. \(f_n\to f\) in \(L^1_{loc}(\bR^2)\text{,}\)
  2. \(Xf_n\to Xf=2\) in \(L^1_{loc}(\bR^2)\text{,}\)
  3. \(f\not\in W^{1,1}_{loc}(\bR^2)\text{.}\)
There is no simple way to characterize in more detail functions in \(D(\overline{X})\text{,}\) all we can say is that
\begin{equation*} D(\overline{X}) = \left\{f\in L^1_{loc}(\bR^2)\,:\,Xf\in L^1_{loc}(\bR^2)\right\}. \end{equation*}
Clearly \(W^{1,1}_{loc}(\bR^2)\subset \overline{X}\text{.}\) The example above shows that \(\overline{X}\) is striclty larger than \(W^{1,1}_{loc}(\bR^2)\text{.}\)
We saw in Example 4.5.8 that
\begin{equation*} \frac{d}{dx}:C^1([0,1])\subset L^2([0,1])\to L^2([0,1]) \end{equation*}
is not closed. We show here that it is closable and find its closure.

It is enough to show that, given any \(f_n\in C^1([0,1])\) such that \(f_n\to0\) and \(f'_n\to g\) in \(L^2([0,1])\text{,}\) then \(g=0\text{.}\) This follows from the fact that, for any \(\phi\in C^\infty_c((0,1))\text{,}\)
\begin{equation*} \langle f'_n,\phi \rangle_{L^2} = - \langle f_n,\phi' \rangle_{L^2}. \end{equation*}
By taking the limit, since the scalar product is continuous, we get that
\begin{equation*} \langle g,\phi \rangle_{L^2} = -\langle 0,\phi' \rangle_{L^2} = 0 \end{equation*}
for every \(\phi\in C^\infty_c((0,1))\text{.}\) If \(g\) where not zero on a set of positive measure, we could choose \(\phi\) so that the integral would be non-zero. Hence, it must be \(g=0\text{.}\)

Who is the closure of this operator? It is the operator \(W\) defined as follows. Its domain \(D(W)\) is the set of all \(f\in L^2([0,1])\) for which there is a \(f_n\in C^1([0,1])\) such that \(f_n\to f\) and \(f''_n\to g\) in \(L^2([0,1])\text{;}\) moreover, \(W(f)=g\text{.}\)

This is precisely the weak derivative operator and \(D(W)=H^1([0,1])\text{.}\) Indeed,
\begin{equation*} \langle f'_n,\phi \rangle_{L^2} = - \langle f_n,\phi' \rangle_{L^2}. \end{equation*}
so that, taking the limit,
\begin{equation*} \langle g,\phi \rangle_{L^2} = - \langle f,\phi' \rangle_{L^2}\text{ for all } \phi\in C^\infty_c((0,1)). \end{equation*}
This is precisely the definition of weak derivative of a function \(f\in L^2([0,1])\text{.}\)

