2
$\begingroup$

Let $X$ be a complex vector space, and let $\{X_n\}_{n=1}^\infty$ be a sequence of vector subspaces such that $X_n \subseteq X_{n+1}$ for all $n$ and $X = \bigcup_n X_n$. Furthermore, suppose that:

  1. Each $X_n$ is a locally convex topological vector space whose topology $\tau_n$ is determined by a separating family of seminorms.

  2. Each $\tau_n$ coincides exactly with the subspace topology that $X_n$ inherits from $X_{n+1}$.

Let $\mathcal{U}$ be the collection of all convex, balanced sets $W \subseteq X$ such that $0 \in W$ and $W \cap X_n \in \tau_n$ for all $n$.

I would like to show that $ \mathcal{B} = \{x + W : x \in X, W \in \mathcal{U}\}$ is a basis for a topology on $X$.

This is the construction for an inductive limit of Frechet spaces in Reed and Simon's Methods of Modern Mathematical Physics, Vol.1.

In order to show that $\mathcal{B}$ is suitable a basis for a topology on $X$, I need to show:

a) That the elements of $\mathcal{B}$ cover $X$,

b) If we are given elements $x_1 + W_1$ and $x_2 + W_2$ in $\mathcal{B}$ with $x \in (x_1 + W_1) \cap (x_2 + W_2)$, then there exists $x_3 + W_3 \in \mathcal{B}$ so that $x \in x_3 + W_3 \subseteq (x_1 + W_1) \cap (x_2 + W_2)$.

Showing a) is easy since $\mathcal{B} \ni 0 + X = X$. But I am having trouble showing b). My thoughts go something like this: if $x_1 + W_1$, $x_2 + W_2$, and $x$ are given as above, choose $n$ large enough so that $x_1, x_2, x \in X_n$. Then $W_1 \cap X_n, W_2 \cap X_n$ are open, balanced, convex sets about $0$ in $X_n$. Because we are now working in the seminorm topology of $X_n$, we should be able to get $W_3$ open (in $X_n$), convex, balanced, and containing $0$ so that

$$x \in x_3 + W_3 \subseteq (x_1 + W_1 \cap X_n) \cap (x_2 + W_2\cap X_n).$$

But I am not sure how to go from here. I'm wondering if there's some way to "lift" this $W_3$ up to the status of an element of $\mathcal{U}$? I am aware aware of is this lemma, and I know its proof. I have tried several times to make this lemma work for me, but have not gotten anywhere.

Hints or solutions are greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ First take $x_3:= x$ then use theorem 1.14 p.12-13 of Rudin's Functional Analysis for $V$ as defined by Luiz Cordeiro. (One finds a balanced subset of $V$, a.k.a. balanced core) $\endgroup$ – Noix07 Mar 18 at 17:14
1
$\begingroup$

EDIT: Let $x_1+W_1$, $x_2+W_2$ in $\mathcal{B}$ and $x\in (x_1+W_1)\cap(x_2+W_2)$. Let's show that there exists $W_3\in\mathcal{U}$ such that $x+W_3\subseteq (x_1+W+1)\cap(x_2+W_2)$ (so that we are considering $x_3=x$).

Let $V=(W_1+x_1-x)\cap (W_2+x_2-x)$. Note that $V$ is convex and contains $0$, so we have to work on balancedness. Let $U=\bigcap_{|\alpha|=1}\alpha\cdot V$. Then $U$ is (an intersection of) convex, balanced sets, so $U$ is also convex and balanced (and contains $0$). Clearly, $x+U\subseteq (x_1+W_1)\cap(x_2+W_2)$.

The last problem we have to deal with is the intersections with $X_n$ being open in $X_n$. Let $\mu_U(x)=\inf\left\{t>0:t^{-1}x\in U\right\}$ be the Minkowski functional (gauge) of $U$ (maybe this is cheating, but it is the only way that I could use to solve this problem), and, finally, $W_3=\left\{x:\mu_U(x)<1\right\}$.

You can prove that $\mu_U$ is a seminorm and $W_3\subseteq U$ (see Theorem 1.35 of Rudin, Functional Analysis), so $W_3$ is also convex and balanced (and contains $0$). Let's show that $W_3\cap X_n\in\tau_n$. By definition of Minkowski functionals, it is easy to see that $W_3=\left\{x\in X_n:\mu_{U\cap X_n}<1\right\}$, and, again, $\mu_{U\cap X_n}$ is a seminorm on $X_n$, so we need only to show that it is continuous. Note that $V\cap X_n$ is a neighbourhood of $0$ in $X_n$, so there exists $A$ open, balanced and convex with $0\in A\subseteq V\cap X_n$, hence $A\subseteq U\cap X_n$, thus $\mu_{U\cap X_n}\leq\mu_A$. Since $\mu_A$ is also a continuous seminorm on $X_n$, then $\mu_{U\cap X_n}$ is continuous. (Continuity of $\mu_A$ follows from $\left\{x\in X_n:\mu_A(x)<1\right\}=A$ being open.)


An alternative approach (which I prefer) is the following: The topology on $X$ we are considering is actually the direct/inductive limit topology (of locally convex spaces) induced by the inclusions $X_n\hookrightarrow X$ (Schaefer, Topological Vector Spaces, II.6), that is, it is the strongest locally convex topology on $X$ for which the inclusions are continuous.

Given $W\in\mathcal{U}$, we can consider its Minkowski functional (gauge) $\mu_W(x)=\inf\left\{t>0:t^{-1}x\in W\right\}$, which happens to be a seminorm in $X$. The inductive limit topology is the one induced by the family $\left\{\mu_W:W\in\mathcal{U}\right\}$ (the construction of a locally convex topology from a family of seminorms can be found in Rudin, Functional Analysis, Theorem 1.37).

$\endgroup$
  • $\begingroup$ Thank you for your answer. The comment about inductive limit topology is insightful. However, I have been unable to verify the property that $x \in x_3 + \widetilde{W}_3 \subseteq (x_1 + W_1) \cap (x_2 + W_2)$ for some $x_3$. Could you provide a few details on your inclusion argument? I also think I have come up with an alternate answer to question, which I will post. $\endgroup$ – JZS Jul 23 '14 at 2:39
  • $\begingroup$ There really was a problem with $\widetilde{W}_3$, because we didn't have enough control of the liftings. I changed the proof (its uglier, but I think that it works now). $\endgroup$ – Luiz Cordeiro Jul 23 '14 at 17:05

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.