Proof of Bounded Linear and Conjugate-Linear Maps on a Dense Subspace of a Complex Hilbert Space Extend Uniquely

Each result cited below is universally quantified over the data in its own statement. Norms and inner products are those of $H$ or of $K$, as the vectors require, and convergence in $H$ and $K$ refers to the norm metrics, as fixed in Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §spaces. By claim 2 of The Induced Norm is a Norm, and Induces a Metric the induced norm is a norm in the sense of Norm on a Complex Vector Space, and its positivity, absolute homogeneity and triangle inequality are used without further mention; in particular $\lVert u-v\rVert=\lVert v-u\rVert$ and $\lVert u\rVert\le\lVert u-v\rVert+\lVert v\rVert$, where $u-v=u+(-1)v$.

Preliminaries. (P1) Let $(u_k)$ and $(v_k)$ converge to $u$ and $v$ in $H$ (or in $K$), and let $c\in\mathbb{C}$. Then $(u_k+v_k)$ converges to $u+v$ and $(cu_k)$ converges to $cu$. Indeed $\lVert(u_k+v_k)-(u+v)\rVert\le\lVert u_k-u\rVert+\lVert v_k-v\rVert$ and $\lVert cu_k-cu\rVert=|c|\,\lVert u_k-u\rVert$. Given $\varepsilon>0$ it therefore suffices to take $k$ so large that $\lVert u_k-u\rVert$ and $\lVert v_k-v\rVert$ are below both $\varepsilon/2$ and $\varepsilon/(|c|+1)$.

(P2) Since $D$ is dense, $\operatorname{cl}_H(D)=H$ by Dense Subset of a Topological Space. So by Sequential Characterization of the Closure in a Metric Space every $\xi\in H$ has an approximating sequence, that is, a sequence $(\xi_k)$ in $D$ converging to $\xi$. For $\xi\in D$ the constant sequence is one. Let $(\xi_k)$ and $(\eta_k)$ approximate $\xi$ and $\eta$, and let $c\in\mathbb{C}$. Then $(\xi_k+\eta_k)$ approximates $\xi+\eta$ and $(c\xi_k)$ approximates $c\xi$, by (P1) and because $D$ is a linear subspace.

(P3) Let $M\ge0$ be real, and let $Q:H\to K$ satisfy $Q(\xi-\eta)=Q\xi-Q\eta$ and $\lVert Q\xi\rVert\le M\lVert\xi\rVert$ for all $\xi,\eta\in H$. Then $\lVert Q\xi-Q\eta\rVert\le M\lVert\xi-\eta\rVert$. Given $\varepsilon>0$, the choice $\delta=\varepsilon/(M+1)$ therefore meets Continuous Map Between Metric Spaces, and $Q$ is continuous on $H$. In particular every $A\in\mathcal{L}(H,K)$ is continuous on $H$: take $M=\lVert A\rVert_{\mathrm{op}}$, which is a bound by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound.

A common construction for clauses 1 and 2. Let $\sigma:\mathbb{C}\to\mathbb{C}$ be either the identity or complex conjugation. Let $R:D\to K$ satisfy $R(\xi+\eta)=R\xi+R\eta$, $R(c\xi)=\sigma(c)R\xi$ and $\lVert R\xi\rVert\le C\lVert\xi\rVert$ for all $\xi,\eta\in D$ and $c\in\mathbb{C}$. In both cases $\sigma(-1)=-1$ and $|\sigma(c)|=|c|$, by claims 1 and 3 of Properties of Complex Conjugation and Modulus. Hence $R(\xi-\eta)=R\xi-R\eta$ for $\xi,\eta\in D$.

(A) Convergence. Let $(\xi_k)$ approximate $\xi\in H$. Given $\varepsilon>0$, choose $N$ with $\lVert\xi_k-\xi\rVert<\varepsilon/(2(C+1))$ for $k\ge N$. For $k,l\ge N$,

So $(R\xi_k)$ is a Cauchy sequence in $K$. It converges because $K$ is complete (Complex Hilbert Space).

(B) Independence of the sequence. Let $(\xi_k)$ and $(\xi'_k)$ both approximate $\xi$, with $R\xi_k\to y$ and $R\xi'_k\to y'$. Since $R\xi_k-R\xi'_k=R(\xi_k-\xi'_k)$, for every $k$

Given $\varepsilon>0$, each term on the right is below $\varepsilon/4$ for $k$ large. So $0\le\lVert y-y'\rVert<\varepsilon$ for every $\varepsilon>0$. Hence $\lVert y-y'\rVert=0$ by Comparison of Real Numbers with Arbitrary Positive Slack §vanishing, and $y=y'$.

Define $\widetilde R\xi$ to be this common limit. This gives the following property (F): $R\xi_k\to\widetilde R\xi$ for every approximating sequence $(\xi_k)$ of $\xi$. Moreover $\widetilde R\xi=R\xi$ for $\xi\in D$: use the constant sequence, and recall that limits are unique by Uniqueness of Limits in a Metric Space.

