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 uH we have

limmn=1mu,enen=n=1u,enen=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=1mu,enen}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 MN such that for all NM we have m=N+1|u,en|2<ϵ

Then for all m>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¯=limmn=1mu,enen

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

uu¯,e=limmun=1mu,enen,e=limm[u,en=1mu,enen,e]=u,eu,e1=0

Thus uu¯,e=0 for all if and only if uu¯=0.

References

References

See Also

Mentions

Mentions

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