• 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?

  • $\begingroup$ It is very simple. Consider the basis $(1,1), (1,0)$ of $\mathbb R^2$ equipped with the usual scalar product $(x, y)\cdot( x', y')=xx'+yy'$. $\endgroup$ 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
    Apr 27, 2018 at 12:47
  • 1
    $\begingroup$ A trivial example: if $\{b_n\}$ is a Hilbert basis, replace $b_1$ by $2b_1$. No longer orthonormal. $\endgroup$
    – Aweygan
    Apr 27, 2018 at 12:57

1 Answer 1


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 $$ \begin{array}{rcl} b_1 & =& (v_1, 0,0,0\ldots) \\ b_2 &=& (v_2, 0 ,0 ,0\ldots)\\ b_3 &=& (0,0,1,0,0\ldots)\\ b_4 &=& (0,0,0,1,0\ldots)\\ &\vdots&\\ \end{array}$$

  • $\begingroup$ Thank you. I wasn't seeing the immediate generalization to $\ell^2$ of what you said in the comments. $\endgroup$
    – giobrach
    Apr 27, 2018 at 13:03
  • $\begingroup$ You are welcome. Glad it helped. $\endgroup$ 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
    Feb 2, 2019 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.