# Question about the basis for the topology on an inductive limit of Frechet spaces

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.

• 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) – Noix07 Mar 18 at 17:14

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).

• 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. – JZS Jul 23 '14 at 2:39
• 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). – Luiz Cordeiro Jul 23 '14 at 17:05