Is $C^\infty\subset L_{1,\text{loc}}(\mathbb{R})$?

Question

Is $C^\infty\subset L_{1,\text{loc}}(\mathbb{R})$? Or, can we connect differentiability to integrability (such as between $C^1$ and $L_{1,\text{loc}}(\mathbb{R})$)? I think this correct, but I don't have any rigorous arguments.

Answer 1

Just use the definition of $L^p_{\operatorname{loc}} (\mathbb R)$, $1 \leq p < \infty$: Let $f \in \mathrm C^\infty(\mathbb R)$. We have to show that $f|_K \in L^p(K)$ for every compact set $K \subseteq \mathbb R$. So let $K \subseteq \mathbb R$ compact. Since $f$ is continuous, we know that $f$ admits it maximum on $K$, i.e. $c := \max_{x \in K} \vert f(x) \vert < \infty$ exists. Thus, \begin{align*} \Vert f \Vert_{L^p(K)}^p = \int_K \vert f(x) \vert^p \, \mathrm d x \leq \int_K c^p \, \mathrm d x = c^p \vert K \vert < \infty, \end{align*} since $K$ is bounded and consequently has finite Lebesgue measure. Since $K$ was arbitrarily chosen, the claim follows. Notice that we didn't even need the differentiability of $f$, just its continuity. Thus, the claim follows even for each $f \in \mathrm C(\mathbb R)$. Finally, the claim also holds for $p = \infty$ obviously. I hope it got clear :)