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Here is the Squeeze Theorem in $\mathbb{R}$:

Let $(a_n)$, $(b_n)$ and $(c_n)$ be sequences taking their values in $\mathbb{R}$. Let $x \in \mathbb{R}$. Assume that:

  • $\forall n \in \mathbb{N}, \ \ a_n \leq b_n \leq c_n$;
  • $\lim \limits_{n \to + \infty} a_n = \lim\limits_{n \to + \infty} c_n = x$.

Then $\lim \limits_{n \to + \infty} b_n = x$.

This theorem is true is one replaces the occurences of $\mathbb{R}$ above by $\overline{\mathbb{R}}$, $\mathbb{R}^n$, $\mathcal{C}_b (\Omega)$ (where $\Omega$ is an open set), $\mathbb{L}^p (\Omega, \mu)$ (where $(\Omega, \mu)$ is a measured space and $p \in [0, + \infty]$), and even in $\mathcal{P} (\mathbb{R})$ (see this related question). However, the proofs I know of these facts have some points in common, but also some individual ingredients.

Is there a general sufficient condition which would ensure that a topological space with a partial order satisfies the Squeeze Theorem, and apply to all examples above? Are there some not too contrived examples of spaces for which the Squeeze Theorem fail?

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D. Thomine
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    how do you formulate the squeeze thoerem for $\mathbb R^n$? – Ittay Weiss Nov 16 '13 at 23:24
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    I'd guess "order topology" is the keyword – Hagen von Eitzen Nov 16 '13 at 23:25
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    @Ittay Weiss: you have to take the partial order "$x \leq y$ if and only if $x_i \leq y_i$ for all $i$. If you take e.g. the lexicographical order, it doesn't work, but this is what I'd call a contrived example (the lexicographcal order is not compatible with the usual topology). – D. Thomine Nov 16 '13 at 23:26
  • @HagenvonEitzen, the usual order topology only applies to total orders. I'm not sure how one might handle other situations. – dfeuer Nov 16 '13 at 23:28
  • If the set is not complete (e.g., the rationals) there could be problems. – marty cohen Nov 16 '13 at 23:48
  • @martycohen, that's not an issue in a LOTS, at least. By the premise given, the squeezing sequences converge. All that remains is to show that the squeezed sequence converges to the same point. If you wanted to weaken the premises, completeness might come into play (e.g., suppose $(\forall n\in \Bbb N)(a_n\le b_n\le c_n)$, every element less than a lower bound of $c_n$ is eventually exceeded by $a_n$, both sequences are bounded appropriately, etc.) – dfeuer Nov 17 '13 at 00:12

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This may not quite answer your question, but it was too long for a comment, and worth pointing out, IMHO :

Many of the function/sequence spaces where this proof works are all Banach lattices. ie. They are ordered normed linear spaces, and the norm respects the order structure.

For such spaces, a functional $f:E \to \mathbb{C}$ is called positive, if $x \geq 0$ implies that $f(x) \geq 0$. In particular, if $a_n \leq b_n \leq c_n$ for all $n$, then $$ f(a_n) \leq f(b_n) \leq f(c_n) \quad\forall f \text{ positive linear functional } $$ Hence, by the squeezing principle in $\mathbb{R}$, it follows that $$ f(b_n) \to f(x) \quad\forall f \text{ positive linear functional } $$ Now, there is a theorem that say that the positive linear functionals generated the (continuous) dual space. In other words, $$ E^{\ast} = \{f-g : f,g \text{ positive linear functionals }\} $$ Hence (by the Banach lattice version of the Hahn-Banach theorem), it follows that $b_n \to x$ in $E$

Prahlad Vaidyanathan
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  • Thanks. It looks like that, even if we don't have a Banach lattice, this method can be tweaked to apply; we just need enough positive functionals to separate the points, which is true with $\mathcal{P} (\mathbb{R})$ (these are nondecreasing bounded functions). This is not exactly what I expected, and I am not sure it really answer the question, but it is definitely a very nice answer. – D. Thomine Nov 20 '13 at 12:09
  • @PrahladVaidyanathan Do you have a reference for the characterization of the dual wrt positive functionals? – tks Oct 15 '15 at 16:14
  • Aliprantis-Burkinshaw's "Positive Operators" should have a proof. Not sure though. – Prahlad Vaidyanathan Oct 16 '15 at 03:13
  • @PrahladVaidyanathan Thanks for the tip! It seems like it one can deduce it from the fact that for a Banach lattice, the norm dual coincides with the order dual. It would be however great to have a reference of a precise statement for people (like me...) who are not too much into lattice theory.... – tks Oct 18 '15 at 15:48
  • Found one: Theorem 23.6, p.228 of the Appendix of Kelley's and Namioka's Linear Topological Spaces. – tks Oct 18 '15 at 15:58
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At least, for metric spaces with a partial order I found some generalizations of the squeeze theorem. They are useful to deduce the cases $\mathbb{R}$, $\mathbb{R}^n$, $\mathcal C_b(\Omega)$ and $\mathcal L_p(\Omega)$.

First, I came up with the following proposition:

Proposition. Let be $(X,d)$ a metric space, $a\in X$, $(x_n)$, $(y_n)$ sequences in $X$ and $A_n\subseteq X$ for $n\in\mathbb{N}$ with the following properties:

  • $x_n,y_n\in A_n\quad\forall n\in\mathbb{N}$
  • $\operatorname{diam} A_n=\sup_{x,y\in A_n} d(x,y)\rightarrow 0$
  • $x_n\rightarrow a$

Then we also have $y_n\rightarrow a$.

One can easily check this by elementary estimations. Now we conclude the general squeeze theorem for metric spaces $(X,d)$ endowed with a partial order $(\preceq, X)$ with respect to the metric $d$, that means we have $d(x,y)\leq d(x,z)$ for all $x\preceq y\preceq z$.

General squeeze theorem. Le be $(X,d)$ a metric space, $(\preceq,X)$ a partial order , $a\in X$ and $(x_n)$, $(y_n)$, $(z_n)$ sequences in $X$ with $$x_n\preceq y_n\preceq z_n$$ for each $n\in\mathbb{N}$ and $x_n,z_n\rightarrow a$. Then it follows that $y_n\rightarrow a$.

Proof. Define $A_n=\{y\in X\ \vert\ x_n\preceq y\preceq z_n\}$. Then we have $x_n,y_n\in A_n$ for each $n\in\mathbb{N}$ and $$\operatorname{diam}A_n=d(x_n,z_n)\rightarrow 0.$$ Thus the theorem follows from the proposition. $\square$

Josef
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