# Linear functionals on a normed vector space

1. Jul 23, 2011

### AxiomOfChoice

I have a question: If $x\in X$ is a normed vector space, $X^*$ is the space of bounded linear functionals on $X$, and $f(x) = 0$ for every $f\in X^*$, is it true that $x = 0$? I'm reasonably confident this has to be the case, but why? The converse is obviously true, but I don't see why there couldn't be an example of a normed space for which all the functionals in $X^*$ are zero at some other value of $x$...

Last edited: Jul 23, 2011
2. Jul 24, 2011

### jambaugh

Start by considering expanding $x$ in a basis, dual some basis of $X^*$. Clearly for the basis $\{f_k\}$ and dual basis $\{e^k\}$(if there is one) of $X$ you have $x = x_k e^k$ where $x_k = f_k(x)$.

The question then becomes whether there exists such a dual basis. You don't have to worry about considering every element of $X^*$ but rather only a basis. Finally the fact that you have a metric space may shed some light on this question.

Last edited: Jul 24, 2011
3. Jul 24, 2011

### micromass

You will need the Hahn-Banach theorem for this. It states exactly what you want:

If V is a normed vector space and if z is in V, then there is a continuous function such that $f(z)=\|z\|$ and such that $\|f\|\leq 1$. ​

The correspond result for metric spaces may fail. For example

$$L^{1/2}(\mathbb{R})=\{f:\mathbb{R}\rightarrow \mathbb{R}~\vert~\int_{-\infty}^{+\infty}{\sqrt{f}}<+\infty\}$$

with metric

$$d(f,g)=\int_{-\infty}^{+\infty}{\sqrt{f-g}}$$

is a metric vector space such that 0 is its only continuous functional!

4. Jul 25, 2011

### AxiomOfChoice

Haha...yep, that's perfect! Thanks a lot!