Is Convolution of continuous function $f$ and $\chi_{[-1,1]}$ differentiable?

Question

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 $\chi$.

Answer 1

It exists $g$ such that $g(b)-g(a)=\int_a^b f(u) du$. Therefore

\begin{align} \lim_{h\to0}\frac{(f*\chi)(x+h)-(f*\chi)(x)}{h}&= \lim_{h\to 0} \frac{\int_{-1}^1f(x+h-t)dt-\int_{-1}^1f(x-t)dt}{h}\\ &= \lim_{h\to 0} \frac{g(x+h+1)-g(x+h-1)-(g(x+1)-g(x-1)}{h}\\ &= \lim_{h\to 0}\frac{g(x+h+1)-g(x+1)}{h} - \lim_{h\to0}\frac{g(x+h-1)-g(x-1)}{h}\\ &= g'(x+1)-g'(x-1) = f(x+1)-f(x-1). \end{align}

Answer 2

$(f*\chi_{[-1,1]})(x)=\int_{-1}^{1} f(x-t)dt=\int_{x-1}^{x+1} f(s)ds$ 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.