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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.

  1. $Cf(x) = k$, where $k \in \{0\}\bigcup\mathbb{N}\bigcup\{\infty\}$.
  2. 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.

  1. $\forall n \in \mathbb{N}$, the pre-image $k^{-1}(\{n\})$ is a finite set.
  2. $\forall x \in $D$, ck(x) \ge k(x)$ but finite and $\forall x \in (0,1)$\D$, ck(x) = \infty$.
  3. $\forall x \in (0,1), Cf(x) = ck(x)$
  4. 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.

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What is $Cf{}$? –  Mariano Suárez-Alvarez Jan 21 '12 at 3:43
    
@Mariano : $Cf(x) = k$, 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 $f$, 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. –  Rajesh D Jan 21 '12 at 4:02
    
@Mariano : Please let me know if you think its a serious problem, I could would try to make a more formal notation. –  Rajesh D Jan 21 '12 at 4:54
2  
No, the problem was that you did not explain anywhere what that notation meant. You should add it the definition of $Cf$ to the body of the question. –  Mariano Suárez-Alvarez Jan 21 '12 at 5:46
2  
If I understand it correctly (which is quite possibly false) then your proof that $S_{pc}$ is closed under addition does not work. If $f$ is exactly $n$-times differentiable at $x$ and $g$ is exactly $m$-times differentiable at $x$ then why should $f + g$ be exactly $min(m,n)$-times differentiable at $x$? Most of the time, it is, but "most" is not good enough. –  Loop Space Jan 25 '12 at 10:05

1 Answer 1

up vote 1 down vote accepted

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

  1. The set of all points where $f_3$ is finitely differentiable is countable and dense in $(0,1)$ and
  2. 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.

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