Notice that \(H^1({[0,1]})\) is the closure of \(C_c^\infty((0,1))\) in the graph norm \(\|f\|_\Gamma=\|f\|_{L^2}+\|f'\|_{L^2}\text{.}\)
Let \(\{e_i\}\) be a basis for \(\ell^2\) and consider the linear functional \(L:\text{span}\{e_i\}\subset\ell^2\to\bR\) defined by
\begin{equation*} Le_i = i. \end{equation*}
This operator is unbounded. In particular, this means that \(L\) is not continuous at 0, so that there is some sequence \(x_n\) in \(\text{span}\{e_i\}\) such that \(x_n\to0\) but \(Lx_n\not\to 0\text{.}\) Hence, there is some subsequence \(x_{n_k}\) and \(\eps>0\) such that \(|Lx_{n_k}|>\eps\) for all \(k=1,2,...\text{.}\) Consider now the sequence \(y_k=\frac{x_{n_k}}{Lx_{n_k}}\text{.}\) Then \(y_k\to0\) but \(Ly_k=1\text{,}\) so \(Ly_k\) converges but not to 0. Hence, \(L\) is not closable.
Consider the linear map
\begin{equation*} \delta_0:C^0([0,1])\to C^0([0,1]) \end{equation*}
given by \(\delta_0(f)=f(0)\text{,}\) where by \(f(0)\) we mean the constant function equal to \(f(0)\) at every point. The map \(\delta_0\) is continuous:
\begin{equation*} \|\delta_0\|_{op} = \sup_{\|f\|_{C^0}=1}f(0)=1. \end{equation*}
Consider now the same map under the \(L^2\) topology:
\begin{equation*} \delta_0:C^0([0,1])\subset L^2([0,1])\to L^2([0,1]). \end{equation*}
Then \(\delta_0\) is neither continuous nor even just closed. To see this, consider the sequence of continuous functions
\begin{equation*} f_n(x)=\max\{1-nx,0\}. \end{equation*}
Then \(\|f_n\|_{L^2}\to0\text{,}\) so that \(f_n\to0\) in \(L^2\text{,}\) but \(\delta_0(f_n)=1\) for every \(n\text{,}\) so that \(\lim\delta_0(f_n)=1\neq\delta_0(0)\text{.}\) Hence, \(\delta_0\) is not closed.
Next proposition shows that, on the contrary, there is no maximal closed extension of a given closable map. A more sophisticated look at the Poisson equation. In Subsection 4.3.1 we discussed the inversion of the operator
\begin{equation*} D(A) = C^2_0([0,1]) = \{u\in C^2: u(0)=u(L)=0\}. \end{equation*}
The compcept of closability and closedness help us having a better understanding of what goes on there. Notice that the \(D(A^{-1}) = R(A) = C^0([0,L])\) but the integral operator representing \(A^{-1}\text{,}\) namely the integral operator
\begin{equation*} A^{-1}(g) = \int_0^Lk(x,s) g(s) ds,\qquad k(x,s) =\chi_{[0,x]}(s) (s-x) - x\frac{(s-L)}{L}, \end{equation*}
is actually well-defined over the whole \(L^2([0,L])\) -- in fact, in principle over the whole \(L^1([0,L])\text{!}\) This extension to \(L^2([0,L])\) of \(A^{-1}\) is a compact operator and so is precisely its closure. In particular, its graph is closed. Notice that it is invertable because one can prove directly that \(\int_0^Lk(x,s) g(s) ds=0\) if and only if \(g=0\text{.}\) Hence, its inverse \(\left(\overline{A^{-1}}\right)^{-1}\) is a closed operator and is clearly an extension of \(A\text{.}\) In fact, it is precisely the closure of \(A\text{:}\)
\begin{equation*} \overline{A} = \left(\overline{A^{-1}}\right)^{-1} \end{equation*}
or, equivalently,
\begin{equation*} \left(\overline{A}\right)^{-1} = \overline{A^{-1}}. \end{equation*}
Indeed, one can verify directly that \(R(\overline{A^{-1}})\subset H^2([0,L])\) and that \(k(0,s)=k(L,s)=0\text{,}\) so that \(R(\overline{A^{-1}})=\{f\in H^2([0,L]) : f(0)=f(L)=0\}\text{.}\)

This provides an alternate way of seen the domain of the closure of an operator: in this setting, such domain is the range of the extended inverse operator.

This situation happens frequently and we can formulate the following general statement:
(1) Since \(R(A)\) is dense and \(A^{-1}\) is bounded, \(A^{-1}\) extends uniquely to a bounded operator \(B:Y\to X\text{.}\) (2) The graph of \(A^{-1}\) is the image of the graph of \(A\) under the continuous invertable map \(\sigma(x,y)=(y,x)\text{.}\) Hence, \(A^{-1}\) is closable if and only if \(A\) is closable. Since \(A^{-1}\) extends to a continuous map, it is closable and so is \(A\text{.}\)

