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I just have a few question about some things in Banach spaces. Let $X$ be a separable, reflexive Banach space with basis $\{e_{i}\}$.

  1. Let $X_{n} = \text{span}\{e_{1},...,e_{n}\}$, then consider the closed ball $\bar{B_{R_{o}}(0)} \subset X_{n}$ for some $R_{o} > 0$. Would it follow immediately that $\bar{B_{R_{o}}(0)}$ is convex and compact in $X$?

  2. Would it follow that $\bar{B_{R_{o}}(0)}$ is homeomorphic to the closed unit ball in $\mathbb{R}^{n}$?

  3. Also if we construct such subspaces $X_{n}$ for each $n \in \mathbb{N}$. Why does it follow that $\cup_{n=1}^{\infty}X_{n}$ is dense in $X$? It seems that $\cup_{n=1}^{\infty}X_{n}$ should immediately be equivalent to $X$.

Thanks for any help.

  • "Basis" means Schauder basis here? – Daniel Fischer May 22 '14 at 22:00
  • @DanielFischer This is from a proof that just stated that 'the basis of $X$ is ${e_{i}}$'. I assumed it was the usual basis. Do you have a good reference for the basics on Schauder basis, definitions and existence proofs? –  May 23 '14 at 14:46
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    There are different concepts that go under the name of "basis". One is the algebraic concept, also called a Hamel basis. No infinite-dimensional Banach space can have a countable Hamel basis. So the meaning cannot be Hamel basis here. For Hilbert spaces, you have Hilbert bases, which are maximal orthonormal subsets. For Banach spaces, the concept of Schauder bases is useful, but not every separable Banach space has a Schauder basis (and non-separable Banach spaces can't have Schauder-bases). There are other concepts of a basis possible, so the question is what "basis" means here. – Daniel Fischer May 23 '14 at 14:58
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    As for references, a not too bad start is wikipedia, and comprehensive books on functional analysis also often treat that concept (not all, Rudin doesn't mention Schauder bases, iirc). E.g. Meise/Vogt, An Introduction to Functional Analysis. – Daniel Fischer May 23 '14 at 15:00
  • @DanielFischer Yes I'm looking at the wikipeida entry but it's not sufficient, will look at the book you recommended. I also saw posts on MSE which deal with separable Banach spaces without a Schauder basis. In the proof I am looking at it states that the Banach space is all of separable and reflexive. Would the extra condition of reflexivity guarantee the existence of the Schauder basis? –  May 23 '14 at 15:06

1 Answers1

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  1. Yes. The space $X_n$ with the norm induced from $X$ is a finite-dimensional normed space. The closed ball is bounded and closed, therefore compact.

  2. Yes. Every convex compact set with nonempty interior in $n$-dimensional space is homeomorphic to the Euclidean closed unit ball. One can follow the proof here.

  3. The [Schauder] basis assumption implies that $\{e_i\}$ has dense linear span in $X$. This linear span is exactly the union of all spaces $X_n$. So the union is dense. To see that it need not be all of $X$, take, for example, $\ell^2$, and observe that $\bigcup_{n=1}^\infty X_n$ does not contain the element $a=(1,1/2,1/3,1/4,\dots)$. Indeed, if $a$ was an element of the union $\bigcup_{n=1}^\infty X_n$, it would have to belong to some $X_n$, which is clearly false.

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  • for 1. Is it always the case that every subset which is compact in a subspace(in this case $X_{n}$) is therefore compact in the whole space? for 3. My understanding about Schauder basis is that any vector can be represented as an infinite linear combination of Schauder basis elements and that as you state it is dense in $X$. Do you have a good reference for the basics of 'Schauder basis'? So ${ e_{i} }$ is not a Hamel basis as well? And how do we know that the space has a Schauder basis? Thanks. –  May 23 '14 at 14:37
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    @Moses 1. Yes. Compactness does not depend on the ambient space. 3. Wikipedia. No, the Schauder basis is never a Hamel basis except in finite dimensional case. The way your question is phrased, the existence of basis $(e_n)$ is assumed. If you want to know for which Banach spaces a Schauder basis exists, ask it in a new question. –  May 23 '14 at 14:59
  • Okay will do. Would I be right in stating that the compactness does of depend on the ambient space since a convergent sequence converges in both the subspace and ambient space, given that it is contained in the subspace of course? But is this the case in a general topological space? –  May 23 '14 at 15:11