Functional Analysis - Dual Space and finite dimensional normed vector spaces

Question

Exercise:

Let $(X,\Vert \cdot \Vert)$ be a normed $\mathbb R$-vector space. Which of the following statements are true and which are wrong?

1.) For each $x \in X$ there exists $f \in X^*$ (dual space) with $\Vert f \Vert \le 1$ so that $\Vert x \Vert = f(x)$

2.) For each $f \in X^*$ there exists $x \in X$ with $\Vert x \Vert \le 1$ so that $f(x) = \Vert f \Vert$

3.) $X$ is finite dimensional if and only if every linear functional is continuous.

4.) If $X$ is finite dimensional, then each hyperplane in $X$ is closed.

My Questions:

For 1.) and 2.) I don't know where to start. It would be great if someone could give me a hint.

For 3.) I know that if X is finite dimensional then every linear functional is continuous. But the other implication does not have to be true. Right?

For 4.) As I know that X is finite dimensional this implies that every linear functional is continuous. There is a proposition which says that if f is continuous than the hyperplane $H=f^{-1}(\{\alpha\})$ is closed. So this should be enough to argue that the statement is true?

Thanks in advance for your help!

Answer 1

1) Hahn-Banach. This is one of the standard implications.

2) Not true in general (it is true if $X$ is reflexive). The standard example is an integral functional on $C([0,1])$.

3) You argument works for the 'only if' part. For the other implication, I am not completely sure. In infinite-dimensional $X$ one can construct discontinuous functionals. The construction requires the axiom of choice. But as (1) requires Hahn-Banach, this may answer the question.

4) Correct

Answer 2

Adding to @daw 's answer.....

Regarding (3), if $X/F$ is infinite dimensional with algebraic basis $B = \{ b_\alpha \}_{\alpha \in A}$ you can define a linear functional $f: X \to F$ by $f(x) = f((\Sigma b_\alpha x_\alpha)) = \Sigma \alpha x_\alpha $ this is unbounded and therefore not continuous. By contrapositive, if all linear functionals are continuous then the space is finite dimensional.

(Axiom of choice is indeed needed for the assumption that $X$ has an algebraic basis)

I guess that one should also specify that $A \subset F$ in order for the functional to make sense as defined.