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Let H be a hilbert space, infinite dimensional. Let $(e_n)$ be an orthonormal basis for H. Can we express any x in H as $x=\sum_{i=1}^{\infty} c_i e_i$ where the constants are (maybe) $<e_i,x>$? Also when is an infinite linear combination of basis vectors like the one above an element in H? I ask because I know on one hand that by definition of Hilbert basis, any element in H can be approximated in the norm by a sequence of finite linear combinations of $e_k$ but have doubts as to whether we can write any element as an infinite linear combination of the one above.. What conditions garantee that such an infinite sum would even be in H? Thanks for any information.

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    $\begingroup$ First, we need to make sure that we are using the same terms and definitions. Page 176, chapter 6 of Linear Operator Theory in Engineering and Science by Naylor and Sell defines (def4.6.1) Given a set $A$ in a linear space $X$, a point $x\in X$ is said to be a linear combination of points in $A$ if there exists a finite set of points $\{x_1,x_2,\dots,x_n\}$ in $A$ and a finite set of scalars $\{\alpha_1,\dots,\alpha_n\}$ such that $x=\alpha_1x_1+\dots+\alpha_nx_n$. It goes on to note that we consider finite linear combinations only. $\endgroup$ – JMoravitz Mar 31 '17 at 1:54
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    $\begingroup$ You can have $\ell_2$ and the linearly independent set $\{e_1,e_2,\dots\}$ where $e_i = (0,0,\dots,0,\underbrace{1}_{i\text{'th term}},0,\dots)$, the element $(1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},\dots)$ cannot be represented as a finite linear combination of these elements, (technically this implies that $\{e_1,e_2,\dots\}$ is not even a hamel basis in the first place, see later in the chapter for more details) $\endgroup$ – JMoravitz Mar 31 '17 at 1:56
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So firstly we give some definitions:

If $K$ is a subset of a Hilbert space H, the set of finite linear combinations of elements of $K$ is called the span of $K$. An orthonormal basis is then an orthonormal subset of $H$ s.t. the closure of its span is all of $H$.

Then we could show that Every Hilbert space has an orthonormal basis. (Given axiom of choice)

Proof. If $B=\{u_{\alpha}\}$ is orthonormal, but not a basis, let $V$ be the closure of the linear span of $B$, i.e. the closure w.r.t. the norm in $H$ of the set of finite linear combinations of elemtns of $B$. Choose $x\in H^{\perp}$, and if we let $B'=B\cup\{\frac{x}{||x||}\}$, then $B'$ is a basis that is strictly bigger than $B$.

It is easy to see that the union of an increasing sequence of orthonormal sets is an orthonormal set, and so there is a maximal one by Zorn's lemma. By preceding paragraph, this maximal orthonormal set must be a basis, for otherwise we could find a larger basis. Q.E.D.

Note: here we implicitly used the fact that the Gram-Schmidt procedure from linear algebra also works in infinitely many dimensions, and this is because we could do the follow step over the linearly independent sequence $\{x_n\}_{n=1}^{\infty}$: $$u_N=x_N-\sum_{i=1}^{n-1}\langle x_N,u_i\rangle u_i,$$ $$u_N=v_N/||v_N||$$

In sum, unfortunately it is not a constructive proof, and we relied on axiom of choice. Thus we really cannot show one such basis that all elements in $H$ could be expressed by it - but as long as you accept axiom of choice, you know there is at least one set of such basis out there.

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$\sum_i c_i e_i$ converges iff $\sum_i c_i^2 <\infty$. For any $x\in H$, $x=\sum_i \langle x,e_i\rangle e_i$ where $\sum_i |\langle x,e_i\rangle|^2 =||x||^2 <\infty$.

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  • $\begingroup$ Thanks, so basically any x in H can be represented as the infinite sum of $<x,e_i>e_i$ ? $\endgroup$ – user172377 Mar 31 '17 at 2:44
  • $\begingroup$ @Socchi Yes, this is essentially the definition of an orthonormal basis. $\endgroup$ – mathematician Mar 31 '17 at 21:31

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