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-2026aA 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.
Each result cited is universally quantified over the data in its own statement. Closures are taken in the topology of the metric indicated, and denotes the closure in , that in . Sums of vectors are finite sums in , and is the span of a tuple, a linear subspace of 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 dense in ). By Hilbert Triples: Standing Notation and Background §separable the metric space is separable, so there is a countable subset with , that is, dense in .
We show that for every and every positive there is with . Let and be given. By Hilbert Triples: Standing Notation and Background §triple the subspace is dense in , so , and claim 3 of Characterization of the Closure in a Metric Space by Open Balls, applied in the metric space to the subset and the point , yields with . Since , the same claim of that theorem, applied in the metric space to the subset and the point , yields with ; and by Hilbert Triples: Standing Notation and Background §triple, because . Hence, since , The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle gives
Applying claim 3 of Characterization of the Closure in a Metric Space by Open Balls in the other direction, in to the subset , we conclude that every lies in ; since , this gives , so is dense in .
Step 2 (an enumeration of ). Taking and in Step 1 produces an element of , so is nonempty. By Countable Set there is therefore a sequence in whose set of terms is ; in particular for every , and every element of is a term of the sequence.
Step 3 (extending an orthonormal tuple inside ). We record the step used repeatedly below. Let , let be an orthonormal -tuple all of whose components lie in , let , and let be the map with of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, which applies because is a real inner product space. Let with . Since is a linear subspace of containing , the smallest such subspace satisfies . By Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace §projection we have and , the orthogonal complement of . Put . Then , because and and is a linear subspace; and , since would give . Hence by Elementary Identities in a Real Inner Product Space §vanishing, and we may set , an element of with by Elementary Identities in a Real Inner Product Space §homogeneity, the scalar being positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. For we have because and ; 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.Loading…
Prerequisites
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