# Does inverse wrapping exist?

function $f$ on $R$ can be made periodic with period 1 by wrapping $$g(x)=\sum_{k=-\infty}^{\infty}f(x+k)$$

Given $g$ with period 1. How to find continuous $f$ in general case?

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What you want is not clear, as far as I can see. Also, the series you wrote does not converge in general... –  Mariano Suárez-Alvarez Mar 5 '11 at 2:34
@Katja: What you write does not really make sense or work for arbitrary $f$. Did you perhaps meant to write something along the lines of "given an $x$ not in [0,1), find an integer $k$ such that $x+k\in[0,1)$ and then define $g(x)$ as $f(x+k)$"? –  Arturo Magidin Mar 5 '11 at 2:50
@Arturo Magidin: It makes perfect sense to me as it stands. It's a series that converges if $f$ decays sufficiently rapidly at infinity. This construction is used e.g. in constructing Bloch functions from Wannier functions in solid state physics: en.wikipedia.org/wiki/Wannier_function#Properties –  joriki Mar 5 '11 at 9:53
It is also used in circular statistics ("wrapped Gaussian" etc). –  Rasmus Mar 5 '11 at 13:34
@joriki: "...for abitrary $f$." Of course, it works for some functions, but there was no restriction placed on $f$, it merely says "function $f$ on $R$" (not even assuming continuity!) –  Arturo Magidin Mar 5 '11 at 20:22

## 1 Answer

The solution to your problem is non-unique. But if you just want a candidate:

assuming $g$ is continuous, let $c$ be the average value of $g$ over $[0,1]$. Then by intermediate value theorem, the function $g - c$ hits 0 at some point. By translating the axis we can assume that $g-c = 0$ at $0$. Define $f_1(x) = g-c$ if $x\in [0,1]$ and $0$ elsewhere. $f_1$ is continuous, has compact support, and its periodicization is $g-c$.

Now if suffices to find a continuous function $f_2$ such that its periodicization is $c$. For this we simply define

$$f_2 = \begin{cases} cx & x\in[0,1]\\ c - c(x-1) & x\in(1,2]\\ 0 & x\notin [0,2] \end{cases}$$

You easily check that $f_2$ is continuous, has compact support, and has periodicization the constant value $c$. So $f = f_1 + f_2$ will "wrap" to $g$.

(By suitable mollification, $f_2$ can clearly be made smooth. By considering derivatives, you can make $f_1$ as smooth as the function $g$.)

Edit:

After thinking about this more, a slightly simpler construction is as follows. Observe that we assumed $g$ to be a continuous, periodic function. Extend $g$ to over $\mathbb{R}$ by periodicity. Then it suffices to construct a function $\psi$, with the following property:

• $\sum_{k = -\infty}^\infty \psi(x+k) = 1$
• $\psi$ is continuous (smooth)

This is because if you let $f(x) = g(x) \psi(x)$, using periodicity of $g$ and the above property of $\psi$, $f$ has the desired properties. Now let $\eta$ be a smooth transition function: $\eta = 0$ if $x < 0$, $\eta = 1$ if $x > 1$, and $\eta$ is smooth. Consider

$$\phi(x) = \frac{1}c \eta(x) (1- \eta(x - k))$$

where $k$ is a positive integer and the constant $c$ is chosen such that the compactly supported function $\phi(x)$ verifies the condition above on its sum. (By construction the periodic sum of $\eta(x)(1- \eta(x-k))$ is some constant integer.)

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To illustrate nonuniqueness, another possible definition of $f_2$ is that $f_2 = cx/2$ if $x\in [0,1]$, $c/2$ if $x\in (1,2)$ and $c/2 - c(x-2)/2$ if $x\in[2,3]$ and $0$ elsewhere. –  Willie Wong Mar 5 '11 at 13:26