# Thread: Dual Space of a Vector Space Question

1. ## 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

2. 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 $\{ x_i\}$.

3. Thank you! I looked at the Hahn Banach theorem and understand what's going on a little better