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I have a question concerning the Lebesgue spaces: Is $C_0^\infty$ dense in $L^p$ ?

And if yes, why?

Thanks!

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Over what set? What is the measure you are working with? –  Siminore Oct 3 '12 at 11:40

3 Answers 3

up vote 1 down vote accepted

Yes. First of all, it is enough to see that any function in $L^p$ with compact support can be approximated in the $L^p$ norm by $C_0^\infty$ functions.

Take $\phi\in C_0^\infty$, $\phi\ge0$ and $\int\phi(x)\,dx=1$, and define for $\epsilon>0$ $\phi_\epsilon(x)=\epsilon^{-1}\phi(x/\epsilon)$. If $f\in L^p$ with compact support, then $\phi_\epsilon\ast f$ has compact support, is of class $C^\infty$ and $\phi_\epsilon\ast f$ converges to $f$ in $L^p$.

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Thank you. And in the case of non-compact support? –  AlexisZorbas Oct 3 '12 at 11:52
    
@AlexisZorbas A function in $C^\infty$ need not be in $L^p$. –  Siminore Oct 3 '12 at 12:05
    
@AlexisZorbas given a function $f$ in $L^p(\mathbb{R}^k)$, you can use the dominated convergence theorem to show that $f_M = f \cdot \chi_{[-M,M]}$ converges to $f$ in $L^p$, so you can do a standard argument by approximating first by a compactly supported function and then approximating the compactly supported function by a $C_0^\infty$ function. –  Chris Janjigian Oct 3 '12 at 12:34

Theorem. Let $\mathcal{L}^k$ be the Lebesgue measure on $\mathbb{R}^k$, $k \geq 1$. For $1 \leq p < \infty$, the set $C_0(\mathbb{R}^k)$ is dense in $L^p(\mathbb{R}^k,\mathcal{L}^k)$.

The proof is based on Lusin's theorem. A more general statement holds true in a locally compact Hausdorff space, provided the measure satisfies some properties.

For all this, you can read W. Rudin, Real and complex analysis, Chapter 3, section Approximation by continuous functions.

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Take a look in the chapeter 4 of this book (Brezis - Functional Analysis, Sobolev Spaces and PED).

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