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Just wondering... Let $E$, $G$ be Banach spaces, let $U\subset E$ be a subset of $E$, and let $f:U\rightarrow G$ be a continuous linear function. Can $f$ be extended to a continuous linear function on $E$, $F:E\rightarrow G$? For which cases does this happen?

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What does it mean for $f$ to be linear if $U$ isn't a subspace? – Qiaochu Yuan Jan 5 '12 at 20:20
(Assuming $U$ to be a subspace.) Sure: Fix a basis $B'$ for $U$ and extend it to a basis $B$ for the whole space $E$. Now define $$F(b) = \begin{cases} f(b), &b \in B', \\ 0, &b \notin B', \end{cases}$$ for $b \in B$; extend by linearity. – Srivatsan Jan 5 '12 at 20:21
@Srivatsan: Such $F$ is not necessarily continuous. – Nate Eldredge Jan 5 '12 at 20:26
@Nate, I see. Thanks for pointing it out. [I guess will just leave that comment around as an attempt that doesn't work.] – Srivatsan Jan 5 '12 at 20:28
This isn't possible in general. For example, suppose $X$ is a Banach space and $A$ is a closed uncomplemented subspace of $X$ (e.g. $c_0$ in $\ell^\infty$). Then the identity map from $A$ to $A$ does not extend to a continuous linear map from $X$ to $A$. – Chris Eagle Jan 5 '12 at 20:47
up vote 8 down vote accepted

Here is a partial answer.

You do want $E$ to be a subspace. A Banach space $X$ is called injective if for every Banach space $Y$ and every subspace $Z$ of $Y$ and every operator $T:Z\rightarrow X$, there is an extension $\overline T: Y\rightarrow X$.

It is a Theorem of R.S. Phillips that $\ell_\infty$ is injective. $L_\infty$ shares the same property. These facts are proved in Diestal's Sequences and Series in Banach spaces. Injectivity properties are gone over in some detail in Wojtaszczyk's Banach Spaces for Analysts. Of course, a google search on "injective Banach space" proves fruitful.

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Will do, thanks @David Mitra – Mario Carrasco Jan 11 '12 at 3:00


A classic and probably the easiest counterexample is $U = G = c_0$, the space of sequences converging to zero and $E = \ell^{\infty}$, all equipped with the supremum norm and $f: c_0 \to c_0$ the identity map. There is no continuous linear map $F: \ell^{\infty} \to c_0$ extending $f$ by a result usually called Phillips's lemma more succinctly: “$c_0$ is not complemented in $\ell^\infty$”.

The proof is not entirely trivial but not too hard either. A nice exposition can be found in Robert Whitley, Projecting $m$ onto $c_0$, The American Mathematical Monthly Vol. 73 (3) (Mar., 1966), pp. 285–286.

Two sufficient conditions:

  • $U$ is a direct sumand of $E$ (trivial).
  • $G$ is a space of the form $C(K)$ with $K$ compact Hausdorff and extremally disconnected. This includes $\ell^{\infty}(S)$ for an arbitrary set $S$ (easy exercise using Hahn-Banach), and $L^{\infty}(\Omega,\mu)$ (less easy see e.g. Benyamini-Lindenstrauss, p. 32).

    One can show furthermore if $F$ can always be found with $\|F\| = \|f\|$ for all pairs $U \subset E$ and all $f: U \to G$ then $G$ is isometrically isomorphic to $C(K)$ as above (Akilov-Goodner-Nachbin-Kelley, '50ies). See Chapters V and  VI of Day's Normed Linear Spaces, 3rd edition, Springer, 1973, for more on this and a proof of the first sentence of this bullet point.

    The spaces having the extension property as $G = C(K)$ above are called injective, spaces with the Hahn-Banch property or spaces with the $P_1$-property in the literature.

Let me mention my answer here where I expand on injectivity (the extension property of $C(K)$-spaces) and flatness of Banach spaces.

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Awesome, thanks @t.b – Mario Carrasco Jan 11 '12 at 3:02
@MarioCarrasco, how you prove the case $U$ is a direct summand of $E$? – Juan Pablo Jul 26 '14 at 22:21

(Assuming $U$ is a subspace.)

$f$ has a unique continuous extension to the closure of $U$, $\overline{U}$. Then an extension exists if $\overline{U}$ is injective or $G$ is injective. For example, see Proposition 2.f.2 on page 105 of Classical Banach spaces by Lindenstrauss and Tzafriri.

This characterizes injectiveness, in the sense that if $\overline{U}$ is not injective then there will exist $E$, $G$, and $f$ such that the extension doesn't exist, and if $G$ is not injective then there will exist $E$, $\overline{U}$ and $f$ such that the extension doesn't exist. (I am only pointing out what the cited theorem shows.)

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Concerning your last paragraph: If $\overline{U}$ is injective then it is complemented in every Banach space containing it (apply injectivity to the identity map of $\overline{U}$ and extend to the containing space to get a projection.) Every Banach space space is isometric to a closed subspace of $\ell^{\infty}(S)$ (e.g. $S$ the unit ball of the dual of $\overline{U}$ and the embedding given by evaluation) and a complemented subspace of an injective space is injective itself, so if $\overline{U}$ is not injective you can take the embedding I described before. – t.b. Jan 5 '12 at 20:56
Thank you both, I'll look into this – Mario Carrasco Jan 11 '12 at 3:01

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