(3)\(B^{-1}\) is closed and extends \(A\text{,}\) namely \(\Gamma_{B}\supset\Gamma_{\overline{A}}\text{.}\) We claim that \(B^{-1}=\overline{A}\text{,}\) namely \(\Gamma_{B}=\Gamma_{\overline{A}}\text{.}\) Indeed, let \((f,g)\in\Gamma_{B^{-1}}\) and let \(g_n\in R(A)\) such that \(g_n\to g\text{.}\) Such sequence always exists since \(R(A)\) is dense in \(Y\text{.}\) Put \(f_n = A^{-1}g_n\text{.}\) Then
\begin{equation*} f_n = Bg_n\to Bg=f\text{ and }Af_n = g_n\to g. \end{equation*}
Hence, \((f,g)\in\overline{\Gamma_A}=\Gamma_{\overline{A}}\text{.}\) So, \(\Gamma_{\overline{A}}=\Gamma_{B}\text{.}\)
Consider the operator
\begin{equation*} \frac{d}{dx}:C^1([0,1])\subset L^2([0,1])\to L^2([0,1]). \end{equation*}
This is not invertable since constant functions are sent to 0. Hence we restrict the derivative and consider the operator \(A=d/dx\) with domain
\begin{equation*} D(A) = \{f\in C^1([0,1])\,:\,f(0)=0\}. \end{equation*}
This is now invertable. Its inverse is given by the antiderivative
\begin{equation*} A^{-1}g = \int_{[0,x]}g(s)ds. \end{equation*}
Clearly this operators is extends to a compact operator \(B\) defined over the whole \(L^2([0,1])\text{.}\) The range of \(B\) is the set of absolutely continuous functions \(f\) whose derivative is \(L^2\) and such that \(f(0)=0\text{,}\) which is precisely the domain of the closure of \(A\text{.}\)
Below we show a simple case of uncountably many closed extensions of a differential operator.

Subsection 4.5.3 An important example: the Lapacian

The Laplacian is arguably the most important differential operator in Functional Analysis and PDE theory. Here we consider this operator under several inequivalent topologies and study its closure.

In general, one would expect that a function \(f\) be at least "two derivatives more regular" than \(\Delta f\text{.}\) The examples below show that this is not necessarily the case. First, though, we present two important estimates that show that this is indeed the case for some class of spaces.

Example 4.5.30 shows that these estimates do not hold for \(\alpha=0\text{.}\)

Example 4.5.32 shows that these estimates do not hold for \(p=1\text{.}\)
Unlike its one-dimensional counterpart, the Laplacian
\begin{equation*} \Delta=\partial_{xx}+\partial_{yy}, \end{equation*}
seen as an operator
\begin{equation*} \Delta:C^2(\bR^2)\subset C^0(\bR^2)\to C^0(\bR^2), \end{equation*}
is not closed under the Fréchet compact-open topology on \(C^0(\bR^2)\text{.}\) Indeed, if it were closed, then \(C^2(\bR^2)\) would be complete under the graph norm. The example below shows that it is not.

The exaple is based on the function
\begin{equation*} f(x,y) = (x^2-y^2)\log|\log(x^2+y^2)|. \end{equation*}
In polar coordinates \(x=r\cos\theta,\,y=r\sin\theta\text{,}\) this is the function
\begin{equation*} F(r,\theta) = f(r\cos\theta,r\sin\theta) = 2\cos(2\theta)r^2\log|\log r|, \end{equation*}
whose Laplacian is
\begin{equation*} \Delta F(r,\theta) = F_{rr}+\frac{F_r}{r}+\frac{F_{\theta\theta}}{r^2}=\frac{4\log r-1}{\log^2 r}\cos(2\theta). \end{equation*}
Hence, \(\Delta f\in C^0(\bR^2)\) (in case you are wondering, \(f\) and \(F\) have the same resularity since they are expressions of the same function in different coordinates). Yet, \(f\not\in C^2(\bR^2)\text{,}\) since
\begin{equation*} F_{rr}(r,\theta) = \cos(2\theta)\left( 2\log|2\log r| + \frac{2}{\log r} + \frac{1}{\log r} - \frac{1}{\log^2 r}\right) \end{equation*}
is not continuous at 0.

