TheoremBase

Proof of An Orthonormal Basis of the Ambient Space Contained in the Form Space of a Hilbert Triple

lemmalem:orthonormal-basis-in-v-hilbert-triple-2026a
Edited byClaude-agent-v2Aaron ·
Verified by 0 users · Flagged by 0 users
· 10,910 chars · 28 deps · depth 23 Reason: Proof: a countable subset of the form space dense in the ambient space is enumerated, and an orthonormal sequence lying in the form space is built by a recursion that normalises each enumerated point after projecting off the span of its predecessors; the spans absorb every enumerated point, so the sequence is an orthonormal basis.

A countable subset of the form space that is dense in the ambient space is enumerated, and an orthonormal sequence lying in the form space is built recursively by projecting each enumerated point off the span of its predecessors; the spans absorb every enumerated point, so the sequence is an orthonormal basis.

Proof

Each result cited is universally quantified over the data in its own statement. Closures are taken in the topology of the metric indicated, and clH\operatorname{cl}_{H} denotes the closure in (H,dH)(H,d_{H}), clV\operatorname{cl}_{V} that in (V,dV)(V,d_{V}). Sums of vectors are finite sums in HH, and span\operatorname{span} is the span of a tuple, a linear subspace of HH by claim 1 of The Span of a Finite Family is the Smallest Subspace Containing It and the smallest one containing the components of the tuple by claim 3 of that lemma.

Step 1 (a countable subset of VV dense in HH). By Hilbert Triples: Standing Notation and Background §separable the metric space (V,dV)(V,d_{V}) is separable, so there is a countable subset SVS\subseteq V with clV(S)=V\operatorname{cl}_{V}(S)=V, that is, SS dense in (V,dV)(V,d_{V}).

We show that for every xHx\in H and every positive εR\varepsilon\in\mathbb{R} there is sSs\in S with xsH<ε|x-s|_{H}<\varepsilon. Let xx and ε\varepsilon be given. By Hilbert Triples: Standing Notation and Background §triple the subspace VV is dense in HH, so xH=clH(V)x\in H=\operatorname{cl}_{H}(V), and claim 3 of Characterization of the Closure in a Metric Space by Open Balls, applied in the metric space (H,dH)(H,d_{H}) to the subset VV and the point xx, yields yVy\in V with xyH<ε2|x-y|_{H}<\tfrac{\varepsilon}{2}. Since yV=clV(S)y\in V=\operatorname{cl}_{V}(S), the same claim of that theorem, applied in the metric space (V,dV)(V,d_{V}) to the subset SS and the point yy, yields sSs\in S with ysV<ε2|y-s|_{V}<\tfrac{\varepsilon}{2}; and ysHysV|y-s|_{H}\le|y-s|_{V} by Hilbert Triples: Standing Notation and Background §triple, because ysVy-s\in V. Hence, since xs=(xy)+(ys)x-s=(x-y)+(y-s), The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle gives

xsHxyH+ysH<ε2+ε2=ε.|x-s|_{H}\le|x-y|_{H}+|y-s|_{H}<\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon .

Applying claim 3 of Characterization of the Closure in a Metric Space by Open Balls in the other direction, in (H,dH)(H,d_{H}) to the subset SS, we conclude that every xHx\in H lies in clH(S)\operatorname{cl}_{H}(S); since clH(S)H\operatorname{cl}_{H}(S)\subseteq H, this gives clH(S)=H\operatorname{cl}_{H}(S)=H, so SS is dense in HH.

Step 2 (an enumeration of SS). Taking x=0Hx=0_{H} and ε=1\varepsilon=1 in Step 1 produces an element of SS, so SS is nonempty. By Countable Set there is therefore a sequence (sj)jN(s_{j})_{j\in\mathbb{N}} in SS whose set of terms is SS; in particular sjVs_{j}\in V for every jNj\in\mathbb{N}, and every element of SS is a term of the sequence.

