Let $k=(k\_1,k\_2) \in \mathbb{Z}^2$. Let $\lambda=(\lambda\_1,\lambda\_2)\in [0,2\pi]^2$ and $F(\lambda)$ be a bounded real function of $\lambda\in [0,2\pi]^2$.


I am interested in the following equation:
$$
\frac{1}{(2\pi)}\int\_{[0,2\pi]}\sum\_{l=-\infty}^\infty \lvert l\rvert \bigg\lvert\frac{1}{2\pi}\int\_{[0,2\pi]} e^{il\lambda\_1 }F(\lambda)d\lambda\_1 \bigg\rvert^2{d\lambda\_2}\\
+\frac{1}{(2\pi)}\int\_{[0,2\pi]}\sum\_{l=-\infty}^\infty \lvert l\rvert \bigg\lvert\frac{1}{2\pi}\int\_{[0,2\pi]} e^{il\lambda\_2 }F(\lambda)d\lambda\_2 \bigg\rvert^2{d\lambda\_1}
\\=\sum\_{k\in \mathbb{Z}^2}|k\_1|\bigg|\frac{1}{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1\lambda\_1+k\_2\lambda\_2)}F(\lambda)d\lambda\bigg|^2\\
\\+\sum\_{k\in \mathbb{Z}^2}|k\_2|\bigg|\frac{1}{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1\lambda\_1+k\_2\lambda\_2)}F(\lambda)d\lambda\bigg|^2
$$
Can anyone give me a brief explanation as to why the statement is wrong or give me a hint for the solution?



2-dimensional Fourier transform
$\newcommand\Z{\mathbb Z}$Suppose that $F$ is a bounded measurable complex-valued function on $[0,2\pi]^2$, so that $F\in L^2([0,2\pi]^2)$. Then
$$F(s,t)=\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{i(ms+nt)} \tag{1}$$
for some complex $c\_{m,n}$'s such that $\sum\_{(m,n)\in\Z^2}|c\_{m,n}|^2<\infty$ (the above double series converges in $L^2([0,2\pi]^2)$).


Now direct calculations (see details below) show that the left-hand side (lhs) of your identity is $\sum\_{(l,n)\in\Z^2}|l|(|c\_{l,n}|^2+|c\_{n,l}|^2)$, whereas the right-hand side (rhs) of your identity is $\sum\_{(k\_1,k\_2)\in\Z^2}(|k\_1|+|k\_2|)||c\_{k\_1,k\_2}|^2$, which is obviously the same as the left-hand side.


Thus, your identity does hold.




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**Details:** By (1),
$$I\_l:=\frac1{2\pi}\int\_0^{2\pi}e^{ils}F(s,t)ds
=\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{int}1(m=l)
=\sum\_{n\in\Z}c\_{l,n}e^{int},$$
and hence
$$|I\_l|^2=I\_l\overline{I\_l}
=\sum\_{(m,n)\in\Z^2}c\_{l,m}\overline{c\_{l,n}}e^{i(m-n)t}$$
and
$$\frac1{2\pi}\int\_0^{2\pi}|I\_l|^2 dt 
=\sum\_{(m,n)\in\Z^2}c\_{l,m}\overline{c\_{l,n}}1(m=n)
=\sum\_{n\in\Z}|c\_{l,n}|^2.$$
So, the first summand on the lhs of your identity is $\sum\_{(l,n)\in\Z^2}|l||c\_{l,n}|^2$ and, similarly, the second summand there is $\sum\_{(l,n)\in\Z^2}|l||c\_{n,l}|^2$, so that the lhs of your identity is $\sum\_{(l,n)\in\Z^2}|l|(|c\_{l,n}|^2+|c\_{n,l}|^2)$.


Next, again by (1),
$$\frac1{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1s+k\_2t)}F(s,t)ds\,dt \\ 
=\frac1{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1s+k\_2t)}\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{i(ms+nt)}ds\,dt \\ 
=\sum\_{(m,n)\in\Z^2}c\_{m,n}1(m=k\_1,n=k\_2)=c\_{k\_1,k\_2}.$$
So, the rhs of your identity is $\sum\_{(k\_1,k\_2)\in\Z^2}(|k\_1|+|k\_2|)|c\_{k\_1,k\_2}|^2$.


