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In this conversation, it was proved that an infinite-dimensional Hilbert space H has a countable orthonormal basis iff H is separable (has a countable dense subset).

Is there any analogous statement for Banach spaces,
(with some generalisation of the notion of orthogonality)?

Michael_1812
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    Well, it wouldn't have an "orthonormal" basis necessarily... – David Raveh Jan 06 '24 at 03:48
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    A normed vector space is separable iff there exists a sequence of linearly independent unit vectors $(u_n)$ such that the union of the vector spaces $\langle u_1, \ldots, u_k \rangle$ is dense in the normed space. I am guessing the vectors $(u_k)$ will play a similar role to the orthogonal basis, however, there is no guarantee (that I am aware of) that the family $(u_k)$ will be a Schauder basis – William M. Jan 06 '24 at 03:53
  • @DavidRaveh Thanks for your point. I have clarified my question. – Michael_1812 Jan 06 '24 at 04:00
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    You are considering now a vector space with two norms, one of which comes from an inner product. If the norms are not equivalent, the normed spaces obtained by these two norms may not be complete simultaneously and convergence of series may fail for one of them while existing for the other one... In other words, your questions is a little too broad. If the norms are equivalent, then they generate the same topology and the orthonormal basis will work for both norms... – William M. Jan 06 '24 at 04:05
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    Based on the edit, which says "in Banach spaces, which have dot products that are not related to the norm", it's not clear what you're asking. Are you concerned only with the original norm (in which case the dot product plays no part, but we need to find a suitable substitute for the notion of orthogonality), or are you concerned with both norms at once, as William's comment asks? Since the original question made no mention of any dot product, I'm inclined to think it's the former. –  Jan 06 '24 at 04:08
  • @jwhite I am concerned with the original norm only (though, yes, the existence of a dot product in fact gives birth to a different, Hilbert-type norm). – Michael_1812 Jan 06 '24 at 04:33
  • @WilliamM. I have edited my question, to explain that I have in mind two different norms, $||x||$ and $\sqrt{(x,x)}$ unrelated to one another. And I believe you have answered my question. Thank you. If you put your explanation not as a comment but as an answer, it will be my pleasure to give you full credit in terms of points. – Michael_1812 Jan 06 '24 at 04:35
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    If the norm induced by the dot product is not related to the original norm, and you're only concerned with the original norm, why are you mentioning the dot product at all? –  Jan 06 '24 at 04:45
  • @jwhite This was a part of my confusion: do we or do we not need to define any kind of orthogonality of the basis? If not, then no dot product is needed at all. Would you recommend me to delete it from my question as redundant? – Michael_1812 Jan 06 '24 at 05:10
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    Since you're only concerned with the original norm, which need not be related to a dot product, the first question is what is the right generalization of the notion of orthogonality, since we do not have such a notion in general Banach spaces. I have offered one such generalization in my answer below (although I could see trying to construct others, but to me that's the most natural). If you're satisfied with biorthogonality being the natural way to generalize (and indeed, orthonormality is a special case), then I would say yes, mentioning an unrelated dot product in the question is redundant. –  Jan 06 '24 at 05:16
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    @jwhite I very much appreciate your time and help, and wish you a happy and successful year. – Michael_1812 Jan 06 '24 at 05:18
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    You probably want to work in the area of reflexive spaces to have any hope of mimicking Hilbert space theory. – Cameron L. Williams Jan 06 '24 at 05:31
  • @CameronWilliams Would it be in your power to explain in short what reflexive spaces are? – Michael_1812 Jan 06 '24 at 06:30
  • @geetha290krm in what sense can a sentence be "very wrong" (as opposed to simply "wrong"/"false")? And exactly what am I saying that is not true? – William M. Jan 07 '24 at 18:07

1 Answers1

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For a Banach space $X$, let $X^*$ denote the dual space (the space of continuous, linear functionals $x^*:X\to \mathbb{K}$, where $\mathbb{K}$ is the scalar field, which is either $\mathbb{R}$ or $\mathbb{C}$). Somewhat suggestively, we denote the action of a functional $x^*$ on the vector $x$ as $\langle x,x^*\rangle$, but we note that $\langle \cdot,\cdot\rangle$ is not an inner product, since the first argument comes from $X$ and the second comes from $X^*$.