Now, consider the sequence
\begin{equation*} f_n(x,y) = (x^2-y^2)\log|\log(x^2+y^2+\frac{1}{n})|\in C^2(\bR^2). \end{equation*}
Pointwise, outside of the unit sphere, this sequence is converging to \(f\text{.}\) In order to avoid the divergence at the unit sphere, we take any \(\phi\in C^\infty_c(\bR^2)\) such that \(\phi=1\) in some neighborhood \(U\) of the origin and \(\phi=0\) for, say, \(x^2+y^2\geq\frac{1}{2}\text{.}\) Then, we have that \(g_n=f_n\cdot\phi\in C^2(\bR^2)\) and that \(g_n|_U=f_n|_U\text{.}\) Notice that all these functions are \(C^\infty\) away from the origin and so it is enough to check what happens in some neighborhood of the origin. Hence, from now on, we will forget about \(\phi\) and refer to just \(f_n\) to make the discussion lighter and easier to follow.

One can prove directly that \(f_n\to f\) and \(\Delta f_n\to\Delta f\) uniformly on \(\overline{U}\text{,}\) so that \(f\in D(\overline{\Delta})\text{.}\) On the other side, as we pointed out above, \(f\not\in C^2(\overline{U})\text{.}\) Hence, \(\Delta\) with \(D(\Delta)=C^2(\bR^2)\) is not a closed operator.

Let us prove that it is closable. Assume that \(f_n\in C^2(\bR^2)\) is such that \(f_n\to0\) and \(\Delta f_n\to g\text{.}\) We must prove that \(g=0\text{.}\) Let \(\phi\in C^\infty_c(\bR^2)\text{.}\) Then, integrating by parts, one sees that
\begin{equation*} \int_{\bR^2} g\phi d\mu = \lim_{n\to\infty}\int_{\bR^2}(\Delta f_n)\phi d\mu = \lim_{n\to\infty}\int_{\bR^2} f_n\Delta\phi d\mu = 0. \end{equation*}
Since this is true for every \(\phi\in C^\infty_c(\bR^2)\text{,}\) we must have \(g=0\text{.}\)

Ultimately, all this shows that \(\Delta\) is closable and that \(D(\overline{\Delta})\) is strictly larger than \(C^2(\bR^2)\text{.}\) A detailed description of \(D(\overline{\Delta})\) is a non-trivial task, the author is not aware of any work in literature where such description is available.

A consequence of Calderon-Zygmund estimates is that, unlike the example below, on spaces of integrable functions the Laplacian behaves just as in dimension one. Below we use a simpler direct proof for the case \(p=2\text{.}\)

The Laplacian
\begin{equation*} \Delta=\partial_{xx}+\partial_{yy}, \end{equation*}
seen as an operator
\begin{equation*} \Delta:C^\infty_c(\bR^2)\subset L^2(\bR^2)\to L^2(\bR^2), \end{equation*}
is not closed in \(L^2(\bR^2)\) for similar reasons to Example 4.5.22.

Let us show first that it is closable. Let \(f_n\in C^\infty_c(\bR^2)\) be such that \(f_n\to0\) and \(\Delta f_n\to g\) in \(L^2(\bR^2)\text{.}\) We must prove that \(g=0\text{.}\) Let \(\phi\in C^\infty_c(\bR^2)\text{.}\) Then, integrating by parts, one sees that
\begin{equation*} \int_{\bR^2} g\phi d\mu = \lim_{n\to\infty}\int_{\bR^2}(\Delta f_n)\phi d\mu = \lim_{n\to\infty}\int_{\bR^2} f_n\Delta\phi d\mu = 0. \end{equation*}
Since this is true for every \(\phi\in C^\infty_c(\bR^2)\text{,}\) we must have \(g=0\text{.}\)

