0
$\begingroup$

I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.

Proposition 5: Let $X$ be a linear space and W a subspace of $X^\#$ (space of linear functionals on X). Then a linear functional $\psi: X \rightarrow \mathbb{R}$ is W-weakly continuous if and only if it belongs to W.

Definition of "$\mathcal{F}$-weakly continuous": If $\mathcal{F}$ is any collection of real-valued functions on a set $X$, the $\mathcal{F}$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $\mathcal{F}$ are continuous. A function that is continuous with respect to the $\mathcal{F}$-weak topology is called $\mathcal{F}$-weakly continuous.

Proof:

By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $\psi: X \rightarrow \mathbb{R}$ is $W$-weakly continuous . By the continuity of $\psi$ at $0$, there is a neighborhood $\mathcal{N}$ of $0$ for which $|\psi(x)| = |\psi(x)- \psi(0)| < 1$ if $x \in \mathcal{N}$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $\mathcal{N}$. Choose $\epsilon > 0$ and $\psi_1, ...,\psi_n$ in $W$ for which $\mathcal{N}_{\epsilon,\psi_1,...,\psi_n} \subset \mathcal{N}$. Thus $$ |\psi(x)| < 1 \text{ if } |\psi_k(x)|<\epsilon \text{ for all } 1\le k \le n$$

By the linearity of $\psi$, and $\psi_k$'s, we have the inclusion $\bigcap_{k=1}^n \ker \psi_k \subset \ker \psi$.

By Proposition 4, $\psi$ is a linear combination of $\psi_k$. Therefore, since W is a linear subspace, $\psi \in W$. $$\tag*{$\Box$}$$

How I understand this is as follows: The step I highlight implies $\bigcap_{k=1}^n \{x: \psi_k(x) < \epsilon\} \subset \{x: \psi(x) < 1\}$. The $\epsilon$ has been fixed, and the $\psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c \in \mathbb{R}$, \begin{align*} \bigcap_{k=1}^n \{x: \psi_k(x) < \epsilon\} \subset \{x: \psi(x) < 1\} \Leftrightarrow& \bigcap_{k=1}^n \{x: \psi_k(cx) < \epsilon\} \subset \{x: \psi(cx) < 1\}\\ \Leftrightarrow& \bigcap_{k=1}^n \{x: \psi_k(x) < \frac{\epsilon}{c}\} \subset \{x: \psi(x) < \frac{1}{c}\} \end{align*}

Taking c arbitrarily large will yield $\bigcap_{k=1}^n \ker \psi_k \subset \ker \psi$ ?

$\endgroup$
  • $\begingroup$ I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $\mathcal{N}_{\epsilon,\psi_1,...,\psi_n}$. I can guess what it should be but it would help other readers if the notation used is defined. $\endgroup$ – Rhys Steele Aug 1 '18 at 11:45
1
$\begingroup$

You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that \begin{align*} \bigcap_{c \in \mathbb{N}} \bigg(\bigcap_{k=1}^n \{x: \psi_k(x) < \frac{\epsilon}{c}\}\bigg) \subset \bigcap_{c \in \mathbb{N}} \bigg(\{x: \psi(x) < \frac{1}{c}\} \bigg) \end{align*} and then convince yourself that the left hand side is $\bigcap_{k=1}^n \ker \psi_k$ and the right hand side is $\ker \psi$. If I'm honest, this argument feels unnecessarily complicated to me.

A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x \in \bigcap_{k=1}^n \ker \psi_k$ and show it's in $\ker \psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |\psi_k(cx)| < \epsilon$ for all $c > 0$, $|\psi(cx)| < 1$ and so $|\psi(x)| < \frac1c$. Since $c>0$ was arbitrary, $|\psi(x)| = 0$ so $x \in \ker \psi$.

$\endgroup$
1
$\begingroup$

Your argument is correct. But (perhaps) a better argument is the following: let $\psi_k(x)=0$ for $1\leq k \leq n$. Let $n$ be any positive integer. Then $\psi_k(Nx)=N\psi_k(x)=0$ for $1\leq k \leq n$. Hence $\psi (Nx) <1$ and $\psi (x) <\frac 1 N$. Since this holds for all $N$ we get $\psi (x) \leq 0$. Changing $x$ to $-x$ we get $\psi (x)\geq 0$. Hence $\psi (x)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.