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?

  • $\begingroup$ The solution of what? $\endgroup$
    – LSpice
    Commented Nov 2, 2021 at 23:21
  • $\begingroup$ My question is how do I get from one side of the equation to the other side of the equation? $\endgroup$
    – HyyFly
    Commented Nov 2, 2021 at 23:49

1 Answer 1


$\newcommand\Z{\mathbb Z}$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.

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 $\sum_{(k_1,k_2)\in\Z^2}(|k_1|+|k_2|)|c_{k_1,k_2}|^2$.

  • 1
    $\begingroup$ @HyyFly : I have added details of the calculation. $\endgroup$ Commented Nov 3, 2021 at 15:45

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