A Hausdorff topological vector space is locally convex if and only if $0$ has a neighborhood base of balanced, convex, absorbing space

functional-analysislocally-convex-spacestopological-vector-spaces

The following is Theorem V.1 in Reed & Simon's book on functional analysis.

Let $V$ be a vector space with a Hausdorff topology in which addition and scalar multiplication are separately continuous. Then $V$ is a locally convex space if and only if $0$ has a neighborhood base of balanced, convex, absorbing sets.

In their proof they reference the following lemma about the Minkowski functional or gauge of an absorbing set $C$ with the additional property that if $x \in C$ and $0 \leq t \leq 1$, then $tx \in C$.

(a) If $t \geq 0$, then $\rho(tx) = t \rho(x)$ for the gauge of any set $C$.

(b) $\rho$ obeys $\rho(x + y) \leq \rho(x) + \rho(y)$ if $C$ is convex.

(c) $\rho$ obeys $\rho(\lambda x) = |\lambda| \rho(x)$ if $C$ is circled/balanced.

(d) $\{x | \rho(x) < 1 \} \subset C \subset \{x | \rho(x) \leq 1\}$.

Their proof of Theorem V.1 is:

Let $\mathscr{U}$ be a neighborhood base at $0$ containing only convex, balanced, absorbing sets; for each $U \in \mathscr{U}$, let $\rho_U$ be the gauge of $U$. By (b) and (c) of the lemma, $\rho$ is a seminorm and by (d) the neighborhoods of $0$ in the original topology are the same as those in the locally convex topology given by the seminorms $\{\rho_U | U \in \mathscr{U}\}$. Since addition is separately continuous in both topologies, the neighborhood about any point are identical in the two topologies.

I have three questions about their proof:

  1. How do we know that such a neighborhood base about $0$ exists?
  2. How does it follow from (d) that the neighborhoods of $0$ in the original topology are the same as those in the locally convex topology given by the seminorms $\{\rho_U | U \in \mathscr{U}\}$?
  3. I do not see how the proof sufficiently proves the forward direction, that if $V$ is locally convex then $0$ has a neighborhood base of balanced, convex, and absorbing sets. It seems to me that we have only proved this for locally convex spaces induced by a particular family of seminorms, but not in full generality.

Best Answer

  1. It is the hypothesis of their proof of "if" (they consider "only if" as obvious, but see point 3).
  2. Because $U$ contains $\{x\mid\rho_U(x) < 1 \}$ and conversely, $\{x\mid\rho_U(x) < 1 \}$ contains $\{x\mid\rho_U(x)\le1/2\}\supset\frac12U.$ If two topologies of vector spaces have respectively B and B' as a base of neighborhoods of $0,$ they are the same iff every element of B contains some element of B' and conversely, because this is equivalent to having the same neighborhoods of $0,$ and in a topological vector space the neighborhoods of $0$ determine (by translation) the neighborhoods of any point, hence the topology.
  3. Let $V$ be a locally convex space, and $(\rho_i)_{i\in I}$ be a family of seminorms inducing its topology. To simplify the notations, assume wlog that this family is saturated, i.e. $\forall i,j\in I\quad\exists k\in I\quad\max\circ(\rho_i,\rho_j)=\rho_k.$ Then, by definition of the topology on $V,$ the "balls" $\{x\mid\rho_i(x) < c\}$ for all $i\in I$ and all $c>0,$ form a base of neighborhoods of $0.$ And the "balls" of a seminorm are always balanced, convex and absorbing.
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