I'm reading a proof of the following claim. The proof contains some claims I am not following, can someone elaborate?


Denote the closed unit balls in $X$ and $X^*$ by $B$ and $B^*$, respectively. Denote $\mathcal{F}(B)$ to be a set of functions of the form $f:B \rightarrow [-1,1]$ with pointwise product topology (for basis, see below). Define the restriction map $R: B^* \rightarrow \mathcal{F}(B)$ by $R(\phi) = \phi|_B$ for $\phi \in B^*$. Show $R(B^*)$ is a closed subset of $\mathcal{F}(B)$.


Let $f: B \rightarrow [-1,1]$ be a point of closure with respect to the pointwise product topology of $R(B^*)$ (defn of topology given below). To show $f \in R(B^*)$ it suffices to show that for all $u, v \in B$ and $\lambda \in \mathbb{R}$ for which $u + v$ and $\lambda u$ also belong to $B$, $$f(u+v)= f(u)+f(v) \text{ and } f(\lambda u)= \lambda f(u)$$ However, for any $\epsilon > 0$, the weak-* neighborhood of $f$, $\mathcal{N}_{\epsilon, u,v,u+v}(f)$ (see below for definition, this is an element of the basis of weak-* topology), contains some $R(\phi_\epsilon)$ and since $\phi_{\epsilon}$ is linear, we have $|f(u + v) - f (u) - f (v)| < 3\epsilon$. Therefore, $f(u+v)= f(u)+f(v)$ holds. The proof of $f(\lambda u)= \lambda f(u)$ is similar.

In the proof above, we denote a basis element of the weak-* topology as: $$\mathcal{N}_{\epsilon,x_1,...x_n}(\psi) =\bigg \{\psi' \in B^* \, \bigg| \, |\psi'(x_i)-\psi(x_i) | < \epsilon \quad \forall i = 1,...n\bigg\} \\ =\bigg \{\psi' \in \mathcal{F}(B) \, \bigg| \, |\psi'(x_i)-\psi(x_i) | < \epsilon \quad \forall i = 1,...n\bigg\}$$ The pointwise product topology (the basis shown in the second equality) and weak-* topology are homeomorphic (there's a bijection between basis elements established by the equality above).


  1. "To show $f \in R(B^*)$ it suffices to show..." -- What are they verifying about $f$ here? Loosely, my understanding is that proofs of closure proceed as follows: suppose $x \in \overline{X}$, check neighborhoods of $x$. In particular, we suppose $x$ exists, there are no "properties" of $x$ to check.

  2. "for any $\epsilon > 0$, the weak-$^*$ neighborhood of $f$, $\mathcal{N}_{\epsilon, u,v,u+v}(f)$ ... contains some $R(\phi_\epsilon)$" -- What is $\phi_{\epsilon}$? Why does there exist $R(\phi_\epsilon) \in \mathcal{N}_{\epsilon, u,v,u+v}(f)$? This is where they are establishing that the $f$ they chose is a limit point, right?

  3. "However... [to the end]" -- What's being verified here? Aren't we done after we show every neighborhood contains a point of the set? (i.e. after step 2).

  • $\begingroup$ Can you please state the source of the proof you're quoting? $\endgroup$ – Nate Eldredge Jun 14 at 18:07
  • $\begingroup$ This is Royden, Chap 15 Section 1. It's part of the proof he writes for Alaoglu's Theorem (thm un-numbered). (I looked for the link in google books, but one cannot search it) $\endgroup$ – yoshi Jun 14 at 18:38
  1. As stated, they are verifying that $f(u+v) = f(u)+f(v)$ and $f(\lambda u) = \lambda f(u)$. That's all.

    Why does this imply $f \in R(B^*)$? You have to show that there exists $\phi \in B^*$ such that $f = \phi|_B$. There is a natural candidate for $\phi$: since $\phi$ is supposed to be linear, it should be determined by its values on the unit sphere, where it is supposed to agree with $f$. So let $\phi(x) = \|x\| f(\frac{x}{\|x\|})$ with $\phi(0)=0$. Given what's been shown about $f$, it should now be a straightforward exercise to show:

    • $f(x) = \phi(x)$ for $x \in B$, i.e. $f = \phi|_B$

    • $\phi$ is linear, i.e. for any $x,y \in X$ and any $\lambda \in \mathbb{R}$, we have $\phi(x+y) = \phi(x) + \phi(y)$ and $\phi(\lambda x) =\lambda \phi(x)$

    • $\|\phi\| \le 1$, i.e. for every $x \in B$ we have $|\phi(x)| \le 1$. (This follows immediately from the fact that $f = \phi|_B$.

