First of all the convolution is defined as
\begin{align}
(f*g) (x) = \int_{-\infty}^{\infty} f(x-t)g(t)dt, \qquad x \in \mathbb{R}
\end{align}
And I know The convolution of two continuous functions $f$ and $g$ is differentiable if any of the functions $f$ and $g$ are differentiable and have a bounded derivative
Then what I want to know is
Suppose that $f$ is continuous on $\mathbb{R}$. Is the function $f* \chi_{[-1,1]}$ differentiable?
i.e., convolution of continuous function and characteristic function is differentiable?
I know $\chi$ is bounded and measurable and $\chi_A$ is continuous on the interior of $A$ and $A^c$ but not at the boundary.
It seems the answer to the above question is no, but I don't know how to explain this with the above information. Naively it seems continuity of $f$ is not enough for a general case but it seems this might be cured due to $\chi$.
Best Answer
$(f*\chi_{[-1,1]})(x)=\int_{-1}^{1} f(x-t)dt=\int_{x-1}^{x+1} f(s)ds$ by the substitution $s=x-t$ Hence, $(f*\chi_{[-1,1]})(x)=F(x+1)-F(x-1)$ where $F$ is an anti-derivative of $f$. Of course, this is differentiable.