Hence, \(\Delta\) has a closure \(\overline{\Delta}\text{.}\) To find the domain of \(\overline\Delta\text{,}\) we use the ideas in Remark 4.5.19. We claim that \(D(\overline{\Delta})=H^2(\bR^2)\text{.}\) It is enough to prove that, for some \(C>0\text{,}\)
\begin{equation*} \|f\|_{H^2(\bR^2)}\leq C\left(\|f\|_{L^2(\bR^2)} + \|\Delta f\|_{L^2(\bR^2)}\right)\text{ for all }f\in C^\infty_c(\bR^2). \end{equation*}
This is a corollary of Theorem 4.5.29. Below we present a simpler direct proof. Recall that
\begin{equation*} \|f\|_{H^2(\bR^2)} = \|f\|_{L^2(\bR^2)} + \|f_x\|_{L^2(\bR^2)} + \|f_y\|_{L^2(\bR^2)} + \|f_{xx}\|_{L^2(\bR^2)} + \|f_{yy}\|_{L^2(\bR^2)} + 2\|f_{xy}\|_{L^2(\bR^2)}. \end{equation*}
Notice first that, by integrating by parts twice,
\begin{align*} \|f_{xy}\|^2_{L^2(\bR^2)} \amp = \phantom{-}\int_{\bR^2}f_{xy}(x,y)f_{xy}(x,y) dxdy =\\ \amp = -\int_{\bR^2}f_{x}(x,y)f_{xyy}(x,y) dxdy =\\ \amp = \phantom{-}\int_{\bR^2}f_{xx}(x,y)f_{yy}(x,y) dxdy, \end{align*}
so that
\begin{equation*} \|f_{xx}\|^2_{L^2(\bR^2)} + \|f_{yy}\|^2_{L^2(\bR^2)} + 2\|f_{xy}\|^2_{L^2(\bR^2)} = \int_{\bR^2}|\Delta f(x,y)|^2dxdy = \|\Delta f\|^2_{L^2(\bR^2)}. \end{equation*}
Moreover, again by integrating by parts,
\begin{align*} \int_{\bR^2} ( f f_{xx} + f f_{yy} ) dxdy \amp = - \int_{\bR^2} (f_x^2+f_y^2) dxdy =\\ \amp = - \|f_x\|^2_{L^2(\bR^2)} - \|f_y\|^2_{L^2(\bR^2)}, \end{align*}
so that
\begin{equation*} \|f_x\|^2_{L^2(\bR^2)} + \|f_y\|^2_{L^2(\bR^2)} \leq \|f\|^2_{L^2(\bR^2)}\|\Delta f\|^2_{L^2(\bR^2)}. \end{equation*}
Hence,
\begin{equation*} \|f\|^2_{H^2(\bR^2)} \leq \|f\|^2_{\Delta}. \end{equation*}
This proves that, as claimed, \(D(\overline{\Delta})=H^2(\bR^2)\text{.}\)
The Laplacian
\begin{equation*} \Delta=\partial_{xx}+\partial_{yy}, \end{equation*}
seen as an operator
\begin{equation*} \Delta:C^2(\bR^2)\subset L^1_{loc}(\bR^2)\to L^1_{loc}(\bR^2), \end{equation*}
is not closed in \(L^1_{loc}(\bR^2)\) for similar reasons to Example 4.5.22.

Assume that \(f_n\in C^2(\bR^2)\) is such that \(f_n\to0\) and \(\Delta f_n\to g\) in \(L^1_{loc}(\bR^2)\text{.}\) We must prove that \(\Delta f_n\to0\text{.}\) Let \(\phi\in C^\infty_c(\bR^2)\text{.}\) Then, by Theorem 3.5.5,
\begin{equation*} \left|\int_{\bR^2}(\Delta f_n-g)\phi d\mu \right|\leq\|\Delta f_n-g\|_{L^1}\cdot\|\phi\|_{L^\infty}\to0 \end{equation*}
and
\begin{equation*} \left|\int_{\bR^2}f_n\Delta\phi d\mu \right|\leq\|f_n\|_{L^1}\cdot\|\Delta\phi\|_{L^\infty}\to0. \end{equation*}
Hence, since
\begin{equation*} \int_{\bR^2}(\Delta f_n)\phi d\mu = \int_{\bR^2} f_n\Delta\phi d\mu, \end{equation*}
we have that
\begin{equation*} \int_{\bR^2}g\phi d\mu = 0\text{ for all }\phi\in C^\infty_c(\bR^2). \end{equation*}
This is possible if and only if \(g=0\) in \(L^1_{loc}\text{.}\)

About the domain of the closure, clearly
\begin{equation*} W^{1,2}_{loc}(\bR^2)\subset D(\overline{\Delta}). \end{equation*}
As in case of Example 4.5.30, we show here that that the inclusion is strict.

