Take $ f \in L^{1} (\mathbb{R})$, and $ g \in L^{\infty}(\mathbb{R})$, with $g$ almost everywhere differentiable and such that $g' \in L^{\infty}(\mathbb{R})$.
Prove or disprove: $(f \ast g) \in C^1({\mathbb{R}})$ $(f\ast g)'=f\ast g'$.
I think this is not true. Take for $g$ the cantor lebesgue function. Extended in the whole real line so $g( x)=0 $ for $x\leq 0$ and $g(x)=1$ for $x \geq1$. And choose $f$ the characteristic function of $[-1,1]$.
The cantor lebesgue function is continuous, and have the property that $g'=0$ a.e.. If the prove or disprove statement was true then $f \ast g$ should be constant, but this is not the case. Since for $g\ast f(10) = 0$ but $g\ast f(0)\neq 0 $.
Am I missing something? Thank you for your time.
Let $$ f(x)=\left\{\begin{array}{l} e^{1/(x^2-1)}&\text{when }|x|\lt1\\ 0&\text{when }|x|\ge1\\ \end{array}\right. $$ and $$ g(x)=\left\{\begin{array}{l} 1&\text{when }|x|\lt1\\ 0&\text{when }|x|\ge1\\ \end{array}\right. $$ $f\in L^1$ and $g\in L^\infty$. Furthermore, $g'(x)=0$ almost everywhere.
It is easy to see that since $$ \begin{align} \int_\mathbb{R}f\ast g(x)\,\mathrm{d}x &=\int_\mathbb{R}f(x)\,\mathrm{d}x\int_\mathbb{R}g(x)\,\mathrm{d}x\\ &\gt0 \end{align} $$ Thus, $(f\ast g)'(x)$ is not identically $0$.
Since $g'(x)=0$ almost everywhere, $f\ast g'(x)=0$ for all $x$.
Note that this result is different if we treat $g'(x)$ as a distribution.