Friedrichs Extension

Every densely defined symmetric operator bounded from below has a distinguished self-adjoint extension obtained by completing its energy form. This is the Friedrichs extension. The construction preserves the optimal lower bound of the seed operator but does not claim that self-adjointness or nonnegative energy alone selects a unique extension.

Helpful background. Quadratic Forms supplies the closed-form representation theorem. Self-Adjoint Extensions treats general extension choices. Semibounded Hamiltonians explains spectral lower bounds.

The Friedrichs extension theorem

Let $S$ be densely defined and symmetric, with

$$\langle\psi,S\psi\rangle\geq c\|\psi\|^2, \qquad \psi\in D(S),$$

for some real $c$. Define its form on the original domain by

$$q_S(\phi,\psi)=\langle\phi,S\psi\rangle.$$

Then $q_S$ is closable. Its closure $\overline q_S$ is a densely defined closed semibounded form, and the associated self-adjoint operator $S_F$ extends $S$. This operator is the Friedrichs extension.

Its form domain $Q_F$ is the completion of $D(S)$ in the shifted form norm

$$\|\psi\|_{q_S}^2 =q_S[\psi]+(1-c)\|\psi\|^2,$$

identified with a subspace of $\mathcal H$. The construction is independent of which sufficiently positive shift is used, because the resulting form norms are equivalent.

The identification with a Hilbert-space subspace is a mathematical step, not an automatic consequence of completing a normed space. It is exactly what closability provides.

Why the seed form is closable

After shifting the operator, it suffices to treat $S\geq I$, so $q_S$ itself is an inner product whose norm dominates the Hilbert norm. Complete $D(S)$ in this norm to an abstract Hilbert space $\widetilde Q$. The original inclusion extends continuously to a map $j:\widetilde Q\to\mathcal H$.

To show $j$ is injective, suppose a $q_S$-Cauchy sequence $\psi_n\in D(S)$ represents $v\in\widetilde Q$ and converges to zero in $\mathcal H$. For every $\phi\in D(S)$, symmetry gives

$$q_S(\phi,\psi_n)=\langle S\phi,\psi_n\rangle\longrightarrow0.$$

On the other hand, convergence in $\widetilde Q$ makes this limit the inner product of $\phi$ with $v$. The original domain is dense in $\widetilde Q$, so $v$ is orthogonal to a dense subspace and must be zero. Consequently $q_S[\psi_n]\to0$, which is the closability criterion.

Thus $\widetilde Q$ embeds injectively in $\mathcal H$ and becomes $Q_F$. The closed-form representation theorem now constructs $S_F$. For $\psi\in D(S)$, continuity extends

$$\overline q_S(\phi,\psi)=\langle\phi,S\psi\rangle$$

from test vectors in $D(S)$ to all $\phi\in Q_F$. Hence $\psi\in D(S_F)$ and $S_F\psi=S\psi$, proving the extension claim.

What is unique and what is preserved?

The closed form $\overline q_S$ has a unique associated self-adjoint operator. More strongly, $S_F$ is the only self-adjoint extension $T$ of $S$ whose operator domain satisfies $D(T)\subseteq Q_F$.

To see this, take $\psi\in D(T)\subset Q_F$. For $\phi\in D(S)$,

$$\overline q_S(\phi,\psi) =\langle S\phi,\psi\rangle =\langle\phi,T\psi\rangle.$$

Continuity extends the identity to every $\phi\in Q_F$, so $T\subseteq S_F$. Two self-adjoint operators cannot be proper extensions of one another; therefore $T=S_F$.

The optimal lower bound is also preserved:

$$\inf\sigma(S_F) =\inf_{\substack{\psi\in D(S)\\\|\psi\|=1}} \langle\psi,S\psi\rangle.$$

Indeed, the seed domain is dense in $Q_F$ in form norm, so its Rayleigh quotients approximate the form-domain infimum. Other self-adjoint extensions may have different form domains and different spectral bottoms.

Form closure is not operator closure

On $L^2(0,L)$ consider $S=-d^2/dx^2$ with domain $C_c^\infty(0,L)$. Its shifted form norm is equivalent to the $H^1$ norm, so

$$Q_F=H_0^1(0,L),\qquad D(S_F)=H^2(0,L)\cap H_0^1(0,L).$$

The associated operator is the Dirichlet Laplacian. Its domain fixes the endpoint values while allowing nonzero endpoint derivatives.

By contrast, the graph norm of the minimal operator is equivalent to the $H^2$ norm on its seed domain. Its operator closure therefore has domain

$$D(\overline S)=H_0^2(0,L) =\{f\in H^2:f(0)=f(L)=f'(0)=f'(L)=0\}.$$

This is a proper subdomain of $D(S_F)$. For example, $\sin(\pi x/L)$ belongs to $D(S_F)$ but not to $D(\overline S)$. The seed is a form core for the Friedrichs operator without being an operator core for it.

The Neumann Laplacian is another nonnegative self-adjoint extension of the same seed. Its form domain is $H^1(0,L)$, not the completion of $C_c^\infty(0,L)$ in the seed form norm. Nonnegativity alone therefore does not select the Friedrichs boundary condition. Physical boundary information must be part of choosing the initial model.

Exercises

1. If $S$ is already self-adjoint and bounded below, show that its Friedrichs extension is $S$ itself.

Solution

The construction gives a self-adjoint extension $S_F$ with $S\subseteq S_F$. Maximality of a self-adjoint operator forces equality. Equivalently, closing the form initially defined on $D(S)$ recovers the spectral form on $D((S-c)^{1/2})$.

2. In the interval example, compare the optimal lower bounds of the seed, its Friedrichs extension, and the Neumann extension.

Solution

The seed and Dirichlet extension have infimum $\pi^2/L^2$, by the Poincaré inequality and form approximation to the first sine mode. The Neumann extension has bottom zero, attained by a constant. Both extensions are nonnegative. Of these two, the Friedrichs extension preserves the seed’s optimal lower bound; the Neumann extension does not.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, Chapter 10, publisher edition — semibounded forms and distinguished extensions.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Section 2.3 and Theorem 2.13, author’s text — Friedrichs extension and uniqueness within the completed form domain.