# Dual Space of a Vector Space Question

• Apr 11th 2011, 08:56 AM
slevvio
Dual Space of a Vector Space Question
Hello everyone I have a question about Dual Spaces. Let $X$ be an infinite dimensional vector space over $\mathbb{C}$, and let
$\{ x_1, \ldots, x_n \}$ be a subset of $X$, which is linearly independent. Then we can define the unique linear functionals $f_i : X \rightarrow \mathbb{C}$ which give $f_i (x_j) = 1$ if $i=j,$ 0 otherwise.

My question is can we choose the $f_i$'s to be continuous? I was asked this question and I am not sure what the answer is. I take it if we have an infinite basis $\{ x_i\}_{i \in I}$ of X then we cannot have every $f_i$ continuous because then surely every linear functional would then be determined by those continuous functions, i.e. it is a linear combination of finitely many of them, hence is continuous. And there exist unbounded linear functionals (?). Any help with this would be appreciated! Thanks
• Apr 11th 2011, 10:26 AM
Opalg
Quote:

Originally Posted by slevvio
Hello everyone I have a question about Dual Spaces. Let $X$ be an infinite dimensional vector space over $\mathbb{C}$, and let
$\{ x_1, \ldots, x_n \}$ be a subset of $X$, which is linearly independent. Then we can define the unique linear functionals $f_i : X \rightarrow \mathbb{C}$ which give $f_i (x_j) = 1$ if $i=j,$ 0 otherwise.

My question is can we choose the $f_i$'s to be continuous? I was asked this question and I am not sure what the answer is. I take it if we have an infinite basis $\{ x_i\}_{i \in I}$ of X then we cannot have every $f_i$ continuous because then surely every linear functional would then be determined by those continuous functions, i.e. it is a linear combination of finitely many of them, hence is continuous. And there exist unbounded linear functionals (?). Any help with this would be appreciated! Thanks

The conditions $f_i (x_j) = 1$ if $i=j,$ 0 otherwise, define $f_i$ uniquely as a continuous linear functional on the finite-dimensional subspace spanned by $x_1, \ldots, x_n$. The functional can then be extended to a linear functional on the whole of X. The extension will not be unique. The functional can in fact be extended to a continuous linear functional on the whole of X, by the Hahn–Banach theorem. This continuous extension will also not be unique.

If the set $\{ x_i\}$ is infinite, then the functionals $f_i$ need not be continuous, even on the subspace spanned by the $\{ x_i\}$.
• Apr 16th 2011, 03:02 AM
slevvio
Thank you! I looked at the Hahn Banach theorem and understand what's going on a little better