The celebrated Tietze extension theorem asserts that any continuous real-valued function defined on a closed subset of a normal space, can be extended to a continuous function on the whole space. Seen as a property of the target space $\mathbb{R}$, this leads to the important notion of absolute neighborhood retract, or AR(normal); in Dugundji's notation, the Tietze extension theorem can thus be rephrased saying that $\mathbb{R}$ is an AR(normal) space.

If in the Tietze theorem we restrict the class of domains from normal to metric spaces, by the Dugundji extension theorem, at least all locally convex topological vector spaces are suitable codomains: any continuous LCTVS-valued function on a closed subset of a metric space can be extended to a continuous function on the whole space.

Of course, this situation in principle allows a wide variety of intermediate situations. The first natural questions, that I would be glad to learn an answer of, are:

Q1. Does Dugundji's theorem hold true for normal spaces, namely, can any continuous LCTVS-valued function on a closed subset of a normal topological space be extended to a continuous function on the whole space?

I guess the answer is no, but I can't imagine a counterexample. In case of a negative (or not known) answer:

Q2. Are Banach spaces absolute retract for Hausdorff compact spaces, namely, can any continuous Banach-valued function on a closed subset of a Hausdorff compact space be extended to a continuous function on the whole space?

edit After Bill Johnson's answer to question 2, and the other useful comments, I would like to focus on the following question, that should have some good reference in the (wide) literature.

Q3. Let $X$ be a Hausdorff compact topological space, $Y\subset X$ a closed set, $E$ a Banach space. Does there always exist a bounded linear extension operator $C(Y,E)\to C(X,E)$?

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    $\begingroup$ Could you clarify what you're really looking for in Q1? Dugundji's theorem, in a slightly generalized form, gives more than a function-wise extension of every $f:A \to Y$ to $F:X \to Y$ where $A \subset X$ and $Y$ is LCTVS -- it asserts a simultaneous extension via a linear map $C(A,Y) \to C(X,Y)$. Are you looking for function-wise extensions or a single simultaneous extension? $\endgroup$ Apr 1, 2015 at 17:59
  • $\begingroup$ Good question. Actually in Q1 I was only wondering about the extension problem for any continuous function, but yes, a linear extension operator would be most welcome, at least for the situation in Q2. $\endgroup$ Apr 1, 2015 at 23:17
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    $\begingroup$ Pietro, the answer to Q3 is also no -- set $X$ = the Michael line (which is Hausdorff) and let $Y$ be the rationals. Then you can't find a bounded linear extension even for $E = \mathbb{R}^n$ $\endgroup$ Apr 3, 2015 at 13:54
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    $\begingroup$ I don't know about "celebrated", but the theorem is by Urysohn, not by Tietze. $\endgroup$ Apr 3, 2015 at 19:20
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    $\begingroup$ @Włodzimierz Holsztyński: thank you for your comment. However, the question is not about the original author of (what is known as) Tietze theorem. As you probably know, most theorems are known after a name which is different from the person who discovered it; often by the author's choice. en.wikipedia.org/wiki/Stigler%27s_law_of_eponymy $\endgroup$ Apr 5, 2015 at 6:37

5 Answers 5


Here is an answer to Q2.

Since compact subsets of Banach spaces are separable, WLOG the target Banach space is separable.

Since all separable infinite dimensional Banach spaces are homeomorphic, WLOG the target Banach space is $c_0$.

Since $c_0$ is a Lipschitz retract of $\ell_\infty$, WLOG the target Banach space is $\ell_\infty$.

The space $\ell_\infty$ clearly has the desired property.

Sorry, Pietro; this being the day after April 1, I could not resist giving this answer.

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    $\begingroup$ Similarly, Q1 has a positive answer for separable Frechet spaces, since they are all homeomorphic to the countable product of lines. $\endgroup$ Apr 2, 2015 at 15:39

This is a comment but the system doesn't regard me as worthy. An important ingredient of Dugundji's theorem is that an extension can be found with values in the closed convex hull of the range. Without this, the fact that the values of the function to be extended are in a lcs is rather artificial as is also evident in Bill Johnson's answer and comment. As regards the stronger version of Q2 mentioned in the comments to the OP, the existence of a continuous, linear extension operator was investigated in some detail (for the scalar case). One could check the publications of Pelczyński, Corson, Lindenstrauss, Semadeni for some results of this nature.

  • $\begingroup$ Thank you for the useful references, that I will check. Do you know of some in particular for a bounded linear extension operator $$C(Y,E)\to C(X,E)$$ in the case of $X$ a compact space, $Y\subset X$ closed, and $E$ Banach space? $\endgroup$ Apr 3, 2015 at 12:04
  • $\begingroup$ If you have such an extension for the scalar case, you can presumably extend to that of Banach space valued functions by taking tensor products but I don't know of any references where this is done explicitly. $\endgroup$
    – report
    Apr 3, 2015 at 13:40
  • $\begingroup$ And I think at least in the case of a separable $E$, one can pass from a linear extensor for the scalar case to one for E-valued mappings, by the argument in Bill Johnson's answer. $\endgroup$ Apr 3, 2015 at 14:40

Bad News

The answer to Q3 as stated is no. Let $X$ be the Michael line, and let $Y$ be the closed subset consisting of all the rationals. Then, there is no bounded linear extension $C(Y,\mathbb{R}) \to C(X,\mathbb{R})$. A proof (I'm not sure if this is the first place where it appeared) may be found in Example 3.3 of the paper

RW Heath and DJ Lutzer, Dugundji extension theorems for linearly ordered spaces, Pacific Journal of Math, 55(2), 419-425 (1974).

