Linear functionals on a normed vector space

[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...

 

[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.

 

[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!

 

[AxiomOfChoice]

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.​

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