关于Hahn-Banach泛函延拓集合的闭性、内部为空性及非紧性的证明咨询

Hey there! You've already nailed the foundational parts—non-emptiness from the Hahn-Banach theorem, boundedness since all extensions share $g$'s norm, and convexity. Let's walk through the remaining three properties you're stuck on, with concrete reasoning and examples:

1. 证明延拓集合是闭的（范数拓扑下） #

Let's denote the set of all Hahn-Banach extensions of $g$ as $E \subset X'$. To show $E$ is closed, take any sequence ${f_n} \subset E$ that converges in norm to some $f \in X'$ (i.e., $|f_n - f| \to 0$):

• First, for every $y \in Y$, $f_n(y) = g(y)$ by definition of $E$. Taking the limit as $n \to \infty$, we get $f(y) = g(y)$, so $f$ is an extension of $g$.
• Next, the norm is lower-semicontinuous, so $|f| \leq \liminf_{n \to \infty} |f_n| = |g|$. But since $f$ extends $g$, we also have $|f| \geq |g|$. Combining these gives $|f| = |g|$.
• Thus, $f \in E$, so $E$ contains all its norm-limits and is closed.

(Note: $E$ is also weak-closed, since weak* convergence means pointwise convergence on $X$, which preserves the extension property and norm bound—this follows similarly.)*

2. 证明延拓集合内部为空 #

Suppose for contradiction that $E$ has non-empty interior. Then there exists some $f_0 \in E$ and $\epsilon > 0$ such that the open ball $B(f_0, \epsilon) = {f \in X' \mid |f - f_0| < \epsilon}$ is entirely contained in $E$:

• If $X = Y$, then $E = {g}$, a singleton, which clearly has empty interior. So assume $X \neq Y$, and pick some $x_0 \in X \setminus Y$.
• Define a linear functional $h$ on $\text{span}(Y \cup {x_0})$ by $h(y + tx_0) = t \cdot \frac{\epsilon}{2} |x_0|$ for all $y \in Y, t \in \mathbb{R}$ (adjust for complex scalars by using modulus). By Hahn-Banach, we can extend $h$ to all of $X$ while preserving its norm: $|h| = \frac{\epsilon}{2} < \epsilon$.
• Now consider $f = f_0 + h$. We have $|f - f_0| = |h| = \frac{\epsilon}{2} < \epsilon$, so $f \in B(f_0, \epsilon)$. But $f(x_0) = f_0(x_0) + \frac{\epsilon}{2} |x_0|$, so $|f| \geq \frac{|f(x_0)|}{|x_0|} = \frac{|f_0(x_0)|}{|x_0|} + \frac{\epsilon}{2}$. Since $|f_0| = |g|$, this means $|f| > |g|$, so $f \notin E$—contradicting the assumption that $B(f_0, \epsilon) \subset E$.
• This shows no point in $E$ can be an interior point, so $\text{int}(E) = \emptyset$.

3. 证明延拓集合不一定是紧的（范数拓扑下） #

We can construct a concrete example where $E$ fails to be compact in the norm topology:

• Let $X = \ell^\infty$ (the space of bounded real sequences, with sup norm), $Y = c_0$ (the subspace of sequences converging to 0).
• Define $g: c_0 \to \mathbb{R}$ by $g((a_n)) = \lim_{n \to \infty} a_n$. This is a bounded linear functional with $|g| = 1$.
• The set $E$ consists of all Banach limits on $\ell^\infty$—linear functionals $F \in (\ell^\infty)'$ such that $F(a) = \lim_{n \to \infty} a_n$ for $a \in c_0$, and $|F| = 1$.
• Now consider the sequence ${F_n} \subset E$ where $F_n((a_k)) = \frac{1}{n} \sum_{k=n}^{2n-1} a_k$. Each $F_n$ is a valid extension: for $a \in c_0$, $\lim_{n \to \infty} F_n(a) = \lim_{k \to \infty} a_k = g(a)$, and $|F_n| = 1$.
• Notice that $|F_n - F_m| = 2$ when $n \neq m$: take a sequence $a$ where $a_k = 1$ for $k \in [n, 2n-1]$ and $a_k = -1$ for $k \in [m, 2m-1]$ (0 elsewhere). Then $F_n(a) = 1$, $F_m(a) = -1$, so $|(F_n - F_m)(a)| = 2$, and $|a|_\infty = 1$. Thus, the sequence ${F_n}$ has no Cauchy (or convergent) subsequences in the norm topology.
• This means $E$ is not compact in the norm topology. (Note: $E$ is weak*-compact by Alaoglu's theorem, since it's a weak*-closed subset of the unit ball of $X'$, but the question refers to norm compactness.)

备注：内容来源于stack exchange，提问作者Babai