Hilbert Triple: a Densely and Continuously Embedded Hilbert Space and Its Form Operator

Let $\mathbb{R}$ be the ordered field of real numbers, with the notation of that item, and let $H$ be a real Hilbert space, with inner product, norm and distance written $\langle\cdot,\cdot\rangle_{H}$, $|\cdot|_{H}$ and $d_{H}$ in the ambient notation, and with zero vector $0_{H}$.

1. (Densely and continuously embedded Hilbert space)(i) Let $V$ be a linear subspace of $H$, which is a vector space over $\mathbb{R}$ under the restricted operations of $H$ by claim 1 of A Linear Subspace is a Vector Space and Inherits an Inner Product, equipped with an inner product $\langle\cdot,\cdot\rangle_{V}$ making $V$ a real Hilbert space with norm $|\cdot|_{V}$ and distance $d_{V}$, and suppose that the following two conditions hold.

(a) (Continuity of the embedding)(ii) $|x|_{H}\le|x|_{V}$ for every $x\in V$. (The constant is normalised to $1$: if one only has $|x|_{H}\le c\,|x|_{V}$ for some real $c>0$, replacing $\langle\cdot,\cdot\rangle_{V}$ by $c^{2}\langle\cdot,\cdot\rangle_{V}$ reduces to this case.)

(b) (Density)(iii) $V$ is dense in $H$.

The data $H$, $V$ with $\langle\cdot,\cdot\rangle_{V}$, and the map $A$ constructed in clause 2 are together called a Hilbert triple and written $(H,V,A)$. This differs from the classical Gelfand triple, whose third entry is the dual space of $V$: here the third entry is the operator $A$, and no dual space is used.

2. (The form operator)(iv) Let $D(A)$ be the set of all $x\in V$ for which there exists $z\in H$ with

$$\langle x,y\rangle_{V}=\langle z,y\rangle_{H}\qquad\text{for every }y\in V .$$

For $x\in D(A)$ such a $z$ is unique: if $z$ and $z'$ both satisfy the display, then $\langle z-z',y\rangle_{H}=0$ for every $y\in V$ by Elementary Identities in a Real Inner Product Space (i), so $z-z'$ lies in the orthogonal complement $V^{\perp}$, which is $\{0_{H}\}$ by Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space (vi) because $V$ is dense; hence $z=z'$. The map $A:D(A)\to H$ assigns to $x\in D(A)$ this unique $z$, written $Ax$, so that $\langle x,y\rangle_{V}=\langle Ax,y\rangle_{H}$ for all $x\in D(A)$ and $y\in V$. It is called the form operator of the triple.