To partition the unity by translating a single function I am trying to show that there exists a (real or complex-valued) function $\psi \in C^\infty(\mathbb{R}^n)$ having the following properties: 


*

*The support of $\psi$ is contained in the unit ball $B(0, 1)$. EDIT As Daniel Fischer points out, this condition cannot be satisfied, at least for $n>4$. The "right" condition is that $\psi$ be supported in the unit cube.

*The sum of all integer translates of $\psi$ is $1$, that is $$\sum_{k \in \mathbb{Z}^n} \psi(x+k)=1,\qquad \forall x \in \mathbb{R}^n.$$

*The sum of the squares of the integer translates of $\psi$ is between two constants: $$0< c \le \sum_{k\in\mathbb{Z}^n} \lvert \psi(x+k)\rvert^2\le C,\qquad \forall x \in \mathbb{R}^n.$$


The existence of a function with those properties is assumed implicitly in the paper that I am reading ${}^{[1]}$, but this does not seem that obvious to me. Can you give me some hint, at least? 

${}^{[1]}$ S. Lee, "On the pointwise convergence of solutions to the Schrödinger equation in $\mathbb{R}^2$", IMRN (2006), pp.1-21. The function $\psi$ appears in the Appendix.

Start with a continuous partition of unity on $\mathbb{R}$ obtained by translation of a single function, say
$$\psi_0(x) = \begin{cases} 1 &, \lvert x\rvert \leqslant \frac14\\ \frac32 - 2\lvert x\rvert &, \frac14 < \lvert x\rvert \leqslant \frac34\\
0 & \lvert x\rvert > \frac34\end{cases}$$
It is easy to check that $\sum_{k\in\mathbb{Z}} \psi_0(x+k) \equiv 1$.
Now take an approximation of the identity $\varphi$ with compact support in $\left[-\frac18, \frac18\right]$ and convolve,
$$\psi_1 := \psi_0 \ast \varphi.$$
The support of $\psi_1$ is contained in $\left[-\frac78,\frac78\right]$, $\psi_1$ is smooth, and
$$1 \equiv \sum_{k \in \mathbb{Z}} \psi_1(x+k) =  \sum_{k \in \mathbb{Z}} (\psi_0 \ast \varphi)(x+k) = \sum_{k \in \mathbb{Z}} \tau_k(\psi_0 \ast \varphi) = \sum_{k \in \mathbb{Z}} (\tau_k \psi_0)\ast\varphi = \left( \sum_{k \in \mathbb{Z}} \tau_k\psi_0\right)\ast \varphi,$$
since the supports of the $\tau_k\psi_0$ are a locally finite family.
Then the function
$$\psi(x_1,\, \dotsc,\, x_n) := \prod_{k=1}^n \psi_1(x_k)$$
has almost the required properties (it support is not contained in the unit ball, but in the cube). However, since "But that is totally not important", let's take that.
