$\int_{\mathbb R^d} (1+ |x-n|^2)^{s} |f(x)|^2 dx = \sum_{k\in \mathbb Z^d}(1+ |k-n|^2)^s \int_{I_k}|f(x)|^2 dx?$

Question

Let $I_k= \prod_{i=1}^d[k_i-1/2, k_i+1/2].$ Then $\mathbb R^d= \cup_{k\in \mathbb Z^d} I_k$ where $k=(k_1,.., k_i,...k_d)\in \mathbb Z^d.$

Let $f\in \mathcal{S}(\mathbb R^d)$ (Schwartz class) and $n\in \mathbb Z^d$ and some $s<0.$

My Question: Can we say that

$$\int_{\mathbb R^d} (1+ |x-n|^2)^{s} |f(x)|^2 dx = \sum_{k\in \mathbb Z^d}(1+ |k-n|^2)^s \int_{I_k}|f(x)|^2 dx?$$

Answer 1

No, let's take a simple case. Suppose $f$ is smooth on $\mathbb R,$ with $f>0$ on $(-1/2,1/2)$ and $f=0$ elsewhere. Let $n\in \mathbb Z.$ Then your equality would be

$$\int_{-1/2}^{1/2}(1+x^2)^sf(x)^2\, dx = \int_{-1/2}^{1/2}f(x)^2\, dx.$$

But $(1+x^2)^s<1$ for $x\in (-1/2,1/2),x\ne 0.$ Thus the integral on the left is less than the integral on the right.