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Let $E$ be a Banach space and $(E)_{\mathcal U}$ be an ultrapower for some ultrafilter $\mathcal U$ on an index set $I$.

It is remarked in a paper that $(E')_{\mathcal U}$ can be naturally embedded into $(E)_{\mathcal U}'$ (where I use $'$ to indicate the normed space dual).

I just want to clarify that this natural embedding is given by the following action.

$$(\varphi_{i})_{\mathcal U}:(x_{i})_{\mathcal U}\mapsto \lim_{i\to\mathcal U}\varphi_{i}(x_{i})$$

Is this correct? (sorry if I chose poor tags)


$\bf{\text{Definition}}$:

$\lim_{i\to \mathcal U}x_i = x$ in a topological space $X$ if for every neighbourhood $U$ of $x, \{i\in I : x_i\in U\}\in \mathcal U$.

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    $\begingroup$ How is this limit defined exactly? Is it guaranteed to exist? $\endgroup$
    – Berci
    Jul 8 '13 at 21:37
  • $\begingroup$ It is guaranteed to exist in any compact space. Since the indexed family is bounded in the scalar field this is enough. Ill add the definition to the question for convenience. $\endgroup$
    – roo
    Jul 10 '13 at 17:37
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Yes, that’s correct. It works because $E_{\mathscr{U}}$ isn’t the full model-theoretic ultrapower: we consider only bounded sequences $x=\langle x_k:k\in\Bbb N\rangle$, so that $\mathscr{U}\text{-}\lim_k\|x_k\|$ exists, and we set $\|x\|_\infty=\mathscr{U}\text{-}\lim_k\|x_k\|$. That already cuts out the infinite elements of the full ultrapower, and in addition we take the quotient by the ideal $\{x:\|x\|_\infty=0\}$, identifying infinitesimally different elements.

Similarly, if $\langle\varphi_k:k\in\Bbb N\rangle_{\mathscr{U}}\in(E')_{\mathscr{U}}$, then by definition there is a $U\in\mathscr{U}$ such that $\{\|\varphi_k\|:k\in U\}$ is bounded. It follows that there is a $U\in\mathscr{U}$ such that $\{\varphi_k(x_k):k\in U\}$ is bounded and hence that $\mathscr{U}\text{-}\lim\varphi_k(x_k)$ exists, and it’s straightforward to verify that the map is a bounded linear functional on $E_{\mathscr{U}}$.

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  • $\begingroup$ @Kyle: You’re welcome! $\endgroup$ Jul 10 '13 at 22:15

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