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Let $X$ be a compact pointed metric subspace of the $d$-dimensional Euclidean space $(\mathbb{R}^d,d_E)$ and let $AE(X)$ denote its Arens-Eells space. Then a result of Nik Weaver shows that for every Lipschitz map $f:X\rightarrow E$ into a separable Banach space, there exists a unique continuous linear extension $F:AE(X)\rightarrow E$ satisfying $$ F\circ \delta = f, $$ where $\delta$ is the canonical isometric embedding of $X$ in $AE(X)$. (See Nik's book for more details).

Question:

Is there a concrete description of what $F$ is or how to explicitly construct it? I would like to use it for computations...

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  • $\begingroup$ Typo in your title (mis-spelling or autocorrect of Eells'sname). $\endgroup$
    – Yemon Choi
    Commented Sep 30, 2019 at 12:32
  • $\begingroup$ Thanks Yemon, for some reason I find this spelling particularly difficult to remember... $\endgroup$
    – ABIM
    Commented Sep 30, 2019 at 12:36
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    $\begingroup$ @YemonChoi: where do you see this result in their paper? I thought it was new when I proved it. $\endgroup$
    – Nik Weaver
    Commented Sep 30, 2019 at 14:03
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    $\begingroup$ @YemonChoi yes, Godefroy and Kalton's paper was 2003 and my book was 1999 (also they only state the result for $X$ a Banach space). $\endgroup$
    – Nik Weaver
    Commented Sep 30, 2019 at 18:23
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    $\begingroup$ @NikWeaver At the risk of adding on to the list of irrelevant references: some variant of the universal property of $AE(X)$ also seems to have been considered in Sec 3.8 of Joe Flood's thesis Free topological vector spaces. $\endgroup$
    – Tobias Fritz
    Commented Jan 30, 2020 at 2:07

2 Answers 2

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$AE(X)$ is the completion of the space of "molecules", i.e., the finitely supported functions $m: X \to \mathbb{R}$ which satisfy $\sum_{p \in X}m(p) = 0$. The extension $F$ of $f: X \to E$ satisfies $F(m) = \sum_{p \in X} m(p)f(p)$. (BTW $E$ need not be separable.)

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  • $\begingroup$ Oh, so in this version you embed $X$ not via the point-evaluation map but through the map $x \mapsto m_{xe}$ (where in your notation e is the base-point...so $F(m_p)=(1_p-1_e)f(p)$...interesting)? Is there a known description when using the description of $AE(X)$ found at the beginning of section 1 of this paper: arxiv.org/pdf/1505.07209.pdf $\endgroup$
    – ABIM
    Commented Sep 30, 2019 at 14:56
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    $\begingroup$ Well, in that paper they construct $AE(X)$ as the closed span of the point evaluations $\delta_p$ in $({\rm Lip}(X))^*$. In this case you would set $F(\delta_p) = f(p)$ and extend linearly. $\endgroup$
    – Nik Weaver
    Commented Sep 30, 2019 at 16:24
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    $\begingroup$ (I notice that they refer to my theorem as "folklore".) $\endgroup$
    – Nik Weaver
    Commented Sep 30, 2019 at 16:25
  • $\begingroup$ The result quoted in the above-mentioned article as folklore had appeared in print in 1976 (and quite probably earlier). $\endgroup$
    – user131781
    Commented Sep 30, 2019 at 20:58
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    $\begingroup$ @NikWeaver I believe that user131781 is referring to this paper: link.springer.com/chapter/10.1007%2FBFb0089276 and "SLN" is an abbreviation for "Springer LNM". $\endgroup$
    – Robert Furber
    Commented Oct 1, 2019 at 17:35
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This has already been answered but I would like to add some points which I hope might be of interest. The clearest expression is, in my opinion, in the general setting of a complete metric space $M$ with a base point $x_0$ and radius $1$. One then defines the Banach space $F$ consisting of the Lipschitz functions which respect the base, i.e., map $x_0$ onto $0$, with the natural norm. Then one can embed the metric space isometrically into a Banach space $E$ with the universal property that every Lipschitz map on $M$ into a Banach space $G$ which respects the base lifts to a unique linear operator on $E$ with the same norm. If one takes $G$ to be one-dimensional, then one sees that the dual of $E$ is the space of Lipschitz functions above. Now the unit ball of the latter has a natural compact topology (pointwise or uniform convergence) and so, by standard duality theory, is a dual space. One can then turn this reasoning on its head and define $E$ to be its predual.

One can see this more clearly if one uses a little terminology from category theory. If we map a Banach onto its unit ball, then we define a functor from the category of Banach spaces (with linear contractions as morphisms) into that of pointed metric spaces with base-point preserving Lipschitz functions, as above, then what we have constructed is just an adjoint functor. That is, the Arens-Eells space can be interpreted as a Free-functor and $AE(X)$ is a free object over $X$.

This is perhaps not really a concrete construction, but it follows from the existence that the space is just the so-called free vector space over $M$ (as a pointed set), completed under a suitable norm (basically the observation of Nik Weaver above). On the other end of the concrete-abstract spectrum, the existence of such an object (often called the free Banach space over $M$) can be deduced from the Freyd adjoint theorem.

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  • $\begingroup$ I added a tiny correction (since the free functor goes has as its sink the category of pointed metric spaces and Lipschitz maps. Since the fact that Met is not complete... I think this can be a little issue.)... Otherwise thanks. This I did know, but since I am looking for a constructive/explicit description of $F$, sadly I can't use it. But I very much like seeing category theory again, like the good-ol-days $\endgroup$
    – ABIM
    Commented Sep 30, 2019 at 15:23
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    $\begingroup$ Well, I meant the category of complete pointed metric spces of radius at most $1$ with base-preserving contractions as morphisms—-this does have products. Could you be more precise about what you mean by „constructive/explicit“? Taking the completion of the free vector space works in some simple cases. Maybe you could give more details on the situation you are interested in. $\endgroup$
    – user131781
    Commented Sep 30, 2019 at 17:36
  • $\begingroup$ Basically, Nik's answer was exactly what I was looking for..a concrete description in terms of input and outputs. $\endgroup$
    – ABIM
    Commented Oct 1, 2019 at 8:26
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    $\begingroup$ Two comments: (1) you don't need radius 1 to do this and (2) all it means to say that $E$ is a predual of $F$ is that $E^* \cong F$, so there's no need to "turn the reasoning on its head". (A predual; uniqueness of the predual is a more subtle question.) $\endgroup$
    – Nik Weaver
    Commented Oct 1, 2019 at 18:43
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    $\begingroup$ There are countless examples of this ( topological spaces into spaces of measures, metric spaces into free Banach spaces, uniform spaces into uniform measures manifolds into distributions, complex manifolds into analytic functionals,...) and the principle is always the same—-you basically construct an adjoint to a forgetful functor. Of course, you don‘t have to do it or think of it in this way—-most people probably wouldn‘t. However, a small minority is of the opinion that so doing sheds some light on the subject. In this sense, it is „folklore“. $\endgroup$
    – user131781
    Commented Oct 2, 2019 at 2:02

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