We prove that the set $S_p$ is closed under the convolution operation. Let $$f_3 = f_1 \ast f_2$$.
In order to prove that $f_3 \in S_p$, we need to prove that
- The set of all points where $f_3$ is finitely differentiable is countable and dense in $(0,1)$ and
The set of all points where $f_3$ is $n \in \mathbb{N}$ times differentiable is a finite set.
Proof for statement 2
Let $D_1,D_2$ are the countable dense sets and $ck_1,ck_2$ are the maps associated with the functions $f_1,f_2$ respectively.
The convolution operation is defined as $$ f_3(\tau) = \int_0^1 f_1(\tau-x) f_2(x) d{x}$$
In convolution we flip $f_1$ about the $y-axis$ and shift it by $\tau$ and place it on the function $f_2$ and multiply pointwise and take a summation to get $f_3(\tau)$.
The minimum value of $n$ = sum of, the number of times $f_1$ is differentiable at $\tau-x$ and the number of times $f_2$ id differentiable at $x$, $\forall x \in (0,1)$.
Let $x_1 \in D_1, x_2 \in D_2$
The points $x_1$, $x_2$ coincide only once for each shift and the point for which they coincide is for $\tau = x_1 - x_2$.
For a point $\tau$ Let $C = \{(x_1,x_2)/ x_1 \in D_1, x_2 \in D_2\}$ be set of all coinciding points for a particular shift $\tau$. Then $$n = \min\limits_{(x_1,x_2) \in C} ck_1(x_1) + ck_2(x_2)$$ where $n$ is the maximum number of times $f_3$ is differentiable at $\tau$. If $C = \phi$ then $f_3$ is infinitely differentiable at $\tau$.
For $f_3$ to be $n$ times differentiable at $\tau$ there should be atleast one coincidence $(x_1,x_2)$ such that $ck1(x_1) + ck2(x_2) = n$. Which means we need coincidence of type $(x_1,x_2)$ with $ck1(x_1) + ck2(x_2) = n$ for points $\tau$ where $f_3$ is $n$ times differentiable. Each coincidence $(x_1,x_2)$ can correspond to only one $\tau$. The set of all coincidences of the form $(x_1,x_2)$ such that $ck1(x_1) + ck2(x_2) = n$ is essentially a subset of $$C_n = \{(x_1,x_2)/x_1 \in D_1, x_2 \in D_2,ck_1(x_1) < n, ck_2(x_2) < n\}$$ The set $C_n$ is a finite set as $f_1,f_2 \in S_p$.
Hence the total number of points where $f_3$ is differentiable $n$ times $\forall n \in \mathbb{N}$ is finite. Also since a countable union of finite sets is countable, the set of all points where $f_3$ is finitely differentiable is countable.
Proof that the set of all points where $f_3$ is finitely differentiable is dense in $(0,1)$.
Consider the interval $(\tau_1,\tau_2)$, as $\tau$ varies from $tau_1$ to $\tau_2$ we need to prove that there is atleast one coinciding point $(x_1,x_2)$ where $x_1 \in D_1$ and $x_2 \in D_2$.
As $D_1$ is dense in $(0,1)$ there is a $x_1 \in (\tau_1,\tau_2)$ and as $D_2$ is dense in $(0,1)$ there is an $x_2 \in (x_1,\tau_2)$.
As we vary the shift $\tau$ from $\tau_1$ to $\tau_2$ the point $x_1$ coincides with the point $x_2$ as $x_2 - x_1 < \tau_2 - \tau_1$ and hence there is point where $f_3$ is finitely differentiable in any in interval $(\tau_1,\tau_2) \subset (0,1)$.
Therefore we have proved that the points where $f_3$ is finitely differentiable is countable and dense in $(0,1)$ and the number of points at which $f_3$ is exactly $n$ times differentiable, for any $n \in \mathbb{N}$ is finite. Hence $f_3 \in S_p$.
Hence the set $S_p$ is closed under the convolution operation.
PS : This proof is intuitive but not in the correct mathematical language. Hence request you to give any suggestions or post the proof in proper mathematical language as a separate answer so that I can award the bounty.
PS2 : Comments from Theo were particularly helpful for this answer.
Best Answer
Well, shifting is enough is the sense that reversing does not really change in an essential way the mathematical object of convolution. But the reason we choose the reversing definition conventionally may be because of several conveniences. For example:
(1) The property of commutativity, that is, $f*g=g*f$, is lost without reversing;
(2) The property that convolution is multiplication on the Fourier side, that is $\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$ where $\mathcal{F}$ denotes the Fourier transform, is lost without reversing;
Etc.