Functional analysis - question about separable dual spaces

[AxiomOfChoice]

Suppose X is a normed space and X*, the space of all continuous linear functionals on X, is separable. My professor claims in our lecture notes that we KNOW that X* contains functionals of arbitrarily large norm. Can someone explain how we know this, please?

 

[micromass]

Isn't that pretty obvious?? If T is a nonzero linear functional, then we can look at nT (for n a positive integer). This has norm $n\|T\|$ and can be made large.

 

[AxiomOfChoice]

Isn't that pretty obvious?? If T is a nonzero linear functional, then we can look at nT (for n a positive integer). This has norm $n\|T\|$ and can be made large.

It certainly is obvious, if we take for granted that there exist nonzero linear functionals! But is that true for an arbitrary normed space?

 

[micromass]

It certainly is obvious, if we take for granted that there exist nonzero linear functionals! But is that true for an arbitrary normed space?

Of course not, since the space could be {0}, in which case there are no nonzero functionals. Other normed spaces do have nonzero linear functionals. This follows from the Hahn-Banach theorem.

 

[AxiomOfChoice]

Of course not, since the space could be {0}, in which case there are no nonzero functionals. Other normed spaces do have nonzero linear functionals. This follows from the Hahn-Banach theorem.

But is this really the only kind of space where we have no nonzero functionals? For example, if $f\in X^*$, we know that the inverse image under $f$ of every open set in $\mathbb C$ is open, so in particular we must have

$$f^{-1} \left( \left\{ |z| < 1 \right\} \right) = \left\{ x\in X : |f(x)| < 1 \right\}.$$

open. But this set is also convex, so our $X$ must contain at least some open convex sets. Doesn't this fail for, say, the nonzero Banach space $\ell^p$ for $0<p<1$?

 

[micromass]

But is this really the only kind of space where we have no nonzero functionals? For example, if $f\in X^*$, we know that the inverse image under $f$ of every open set in $\mathbb C$ is open, so in particular we must have

$$f^{-1} \left( \left\{ |z| < 1 \right\} \right) = \left\{ x\in X : |f(x)| < 1 \right\}.$$

open. But this set is also convex, so our $X$ must contain at least some open convex sets.

Indeed. You have now discovered a more general criterion for the existence of nonzero functionals: there must be "enough" convex sets. Enough means here that at any point x in X, there must be a neighborhood basis consisting out of convex sets. A topological vector space satisfying this criterion is called locally convex. By (a more general form of) the Hahn-Banach theorem, we can show that any locally convex topological space has nonzero functionals.

Doesn't this fail for, say, the nonzero Banach space $\ell^p$ for $0<p<1$?

Sure, this fails. But $\ell^p$ for $0<p<1$ are not Banach spaces! There is no way to put a norm on $\ell^p$. The usual norm

$$\|(x_n)_n\|_p=\sqrt[p]{\sum |x_k|^p}$$

doesn't satisfy the triangle inequality. The best we can do is put a metric on the space by

$$d_p((x_n)_n,(y_n)_n)=\sum |x_k-y_k|^p$$

Under this metric, the space is indeed complete. But this metric does not come from a norm, and is no Banach space (nor locally convex topological vector space).

 

[AxiomOfChoice]

Sure, this fails. But $\ell^p$ for $0<p<1$ are not Banach spaces! There is no way to put a norm on $\ell^p$. The usual norm

$$\|(x_n)_n\|_p=\sqrt[p]{\sum |x_k|^p}$$

doesn't satisfy the triangle inequality. The best we can do is put a metric on the space by

$$d_p((x_n)_n,(y_n)_n)=\sum |x_k-y_k|^p$$

Under this metric, the space is indeed complete. But this metric does not come from a norm, and is no Banach space (nor locally convex topological vector space).

Interesting! I can see why the "standard" p-norm doesn't work, but how does one go about proving that we can't put a norm (of any sort) on this space?

 

[micromass]

Interesting! I can see why the "standard" p-norm doesn't work, but how does one go about proving that we can't put a norm (of any sort) on this space?

Well, it seems that you already know that $\ell^p$ (for 0<p<1) has no nonzero functionals. That would already show it, since any normed space admits nonzero continuous functionals.

 

[AxiomOfChoice]

Of course not, since the space could be {0}, in which case there are no nonzero functionals. Other normed spaces do have nonzero linear functionals. This follows from the Hahn-Banach theorem.

I'm not quite sure how to get this result from the (complex) Hahn-Banach theorem...can you help me with this? Maybe I just have the "wrong" formulation of the theorem; the one I'm looking at talks about defining a real-valued function $p(x)$ such that $p(\alpha x + \beta y) \leq |\alpha|p(x) + |\beta|p(y)$ on some subspace of $X$. It's certainly not hard to think of such a function (try $p(x) = \|x\|$) or such a subspace (just try $\{0\}$)...but you then need to have a (nonzero) functional $\lambda$ that satisfies $|\lambda(x)| \leq p(x)$ to move any further, and I can't in general think of one!

 

[micromass]

I'm not quite sure how to get this result from the (complex) Hahn-Banach theorem...can you help me with this? Maybe I just have the "wrong" formulation of the theorem; the one I'm looking at talks about defining a real-valued function $p(x)$ such that $p(\alpha x + \beta y) \leq |\alpha|p(x) + |\beta|p(y)$ on some subspace of $X$. It's certainly not hard to think of such a function (try $p(x) = \|x\|$) or such a subspace (just try $\{0\}$)...but you then need to have a (nonzero) functional $\lambda$ that satisfies $|\lambda(x)| \leq p(x)$ to move any further, and I can't in general think of one!

OK. So take $p(x)=\|x\|$.

Now, take $x_0$ an arbitrary nonzero vector. Then $span(x_0)$ is a closed subspace of our normed space (closed because finite-dimensional). Define a functional

$$f:span(x_0)\rightarrow \mathbb{R}:tx_0\rightarrow t\|x_0\|$$

now apply Hahn-Banach.

 

[AxiomOfChoice]

OK. So take $p(x)=\|x\|$.

Now, take $x_0$ an arbitrary nonzero vector. Then $span(x_0)$ is a closed subspace of our normed space (closed because finite-dimensional). Define a functional

$$f:span(x_0)\rightarrow \mathbb{R}:tx_0\rightarrow t\|x_0\|$$

now apply Hahn-Banach.

Thanks a lot for your help, micromass!

So, just to recap: It is not enough to say that one has "functionals of arbitrarily large norm" in an arbitrary normed space; we must specify that the normed space is nonzero. But, once we have done that, we automatically have nonzero functionals, since every normed space is locally convex.