    And the latter two statements are exactly the definition of $\phi \in B^*$.

  2. By assumption, $f$ is in the closure of $R(B^*)$. Since $\mathcal{N}_{\epsilon, u,v,u+v}$ is a (weak-*) neighborhood of $f$, it must therefore contain some element of $R(B^*)$; call it $g$. But saying $g \in R(B^*)$ means there exists some element of $B^*$, call it $\phi_\epsilon$, such that $g = R(\phi_\epsilon)$. That is to say, there exists $\phi_\epsilon$ such that $R(\phi_\epsilon) \in \mathcal{N}_{\epsilon, u,v,u+v}$. Their wording "$\mathcal{N}$ contains some $R(\phi_\epsilon)$" is just a more abbreviated way to say it.

  3. The paragraph starting with "However..." is where they actually verify that $f(u+v) = f(u)+f(v)$ and $f(\lambda u) = \lambda f(u)$.

  • $\begingroup$ Thanks! okay, at a high level then: To show $R(B^*)$ is a closed subset of $\mathcal{F}(B)$, they suppose $f$ is in the closure of $R(B^*)$. Then they have to show that there's an element of $B^*$ that maps to it. How does this show $R(B^*)$ is a closed subset of $\mathcal{F}(B)$? $\endgroup$ – yoshi Jun 14 at 18:56
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    $\begingroup$ @yoshi: It shows that every element of the closure of $R(B^*)$ is actually in $R(B^*)$. In other words, $\overline{R(B^*)} \subseteq R(B^*)$. Since the reverse containment is trivial, this shows $R(B^*)$ is equal to its own closure and therefore is closed. $\endgroup$ – Nate Eldredge Jun 14 at 18:58
  • $\begingroup$ I'm having trouble seeing that we showed that every element is in the closure. The next comment contains my (probably incorrect) reasoning. $\endgroup$ – yoshi Jun 14 at 19:23
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    $\begingroup$ @yoshi: We select an arbitrary element $f$ of $\overline{R(B^*)}$, and try to show that it's in $R(B^*)$. Please reread what I wrote in #1. To do this, we begin by verifying the linearity properties $f(u+v) = f(u)+f(v)$, $f(\lambda u) = \lambda f(u)$ (call these facts (*)). We then define a map $\phi : X \to \mathbb{R}$ via $\phi(x) = \|x\| \phi(\frac{x}{\|x\|})$, verify that $\phi|_B = f$, and verify that $\phi \in B^*$. The latter is not based on directly appealing to the original assumption that $f \in \overline{R(B^*)}$, but on the statement (*). $\endgroup$ – Nate Eldredge Jun 14 at 19:28
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    $\begingroup$ By the way, note that the computation $\|x\| f(\frac{x}{\|x\|}) = f(x)$ is only valid for $x \in B$ (after we have shown that $f(\lambda x) = \lambda f(x)$). It's not valid for $x \in X \setminus B$ because $f(x)$ is not defined in that case. The purpose of introducing $\phi$ is to extend $f$, in a linear fashion, from $B$ to $X$. $\endgroup$ – Nate Eldredge Jun 14 at 19:32

The weak$^\ast$ topology on $B^\ast$ is just the subspace topology induced from seeing all linear maps on $X^\ast$ as a subset of the set of all functions from $X^\ast$ to $\Bbb R$. This set of functions has the base as described (an alternative base for this product topology). And $\phi \in B^\ast$ iff any $x \in B$ maps into $[-1,1]$, hence that the map $R$ is well-defined. So $R(B^\ast)$ is a subset of $\prod_{x \in B} [-1,1]_x$, where $[-1,1]_x$ is a copy of $[-1,1]$.

The thing to verigy is that the closure of a set of linear functions only contains linear functions, in the product topology. I would personally use nets for that (it's more natural) but his way is also possible. So $f \in \overline{R(B^\ast)}$ must imply that $f$ is linear on $B$ and so in $B^\ast$.


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