Is this class of periodic functions closed under the (circular) convolution operation ? Help in proving. I am studying the properties of a particular class of functions, and I'd appreciate some help in proving a property of that class. I started with a class of functions and made some modifications to show that it forms a vector space and to show that it is closed under convolution. I have given the entire description below, and require help in proving that the set is closed under the operation of convolution (circular). I have also mentioned what methods I have used to prove that it forms a vector space, so that i can get cleared of the difficulty i am facing in arriving at the result of closure under convolution. ( in case you want to avoid reading the entire post linearly, the property needs to be proved on the set $S_{pc}$ and it is described in the section "closure under addition"). 
PS : Please do let me know if you want to know the motivation for such a study. I do not have any concrete reasons which can be expressed in mathematically precise terms, behind the motivation, but I have an intuition behind it.
PS 2 : A construction method and proof of non-emptiness of this class of functions is given in this Q&A.
Definition
Let $f \colon \mathbb{R} \to \mathbb{R}$ be a function, consider a point $x \in \mathbb{R}$ where $f$ is continuous, then the following two statements are equivalent.


*

*$Cf(x) = k$, where $k \in \{0\}\bigcup\mathbb{N}\bigcup\{\infty\}$.

*The function is continuous at $x$ and the maximum number of times $f$ is differentiable at $x$ is $k$.



Clarification (Added)
$Cf(x)=k$, where we know that $f$ is a function and $x$ is the point in its domain, then it means that the maximum number of times $f$ is differentiable at $x$ is $k$. This is not a good notation as someone could think that $Cf$ itself is one function different from f and not associated with it, but here I intended that $C$ tells about a proprty of the function at a point in its domain.....not a convincing notation......but i'd like know any ideas to make a good notation for it.

Definition of a class of functions
The set $S$ consists of functions $f \colon (0,1) \to \mathbb{R}$, which satisfy the following properties.
Given any $f \in S$ there exist a countable dense subset $D \subset (0,1)$ and maps $k,ck$ defined as $k \colon D \to \mathbb{N}$ and $ck \colon (0,1) \to \mathbb{N}\bigcup\{\infty\}$ which satisfy the following properties.


*

*$\forall n \in \mathbb{N}$, the pre-image $k^{-1}(\{n\})$ is a finite set.

*$\forall x \in $D$, ck(x) \ge k(x)$ but finite and $\forall x \in (0,1)$\D$, ck(x) = \infty$.

*$\forall x \in (0,1), Cf(x) = ck(x)$

*Whenever $Cf(x) = k$ is finite, for the $(k+1)^{nth}$ derivative of $f$ at $x$, the left and right limits exist and are not equal. (the left and right limits do not diverge).


Let $S_p$ be the set of all functions $f_p$ which are periodic versions of the functions $f \in S$.
The periodic version $f_p$ of the function $f \in S$ is defined as $f_p(x) = f(x) \forall x \in (0,1)$, $f_p(0) = f(0+)$ and $f_p(1) = f_p(0)$ and $ \forall x \in \mathbb{R}, f_p(x) = f_p(x+1)$.

Closure property under addition
EDIT : (This argument is false and the set $S_{pc}$ is not closed under addition) (see comment by Andrew)
Let $f_1,f_2 \in S_p$ such that $f_1$ is not same as $-f_2$. I am able to prove that if $f_3 = f_1 + f_2$, then $f_3 \in S_p$, by making the following considerations.
Let $D_1,k_1,ck_1$ be the required set and maps respectively for the function $f_1$ as per the definitions given above. Let $D_2,k_2,ck_2$ be the required set and maps of $f_2$.
Let $D_3,k_3,ck_3$ be defined as below.
$D_3 = D_1 \bigcup D_2$.
Let $l_1 \colon (0,1) \to \mathbb{N}\bigcup\{\infty\}$ and $l_2 \colon (0,1) \to \mathbb{N}\bigcup\{\infty\}$ are defined as
$l_1(x) = k_1(x)$ if $x \in D_1$ otherwise $\infty$.
$l_2(x) = k_2(x)$ if $x \in D_2$ otherwise $\infty$.
$k_3(x)$ is assigned as $k_3(x) = \min\{l_1(x),l_2(x)\}$
and 
Let $m_1 \colon (0,1) \to \mathbb{N}\bigcup\{\infty\}$ and $m_2 \colon (0,1) \to \mathbb{N}\bigcup\{\infty\}$ are defined as
$m_1(x) = ck_1(x)$ if $x \in D_1$ otherwise $\infty$.
$m_2(x) = ck_2(x)$ if $x \in D_2$ otherwise $\infty$.
$ck_3(x)$ is assigned as $ck_3(x) = \min\{m_1(x),m_2(x)\}$.
By assigning $D_3,k_3,ck_3$ as mentioned above I am able to prove that $f_3 \in S_p$.
To get the closure property under addition, we can add the set of all constant functions $K$ to the set $S_p$ to form a new set $S_{pc}$.
Hence I am able to prove that the set $S_{pc}$ as defined above is closed under the addition operation (and it easily follows that the set $S_{pc}$ is closed under the multiplication operation as well).
There by I am able to show that the set $S_{pc}$ is indeed a vector space. (as it cab be easily seen that the set $S_{pc}$ is closed under scalar multiplication).

The Question
Where I need some help is to show that the set $S_{pc}$ is closed under the binary operation of circular convolution. Here by circular convolution, i mean the convolution operation with the convolution integral summed only over one period i.e., on $[0,1]$.
Specifically How should i make the choice of $D_3,k_3,ck_3$, the required set and maps for the resultant function of a convolution, to show that it belongs to $S_{pc}$.

What I know
Let $f_1$ is periodic with period $1$ and is smooth (within one period) except at $x = a_1 \in [0,1]$, where it is only $n_1$ times differentiable and Let $f_2$ is periodic with period $1$ and is smooth (within one period) except at $x = a_2 \in [0,1]$, where it is only $n_2$ times differentiable.
Now the function $f_3 = f_1 \star f_2$ is smooth (within one period) except at $x = a_1 - a_2$ where it is only $n_1 + n_2$ times differentiable.
But I am confused as to how to use this fact to arrive at the desired result.
 A: 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.