(C) Algebra. Let $(\xi_k)$ and $(\eta_k)$ approximate $\xi,\eta\in H$, and let $c\in\mathbb{C}$. By (P2) and (F), $R(\xi_k+\eta_k)\to\widetilde R(\xi+\eta)$ and $R(c\xi_k)\to\widetilde R(c\xi)$. By (P1) and (F), $R(\xi_k+\eta_k)=R\xi_k+R\eta_k\to\widetilde R\xi+\widetilde R\eta$ and $R(c\xi_k)=\sigma(c)R\xi_k\to\sigma(c)\widetilde R\xi$. By Uniqueness of Limits in a Metric Space, $\widetilde R(\xi+\eta)=\widetilde R\xi+\widetilde R\eta$ and $\widetilde R(c\xi)=\sigma(c)\widetilde R\xi$. In particular $\widetilde R(\xi-\eta)=\widetilde R\xi-\widetilde R\eta$.

(D) Bounds. Let $C'\ge0$ be real with $\lVert R\xi\rVert\le C'\lVert\xi\rVert$ for every $\xi\in D$. Let $\xi\in H$, with approximating sequence $(\xi_k)$. For every $k$,

Given $\varepsilon>0$, the first two terms on the right are each below $\varepsilon/2$ for $k$ large. So $\lVert\widetilde R\xi\rVert\le C'\lVert\xi\rVert+\varepsilon$ for every $\varepsilon>0$, and $\lVert\widetilde R\xi\rVert\le C'\lVert\xi\rVert$ by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above. With $C'=C$ this gives $\lVert\widetilde R\xi\rVert\le C\lVert\xi\rVert$ on $H$.

Suppose moreover that $\lVert R\xi\rVert=\lVert\xi\rVert$ on $D$. Then $C'=1$ gives $\lVert\widetilde R\xi\rVert\le\lVert\xi\rVert$. Conversely, for every $k$,

which gives $\lVert\xi\rVert\le\lVert\widetilde R\xi\rVert$ in the same way. So $\lVert\widetilde R\xi\rVert=\lVert\xi\rVert$.

(E) Continuity and uniqueness. By (C), (D) and (P3), $\widetilde R$ is continuous on $H$. Let $Q:H\to K$ be any map that is continuous on $H$ and satisfies $Q\xi=R\xi$ for $\xi\in D$. Let $\xi\in H$, with approximating sequence $(\xi_k)$. Then Continuity Between Metric Spaces is Equivalent to Sequential Continuity §sequential, applied with $A=H$, gives $R\xi_k=Q\xi_k\to Q\xi$, while $R\xi_k\to\widetilde R\xi$ by (F). So $Q\xi=\widetilde R\xi$ by Uniqueness of Limits in a Metric Space, that is, $Q=\widetilde R$.

Proof of clause 1 (Linear maps). Take $\sigma$ to be the identity and $R=T$; the hypotheses hold because $T$ is linear with bound $C$, and set $\widetilde T=\widetilde R$. By (C), $\widetilde T$ is linear. By (D), $C$ is a bound for it, so $\widetilde T\in\mathcal{L}(H,K)$ by Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §bounded, and $\lVert\widetilde T\rVert_{\mathrm{op}}\le C$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. By (F), $\widetilde T\xi=T\xi$ for $\xi\in D$. By (D), $\widetilde T$ is isometric when $T$ is. For uniqueness, let $A\in\mathcal{L}(H,K)$ agree with $T$ on $D$. Then $A$ is continuous on $H$ by (P3), so $A=\widetilde T$ by (E).

Proof of clause 2 (Conjugate-linear maps). Take $\sigma$ to be complex conjugation and $R=S$, and set $\widetilde S=\widetilde R$. By (E), $\widetilde S$ is continuous and is the only continuous map $H\to K$ that agrees with $S$ on $D$; it does agree with $S$ on $D$ by (F). Additivity and $\widetilde S(c\xi)=\overline{c}\,\widetilde S\xi$ hold by (C). The bound, and the isometry when $S$ is isometric, hold by (D).

Proof of clause 3 (Equality of operators). First let $\xi\in D$ and put $w=A\xi-B\xi$. For $\eta\in E$, conditions 2 and 3 of Complex Inner Product Space give $\langle\eta,w\rangle=\langle\eta,A\xi\rangle-\langle\eta,B\xi\rangle=0$. Since $E$ is dense in $K$, Dense Subset of a Topological Space and Sequential Characterization of the Closure in a Metric Space give a sequence $(\eta_k)$ in $E$ converging to $w$. Now $w=(w-\eta_k)+\eta_k$, so claim 1 of Elementary Properties of a Complex Inner Product and claim 1 of The Induced Norm is a Norm, and Induces a Metric give, for every $k$,

Given $\varepsilon>0$, the right-hand side is below $\varepsilon$ for $k$ large. Since $\langle w,w\rangle$ is real and nonnegative, it equals $|\langle w,w\rangle|$ by claim 8 of Properties of Complex Conjugation and Modulus. Hence $0\le\langle w,w\rangle<\varepsilon$ for every $\varepsilon>0$, so $\langle w,w\rangle=0$ by Comparison of Real Numbers with Arbitrary Positive Slack §vanishing, and $w=0$ by condition 4 of Complex Inner Product Space. Thus $A\xi=B\xi$ for every $\xi\in D$.

The restriction of $A$ to $D$ is a linear map $D\to K$ with bound $\lVert A\rVert_{\mathrm{op}}$ (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound). Both $A$ and $B$ are elements of $\mathcal{L}(H,K)$ agreeing with it on $D$, so $A=B$ by the uniqueness in clause 1. The same argument proves the final sentence of the clause, whose hypothesis is exactly the agreement $A\xi=B\xi$ on $D$ used here.