all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements

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Theorem

If {en} is an orthonormal basis of a hilbert space H, then for all u∈H we have

limm→∞∑n=1m⟨u,en⟩en=∑n=1∞⟨u,en⟩en=u
Note

Thus if we have an orthonormal basis, every element can be expanded in this series in terms of the basis elements (called a Bessel-Fourier series). And thus every separable Hilbert space has an orthonormal basis.

Note

ie like in finite-dimensional linear algebra, we can write any element as a linear combination of the basis elements. But in this space, there might be an infinite number of such elements.

Proof (via Bessel's inequality)

First, we prove {∑n=1m⟨u,en⟩en}m is Cauchy. Let ϵ>0. Then by Bessel's inequality, we have

\sum_{n=1}^\infty \lvert \langle u, e_{n} \rangle \rvert { #2} \leq \lvert \lvert u \rvert \rvert { #2} < \infty

Thus, for all M∈N such that for all N≥M we have ∑m=N+1∞|⟨u,en⟩|2<ϵ

Then for all m>ℓ≥M we compute

\begin{align} \left\lvert \left\lvert \sum_{n=1}^m \langle u, e_{n} \rangle e_{n} - \sum_{n=1}^\ell \langle u, e_{n} \rangle e_{n} \right\rvert \right\rvert &= \sum_{n=\ell+1}^m \lvert \langle u, e_{n} \rangle \rvert { #2} \\ &\leq \sum_{\ell+1}^\infty \lvert \langle u, e_{n} \rangle \rvert { #2} \\ &< \epsilon^2 \end{align}

thus the sequence is indeed Cauchy. Since H is complete, there exists some u¯∈H where

u¯=limm→∞∑n=1m⟨u,en⟩en

By continuity of inner product, we know that for all ℓ∈N

⟨u−u¯,eℓ⟩=limm→∞⟨u−∑n=1m⟨u,en⟩en,eℓ⟩=limm→∞[⟨u,eℓ⟩−∑n=1m⟨u,en⟩⟨en,eℓ⟩]=⟨u,eℓ⟩−⟨u,eℓ⋅1⟩=0

Thus ⟨u−u¯,eℓ⟩=0 for all ℓ if and only if u−u¯=0.

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References

References

See Also

Mentions

Mentions

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Created 2025-07-15 Last Modified 2025-07-15