When reading a proof of this theorem (which states every uniformly convex Banach space is reflexive), I came across the following claim:

In a infinite-dimensional Banach space $X$, if $C$ is a weak-$\ast$ compact subset of $X^{\ast\ast}$ and $z$ is a point disjoint from $C$, then we can find $g \in X^{\ast}$ such that $g$ strictly separates them.

My questions are:

  1. By Hahn-Banach Separation Theorem we could find $\phi \in X^{\ast\ast\ast}$ and $\alpha \in \mathbb{R}$ such that $\sup_{x \in C}\operatorname{Re}\phi(c) < \alpha < \operatorname{Re}\phi(z)$. Can we find such $g \in X^{\ast}$ that can replace $\phi$? I suppose the author wants to cite Goldstine's Theorem but in this case the set $\{\psi \in X^{\ast\ast\ast}\,\vert\,\vert\psi(x) - \phi(x)\vert < \epsilon\,\forall\,x \in C\} = \bigcap_{x \in C}\{\psi \in X^{\ast\ast\ast}\,\vert\,\vert\psi(x) - \phi(x)\vert < \epsilon\}$ may not be open and how can we find $g \in X^{\ast}$ that is close enough to $\phi$ on every point in $C$?
  2. Given a infinite-dimensional Banach space $X$, in $X^{\ast}$ equipped with the weak-$\ast$ topology are there any sufficient conditions to finding an injective bounded linear functional (an example will be greatly appreciated)? How about in a general topological vector space? (I suppose a nice mapping between the basis in an neighborhood of $\overset{\rightarrow}{0}$ and a bounded set in $\mathbb{R}$ always exists but hope to a see some details).

1 Answer 1

  1. Yes. In general, let $Y$ be a normed space and endow $Y^*$ with the weak-star topology, which is a locally convex topology. If $C\subset Y^*$ is a convex, weak-* closed set and $z\in Y^*\setminus C$, then by the Hahn-Banach theorem we can find a real number $t$ and a functional $\phi:Y^*\to\mathbb{C}$ that is continuous $\textbf{with respect to the weak-star topology}$ such that $$\text{Re}(\phi(c))<t<\text{Re}\phi(z)$$ for all $c\in C$. I claim that the functionals $\phi:Y^*\to\mathbb{C}$ that are continuous with respect to the weak-star topology are precisely the evaluation functionals of points of $Y$, i.e. precisely the functionals of the form $\text{ev}_y:Y^*\to\mathbb{C}$, $f\mapsto\text{ev}_y(f):=f(y)$.

To see this, let $\phi:Y^*\to\mathbb{C}$ be continuous for the weak-star topology. Since the weak-star topology is generated by the seminorms $\{p_y\}_{y\in Y}$ where $p_y(f)=|f(y)|$, we can find $M>0$ and $y_1,\dots,y_n\in Y$ such that $|\phi(f)|\leq M\cdot\max_{1\leq i\leq n}|p_{y_i}(f)|$ for all $f\in Y^*$, i.e. $|\phi(f)|\leq M\max_{1\leq i\leq n}|f(y_i)|$ for all $f\in Y^*$. But this shows that $\bigcap_{i=1}^n\ker(\text{ev}_{y_i})\subset\ker(\phi)$. By elementary linear algebra this shows that $\phi$ is a linear combination of $\text{ev}_{y_i}$, thus of the form $\text{ev}_y$, where $y$ is a linear combination of the $y_i$.

Therefore, we can restate the application of Hahn-Banach as "we find a real number $t$ and a point $y\in Y$ such that $$\text{Re}(c(y))<t<\text{Re}(z(y))$$ for all $c\in C$.

  1. Let $X$ be a vector space in general. If we have an injective linear map $T:X\to\mathbb{C}$, then $X$ embeds as a vector space in $\mathbb{C}$, so $\dim(X)\leq1$. The condition you are looking for is this: $X$ admits an injective linear functional if and only if $X$ is trivial or one-dimensional.

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