Every Banach limit on $l_{\mathbb{C}}^{\infty}(\mathbb{N})$ is an extension of some Banach limit on $l_{\mathbb{R}}^{\infty}(\mathbb{N})$

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.

• What's your definition of a Banach limit? If you use Wikipedia's definition, for instance, this follows immediately from the positivity assumption. – Eric Wofsey Mar 15 at 4:27
• 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! – vxnture Mar 15 at 8:58