Let $H$ be a Hilbert space with a countable basis $B$. Does it mean that any vector $x\in H$ can be expressed as a finite linear combination of elements from $x$, or as an infinite linear combination?

Thanks in advance


Take $\ell^2$ for example, i.e. the square-summable sequences of complex numbers with inner product $$\langle x,y\rangle = \sum_{n=1}^\infty x_n\overline{y_n}. $$ This has the countable orthonormal basis $$\{(1,0,0,\ldots), (0,1,0,\ldots), (0,0,1,0,\ldots),\ldots\}. $$ As $$\sum_{n=1}^\infty 2^{-n} = 1<\infty, $$ we have $$(2^{-1}, 2^{-2}, 2^{-3},\ldots)\in\ell^2, $$ and it is clear that this element cannot be written as a finite linear combination of the basis elements.

  • 3
    $\begingroup$ It's also important to distinguish between Hamel and Hilbert bases. $\endgroup$ – anon Jan 23 '15 at 15:11

In Hilbert space theory and Functional analysis the term 'basis' has a different definition than in linear algebra. For this reason one often encounters the term 'Hamel basis' for the latter and 'Hilbert space basis' in the first case.

In the Hilbert space case the requirement is that each vector can be represented as in infinite sum, equality being understood in the sense of convergence in the Hilbert space norm. The term 'Schauder basis' is also used in B spaces for which infinite sums are allowed.

See e.g. http://en.wikipedia.org/wiki/Schauder_basis


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.