$ f\in C_{0}^{\infty} \Rightarrow f\in L^m $? Let $ C_{0}^{\infty} $ be the subspace of $ C^{\infty} $ functions with compact support in $ R^n $. It's true that if $ f \in C_{0}^{\infty}, $ then $ f \in L^{m} \cap H^{s} $ where $ 1\leq m \leq 2$ and $ s\geq 0 $ is an integer number?
Can anyone can me explain why, please?
 A: Let me first prove that $C_0^\infty \subset L^p$. Let $f\in C_0^\infty$, then its support is contained in some ball $B(R)$ of radius $R$. However, then we get
$$ \Vert f \Vert_p = \left( \int \vert f (x) \vert^p dx \right)^\frac{1}{p}
= \left( \int_{B(R)} \vert f (x) \vert^p dx \right)^\frac{1}{p}  
\leq \Vert f \Vert_\infty \vert B(R)\vert <\infty$$
where $\vert B(R)\vert$ denote the measure of the ball $B(R)$. Thus, $f\in L^p$. As you see, we did not need the smoothness for this argument. We only used the continuity to conclude that our function is bounded.
Next we show the inclusion in the Sobolev spaces. Compactly supported smooth functions are in Schwartz space. One can show that the Fourier transform of a Schwartz function is again a Schwartz function. If you explicitely write down the $H^s$ norm of $f$, then you see that you just take the $L^2$ norm of the Fourier transform of $(1+|⋅|)^s \hat{f}$. However, using that $\hat{f}$ is a Schwartz function, one gets that this norm is finite. Thus $f$ is in any $H^s$ for $s\geq 0$. The main trick here is the following: Let $m\geq s$ be an integer. Then we get
$$ (1+ \vert x \vert)^{s} \vert \hat{f}(x) \vert 
\leq \frac{1}{(1+\vert x \vert)^m} \cdot (1+ \vert x \vert)^{2m} \vert \hat{f}(x) \vert \leq \frac{C_m}{(1+\vert x \vert)^m} $$
where $C_m$ is a constant independent of $x$ (this follows from the defintion of the Schwartz space). Chosing $m$ sufficiently large, we get that
$$ \frac{C_m}{(1+\vert x \vert)^m} $$
is in $L^2$. Therefore, we get that $(1+|⋅|)^s \hat{f}$ is in $L^2$ and hence $f\in H^s$.
