6

So I know that a completion of $X$ is a Banach space $Y$ such that $X$ is isometrically isomorphic to a dense subset of $Y$, say $A$.

So we need to prove that we can always find a $T \in L(X,A)$ such that T is bounded and $\|T\| = 1$. How do we construct this?

And if we have more completions, are these completions isomorphic as well?

Kees

Martin Sleziak
  • 53,687
Kees Til
  • 1,958
  • I think you construct the completion by taking the quotient of Cauchy successions... – Riccardo Orlando May 02 '16 at 12:04
  • Completions are unique up to isometric isomorphism. There are several constructions of a completion of a normed space, but the isometric embedding of $X$ into the completion is part of the construction in every case. – Daniel Fischer May 02 '16 at 12:08
  • 1
    See section 6 of the following notes for some contructions: https://math.mit.edu/~rbm/18-102-S14/Chapter1.pdf – Mathematician 42 May 02 '16 at 12:09
  • @DanielFischer : what do you mean ? in every case the limits of Cauchy sequences are in the obtained complete space, and nothing else ? – reuns May 02 '16 at 12:11
  • 1
    @user1952009 "The construction" is not just the constructed space, I used it in the sense of the process of constructing the completion. In any case, however, a completion of a normed space consists of both, a Banach space, and an isometric embedding, by definition of completion. – Daniel Fischer May 02 '16 at 12:16
  • For example, with the Cauchy sequence construction, $T$ should map the vector $x \in X$ to the equivalence class of the constant sequence $(x,x,x,x,\cdots)$ in the completion $Y$. – GEdgar May 02 '16 at 12:51
  • The definition of a completion includes the existence of your map $T$. The definition of the completion of a normed space $(X,|\cdot|{X})$ is a complete normed linear space $(Y,|\cdot|{Y})$ and a linear map $T : X\rightarrow Y$ such that (a) $|Tx|{Y}=|x|{X}$ and (b) $TX$ is dense in $Y$. – Disintegrating By Parts May 02 '16 at 12:57

1 Answers1

15

You can construct the completion of a normed vector space $X$ by using a construction of the completion of a general metric space and then defining a normed vector space structure on that completion.

But there's a better way, or at least a way that a lot of people might regard as simpler and more elegant: If $X$ is a normed vector space then the dual $X^*$ is a Banach space, and $X\subset X^{**}$ just as if $X$ were a Banach space; the completion of $X$ is just its closure in $X^{**}$.

(More precisely, the completion of $X$ is the closure of $j(X)$, where $j:X\to X^{**}$ is the canonical injection; so $j$ is exactly the $T$ you ask for.)

David C. Ullrich
  • 89,985