So I am currently working on the proof of Grothendieck's Lemma :

Let S $ \subset L^{\infty}(X) $, of finite measure, be a closed vector subspace of $L^p $ for a certain p such that $ S \subset L^{\infty} $

We want to show that S is finite dimensional.

Suppose that the embedding $ L^2 \subset L^{\infty} $ is continuous.

We give an orthonormal family of vectors of S.

$ \forall c = (c_1, ..., c_n) \in \mathbb{Q}^n $ (countable and dense in $ \mathbb{R}^n $ ) $ \exists $ $ X_c \subset X $ of full measure, such that

$ \forall x \in X_C $ , $ | \sum_i c_i f_i (x) | \leq M || \sum_i c_i f_i ||_{\infty} \leq M || \sum_i c_i f_i ||_{2} $

We also have $ \forall c \in \mathbb{Q}^n \: \: \forall x \in \cup_{c \in \mathbb{Q}^n } X_c $ , $ | \sum_i c_i f_i (x) | \leq M || \sum_i c_i f_i ||_{\infty} \leq M || \sum_i c_i f_i ||_{2} $

So as $ \mathbb{Q}^n $ is countable and dense in $ \mathbb{R}^n $ we get the same result for all $ c \in \mathbb{R}^n $

In particular for $ c_i = f_i (x) $ we get $ \sum_i (f_i (x))^2 \leq M \sqrt{\sum_i f_i ^2} $ so $ \sum_i (f_i (x))^2 \leq M^2 $

By integration on $ \cup_{c \in \mathbb{Q}^n } X_c $ we deduce that : $ n \leq M^2 $

  • my question : Wy did we take $ \mathbb{Q}^n $ countable and dense ?

Apparently as $ X_c $ depends on c we would get an issue trying that integration.

We also need to use the definition of $ || . ||_{\infty} $ I suppose.

But is there something to do with a Quantifier inversion? Why specificaly a countable set?


1 Answer 1


The simple reason for varying $c$ over a countable set is that an uncountable union of sets of measure $0$ need not have measure $0$. [ Like the singleton sets in $\mathbb R$]. In the line next to the one where $X_c$ is introduced we want the inequality to hold for $x$ outside one null set $X_0$ for all $c$. But the union of the sets $X_c^{c}$ need not have measure $0$ if we vary $c$ over an uncountable set. So we first vary $c$ over $\mathbb Q^{n}$ and then observe that if the inequality we get holds for $c$ in a dense set of $c$'s it holds for all $c$ by a continuity argument.


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