# The restriction of unit norm functionals to the ball are a closed subset of functionals defined over a ball

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

Claim:

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 $$\mathcal{F}(B)$$.

Proof:

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).

Questions:

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).

• Can you please state the source of the proof you're quoting? – Nate Eldredge Jun 14 at 18:07
• 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) – 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 $$f(\lambda u) = \lambda f(u)$$.

• 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 $\mathcal{F}(B)$? – yoshi Jun 14 at 18:56
• @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. – Nate Eldredge Jun 14 at 18:58
• I'm having trouble seeing that we showed that every element is in the closure. The next comment contains my (probably incorrect) reasoning. – yoshi Jun 14 at 19:23
• @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 (*). – Nate Eldredge Jun 14 at 19:28
• 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 $X$. – 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 $$B^\ast$$.