Construct a special $C_0^\infty$ function

Question

I'am quite curious about how to construct a nonnegative $C_0^\infty$ function $f(x)$ satisfying for $1<p<\infty$ $$ f(0)=1~~~~~\text{and}~~~~~\sum_{k\in \mathbb Z^n}f(x+k)^p=1. $$

It seems easy to construct such a function in $\mathbb R^1$.

We first construct a nonnegative radial function $f\in C_0^\infty(\mathbb R^1)$ and $\operatorname{supp}f\subset (-1,1)$. $f$ also satisfies $f(0)=1$ and $f(\tfrac 12)=\tfrac 12$. (We can ask for such $f$ equals to 1 in a small neighborhood of $0$.)

Then we ask that $f(x)+f(x-1)=1$ for $x\in [0,1)$. Note that this can be done if we first aske for the smoothness in $[0,1/2)$ and just by let $f(1-x)=1-f(x)$ for $x\in (0,1/2)$.

And just by letting $g(x)=f(x)^{1/p}$, we finish this.

My question is:

(1) We know that the lattices contained in the unit ball is related to the dimension, and how to extending the construction to $\mathbb R^n$ (It can not be just rotation to get a $\mathbb R^n$ function.)

(2) I think there are some other convenient way to constrcut such a funcion.

(3) I think the construction has nothing to do with $p$, is it right?

Thanks in advance. I hope to get some help from you.

Answer 1

The answer is rather simple. First, let $h(x) = e^{-1/x}$ for $x>0$, otherwise $0$. And let $f(x)=\frac{h(1-|x|)} {h(1-|x|) + h(|x|)}$ for $x \in \mathbb{R}$.

Also, constructing such a function in $\mathbb{R}^n$ is not difficult: $$F_n(x) = \prod_{j=1}^{n} f(x_j)$$ Where $x = (x_1,x_2, \cdots, x_n)$.

We can prove that $$ \sum_{k \in \mathbb{Z}^n} F_n(x+k) = \sum_{k_1, \cdots, k_n \in \mathbb{Z}}\prod_{j=1}^{n} f(x_j+k_j) = \prod_{j=1}^{n} \sum_{k_j \in \mathbb{Z}}f(x_j+k_j) = \prod_{j=1}^{n}1=1$$ Then let $G_n = F_n^{1/p}$.

Does this solve your problem? I think this construction is convenient enough.

Answer 2

General process of creating of a partition of unity does the job when we let the inherent symmetry manifest.

1. Take a covering: $G_k := Q(k,1)$ (cubes). Let us skip this step.

2. Take some function supported by $G_k$, or, in our case, we just take some functions: Let $h_0$ be a any smooth non-negative function which is positive on the closure of the unit cube $Q(0,1/2)$ and whose support is bounded. Let $h_k(x) = h_0(x -k)$, for all $k\in \mathbb Z^n$.

3. We normalise the sum to 1: Let $g_k(x) = h_k(x) / \sum_l h_l(x)$.

Now we have a partition of unity. The formula preserves the relation $g_k(x) = g_0(x-k)$. Hence if $f=g_0^{1/p}$, we have $\sum_{k\in \mathbb Z^n}f(x+k)^p=1$.

Now what is $f(0)$ ? This is some number $p$ in $[0,1]$, and if you review steps 1 and 2, you see you get any of that numbers. For $p=1$, we find $g_0$ which satisfies $g_0(1)=1$, which is positive on the closure of the unit cube $Q(0,1/2)$ and whose support is contained in $Q(0,1)$. For other value $p\in [0,1]$, instead of messing with $g_0$, let us chose $a$ such that $g_0(a)=p$ and shift the function sideways, $f(x)=g_0(x-a)^{1/p}$

The next question is if we can get $f(0)=p$ for other $p$ then in $[0,1]$.