If $f \in \mathcal C^k (\mathbb R^n)$ and $\alpha = (\alpha_1, \ldots, \alpha_n)$ is a multi-index of length $|\alpha | = \alpha_1 + \cdots + \alpha_n \le k$, then we write $$ D^\alpha f := D_1^{\alpha_1} D_2^{\alpha_2} \ldots D_n^{\alpha_n} f \quad \text{where} \quad D_i^{\alpha_i} := \frac{\partial^{\alpha_i}}{\partial x_i^{\alpha_i}}. $$
I'm trying to prove Proposition 4.20. in Brezis's Functional Analysis, i.e.,
Let $k \ge 1,f \in \mathcal C^k_c (\mathbb R^n)$, and $g \in L_{\text{loc}}^1 (\mathbb R^n)$. Then $(f*g) (x)$ is well-defined for all $x \in \mathbb R^n$. Also, $(f * g) \in \mathcal C^k (\mathbb R^n)$ and $$ D^\alpha (f*g) = (D^\alpha f) *g \quad \forall \alpha \in \mathbb N^n \text{ s.t. } |\alpha| \le k. $$ In particular, if $f \in \mathcal C^\infty_c (\mathbb R^n)$ and $g \in L_{\text{loc}}^1 (\mathbb R^n)$, then $(f*g) \in \mathcal C^\infty (\mathbb R^n)$.
Could you have a check on my below attempt?
Proof We need the following lemma, i.e.,
Lemma Let $f \in \mathcal C_c (\mathbb R^n)$ and $g \in L_{\text{loc}}^1 (\mathbb R^n)$. Then $(f*g) (x)$ is well-defined for all $x \in \mathbb R^n$ and $(f * g) \in \mathcal C (\mathbb R^n)$.
By above Lemma, $(f*g) (x)$ is well-defined for all $x \in \mathbb R^n$.
First we prove for the case $k=1$. Fix $x \in \mathbb R^n$, we claim that $f*g$ is differentiable at $x$ and $\nabla (f*g) (x) = (\nabla f )*g (x)$. Let $h \in \mathbb R^n$ with $|h|<1$. By mean value theorem (in integral form), for all $y \in \mathbb R^n$, $$ \begin{align} & |f(x-y+h)-f(x-y) - h \cdot \nabla f (x-y)| \\ = & \bigg | \int_0^1 h \cdot (\nabla f(x-y + sh) - \nabla f (x-y)) \ \mathrm d s \bigg| \\ \le & |h| \varepsilon (|h|) \end{align} $$ with $\varepsilon (|h|) \to 0$ as $|h| \to 0$ (since $\nabla f$ is uniformly continuous).
Let $K := \operatorname{supp} f$. Then $\operatorname{supp} (\nabla f) \subset K$. Let $K'$ be a compact set that contains $x+ B(0, 1)-K$. Then $$ \begin{align} & \bigg |\frac{(f*g) (x+h) - (f*g) (x) - h \cdot ((\nabla f )*g (x))}{h} \bigg | \\ = & \frac{1}{h}\bigg | \int_{\mathbb R^n} f(x+h-y) g(y) \ \mathrm d y - \int_{\mathbb R^n} f(x-y) g(y) \ \mathrm d y - h \cdot \int_{\mathbb R^n} \nabla f(x-y) g(y) \ \mathrm d y \bigg | \\ = & \frac{1}{h}\bigg | \int_{\mathbb R^n} [ f(x+h-y) - f(x-y) - h \cdot \nabla f(x-y) ] g(y) \ \mathrm d y \bigg | \\ \le &\varepsilon (|h|) \int_{\mathbb R^n} g(y) 1_{K'} (y) \ \mathrm d y. \end{align} $$
Notice that $\int_{\mathbb R^n} g(y) 1_{K'} (y) \ \mathrm d y$ is finite because $g \in L_{\text{loc}}^1 (\mathbb R^n)$. The claim follows by taking limit $h \to 0$. It follows from $f \in \mathcal C^1_c (\mathbb R^n)$ that $\nabla f \in \mathcal C_c (\mathbb R^n)$. Then $(\nabla f )*g \in \mathcal C (\mathbb R^n)$ by Lemma. Because $\nabla (f*g) = (\nabla f )*g$, we get $\nabla (f*g) \in \mathcal C (\mathbb R^n)$ and thus $f*g \in \mathcal C^1 (\mathbb R^n)$.
Now we prove for general $k \ge 1$ by induction. Assume the statement holds for $k$. We will prove that it holds for $k+1$. Let $\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_n)$ be a multi-index such that $|\alpha| = k+1$. WLOG, we assume $\alpha_1 \ge 1$. Let $\beta = (\alpha_1-1, \alpha_2, \ldots, \alpha_n)$. Then $|\beta| =k$. By inductive hypothesis,
- $(f * g) \in \mathcal C^k (\mathbb R^n)$, and
- $$ D^\beta (f*g) = (D^\beta f) *g. $$
Clearly, $D^\beta f \in \mathcal C^1_c (\mathbb R^n)$. The base case $k=1$ implies
- $(D^\beta f) *g \in \mathcal C^1 (\mathbb R^n)$ and thus $D^\beta (f*g) \in \mathcal C^1 (\mathbb R^n)$.
- $$ D_1^1 ((D^\beta f) *g) = (D_1^1 D^\beta f ) *g = (D^\alpha f) * g. $$
Hence
- $(f*g) \in \mathcal C^{k+1} (\mathbb R^n)$, and
- $$ D^\alpha (f*g) = D_1^1 D^\beta (f*g) \in \mathcal C (\mathbb R^n). $$
This completes the proof.