Differentiability of convolution without compact support

Question

Let $g \in C^1(\mathbb R), f, g, g' \in L^1(\mathbb R)$. How do I then show $f * g$ is differentiable? Variations of this question have been asked where $g$ or $g'$ have compact support, but these assumption make it easier since that implies boundedness due to continuity.

My work so far: Compute $$\lim_{h \to 0} \frac{(f * g)(x+h) - (f * g)(x)}{h} = \lim_{h \to 0} \int_{\mathbb{R}} f(y) \frac{g(x+h-y) - g(x-y)}{h} dy.$$

Because of dominated convergence, we can swap the limits as we wish to if we are able to exhibit an integrable bound $H \in L^1(\mathbb R)$ with $$|f(y) (g(x-y + 1/n)-g(x-y)) n| \le H(y)$$ for all $n$. If you restrict yourself to $H : (-a,a) \to \mathbb R$, this easily follows by the mean value theorem, indeed you can choose $H$ constant in this case. However, $g'$ does not have to be bounded over $\mathbb R$, so there's no obvious way of recovering a similar result.

I also know $g$ is bounded due to the integrability of $g'$, but of course, that's not quite sufficient.

Any ideas? In particular, I'd like to avoid too much functional analysis, this is for a relatively basic analysis class.

Answer 1

Your assumption implies that $f \in L^1 (\mathbb R)$ and $g \in W^{1, 1} (\mathbb R)$. Then we can apply the following result from Brezis' Functional Analysis:

Lemma 8.4. Let $\rho \in L^1(\mathbb{R})$ and $v \in W^{1, p}(\mathbb{R})$ with $1 \leq p \leq \infty$. Then $\rho * v \in$ $W^{1, p}(\mathbb{R})$ and $(\rho * v)'=\rho * v'$.

Notice that a function in $W^{1, 1} (\mathbb R)$ is locally absolutely continuous and thus differentiable almost everywhere.