$C_c^\infty(\Omega)\subseteq L^p(\Omega)$ for any open $\Omega$?

Question

Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$. Can we show that $$C_c^\infty(\Omega)\subseteq L^p(\Omega)\tag 1$$ for all $p\in [1,\infty]$? It's clear that $(1)$ holds if $\Omega$ has finite Lebesgue measure. And it's clear that $(1)$ holds for $p=\infty$.

Answer 1

Let $f\in C^\infty_c(\Omega)$. Then $f$ is supported in a compact set $K$ and $|f|$ attains a maximum $C$ in this $K$. Thus

$$\int_{\Omega} |f|^p dx = \int_K |f|^p dx \le \int_K C^p dx = \text{Vol}(K) C^p.$$

Thus $f\in L^p$ for all $p$. Indeed $C^\infty_c(\Omega)$ is dense in $L^p$ for all $1\le p <\infty$.

Answer 2

Well yeah it has compact support, hence outside of a compact set $K$ is zero. It's continuos on a bounded, closed set hence its integral must be smaller than $M\cdot m(K)$, where $M$ is the maximum of the function on $K$ and $m$ is the lebesgue measure

Answer 3

I'm assuming that $C^\infty_c(\Omega)$ is the space of $C^\infty$ with compact support contained in $\Omega$. If that's the case, then you don't care whether or not $\Omega$ has finite measure, since

$$ \int_\Omega |u|^pdx\leq |spt(u)|\cdot\max |u|^p<\infty. $$

where $spt(u)$ is the support of $u$, that is defined as

$$ spt(u)=\overline{\{x\in \Omega : u(x)\neq 0\}}. $$

Since the support is compact, it is boudned, so $|spt(u)| < \infty$.

Answer 4

Not sure this is the fastest and most elegant solution, but this fact is rather a standard fact from distribution theory.

$\textbf{First case}:$ If $\Omega$ is bounded than it is obvious (as you wrote), since set has finite Lebesgue measure.

$\textbf{Second case}:$ Set $\Omega$ is open and unbounded. This is not trivial fact (but a good excercise) to show that in $R^d$ every open set can be represented as a countable union of open balls - therefore it is true for $\Omega$. Therefore it can be shown that it is possible to build an increasing sequence of compact sets $\{K_n\}, \, K_i\subset \Omega$ $$ K_1 \subset K_2 \dots \subset K_n\dots, \, \bigcup_i K_i = \Omega. $$ Having such sequence it is possible to build $L^p$ approximations of any function $f\in L^p(\Omega)$ by setting $f_n(x) = f(x)\cdot I_{K_n}(x)$. By the Lebesgue Theorem for limits of function sequences, this $\{f_n(x)\}$ will be indeed an approximation in $L^p$ - norm (also as an exercise to show it). The technical point then is to choose the $K_i$ such that the support of $f_n(x)$ will be $\delta$-inside the $\Omega$ (it is not difficult to do, knowing that $\Omega$ is a countable union of open balls). The last we do - take some function $\phi\in C^{\infty}_c(\Omega)$ from the ``delta-sequence'' with support in the $\delta/2$-ball in $R^d$. Taking a convolution $\phi*f_n$ will give you a function from $C^{\infty}_c(\Omega)$ with desired accuracy in $L^p$ norm.

p.s. Sorry that I didn't wrote what is delta-sequence, but it is rather standard object from distribution theory, same as idea of smoothing through convolution.