Let $X$ be a topological vector space (over $\mathbb{C}$) whose topology is defined by a family of separating seminorms $\{\rho_\alpha\}$. Let $X_1$ be a vector subspace of $X$ whose topology is the relative topology inherited from $X$ (it can be show, then, that the topology on $X_1$ is simply determined by the seminorms restricted to $X_1$). Suppose that $V$ is an open, convex, balanced neighborhood of $0$ in $X_1$ ($V$ being balanced means that $y \in Y, \lambda \in \mathbb{C}, |\lambda| = 1$ implies $\lambda y \in Y$).

I would like to show that there exists open, convex, balanced, $Z$ in $X$ such that $Z\cap X_1 = V$.

This is a lemma stated just after Theorem V.15 in Reed and Simon's Methods of Modern Mathematical Physics Vol. 1. This theorem is all about constructing an inductive limit of Frechet spaces.

A proof of the lemma is given, but I am having trouble understanding the proof. I will summarize the proof below and identify the step that I do not understand:

  1. Since $X_1$ has the relative topology, there exists open $O$ in $X$ so that $O \cap X_1 = V$.

  2. Because $O$ is open in the locally convex topology on $X$, and contains $0$, there is an open, convex, balanced set $O_1 \subseteq O$ with $0 \in O_1.$

  3. Define $Z = \{ \alpha x + \beta y : x \in O_1, y \in V, |\alpha| + |\beta| = 1\} = \bigcup_{y \in V,|\alpha| + |\beta| = 1, \alpha \neq 0}(\beta y + \alpha O_1)$.

  4. This set $Z$ is open, balanced, and convex with $Z \cap X_1 = V$.

The portion of the proof I do not understand is why $Z$ is a convex set.

I have taken a look at this previous question, and it seems that there may be a mistake in the proof of the lemma. That is, $Z$ may not be convex : usually, one constructs the convex hull of a set by taking finite convex combinations of all sizes (not just convex combinations of two elements).

I am wondering if there is a way to adjust the definition of $Z$ so that it remains open and balanced, but is also convex?

Hints or solutions are greatly appreciated.


1 Answer 1


Starting with step 3 in the question above, we need to modify things slightly.

First, change the definition of $Z$ slightly from the one above. $$Z = \{\alpha x + \beta y : x \in O_1, y \in V, |\alpha| + |\beta| \le 1\} $$

So we are using the condition $|\alpha| + |\beta| \le 1$ whereas, above we use $|\alpha| + |\beta| = 1.$

It's easy to see:

$$Z = \bigcup_{y \in V, |\alpha| + |\beta| \le 1, \alpha \neq 0} \beta y + \alpha O_1.$$

So $Z$ is a union of open sets, and is therefore open.

Next, we still have $Z \cap X_1 = V$ as before: showing $V \subseteq Z \cap X_1$ is trivial, so let's just work on the other containment:

Suppose $x_1 \in Z \cap X_1$. So then $x_1 = \alpha x + \beta y$ where $x \in O_1$, $y \in V$, $\alpha \neq 0$, $|\alpha| + |\beta| \le 1$, and $x_1 \in X_1.$ But then $x = \alpha^{-1}(x_1 - \beta y) \in X_1$. Then $x \in O_1 \cap X_1 \subseteq O \cap X_1 = V$. Therefore, $ \alpha x + \beta y$ is a balanced-convex combination. That is, $ \alpha x + \beta y$ of the form $\sum_{i=1}^N \eta_i y_i$, where $\sum_i |\eta_i| \le 1$ and each $y_i \in V$. It is a short exercise to show that any balanced, convex set containing $0$ contains all of its balanced-convex combinations. Hence $ \alpha x + \beta y \in V$ and we have completed the argument that $Z \cap X_1 = V$.

It remains to show, then, that $Z$ is convex and balanced. Well, let $W$ be the balanced, convex hull of $O_1 \cup V$. That is: $$W = \{\text{finite sums }\sum_{i=1}^N \eta_i v_i : \sum_{i=1} |\eta_i| \le 1, \text{and each } v_i \in O_1 \cup V\}$$

It's easy to see that $W$ is convex, balanced, and that $Z \subseteq W$. So all that's left to see is that $W \subseteq Z$. Well, let $\sum_{i=1}^N \eta_i v_i \in W$ (WLOG, we can assume all $\eta_i \neq 0$). Since the $v_i$ all belong to $O_1 \cup V$, we can of course split the finite sum into two terms:

$$\sum_{i=1}^N \eta_i v_i =\sum_{i=1}^M \sigma_i x_i + \sum_{j=1}^L \gamma_j y_j, $$

where each $x_i \in O_1$, each $y_j \in V$, $M + L = N$, and $\sum_{i=1}^M |\sigma_i|+\sum_{j=1}^L |\gamma_j| \le 1$.

Let's observe how we can rewrite the first term:

$$\sum_{i=1}^M \sigma_i x_i = \sum_{i=1}^M |\sigma_i|\left(\sum_{i=1}^M \frac{\sigma_i} {\sum_{i=1}^M |\sigma_i|}x_i\right). $$

The term in parenthesis (call it $\bar{x}$) is a balanced-convex combination of elements in $O_1$ (hence $\bar{x} \in O_1$ because $O_1$ is convex and balanced). Similarly we can show that the term $\sum_{j=1}^L \gamma_j y_j = \sum_{j=1}^L |\gamma_j| \bar{y}$ for some $\bar{y} \in V$. All of this demonstrates that

$$\sum_{i=1}^N \eta_i v_i = \sum_{i=1}^M |\sigma_i| \bar{x} + \sum_{j=1}^L |\gamma_j| \bar{y} \in Z.$$

And this completes the proof of the lemma.


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