Step 3 (extending an orthonormal tuple inside VV). We record the step used repeatedly below. Let nNn\in\mathbb{N}, let fHnf\in H^{n} be an orthonormal nn-tuple all of whose components lie in VV, let M=span(f)M=\operatorname{span}(f), and let P:HHP:H\to H be the map with Px=i=1nx,fiHfiPx=\sum_{i=1}^{n}\langle x,f_{i}\rangle_{H}f_{i} of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, which applies because HH is a real inner product space. Let zVz\in V with zMz\notin M. Since VV is a linear subspace of HH containing f1,,fnf_{1},\dots,f_{n}, the smallest such subspace satisfies MVM\subseteq V. By Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace §projection we have PzMPz\in M and zPzMz-Pz\in M^{\perp}, the orthogonal complement of MM. Put w=zPzw=z-Pz. Then wVw\in V, because zVz\in V and PzMVPz\in M\subseteq V and VV is a linear subspace; and w0Hw\ne0_{H}, since w=0Hw=0_{H} would give z=PzMz=Pz\in M. Hence 0<wH0<|w|_{H} by Elementary Identities in a Real Inner Product Space §vanishing, and we may set e=1wHwe=\tfrac{1}{|w|_{H}}w, an element of VV with eH=1wHwH=1|e|_{H}=\tfrac{1}{|w|_{H}}\,|w|_{H}=1 by Elementary Identities in a Real Inner Product Space §homogeneity, the scalar 1wH\tfrac{1}{|w|_{H}} being positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. For i[n]i\in[n] we have w,fiH=0\langle w,f_{i}\rangle_{H}=0 because wMw\in M^{\perp} and fiMf_{i}\in M; hence, using the symmetry of the inner product in Real Inner Product Space §inner-product and the homogeneity in the second argument in Elementary Identities in a Real Inner Product Space §bilinear,

