Is $C_0^\infty$ dense in $L^p$?

Question

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

And if yes, why?

Thanks!

Answer 1

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 $L^p$.

Answer 2

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

Answer 3

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.