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I'm wondering about the following:

Let$\ X $ be a topological vector space. Then one could pick balanced neighborhoods$\ W $ and$\ U $ of$\ 0 $ such that

$\ \overline{U} + \overline{U} \subset W $, where $\ U+U:=\{u_1 + u_2 | u_1,u_2 \in U \} $

I was faced this question while reading Rudin's "Functional Analysis". I'm able to prove this, but I think there should be a more elegant and easier way to do it.
Listed are the properties I used:

  • using continuity of$\ + $ one can easily show that there exist $\ U, W $ as above such that $\ U+U \subset W $
  • Use that every topological vector space has a balanced local base (local means here at$\ 0 $.)
  • If $\ \mathcal{B} $ is a local base (in the above sense) for a topological vector space$\ X $ then every member of$\ \mathcal{B} $ contains the closure of some member of$\ \mathcal{B} $.
  • and the last property I used was:$\ \overline{U_1} + \overline{U_2} \subset \overline{U_1+U_2} $ where$\ U_1,U_2 \subset X $

As you can see I need a lot of theory / basic properties about topological vector spaces and I'm just wondering if there's not an easier way. Thx for suggestions.

cheers

math

  • 0
    Also observe that the result you ask about directly implies the first three bullet points at the end of your question, so some manipulations will have to enter the argument.2011-09-18

1 Answers 1

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I have to agree with Theo that there is no easier way to prove this, than from the first principles of the definition of topological vector spaces. You can find it as a part of theorem 3.1 in the book

  • Francois Treves: "Topological Vector Spaces, Distributions and Kernels"

(At least in the edition from 1967 that I have. I see that the content of chapter 3 is different on google books.) Anyway, if you feel the need to crosscheck your own proof, you can look it up there.

It is the first nontrivial theorem based on the definition of TVS.