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
From the definition of convolution, we have$$ y(t)=(x∗h)(t)=\int_{-∞}^∞ x(τ)h(t−τ)\,\mathrm{d}τ. $$ Suppose that $x(t)$ is periodic with period $T$, then we have $x (t)=x(t+T)$ and we can rewrite the above integral as$$ y(t)=\int_{-∞}^∞ x(τ+T)h(t−τ)\,\mathrm{d}τ. $$ Now, we employ a change of variable. Let $λ=τ+T$ so that $τ=λ−T$ and $\mathrm{d}λ=\mathrm{d}τ$. This yields\begin{align*} y(t)&=\int_{-∞}^∞ x(λ)h(t−(λ−T)) \,\mathrm{d}λ\\ &=\int_{-∞}^∞ x(λ)h(t+T−λ) \,\mathrm{d}λ\\ &=\int_{-∞}^∞ x(λ)h((t+T)−λ) \,\mathrm{d}λ\\ &=(x∗h)\bigr|_{v=t+T}=y(v)\bigr|_{v=t+T}=y(t+T). \end{align*} Therefore, $y(t)$ is periodic with period $T$.