Lack of a basis for a Dual Space of an infinite dimensional vector space

Question

I understand that the dual basis of a finite dimensional vector space is a basis for the dual space. Conceptually, I understand that this is not true in the infinite dimensional case, but my understanding is that if I begin calculating a “basis” as if I were seeking a dual basis in the finite case (defined in the usual way using the Kronecker symbol), the elements should still be linearly independent. Is this correct? Could someone demonstrate why?

Answer 1

The dual space consists of continuous functionals. You can use Zorn's lemma to construct a "Hamel basis" $H$ for any vector space $V$. Then you can define, for $h\in H$, the dual element $h^*$ in the same way you define it for finite-dimensional vector spaces. But $h^*$ generally won't be continuous.

To see an example that looks like the finite dimensional case, you can read about Hilbert spaces. The basic example of a Hilbert space is $\ell^2$, the set of complex sequences $c$ such that $\sum_n |c_n|^2 < \infty$. We say that $\ell^2$ has a "basis" consisting of the sequences $e_1 = (1, 0, 0, 0, \ldots)$, $e_2 = (0, 1, 0, 0, \ldots)$, et cetera. But this is not a Hamel basis. Its linear combinations are merely dense in $\ell^2$.