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?