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Why $C_0^\infty$ is dense in $L^p$? Would you give me a simple proof or the outline of the proof?

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Another approach is the use of Hahn-Banach separation theorem: because of this theorem and Riesz's Representation Theorem, every continuous linear functional on $L^p$ with $1\le p<\infty$ that vanishes on $C_o^\infty$ is identically zero. – Giuseppe Negro Jun 10 '12 at 13:14
up vote 6 down vote accepted

The outline (the proof isn't simple, at least not according to my understanding of simple):

Let $f \in L^p$. Then there is a sequence of simple functions $s_n \in L^p$ that converges pointwise to $f$. Then you show that they converge in norm.

These $s_n$ have finite support. You can therefore apply Lusin's theorem:

(Lusin) Let $\Omega$ be a locally compact Hausdorff space and $\mu$ a Radon measure and $f: \Omega \to \mathbb R$ a $\mu$-measurable function with finite support $E$. Then for every $\delta > 0$ there exists a closed set $K \subset E$ such that $\mu(E \setminus K) < \delta$ and $f$ is continuous on $K$.

to get a sequence of continuous functions.

Then you apply Tietze's theorem:

(Tietze) If $\Omega$ is a locally compact Hausdorff space and $K \subset \Omega$ compact then any $f \in C(K, \mathbb R)$ can be extended to a bounded function in $C(\Omega, \mathbb R)$.

to get a sequence of compactly supported functions.

I did this here for $p=1$.

Once you have $C_c$ dense in $L^p$ you mollify. Which I did here. Hope this helps.

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I'm assuming you are using $C_0$ to mean compactly supported functions. This might of course not be right since you could be using $C_0$ to denote functions vanishing at infinity. – Rudy the Reindeer Jun 10 '12 at 12:14
@ Matt N. Really thanks! – Misaj Jun 10 '12 at 12:20
The approach with Lusin's theorem is also explained in detail in the chapter about $L^p$ spaces in Rudins "Real and Complex Analysis". – Listing Jun 10 '12 at 12:29

Maybe I am missing something, but rather than using convolution and Lusins Lemma to approximate continuous finitely supported functions I would use the functions

$$ f(x) := \begin{cases} e^{1-\frac{1}{1-x^2}} & \text{ for } \vert x\vert <1 \\ 0,& otherwise \end{cases} $$


$$ g_n(x) := \frac{1}{n}\tan(\frac{x ~\pi }{2}) $$

to approximate indicator functions $\chi_{[a,b]}(x)$ by the functions $\frac{a+b}{2}+ \frac{b-a}{2} f\circ g_n (x)$.

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