Proving Non-Separability of $\ell^{\infty}$ and Duality with $\ell^1$

[benorin]

I would like to prove that $\ell^{\infty}$, namely the Banach space whose elements are sequences of complex numbers that have a fininte infinity-norm (a.k.a. the supremum-norm,) that is for [itex]\alpha = \left\{ \alpha_k \right\}_{k=1}^{\infty},[/tex]<br /> <br /> $$\ell^{\infty}=\left\{\alpha : \sup\left\{ \left| \alpha_n \right| : n\in\mathbb{N}\right\}<\infty \right\},$$ normed by $$\| \alpha\|_{\infty}=\sup\left\{ \left| \alpha_n \right| : n\in\mathbb{N}\right\}$$<br /> <br /> is not separable, that is that it does not possesses a countable dense basis. I do not well understand what it means for a space to be separable <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /> : does it mean that any element (e.g. any sequence, vector, function, ...) of that space can be expressed as either a linear combination of the elements (say, functions) of some countable (basis?) set or a limit thereof?<br /> <br /> Would someone please clearly explain this topic that I might more fully understand it, and, perhaps, the concept of dual spaces: specifically, why is $\ell^{\infty}=(\ell^{1})^*$? where X* denotes the dual space of X and where $\ell^{1}$ is the Banach space of sequences of complex numbers defined by<br /> <br /> $$\ell^{1}=\left\{\alpha : \sum_{k=1}^{\infty} \left| \alpha_k \right| <\infty \right\},$$ normed by $$\| \alpha\|_{1}=\sum_{k=1}^{\infty} \left| \alpha_k \right|$$<br /> <br /> And, why, despite this relationship, is $(\ell^{\infty})^*=(\ell^{1})^{**}\neq \ell^{1}$? I realize that I have asked a lot, but I would rather that sufficient information be put forth that I could join-in the discussion.<br /> <br /> Thanks,<br /> --Ben[/itex]

 

[matt grime]

Let me do the second one first, and get it out the way.

l^infinity is the dual space of l^1 'because it is'. you can work out what the linear functionals on l^1 are and they are precisely the space of bounded sequences.

the reflexive property fails 'because it does', ie you can find a linear functional on l^infinity that is not in l^1, obviously l^1 is contained in the dual space of l^infinity but it is not all of it.

you should try to prove these things to your self. but do not try to work out what the dual space of l^infinity actually is, because it will give you a headache.

 

[benorin]

I'm a little confused...

Regarding functionals on the dual space of a Banach space, I'm a little confused...

So let X be a Banach space, a functional on X is a mapping of vectors in X to complex numbers (or whatever the scalar field is,) correct? Then a functional on X* is what? a vector in X that maps functionals in X* to complex numbers?

 

[matt grime]

So let X be a Banach space, a functional on X is a mapping of vectors in X to complex numbers (or whatever the scalar field is,) correct?

add in the word linear and yes

Then a functional on X* is what?

why don't you put X* in the sentence above, since that is its definition?

Incidentally there is an existential proof that the double dual of l^1 is not l^1.

Thm: Let X be a banach space, then if X^* is separable so is X.

Cor. l^infinity dual is not l^1.

Proof of cor: if l^infinity dual were l^1(which is separable) then applying the theorem l^infinity would be separable, but we know it isn't.