Consider the function
\begin{equation*} f(x,y) = \log|\log(x^2+y^2)| = 2\log|\log r|. \end{equation*}
A direct calculations shows that \(f\in W^{1,1}_{loc}(\bR^2)\) and that \(f\not\in W^{2,1}_{loc}(\bR^2)\text{.}\) On the other side,
\begin{equation*} \Delta(2\log|\log r|)=-\frac{2}{r^2\log^2 r}, \end{equation*}
so \(\Delta f\in L^1_{loc}(\bR^2)\text{.}\)

Now, consider the sequence
\begin{equation*} f_n(x,y) = \log|\log(x^2+y^2+\frac{1}{n})| \end{equation*}
and let \(\phi\in C^\infty_c(\bR^2)\) be such that \(\phi=1\) in some neighborhood of the origin and \(\phi=0\) for \(x^2+y^2\geq1/4\text{.}\) Then, for each \(n\text{,}\) \(g_n=f_n\cdot\phi\in C^2(\bR^2)\text{.}\) One can prove that \(g_n\to g\) and \(\Delta g_n\to\Delta g\) in \(L^1_{loc}(\bR^2)\text{.}\) Since \(f\not\in W^{2,1}_{loc}(\bR^2)\text{,}\) this shows that the domain of \(\overline{\Delta}\) is strictly larger than \(W^{2,1}_{loc}(\bR^2)\text{.}\)

A similar counterexample can be found in any dimension \(n\geq2\) by selecting as \(f\) a suitable antiderivative of \(r^{1-n}/\log r\text{.}\)

Subsection 4.5.4 An interesting example: uncountably many closed extensions of a differential operator

Let \(p\geq 1\) and denote by \(A\) the first-derivative operator
\begin{equation*} \frac{d}{dx}:C_c^\infty((0,1))\subset L^p([0,1])\to L^p([0,1]). \end{equation*}
Arguments similar to those used in the examples of Section 4.5 show that \(A\) is not closed but it is closable. The domain of \(\overline{A}\) is the closure of \(C_c^\infty((0,1))\) under the graph norm. To find this closure, we proceed in steps:

\(W^{1,p}([0,1]) = AC([0,1])\text{.}\)

Recall that \(AC([0,1])\) denotes the absolutely continuous functions on \([0,1]\text{.}\) These are functions \(f\in C^0([0,1])\) such that, for every \(\eps>0\text{,}\) there is a \(\delta>0\) such that, for every finite collection of disjoint subintervals \((x_i,y_i)\subset[0,1]\text{,}\) \(i=1,\dots,n\text{,}\)
\begin{equation*} \sum_{i=1}^n|x_i-y_i|<\delta\;\implies\;\sum_{i=1}^n|f(x_i)-f(y_i)|<\eps. \end{equation*}
A standard analysis result is that \(AC([0,1])\) is the set of all \(f\in C^0([0,1])\) for which there is a \(g\in L^1([0,1])\) such that
\begin{equation*} f(x) = f(0) + \int_{[0,x]} g(s)\,ds. \end{equation*}
Now, let \(f\in W^{1,p}([0,1])\text{.}\) Then \(f\) has a weak derivative \(f'\in L^p([0,1])\subset L^1([0,1])\text{,}\) so that \(h(x)=\int_{[0,x]} f'(s)\,ds\) is absolutely continuous. We want to prove that \(h = f\) outside of a set of measure 0. Let \(\phi\in C^\infty_c((0,1))\text{.}\) Then
\begin{equation*} \int_{[0,1]} (f(s)-h(s))\phi'(s)\,ds = \int_{[0,1]} h'(s)\phi(s)\,ds - \int_{[0,1]} f'(s)\phi(s)\,ds = 0, \end{equation*}
namely \(f\) coincides almost everywhere with the absolutely continuous function \(h\text{.}\)

