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

[[concept]]

Topics

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Theorem

If $\{e_n\}$ is an orthonormal basis of a hilbert space $H$, then for all $u\in H$ we have ${\lim }_{m\to \infty }{\sum }_{n=1}^m⟨u,e_n⟩e_n={\sum }_{n=1}^{\infty }⟨u,e_n⟩e_n=u$

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 ${\left\{{\sum }_{n=1}^m⟨u,e_n⟩e_n\right\}}_m$ is Cauchy. Let $ϵ>0$. Then by Bessel’s inequality, we have ${\sum }_{n=1}^{\infty }|⟨u,e_n⟩{|}^2\le ||u|{|}^2<\infty$ Thus, for all $M\in \mathbb{N}$ such that for all $N\ge M$ we have ${\sum }_{m=N+1}^{\infty }|⟨u,e_n⟩{|}^2<ϵ$

Then for all $m>\ell \ge M$ we compute

\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}$$ thus the sequence is indeed Cauchy. Since $H$ is complete, there exists some $\bar{u}\in H$ where $$\bar{u} = \lim_{ m \to \infty } \sum_{n=1}^m \langle u, e_{n} \rangle e_{n}$$ By [[continuity of inner product]], we know that for all $\ell \in \mathbb{N}$ $$\begin{align} \langle u-\bar{u}, e_{\ell} \rangle &= \lim_{ m \to \infty } \left\langle u - \sum_{n=1}^m \langle u, e_{n} \rangle e_{n}, e_{\ell} \right\rangle \\ &= \lim_{ m \to \infty } \left[ \langle u, e_{\ell} \rangle - \sum_{n=1}^m \langle u, e_{n} \rangle \langle e_{n}, e_{\ell} \rangle \right] \\ &= \langle u, e_{\ell} \rangle - \langle u, e_{\ell}\cdot 1 \rangle \\ &= 0 \end{align}$$ Thus $\langle u-\bar{u}, e_{\ell} \rangle = 0$ for all $\ell$ if and only if $u - \bar{u} = 0$. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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