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Let $f: \mathbb{R} \rightarrow\mathbb{R}$ be defined as:

$$f(x) = \begin{cases}\frac{1+\cos x}{2} \text{, for }-\pi\leq x\leq \pi \\ 0 \text{, otherwise} \end{cases}$$

For each integer $m\geq 0$ set $\phi_{2m+1}(x) = f(x-m\pi)$ and for each integer $m\geq 1$, set $\phi_{2m}(x) = f(x+m\pi)$. Prove that $\left\{\phi_m\right\}$ forms a partition of unity for $\mathbb{R}$.

So far I proved that the supports are compact and contained $\mathbb{R}$, also that the partition is locally finite. It is trivial to show that $\phi_k(x)\geq 0$ and that $\phi_k$ is $C^{\infty}$. However I'm somewhat stuck in showing that $\sum \phi_{m}(x) = 1$ for all $x\in\mathbb{R}$.

Here's a hint someone gave me: Consider the function $f_m = f(x-m\pi)$ (for every integer, not just the positive ones) and calculate $\sum_{m\in\mathbb{Z}} f_{2m}(x)$ and$\sum_{m\in\mathbb{Z}} f_{2m+1}(x)$.

The first sum is $\frac{1+\cos x}{2}$ while the second is $\frac{1-\cos x}{2}$, clearly, $\sum_{m\in\mathbb{Z}}f_m(x)$ is then $1$ and I know intuitively that this proves what I want. Still, I can't quite formalise the intuition, any hints?

Thanks a lot.

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    $\begingroup$ Note that$\phi_1=\phi_0$. You probably want to restrict to only incude one of them, because the sum of the two is $1+\cos x$, which means you'd need negative values to correct for the times when $\cos x>0.$ $\endgroup$ – Thomas Andrews Apr 15 '18 at 23:33
  • $\begingroup$ Right, actually missed something when typing the exercise, will edit. $\endgroup$ – user371816 Apr 15 '18 at 23:36
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First of all, please differentiate these functions at their boundaries. You will find these functions are $C^1$, not $C^\infty$.

Consider $x\in[0,\pi]$ and handle others by translation invariance. There are two positive contributions: from $\phi_1$ and $\phi_3$. They are $(1+cos x)/2$ and $(1+cos(x-\pi))/2$. Adding them gives $1+\frac{cos(x)+cos(x-\pi)}{2}$. But the cosine of an angle and the same angle plus \pi is always zero (elementary property of sine and cosine). So the fraction in the last expression is 0, and the whole expression simplifies to 1. So you do have a partition of unity, but it's $C^1$ only. In particular, it's not $C^2$ so it won't be adequate for situations where you need to work with $C^2$ functions.

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  • $\begingroup$ I should add for the interval $[\pi,2\pi]$ you would use $\phi_3$ and $\phi_5$, and for n>=2 for $[n\pi,(n+1)\pi]$ $\phi_{2n+1}$ and $\phi_{2n+3}$. Formalize that and the same for the intervals containing negative x and you will be close to formalizing your proof. $\endgroup$ – C Monsour Apr 15 '18 at 23:55
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Observe that $1 = \sum_{n\in \mathbb Z} \mathbf 1_{[(-2n-1)\pi, (-2n+1)\pi[} (x)$ $$\sum_{m\in\mathbb Z} f_{2m}(x) = \sum_{m\in \mathbb Z}\sum_{n\in\mathbb Z} \underbrace{f_{2m}(x)\mathbf 1_{[(-2n-1)\pi, (-2n+1)\pi[} (x)}_{=\left\{\begin{array}{cc} 0 & \text{if $n\neq m$} \\ \frac{1+\cos(x)}{2}\mathbf 1_{[(-2n-1)\pi, (-2n+1)\pi[} (x) & \text{otherwise}\end{array}\right.} = \sum_{n\in\mathbb Z} \frac{1+\cos x}{2}\mathbf 1_{[(-2n-1)\pi, (-2n+1)\pi[} (x) = \frac{1+\cos x}2 $$

You do the same thing for the other sum and you have what you are looking for.

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