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Proof of The Subspaces Spanned by an Orthonormal Sequence and Exhausting Sequences

lemmalem:orthonormal-basis-exhausting-2026b
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The spanned subspaces are identified with the ranges of the finite-dimensional projections, and the exhausting property is matched against the expansion of an arbitrary vector in one direction and against the nearest-point property of the projections in the other.

Proof

For each nNn\in\mathbb{N} the tuple e(n)e^{(n)} is orthonormal, since its components are components of the orthonormal sequence (ek)(e_{k}) and distinct indices in [n][n] are distinct in N\mathbb{N}. Write P(n)P^{(n)} for the projection onto Hn=span(e(n))H_{n}=\operatorname{span}(e^{(n)}) of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, so that P(n)x=k=1nx,ekekP^{(n)}x=\sum_{k=1}^{n}\langle x,e_{k}\rangle e_{k}.

Claim 1. By The Span of a Finite Family is the Smallest Subspace Containing It the set HnH_{n} is a linear subspace of HH, and it is closed in HH by claim 4 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, hence a closed linear subspace. Since e1=1|e_{1}|=1 we have e10He_{1}\ne0_{H}, so Hn{0H}H_{n}\ne\{0_{H}\}; apply claim 1 of Gram-Schmidt Orthonormalisation in a Real Inner Product Space to the tuple e(n)e^{(n)}, whose span is HnH_{n}. Since e(n)e^{(n)} is itself an orthonormal nn-tuple with span(e(n))=Hn\operatorname{span}(e^{(n)})=H_{n}, it is a tuple of the kind to which the second sentence of that claim refers, so the corresponding tuple in HnH_{n} is a basis of HnH_{n}; hence HnH_{n} is finite-dimensional.

For the inclusion, let uHnu\in H_{n}, say u=k=1nckeku=\sum_{k=1}^{n}c_{k}e_{k} with cRnc\in\mathbb{R}^{n}. Let cRn+1c'\in\mathbb{R}^{n+1} have components ck=ckc'_{k}=c_{k} for k[n]k\in[n] and cn+1=0c'_{n+1}=0. By the recursion of Finite Sum Notation in a Vector Space and claim 1 of Properties of Finite Sums of Vectors,

k=1n+1ckek=(k=1nckek)+0en+1=u+0H=u,\sum_{k=1}^{n+1}c'_{k}e_{k}=\Bigl(\sum_{k=1}^{n}c_{k}e_{k}\Bigr)+0\,e_{n+1}=u+0_{H}=u,

using claim 3 of Elementary Identities in a Vector Space for 0en+1=0H0\,e_{n+1}=0_{H}. Hence uHn+1u\in H_{n+1} and HnHn+1H_{n}\subseteq H_{n+1}.

Finally, HH is a real Hilbert space and HnH_{n} is closed, so by claim 5 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace the map P(n)P^{(n)} is the orthogonal projection PHnP_{H_{n}}, which is the asserted formula.

Claim 2. Suppose first that (ek)kN(e_{k})_{k\in\mathbb{N}} is an orthonormal basis of HH. Conditions 1 and 2 of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space hold by claim 1. For condition 3, let xHx\in H and let ε\varepsilon be positive. By Orthonormal Expansions in a Real Hilbert Space §expansion the series k=1x,ekek\sum_{k=1}^{\infty}\langle x,e_{k}\rangle e_{k} converges to xx, so there is nNn\in\mathbb{N} with xPHnx<ε|x-P_{H_{n}}x|<\varepsilon, and PHnxHnP_{H_{n}}x\in H_{n}. Hence (Hn)nN(H_{n})_{n\in\mathbb{N}} is an exhausting sequence for HH.

Conversely, suppose (Hn)nN(H_{n})_{n\in\mathbb{N}} is an exhausting sequence for HH, and let xHx\in H satisfy x,ek=0\langle x,e_{k}\rangle=0 for every kNk\in\mathbb{N}. By claim 3 of Properties of Finite Sums of Vectors, applied with the factor 00, and claim 1 above,

PHnx=k=1n0ek=0Hfor every nN.P_{H_{n}}x=\sum_{k=1}^{n}0\,e_{k}=0_{H}\qquad\text{for every }n\in\mathbb{N}.

Let ε\varepsilon be positive. By condition 3 of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space there are nNn\in\mathbb{N} and yHny\in H_{n} with xy<ε|x-y|<\varepsilon. By claim 5 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace the point PHnxP_{H_{n}}x is a nearest point of HnH_{n} to xx, so

x=x0H=xPHnxxy<ε.|x|=|x-0_{H}|=|x-P_{H_{n}}x|\le|x-y|<\varepsilon .

As ε\varepsilon was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack gives x0|x|\le0; since 0x0\le|x|, we get x=0|x|=0 and hence x,x=0\langle x,x\rangle=0, so x=0Hx=0_{H} by condition (d) of Real Inner Product Space §inner-product. Thus (ek)kN(e_{k})_{k\in\mathbb{N}} is an orthonormal basis of HH.

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