# 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 L$^2$ space is a non-negative self-adjoint operator.
Example. Let U be an open set in R$^n$. On L$^2$(U) we consider differential operators of the form
    $[T\phi ](x)=-\sum _{i,j}\partial _{x_{i}}\{a_{ij}(x)\partial _{x_{j}}\phi (x)\}\quad x\in U,\phi \in \operatorname {C} _{c}^{\infty }(U),$
where the functions a$_{ij}$ 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 × n matrix
    ${\begin{bmatrix}a_{11}(x)&a_{12}(x)&\cdots &a_{1n}(x)\\a_{21}(x)&a_{22}(x)&\cdots &a_{2n}(x)\\\vdots &\vdots &\ddots &\vdots \\a_{n1}(x)&a_{n2}(x)&\cdots &a_{nn}(x)\end{bmatrix}}$
is non-negative semi-definite, then T is a non-negative operator. This means (a) that the matrix is hermitian and
    $\sum _{i,j}a_{ij}(x)c_{i}{\overline {c_{j}}}\geq 0$
for every choice of complex numbers c$_1$, ..., c$_n$. 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 H$_1$ be the completion of dom T with respect to Q. H$_1$ 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 H$_1$ 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 H$_1$ → H. We regard H$_1$ 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 H$_1$ inherited from H. By the Riesz representation theorem applied to the linear functional φ$_ξ$ 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 T$_1$=A - I extends T.
T$_1$ 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 T$_{min}$ and T$_{max}$ 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 T$_{min}$ and T$_{max}$, i.e.
    $T_{\mathrm {min} }\leq S\leq T_{\mathrm {max} }.$