Lipschitz functions dense in C0M

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curtdbz
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I'm working on Pugh's book on analysis and there's this problem that should be very easy to solve. It's asking to show that the set of continuous functions, [tex]f:M \rightarrow R, f\in C^{Lip}[/tex] obeying the Lipschitz condition (where M is a compact metric space):

[tex]|f(a) - f(b)| \leq L d(a,b)[/tex] for some L, for every a and b belonging to to M.

Show that the above is dense in [tex]C^{0}(M,R)[/tex]. My attempt is to use the Stone-Weirestrass theorem. That is to show that the set [tex]C^{Lip}[/tex] vanishes nowhere and separates points. The latter is easy for me, I just showed how the above equation implies that f(a) does not equal f(b) if a does not equal b. However, showing the vanishing property is proving difficult. Is there some trick I'm supposed to use? Hm...

Also, I assume I'll have to actually show that [tex]C^{Lip}[/tex] is infact a function algebra; that is, it obeys the 3 properties that makes something that (closed under addition, constant multiples, and multiplication), but I can't seem to manipulate the equations in such a way that shows the Lipschitz property implies those.

Any help is appreciated. Thank you.
 
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The Stone-Weierstrass theorem is applicable to subalgebras, so you must show first that the LIpschitz functions are indeed a subalgebra of [itex]C^{0}\left(M,\mathbb R \right)[/itex]. This is not difficult, for the multiplication, start with:

[tex]\left|f\left(a\right)g\left(a\right)-f\left(b\right)g\left(b\right)\right|[/tex]

Then add and subtract [itex]f\left(a\right)g\left(b\right)[/itex], expand and use the fact that the functions are continuous and defined in a compact space.

Now, there are some things I don't understand in your question:

That is to show that the set vanishes nowhere ...

This is indeed necessary for the locally compact version of the theorem but, as you have a compact metric space, you only need to show that your set contains the constant functions.

The latter is easy for me, I just showed how the above equation implies that f(a) does not equal f(b) if a does not equal b.

How did you prove that?