In general Banach spaces, the appropriate notion is the notion of biorthogonality. Let $X$ be a Banach space, $A$ a set, $\alpha:A\to X$, and $\beta:A\to X^*$ functions. We say the collection $(\alpha(a), \beta(a))_{a\in A}\in \prod_{a\in A}X\times X^*$ is biorthogonal if for $a,b\in A$, $$\langle \alpha(a),\beta(b)\rangle = \left\{\begin{array}{ll} 1 & : a=b \\ 0 & : a\neq b.\end{array}\right.$$

Suppose we have a Hilbert space with inner product $(\cdot,\cdot):H\times H\to \mathbb{K}$. By (one version of) the Riesz Representation Theorem, there is a canonical conjugate-linear isometric surjection $\phi:H\to H^*$. Then for any set $A$ and any function $\alpha:A\to H$, we can naturally build $\beta:A\to H^*$ as $\beta(a)=\phi(\alpha(a))$. The collection $(\alpha(a))_{a\in A}$ is orthonormal if and only if $(\alpha(a),\beta(a))_{a\in A}=(\alpha(a),\phi(\alpha(a)))_{a\in A}$ is biorthogonal. It is for this reason that I view biorthogonality as the appropriate generalization of the notion of orthonormality. However, it is no longer the case that biorthogonality has some relation to the norm, whereas orthonormality implies the vectors $\alpha(a)$ each have norm $1$.

Let $X$ be a Banach space and let $X^*$ be the dual space (the space of continuous, linear functionals on $X$). A Markushevich basis (or $M$-basis) is a set $A$ together with a pair of functions $\alpha:A\to X$, $\beta:A\to X^*$ such that

  1. $\text{span}\{\alpha(a):a\in A\}$ is dense in $X$,
  2. $(\alpha(a),\beta(a))_{a\in A}$ is biorthogonal, and
  3. $\bigcap_{a\in A}\text{ker}(\alpha(a))=\{0\}$.

Conditions 1,2 yield a biorthogonal system whose first components have dense span in $X$. Condition 3 says that, although the second components need not have dense span in $X^*$, there are still sufficiently many functionals in $\{\beta(a):a\in A\}$ to separate points. We typically use the notation $x_a$ instead of $\alpha(a)$ and $x^*_a$ instead of $\beta(a)$, so the Markushevish basis is denoted $(x_a,x^*_a)_{a\in A}$. We have the following.

A Banach space $X$ admits an $M$-basis $(x_n,x^*_n)_{n\in \mathbb{N}}$ iff $X$ is separable.

There is also the notion of a Schauder basis for $X$. A Schauder basis is a sequence $(x_n)_{n=1}^\infty$ such that for each $x\in X$, there exists a unique scalar sequence $(a_n)_{n=1}^\infty$ such that $x=\sum_{n=1}^\infty a_nx_n:=\lim_N\sum_{n=1}^N a_nx_n$. This implies a property called the approximation property (and even stronger properties). Per Enflo gave the first example of a separable Banach space which lacked the approximation property, and therefore does not admit any Schauder basis. So the $M$-basis result above seems to be the full extent to which a general analogue holds.

As a side note, if we're only concerned with the original norm, and not necessarily a norm which comes from an inner product, we can note that for any Banach space $(X, \|\cdot\|)$, there is an inner product $(\cdot,\cdot)$ on $X$ such that, with $|x|=\sqrt{(x,x)}$ the norm induced by the inner product, both $(X, \|\cdot\|)$ and $(X, |\cdot|)$ are complete. However, there need not be any real connection between $\|\cdot\|$ and $|\cdot|$, so the fact that this dot product exists does nothing to address the original question about $(X, \|\cdot\|)$. That's because neither orthogonality nor normality with respect to $(\cdot,\cdot)$ and $|\cdot|$ have any connection to $\|\cdot\|$. It does not induce any interpretation of "orthogonality" with respect to the original norm $\|\cdot\|$, and $\|\cdot\|$-normality need not be connected to $|\cdot|$ normality. In fact, in the infinite dimensional case, you can guarantee that $\{\|x\|/|x|:0\neq x\in X\}=(0,\infty)$.