Friedrichs extension

In functional analysis, the Friedrichs extension is a canonical self-adjoint extension of a non-negative densely defined symmetric operator. It is named after the mathematician Kurt Friedrichs. This extension is particularly useful in situations where an operator may fail to be essentially self-adjoint or whose essential self-adjointness is difficult to show.

An operator T is non-negative if


 * $$ \langle \xi \mid T \xi \rangle \geq 0 \quad \xi \in \operatorname{dom}\ T $$

Examples
Example. Multiplication by a non-negative function on an L2 space is a non-negative self-adjoint operator.

Example. Let U be an open set in Rn. On L2(U) we consider differential operators of the form


 * $$ [T \phi](x) = -\sum_{i,j} \partial_{x_i} \{a_{i j}(x) \partial_{x_j} \phi(x)\} \quad x \in U, \phi \in \operatorname{C}_c^\infty(U), $$

where the functions ai j are infinitely differentiable real-valued functions on U. We consider T acting on the dense subspace of infinitely differentiable complex-valued functions of compact support, in symbols


 * $$ \operatorname{C}_c^\infty(U) \subseteq L^2(U). $$

If for each x ∈ U the n &times; n matrix


 * $$ \begin{bmatrix} a_{1 1}(x) & a_{1 2}(x) & \cdots & a_{1 n}(x) \\ a_{2 1}(x) & a_{2 2} (x) & \cdots & a_{2 n}(x) \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1}(x) & a_{n 2}(x) & \cdots & a_{n n}(x) \end{bmatrix} $$

is non-negative semi-definite, then T is a non-negative operator. This means (a) that the matrix is hermitian and


 * $$ \sum_{i, j} a_{i j }(x) c_i \overline{c_j} \geq 0 $$

for every choice of complex numbers c1, ..., cn. This is proved using integration by parts.

These operators are elliptic although in general elliptic operators may not be non-negative. They are however bounded from below.

Definition of Friedrichs extension
The definition of the Friedrichs extension is based on the theory of closed positive forms on Hilbert spaces. If T is non-negative, then


 * $$ \operatorname{Q}(\xi, \eta) = \langle \xi \mid T \eta \rangle + \langle \xi \mid \eta \rangle $$

is a sesquilinear form on dom T and


 * $$ \operatorname{Q}(\xi, \xi) = \langle \xi \mid T \xi\rangle + \langle \xi \mid \xi \rangle \geq \|\xi\|^2.$$

Thus Q defines an inner product on dom T. Let H1 be the completion of dom T with respect to Q. H1 is an abstractly defined space; for instance its elements can be represented as equivalence classes of Cauchy sequences of elements of dom T. It is not obvious that all elements in H1 can be identified with elements of H. However, the following can be proved:

The canonical inclusion


 * $$ \operatorname{dom} T \rightarrow H $$

extends to an injective continuous map H1 → H. We regard H1 as a subspace of H.

Define an operator A by


 * $$ \operatorname{dom}\ A = \{\xi \in H_1: \phi_\xi: \eta \mapsto \operatorname{Q}(\xi, \eta) \mbox{ is bounded linear.} \} $$

In the above formula, bounded is relative to the topology on H1 inherited from H. By the Riesz representation theorem applied to the linear functional φξ extended to H, there is a unique A ξ ∈ H such that


 * $$ \operatorname{Q}(\xi,\eta) = \langle A \xi \mid  \eta \rangle \quad \eta \in H_1 $$

Theorem. A is a non-negative self-adjoint operator such that T1=A - I extends T.

T1 is the Friedrichs extension of T.

Another way to obtain this extension is as follows. Let :$$ L:H_1\rightarrow H $$ be the bounded inclusion operator. The inclusion is a bounded injective with dense image. Hence $$ LL^*:H\rightarrow H $$ is a bounded injective operator with dense image, where $$ L^* $$ is the adjoint of  $$  L $$ as an operator between abstract Hilbert spaces. Therefore the operator $$ A:=(LL^*)^{-1} $$ is a non-negative self-adjoint operator whose domain is the image of $$  LL^* $$. Then $$ A-I $$ extends T.

Krein's theorem on non-negative self-adjoint extensions
M. G. Krein has given an elegant characterization of all non-negative self-adjoint extensions of a non-negative symmetric operator T.

If T, S are non-negative self-adjoint operators, write


 * $$ T \leq S $$

if, and only if,


 * $$ \operatorname{dom}(S^{1/2}) \subseteq \operatorname{dom}(T^{1/2}) $$
 * $$ \langle T^{1/2} \xi \mid T^{1/2} \xi \rangle \leq  \langle S^{1/2} \xi \mid S^{1/2} \xi \rangle \quad \forall \xi \in \operatorname{dom}(S^{1/2}) $$

Theorem. There are unique self-adjoint extensions Tmin and Tmax of any non-negative symmetric operator T such that


 * $$ T_{\mathrm{min}} \leq T_{\mathrm{max}}, $$

and every non-negative self-adjoint extension S of T is between Tmin and Tmax, i.e.


 * $$ T_{\mathrm{min}} \leq S \leq T_{\mathrm{max}}. $$