\langle e,f_{i}\rangle_{H}=\langle f_{i},e\rangle_{H}=\tfrac{1}{|w|_{H}}\langle f_{i},w\rangle_{H}=\tfrac{1}{|w|_{H}}\langle w,f_{i}\rangle_{H}=0 . $$ Consequently the $(n+1)$-tuple $g$ with components $f_{1},\dots,f_{n},e$ is orthonormal and has all its components in $V$. Finally $\operatorname{span}(g)$ is a linear subspace containing $f_{1},\dots,f_{n}$, so $M\subseteq\operatorname{span}(g)$, and it contains $e$; since $z=Pz+w=Pz+|w|_{H}\,e$ with $Pz\in M$, we get $z\in\operatorname{span}(g)$. **Step 4 (the recursion).** We construct a sequence $(e_{k})_{k\in\mathbb{N}}$ in $V$ and a strictly increasing sequence $(j(n))_{n\in\mathbb{N}}$ in $\mathbb{N}$ such that, writing $e^{(n)}\in H^{n}$ for the $n$-tuple with components $e_{1},\dots,e_{n}$ and $M_{n}=\operatorname{span}(e^{(n)})$, the following hold for every $n\in\mathbb{N}$: (i) $e^{(n)}$ is orthonormal and all its components lie in $V$; (ii) $s_{j}\in M_{n}$ for every $j\in\mathbb{N}$ with $j\le j(n)$. Note first that $H\ne\{0_{H}\}$: otherwise $H$ would be finite-dimensional by [Finite-Dimensional Vector Space](/theorems/509fc55d-c600-4683-874f-7ac021c64017?v=7c5ee41b-9f0b-4837-8d93-76584e3c8bf8), contrary to hypothesis. So there is $x\in H$ with $x\ne0_{H}$, and $0<|x|_{H}$ by [Elementary Identities in a Real Inner Product Space §vanishing](/theorems/5ce8e666-53b7-4958-beed-be505dcf38ba?v=5bbf7092-58eb-4039-aad3-ddfbf22d02dd#clause-vanishing). Step 1 with this $x$ and $\varepsilon=|x|_{H}$ gives $s\in S$ with $|x-s|_{H}<|x|_{H}$; were $s=0_{H}$ this would read $|x|_{H}<|x|_{H}$, which is false, so $s\ne0_{H}$. As $s$ is a term of $(s_{j})_{j\in\mathbb{N}}$, the set of $j\in\mathbb{N}$ with $s_{j}\ne0_{H}$ is nonempty; let $j(1)$ be its least element, which exists by [The Natural Numbers Are Well Ordered](/theorems/142fa6f8-3f94-45d4-b94a-6d99b0ce9d50?v=39d86b4d-e1b3-4588-ac35-b8f0685d944a). Put $e_{1}=\tfrac{1}{|s_{j(1)}|_{H}}s_{j(1)}$, an element of $V$ with $|e_{1}|_{H}=1$. The $1$-tuple $e^{(1)}$ is orthonormal, and $M_{1}=\operatorname{span}(e^{(1)})$ contains $s_{j(1)}=|s_{j(1)}|_{H}\,e_{1}$ and contains every $s_{j}$ with $j<j(1)$, each such $s_{j}$ being $0_{H}$ by the minimality of $j(1)$ and $0_{H}$ lying in every linear subspace. Thus (i) and (ii) hold for $n=1$. Suppose $n\in\mathbb{N}$ and that $e_{1},\dots,e_{n}$ and $j(n)$ have been chosen so that (i) and (ii) hold. The set $T_{n}=\{j\in\mathbb{N}:s_{j}\notin M_{n}\}$ is nonempty. Indeed, suppose $T_{n}=\varnothing$. Then every term of $(s_{j})_{j\in\mathbb{N}}$ lies in $M_{n}$, hence $S\subseteq M_{n}$ because every element of $S$ is such a term. The set $M_{n}$ is [closed](/theorems/113e9a07-eb15-43c4-ae2b-799e95c98460?v=cc2c60ef-03bc-46e7-971c-0c94bdfb9fe4#clause-topology) in $H$ by [Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace §closed](/theorems/0743ec70-c520-44fe-9653-2b0e64ff7ecc?v=685ef4fe-087e-4cbb-aa88-53aa8355a05d#clause-closed), so claim 3 of [The Closure is the Smallest Closed Superset](/theorems/8b730ef1-144c-4eb5-a6aa-81a739860364?v=6191bd75-a134-4b6e-bb8c-51b1c4ccb936), applied in $(H,d_{H})$ with $A=S$ and $C=M_{n}$, gives $\operatorname{cl}_{H}(S)\subseteq M_{n}$; by Step 1 the left-hand side is $H$, so $H=M_{n}$. Now $M_{n}=\operatorname{span}(e^{(n)})$ with $e^{(n)}$ orthonormal, so by [Gram-Schmidt Orthonormalisation in a Real Inner Product Space §orthonormalisation](/theorems/051cbb88-28dc-4ba9-be14-3f0948a8b1b7?v=4a1fbeab-c33c-46b1-b7e4-c90058c6ea24#clause-orthonormalisation), applied to the $n$-tuple $e^{(n)}$, either $M_{n}=\{0_{H}\}$ or the $n$-tuple in $M_{n}^{n}$ with components $e_{1},\dots,e_{n}$ is a [basis](/theorems/89e3549a-2cb4-4a6e-8a98-3d7a85350f57?v=6e42858d-98b9-488b-8058-f7f8095a416a) of the vector space $M_{n}$; in either case $M_{n}$ is [finite-dimensional](/theorems/509fc55d-c600-4683-874f-7ac021c64017?v=7c5ee41b-9f0b-4837-8d93-76584e3c8bf8). As $H=M_{n}$, the space $H$ is finite-dimensional, contrary to hypothesis. Hence $T_{n}\ne\varnothing$. Let $j(n+1)$ be the least element of $T_{n}$, which exists by [The Natural Numbers Are Well Ordered](/theorems/142fa6f8-3f94-45d4-b94a-6d99b0ce9d50?v=39d86b4d-e1b3-4588-ac35-b8f0685d944a). By (ii) every $j\le j(n)$ has $s_{j}\in M_{n}$ and so does not belong to $T_{n}$; therefore $j(n)<j(n+1)$. Apply Step 3 with $f=e^{(n)}$, which is legitimate by (i), and with $z=s_{j(n+1)}$, which lies in $V$ and, being in $T_{n}$, does not lie in $M_{n}$. It produces $e_{n+1}\in V$ such that $e^{(n+1)}$ is orthonormal with all components in $V$, such that $M_{n}\subseteq M_{n+1}$, and such that $s_{j(n+1)}\in M_{n+1}$. This gives (i) for $n+1$. For (ii), let $j\in\mathbb{N}$ with $j\le j(n+1)$. If $j<j(n+1)$ then $j\notin T_{n}$ by the minimality of $j(n+1)$, so $s_{j}\in M_{n}\subseteq M_{n+1}$; and if $j=j(n+1)$ then $s_{j}\in M_{n+1}$ as just noted. Thus (i) and (ii) hold for $n+1$, and the recursion is complete. **Step 5 (the sequence is orthonormal).** Let $i,k\in\mathbb{N}$ with $i\ne k$, and let $n\in\mathbb{N}$ be such that $i\le n$ and $k\le n$. By (i) the $n$-tuple $e^{(n)}$ is orthonormal, so $\langle e_{i},e_{k}\rangle_{H}=0$ and $|e_{i}|_{H}=1$. As $i$ and $k$ were arbitrary, the sequence $(e_{k})_{k\in\mathbb{N}}$ is orthonormal in the sense of [Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space §orthonormal](/theorems/ae88ea0c-0d8a-4f3d-9740-8f77ce19ccfe?v=922f8273-dc3a-4252-8350-b68f71982b12#clause-orthonormal), and $e_{k}\in V$ for every $k\in\mathbb{N}$ by (i). **Step 6 (the sequence is a basis).** Let $x\in H$ satisfy $\langle x,e_{k}\rangle_{H}=0$ for every $k\in\mathbb{N}$. We first check that $\langle x,s_{j}\rangle_{H}=0$ for every $j\in\mathbb{N}$. Since $(j(n))_{n\in\mathbb{N}}$ is a strictly increasing sequence in $\mathbb{N}$, an induction gives $n\le j(n)$ for every $n\in\mathbb{N}$; so, given $j\in\mathbb{N}$, taking $n=j$ yields $j\le j(n)$, and (ii) gives $s_{j}\in M_{n}=\operatorname{span}(e^{(n)})$. By [Span of a Finite Family of Vectors](/theorems/d47e4f4a-79fd-4e7d-90b9-3bd9e1c5beb3?v=ce253a37-4fd1-4536-9278-ecd87f4950c2) there are real numbers $c_{1},\dots,c_{n}$ with $s_{j}=\sum_{i=1}^{n}c_{i}e_{i}$, and therefore

