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  • Given an infinite dimensional Banach space $(V,\|\cdot\|)$ over the field $\Bbb K=\Bbb C$ or $\Bbb R$, a countable ordered set $B:=\{b_n\}_{n\in\Bbb N}⊂V$ is called Schauder basis, if every $v\in V$ can be uniquely decomposed as: $$ v=\sum_{n\in\mathbb N}c_nb_n \tag1$$ for a set (generally infinite) of numbers $c_n\in\mathbb K$ depending on $v$, where the convergence of the sum is referred both to the Banach space topology and to the order used in labelling $B$. Identity $(1)$ is then taken to be equivalent to: $$\lim_{m\to\infty}\left\|v−\sum_{n=1}^mc_nb_n\right\|=0$$
  • Given an infinite dimensional Hilbert space $(V,\langle\cdot | \cdot\rangle)$ over the field $\mathbb K=\mathbb C$ or $\mathbb R$, a set $B⊂V$ is called Hilbert basis, or complete orthonormal system, if the following conditions are true:
    1. $⟨z|z⟩=1$ and $⟨z|z′⟩=0$ if $z,z'∈B$ and $z≠z'$, i.e. $B$ is an orthonormal system;
    2. if $x \in V$ and $⟨x|z⟩=0$ for all $z\in B$ then $x=0$ (i.e. $B$ is maximal with respect to the orthogonality requirment).

If $(V,\langle\cdot | \cdot\rangle)$ is separable, i.e. it contains a dense countable subset, then every Hilbert basis is also a Schauder basis with respect to the norm induced by the Hilbert scalar product. However, the converse is not generally true. Are there any explicit examples of Schauder bases of infinite-dimensional, separable Hilbert spaces that are not Hilbert bases?

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  • $\begingroup$ It is very simple. Consider the basis $(1,1), (1,0)$ of $\mathbb R^2$ equipped with the usual scalar product $(x, y)\cdot( x', y')=xx'+yy'$. $\endgroup$ Commented Apr 27, 2018 at 12:45
  • $\begingroup$ @GiuseppeNegro My question is in the context of infinite dimensional vector spaces. I will specify this in an edit $\endgroup$
    – giobrach
    Commented Apr 27, 2018 at 12:47
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    $\begingroup$ A trivial example: if $\{b_n\}$ is a Hilbert basis, replace $b_1$ by $2b_1$. No longer orthonormal. $\endgroup$
    – Aweygan
    Commented Apr 27, 2018 at 12:57

1 Answer 1

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There is no need to go to the infinite dimensional case. Consider the immediate $\mathbb R^2$ example $v_1=(1,1), v_2=(1,0)$. Once you have that, you trivially construct an infinite-dimensional example in $\ell^2$ by setting $$ \begin{array}{rcl} b_1 & =& (v_1, 0,0,0\ldots) \\ b_2 &=& (v_2, 0 ,0 ,0\ldots)\\ b_3 &=& (0,0,1,0,0\ldots)\\ b_4 &=& (0,0,0,1,0\ldots)\\ &\vdots&\\ \end{array}$$

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  • $\begingroup$ Thank you. I wasn't seeing the immediate generalization to $\ell^2$ of what you said in the comments. $\endgroup$
    – giobrach
    Commented Apr 27, 2018 at 13:03
  • $\begingroup$ You are welcome. Glad it helped. $\endgroup$ Commented Apr 27, 2018 at 13:10
  • $\begingroup$ Heyy, really like your answer, if you could help with an exercise, would really appreciate it :) math.stackexchange.com/questions/3096022/… $\endgroup$
    – Homaniac
    Commented Feb 2, 2019 at 5:53

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