Hilbert completion, null vectors and bounded extension

This preceding programme proof supplies the complete Hilbert-completion contract used in the bounded comparison of weights. It applies to complex spaces of arbitrary dimension. It uses the Cauchy–Schwarz, Cauchy-sequence completion and dense-extension constructions proved in BK-01.

The complete construction

Let $E$ be a complex vector space with a positive semidefinite sesquilinear form $q$, linear in its first variable. Define $p(x)=q(x,x)^{1/2}$. The positive-form Cauchy–Schwarz inequality gives $|q(x,y)|\leq p(x)p(y).$ In particular, a vector with $p(x)=0$ pairs to zero with every vector. The triangle inequality for $p$ follows by expanding $q(x+y,x+y)$ and using this bound on the two mixed terms. Thus $N=\{x:p(x)=0\}$ is a linear subspace. Changing either argument by an element of $N$ does not change the pairing, so $\langle[x],[y]\rangle=q(x,y)$ is a well-defined positive definite inner product on $E/N$.

Apply BK-01's completion construction to $E/N$. Explicitly, take its Cauchy sequences modulo pairs whose difference tends to zero, perform the vector operations termwise, and pair two classes by the limit of the original pairings. Cauchy–Schwarz makes that limit existent and independent of representatives. The diagonal approximation proof in BK-01 proves completeness. Constant sequences give a dense linear isometry from $E/N$ into the resulting Hilbert space $H$. Composing it with $E\to E/N$ gives a map $i:E\to H$, with $\langle i(x),i(y)\rangle=q(x,y),\qquad \|i(x)\|=p(x),\qquad \ker i=N.$ If $q$ is positive definite, $N=0$, so $i$ itself is a dense linear isometry.

Now let $K$ be any complex Hilbert space and let $T:E\to K$ be linear with $\|Tx\|\leq C p(x)$. It vanishes on $N$, and hence defines a bounded map $T_0:i(E)\to K$ by $T_0i(x)=Tx$. For $h\in H$, choose $i(x_n)\to h$. The bound makes $Tx_n$ Cauchy, so completeness of $K$ supplies its limit. If $i(y_n)\to h$ as well, then $\|Tx_n-Ty_n\|\leq C p(x_n-y_n)\longrightarrow0.$ The limit is therefore independent of the chosen sequence. Define $\overline T h=\lim_n Tx_n$. Taking limits proves linearity and $\|\overline T h\|\leq C\|h\|$. Density proves uniqueness of the bounded extension. Its norm is exactly the seminorm-operator norm of $T$: the extension bound gives one inequality, and restriction to $i(E)$ gives the reverse inequality. If $E/N=0$, both operators are zero and both norms are zero.

Finally, if $i':E\to H'$ is another dense realization with the same pairings, the assignment $i(x)\mapsto i'(x)$ is well defined and isometric. The extension just proved gives an isometry $U:H\to H'$. Its image is closed, since $H$ is complete, and dense, since it contains $i'(E)$; thus it is onto. Density also proves uniqueness. This proves uniqueness of the completion up to its canonical unitary, without any dimension or separability restriction.