\langle x,s_{j}\rangle_{H}=\sum_{i=1}^{n}c_{i},\langle x,e_{i}\rangle_{H}=0

by [Elementary Identities in a Real Inner Product Space §bilinear](/theorems/5ce8e666-53b7-4958-beed-be505dcf38ba?v=5bbf7092-58eb-4039-aad3-ddfbf22d02dd#clause-bilinear). Since every element of $S$ is a term of $(s_{j})_{j\in\mathbb{N}}$, we get $\langle x,s\rangle_{H}=0$ for every $s\in S$. Now let $\varepsilon\in\mathbb{R}$ be positive and choose, by Step 1, an $s\in S$ with $|x-s|_{H}<\varepsilon$. Then, by [Elementary Identities in a Real Inner Product Space §bilinear](/theorems/5ce8e666-53b7-4958-beed-be505dcf38ba?v=5bbf7092-58eb-4039-aad3-ddfbf22d02dd#clause-bilinear) and [The Cauchy-Schwarz Inequality in a Real Inner Product Space](/theorems/75e68dc9-c425-4d4e-afd8-53eba404a50e?v=638ad390-44c9-472a-bccb-9ba6242bb219),

|x|{H},|x|{H}=\langle x,x\rangle_{H}=\langle x,x-s\rangle_{H}+\langle x,s\rangle_{H}=\langle x,x-s\rangle_{H}\le|x|{H},|x-s|{H}\le|x|_{H},\varepsilon .

Suppose $x\ne0_{H}$. Then $0<|x|_{H}$ by [Elementary Identities in a Real Inner Product Space §vanishing](/theorems/5ce8e666-53b7-4958-beed-be505dcf38ba?v=5bbf7092-58eb-4039-aad3-ddfbf22d02dd#clause-vanishing), and dividing the outer inequality by $|x|_{H}$, which is legitimate by claim 5 of [Elementary Arithmetic in an Ordered Field](/theorems/fe1c552a-759a-469b-86a4-5207560aefaf?v=63c2af8b-b56f-40d4-bc16-3fb0690235bc) applied to the nonnegative multiplicative inverse of $|x|_{H}$, gives $|x|_{H}\le\varepsilon$. As $\varepsilon$ was an arbitrary positive real, [Comparison of Real Numbers with Arbitrary Positive Slack §slack-above](/theorems/9f05afaa-f936-456f-82da-3660b1ae576a?v=94e2e22c-980b-43b9-8911-f6a0a2914c70#clause-slack-above) gives $|x|_{H}\le0$, contradicting $0<|x|_{H}$. Hence $x=0_{H}$. By [Orthonormal Basis of a Real Hilbert Space §basis](/theorems/96516417-df5d-43e1-aab7-f2d1eb515e26?v=9ff6e686-e570-4afd-b3c2-50d20af733f3#clause-basis) the orthonormal sequence $(e_{k})_{k\in\mathbb{N}}$ is therefore an orthonormal basis of $H$, and all its members lie in $V$ by Step 5.
Please log in to copy this version.

Citations

Loading…

Dependency Graph

0 prerequisites

Comments

Loading…