I've noted, that the following fact can be proven in a few lines using $C^*$-algebra theory. I wonder if it has a simple elementary proof or not. Probably you can give me a reference.

Suppose $X$ is a compact Hausdorff space, $V\subset X$ is a closed subset, $f\colon V\to \mathbb{R}$ is a continuous function. Then there exists a continuous function $g\colon X\to \mathbb{R}$, whose restriction on $V$ is $f$.

C*-algebraic proof is the following.

First note, that it is enough to find $g\colon X\to \mathbb{C}$ (then we can take its real part). Consider restriction map $\phi\colon C(X)\to C(V)$. It is a * -homomorphism, and therefore its image $\phi(C(X))$ is a commutative $C^*$-algebra. Moreover $\phi(C(X))$ separates points of $V$ (because $C(X)$ does). By Stone-Weierstrass theorem $\phi(C(X))=C(V)$.

  • $\begingroup$ I would be a little surprised if the C*-algebra proof was really much quicker once one put back in all the necessary lemmas which one takes for granted. What is your argument? $\endgroup$ – Yemon Choi Oct 27 '10 at 5:16
  • $\begingroup$ You mean that if we will add a piece of $C^*$-algebra theory to the proof, it will not be short. Yes, of course. I think if we will leave only piece of $C^*$-theory essential for this proof we will still end up with about 6 pages. $\endgroup$ – Fiktor Oct 27 '10 at 5:41
  • $\begingroup$ But it doesn't use any "specific" lemmas, only general facts of $C^*$-algebra theory, which can be found for instance in a textbook by Murphy. $\endgroup$ – Fiktor Oct 27 '10 at 5:43
  • $\begingroup$ nice proof, anyway. $\endgroup$ – Pietro Majer Oct 27 '10 at 7:53
  • $\begingroup$ BUT ... the real question is whether this exact theorem is used in the elementary theory of $C^*$-algebras. $\endgroup$ – Gerald Edgar Oct 27 '10 at 13:24

I think this is a special case of the Tietze extension theorem (since any compact Hausdorff space is normal). (Here is one proof.)

  • $\begingroup$ This is indeed the Tietze extension theorem (which I think needs Urysohn's lemma?) $\endgroup$ – Yemon Choi Oct 27 '10 at 5:17
  • $\begingroup$ Dear Yemon, I believe so. (It's a long time since I studied the proof, but Urysohn's lemma is an ingredient in the PlanetMath proof that I linked.) $\endgroup$ – Emerton Oct 27 '10 at 5:22

If you like to see the Tietze theorem from a Functional Analysis viewpoint, another possibility is by means of the following basic lemma, stating that a certain "approximate surjectivity" actually implies "surjectivity"; this allows to prove a kind of dual result too.

Iteration Lemma. Let $X$ and $Y$ be Banach spaces, $T$ a linear bounded operator, $B_X$ the unit closed ball of $X$, $U$ a bounded nbd of $0\in Y$, $0\le t <1$ such that $$U\subset > TB_X+tU.$$ Then, $T$ is onto and in fact $$(1-t)U\subset TB_X\\ .$$

You may have nice time applying it to prove the following results, in which we assume wlog that the functions are bounded, so that we can work with the Banach space $C_b$.

  • Tietze-Dugundji extension theorem. Let $M$ be a metric space, $V\subset M$ closed, $E$ a Banach space. Then the restriction operator $C_b(M,E)\to C_b(V,E)$ is surjective.

  • Bartles-Groves lifting theorem. Let $L:E\to F$ a bounded linear surjective operator; $M$ a metric space. Then the composition operator $C_b(M,E)\to C_b(M,F)$, such that $f\mapsto L\circ f,$ is surjective.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.