\(\overline{C^\infty_c((0,1))}^{W^{1,p}} = \{ f\in AC([0,1])\,:\,f(0)=f(1)=0\}.\)

By what we wrote above, we know already that \(\overline{C^\infty_c((0,1))}^{W^{1,p}}\subset AC([0,1])\text{.}\) Moreover, since
\begin{equation*} |f(0)| \leq |f(x)| + \int_{[0,x]} |f'(s)|\,ds\text{ for all }x\in[0,1], \end{equation*}
we see that there are some \(C,C'>0\) such that
\begin{equation*} |f(0)| \leq \int_{[0,1]} |f(x)|dx + \int_{[0,1]}\int_{[0,x]} |f'(s)|\,dsdx \leq \|f\|_{L^1} + C \|f'|\_{L^p} \leq C'\|f\|_{W^{1,p}} \end{equation*}
and similarly for \(|f(1)|\text{.}\) Now, let \(\phi_n\in C^\infty_c((0,1))\) be such that \(\phi_n\to f\) in \(W^{1,p}([0,1])\text{.}\) Then \(\phi_n(0)=\phi_n(1)=0\) for all \(n\) and so
\begin{equation*} |f(0)| = \lim|f(0)-\phi_n(0)|\leq\|\phi_n-f\|_{W^{1,p}}=0 \end{equation*}
and simiarly for \(|f(1)|\text{.}\) Finally, via mollifiers one can prove that all functions in \(\{ f\in AC([0,1])\,:\,f(0)=f(1)=0\}\) can be approximated by smooth compactly supported functions, so that the two sets are indeed identical. We denote this set by \(W^{1,p}_0([0,1])\text{.}\)

Hence, so far we saw that \(D(\overline{A})=W^{1,p}_0([0,1])\text{.}\) Denote now by \(W\) the weak derivative operator on the whole \(W^{1,p}([0,1])\text{.}\) From arguments that we already used in examples above, we know that \(W\) is a closed operator, and clearly extends \(\overline{A}\) and, consequently, extends \(A\text{.}\) We present below an uncountable family of closed extensions of \(\overline{A}\) (and so of \(A\)) that are "in the middle" between \(\overline{A}\) and \(W\text{.}\)

A galore of closed extensions. Let \(\lambda\in\bR\) and set
\begin{equation*} D_\lambda = \{f\in W^{1,p}([0,1])\,|\,f(1)=\lambda f(0)\}. \end{equation*}
Notice that, for every \(\lambda\in\bR\text{,}\) \(D_\lambda\) is a closed linear subspace of \(W^{1,p}([0,1])\) and
\begin{equation*} W^{1,p}_0([0,1])\subset D_\lambda\subset W^{1,p}([0,1]). \end{equation*}
Moreover, \(D_\lambda\cap D_\lambda'=W^{1,p}_0([0,1])\) for \(\lambda\neq\lambda'\text{.}\)

Hence, for each \(D_\lambda\) we get a distinct extension \(A_\lambda\) of \(A\) by setting \(A_\lambda=W|_{D_\lamdba}\text{.}\) We claim that all \(A_\lambda\) are closed. Indeed, take a sequence \(f_n\in D_\lambda\) such that \(f_n\to f\) in \(L^p([0,1])\) and \(f'_n\to g\) in \(L^p([0,1])\text{.}\) Then \(f\in W^{1,p}([0,1])\) and \(f'=g\text{.}\) Moreover, since \(\int_0^1 f_n'(t)dt = f_n(1)-f_n(0)=(\mu-1)f_n(0)\) for all \(n\text{,}\) then \(\int_0^1 f'(t)dt = f(1)-f(0)=(\mu-1)f(0)\) and so \(f(1)=\mu f(0)\text{,}\) namely \(f\in D_\lambda\text{.}\) Hence, for each \(\lambda\text{,}\) the weak derivative acting on \(D_\lambda\) is a closed proper extension of \(A\text{.}\)