# Functional analysis - question about separable dual spaces

1. Oct 1, 2012

### 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?

2. Oct 1, 2012

### micromass

Staff Emeritus
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.

3. Oct 3, 2012

### AxiomOfChoice

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

4. Oct 3, 2012

### micromass

Staff Emeritus
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.

5. Oct 8, 2012

### AxiomOfChoice

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 $0<p<1$?

6. Oct 8, 2012

### micromass

Staff Emeritus
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.

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).

7. Oct 8, 2012

### AxiomOfChoice

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?

8. Oct 8, 2012

### micromass

Staff Emeritus
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.

9. Oct 9, 2012

### AxiomOfChoice

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!

10. Oct 9, 2012

### micromass

Staff Emeritus
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.

11. Oct 9, 2012

### AxiomOfChoice

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.