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.
  • $\begingroup$ @AnneBauval Thank you for your answer. I am having some trouble seeing how if $U$ contains $\{x\mid\rho_U(x) < 1 \}$ or how the converse ($\{x\mid\rho_U(x) < 1 \}$ containing $\{x\mid\rho_U(x)\le1/2\}\supset\frac12U$) implies that the two topologies are the same. $\endgroup$
    – CBBAM
    May 29 at 8:14
  • $\begingroup$ Also for (3), how do the sets $\{x\mid\rho(x) < 1 \}$ form a neighborhood base for 0? And how do we know that these sets are necessarily balanced, convex, and absorbing? I apologize for all the questions, I am new to seminorms and locally convex spaces. $\endgroup$
    – CBBAM
    May 29 at 8:16
  • 1
    $\begingroup$ @AnneBauval Reed & Simon define a locally convex space to be a vector space with a family of seminorms separating points. I am still a little confused, but I will think about this proof some more and return to this question later today. Thank you for all your help so far. $\endgroup$
    – CBBAM
    May 29 at 8:29

1 Answer 1

  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.
  • $\begingroup$ Thank you very much! Everything is clear now. $\endgroup$
    – CBBAM
    May 29 at 16:19

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