On the other hand, the authors provide an extension theorem for what they call linearly ordered spaces. There is a zoo of such conditions which are related to -- but strictly stronger than -- normality under which you can find a simultaneous extension. It would be quite a painful task to try and list all of them anywhere. There was some work on the class of linearly stratifiable spaces, I think going back to work of CJR Borges in the mid-70s, but people found some gaps and some counterexamples, so I'm not sure where things stand with all that right now.

Good News

Here is one rather typical example where Dugundji extension certainly works - I'll call this assumption $N^+$. A space $X$ satisfies $N^+$ if it admits a distinguished collection of open sets $\{W(n,x) \subset X ~ \mid ~ x \in X \text{ and } n \in \omega\}$ so that

  1. $x \in W(n,x)$ for each $n$,
  2. $W(n,x) \subset W(n+1,x)$ for each $x$, and
  3. For any open $U \subset X$ with $x \in U$, there exists an open $V \subset X$ so that for each $y \in V$ there is some $n$ with $x \in W(n,y) \subset U$.

By the way, if $X$ is $T_1$ then this condition is equivalent to metrizability. Here is the result you want:

Theorem Let $Y$ be a closed subspace of a topological space $X$ satisfying $N^+$. If $L$ is any locally convex topological vector space, then there exists a linear map $C(Y,L) \to C(X,L)$ which produces Dugundji extensions (and in particular, satisfies the convex hull condition).

You can find the proof in a nice and short paper:

IC Starc, Concerning the Dugundji extension property, Topology and its Applications 63(2), 165–172 (1995).

  • $\begingroup$ Thanks this is a fantastic answer! $\endgroup$
    – ABIM
    Aug 4, 2019 at 21:53
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    $\begingroup$ Why does this answer Q3? The Michael line is centainly not compact. And for general topological spaces $X$, what should boundedness of an extension operator mean? Either one considers other locally convex topologies on $C(X,E)$ like uniform convergence on compact sets or, to stay in the category of Banach spaces, one considers the space $CB(X,E)$ of bounded continuous functions with the supremum norm. $\endgroup$ Oct 4, 2022 at 7:10

The following is a rather well-known theorem of Haydon that might be useful for your purpose.

Theorem: The following are equivalent for a compact Hausdorff space $Y$:

  1. ($Y$ is a Dugundji space): For every compact $X$ and embedding $e:Y \to X $ there is a bounded linear extension operator $T:C(Y) \to C(X)$ such that $T(1)=1$ and $T(f)\geq0$ whenever $f\geq0$.

  2. ($Y$ is $\mathrm{AE}(0)$): For every compact zero-dimensional $Z$, every closed $A \subseteq Z$ and every map $f:A \to Y$ there is an extension $f:Z \to Y$.

  3. $Y$ is the inverse limit of a Haydon system.

See for example Todorcevic´s book "Topics in Topology" for a proof of this theorem (and the definition of Haydon system).

For example any product of compact metrizable spaces is Dugundji and compact topological groups are Dugundji.

  • $\begingroup$ What is AE(0)? Arens-Eells space? $\endgroup$
    – ABIM
    Aug 4, 2019 at 21:51
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    $\begingroup$ @AIM_BLB, $AE(0)=$"Absolute Extensor on dimension $0$" and it means what is written in 2. $\endgroup$ Aug 5, 2019 at 14:20
  • $\begingroup$ Ah okay, thanks Ramiro $\endgroup$
    – ABIM
    Aug 5, 2019 at 14:21

I think that question Q3 has nothing to do with Banach space valued functions: If $Y$ is a closed subset of a compact space $X$ such that there is a continuous linear extension operator $F:C(Y)\to C(X)$ then one can use the injective tensor product to get an extension operator $$F\otimes id_E: C(Y,E) \cong C(Y)\hat\otimes_\varepsilon E \to C(X)\hat\otimes_\varepsilon E \cong C(X,E)$$ even for every complete locally convex space $E$. The isomorphism $C(X)\hat\otimes_\varepsilon E\cong C(X,E)$ is desribed, e.g., in Jarchow's book Locally Convex Spaces, chapter 16.

I would be very surprised if it would be unknown whether the restriction operator $C(X)\to C(Y)$ always has a continuous linear right inverse.

Edit. I just learned from Tomasz Kania's answer to this question Nonseparable counterexamples in analysis that an example is $X=\beta\mathbb N$, the Stone-Cech compactification of $\mathbb N$, and $Y=\beta\mathbb N\setminus\mathbb N$: If there were an extension operator $C(Y)\to C(X)$ then the kernel of the restriction operator would be complemented in $C(X)=C(\beta\mathbb N)=\ell^\infty(\mathbb N)$ but for the remainder $Y$ this kernel is $c_0$ which is not complemented (the theorem of Phillips from 1940).

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    $\begingroup$ Pełczyński calls a compact Hausdorff space $X$ such that for all closed $Y \subseteq X$ there exists a bounded linear extension map $C(Y) \rightarrow C(X)$ an almost Dugundji space. In Corollary 8.14 on p. 47 he states that a non-metrizable compact Hausdorff space is not almost Dugundji if it is extremally disconnected, scattered, or has uncountable cellularity. So $\beta \mathbb{N}$ is covered by this, but also e.g. the one-point compactification of an uncountable set. $\endgroup$ Oct 4, 2022 at 13:54

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