Density of smooth functions in $L^p$ space on Cantor Set

Question

Let $\mu$ be the Cantor measure on the Cantor set $C$ in $[0,1]$. Is the space of functions obtained by restricting smooth compactly supported functions to $C$ dense in $L^p(\mu)$ for $1 \leq p < \infty$?

Answer 1

Let $\mu$ be a regular Borel measure on $\mathbf{R}^d$. Then the Riesz Kakutani representation theorem implies that $C_c(\mathbf{R}^d)$ is dense in $L^p(\mu)$ for $1 \leq p < \infty$. If $\{ \phi_\varepsilon \}$ is a mollifier on $\mathbf{R}^d$ and $f \in C_c(\mathbf{R}^d)$, then a simple application of uniform continuity shows that the smooth, compactly supported functions $f * \phi_\varepsilon$ converge to $f$ uniformly.

If $\mu$ is a finite measure, this implies that $f * \phi_\varepsilon$ converges to $f$ in $L^p(\mu)$ for all $1 \leq p \leq \infty$. On the other hand, since we may choose $\phi_\varepsilon$ to be supported on a ball of radius $\varepsilon$, all the functions $f * \phi_\varepsilon$ are supported on $\text{supp}(f) + B$, where $B$ is the unit ball at the origin, and the regularity of $\mu$ implies $\mu|_B$ is a finite regular Borel measure, so there is no loss of generality in assuming that $\mu$ is finite.

Answer 2

Hint: There are two points to consider.

(1) $C([0,1])$ is dense in $L^{p}(\mu)$ for all $1 \leq p < \infty$. This is true whenever $\mu$ is a finite Borel measure on $[0,1]$.

(2) In the case of the Cantor measure, $\mu(\{0\}) = \mu(\{1\}) = 0$.

(Also, in a sense, restricting the function to $C$ is only going to confuse your intuition slightly since $f \restriction_{C}$ and $f$ look the same to the Cantor measure.)