I have a question concerning the proof of the Goldstine theorem, that the image of the unit ball $B_X$ of some Banach space $X$ is dense in the unit ball $B_{X^{**}}$ of its bidual under the canonical embedding of $X$ into $X^{**}$. The proof I am referring to is taken verbatim from the English Wikipedia:

Given an $x^{**} \in B_{X^{**}}$, a tuple $(\phi_1, \dots, \phi_n)$ of linearly independent elements of $X^*$ and a $\delta>0$ we shall find an $x \in (1+\delta) B_{X}$ such that $\phi_i(x)=x^{**}(\phi_i)$ for every $i=1,\dots,n$. If the requirement $||x|| \leq 1+\delta$ is dropped, the existence of such an $x$ follows from the surjectivity of $\Phi : X \to \mathbb{C}^{n}, x \mapsto (\phi_1(x), \dots, \phi_n(x)).$ Let now $Y = \cap_{i} \ker \phi_i = \ker \Phi$. Every element of $x+Y \cap (1+\delta) B_{X}$ has the required property, so that it suffices to show that the latter set is not empty. Assume that it is empty. Then $\mathrm{dist}(x,Y) \geq 1+\delta$ and by the Hahn-Banach theorem there exists a linear form $\phi \in X^*$ such that $\phi|_{Y}=0$, $\phi(x) \geq 1+\delta$ and $||\phi||_{X^*}=1$. Then $\phi \in \mathrm{span}(\phi_1, \dots, \phi_n)$ and therefore $$1+\delta \leq \phi(x) = x^{**}(\phi) \leq ||\phi||_{X^*} ||x^{**}||_{X^{**}} \leq 1,$$

I agree with most of the proof, however, I have a problem with one of last steps, namely why or how it follows that $\phi \in \mathrm{span}(\phi_1, \dots, \phi_n)$. There will be many functionals that fulfill the previous conditions $\phi|_{Y}=0$, $\phi(x) \geq 1+\delta$ and $||\phi||_{X^*}=1$, but are not in $\mathrm{span}(\phi_1, \dots, \phi_n)$. E.g. I think I could construct by Hahn-Banach also a functional $\psi$ that has a kernel $\ker \psi\supsetneq\ker \Phi$ and fulfills $\psi(x) \geq 1+\delta$ and $||\psi||_{X^*}=1$, but cannot be in $\mathrm{span}(\phi_1, \dots, \phi_n)$, since the kernel is a strict superset. Is there something missing in the argument or am I mistaken here?


1 Answer 1


$\phi|_Y=0$ iff $\phi \in \mathrm{span}\left( \phi_1 , \ldots \phi_n \right)$:

The "if" part is obvious.

If $\phi|_Y=0$, then $\phi$ defines a linear form on $X/Y$, which is a vector space of dimension $\leq n$ (using your $\Phi$), and $( \phi_1, \ldots, \phi_n)$ is a linearly independent family of elements of $\left( X/Y \right)^*$, so $\dim X/Y = n$ and $( \phi_1, \ldots, \phi_n)$ is a basis of $\left( X/Y \right)^*$.

  • $\begingroup$ Yes, thanks a lot. This helped me much to think about these things in terms of quotient spaces and duals thereof, which leads to really elegant arguments. Now, I see also where my argument went totally wrong. $\endgroup$ May 9, 2011 at 12:26

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