Let $l_{\mathbb{C}}^{\infty}(\mathbb{N})$ be the space of bounded complex-valued sequences, $l_{\mathbb{R}}^{\infty}(\mathbb{N})$ the subspace of real-valued sequences. Given any Banach limit $L_1: l_{\mathbb{R}}^{\infty}(\mathbb{N}) \to \mathbb{R}$, you can define a Banach limit $L: l_{\mathbb{C}}^{\infty}(\mathbb{N}) \to \mathbb{C}$ by letting $L((a_n + i b_n)) = L_1((a_n)) + i.L_1((b_n))$.

However, the converse is also true - that any Banach limit on $l_{\mathbb{C}}^{\infty}(\mathbb{N})$ is an extension of some Banach limit on $l_{\mathbb{R}}^{\infty}(\mathbb{N})$. This is equivalent to saying that any Banach limit on $l_{\mathbb{C}}^{\infty}(\mathbb{N})$ takes only real values on $l_{\mathbb{R}}^{\infty}(\mathbb{N})$.

The proof that I know uses contradiction to show that if a Banach limit takes the value $i$ at some bounded real sequence, then it cannot have operator norm $1$, so it is more technical than intuitive. I was wondering if there are any other intuitive proofs, or even some reasoning for why this should be true?

EDIT: Say we define a Banach limit as a linear functional $L$ with the following properties:

  1. $||L||_{op}=1$
  2. $ker(L) \supset M = \{x-Sx: x \in l_{\mathbb{C}}^{\infty}(\mathbb{N})\}$, where $Sx$ is the left shift $S(x_1,x_2, ...) = (x_2, x_3, ...)$
  3. $L(\mathbb{1}) = 1$

In the real-valued case, the usual definition of a Banach limit follows from these three properties. In particular, positivity is proved by scaling the sequence $x$ to lie between $0$ and $1$, but the proof that $L(\frac{x}{||x||_{\infty}}) = 1 - L(\mathbb{1}-\frac{x}{||x||_{\infty}}) \geq 0$ seems to use the fact that $L$ is real-valued. So what is it about these three properties that forces the Banach limit to take real values on $l_{\mathbb{R}}^{\infty}(\mathbb{N})$?

I hope this question is not too vague! I'm just curious about the differences in the theory of real and complex vector spaces in functional analysis.

  • $\begingroup$ What's your definition of a Banach limit? If you use Wikipedia's definition, for instance, this follows immediately from the positivity assumption. $\endgroup$ – Eric Wofsey Mar 15 at 4:27
  • $\begingroup$ Ah, that's true! However, our construction of a Banach limit in the real-valued case prescribes a linear functional with three properties that seemingly generalise to the complex case, and positivity in the real case follows from these. I've updated the question to include a definition. Thanks for catching that! $\endgroup$ – nammie Mar 15 at 8:58

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.