$\|f\|_{L^{p}}=\|g\|_{L^p} \implies \|f\|_{L_w^{p}}=\|g\|_{L_w^p} $

Question

Let $f, g: \mathbb C \to \mathbb C$ be functions such that it's $L^{p}-$norm ($1\leq p < \infty$) is the same, that is,

$$\int_{\mathbb C} |f(z)|^{p} dz= \int_{\mathbb C} |g(z)|^p dz.$$

Put $w(z)= \left( 1+ \left|\frac{z-\bar{z}}{2}\right|^2 \right)^{s/2} \ (s>0).$

My Question is: Can we expect $$\int_{\mathbb C} |f(z) w(z)|^{p} dz= \int_{\mathbb C} |g(z) w(z)|^pdz?$$

Answer 1

No. Let $f(z)$ be the characteristic function of the square $[0,1]\times[0,1]$, and $g(z)$ the characteristic function of the square $[0,1]\times[5,6]$. Then $$ \int_{\mathbb C}|f|^p=\int_{\mathbb C}|g|^p, $$ $$ \int_{\mathbb C}|fw|^p=\int_{[0,1]\times[0,1]}|w|^p\leq\left(\frac54\right)^{sp/2}, $$$$ \int_{\mathbb C}|gw|^p=\int_{[0,1]\times[5,6]}|w|^p\geq\left(\frac{27}4\right)^{sp/2}. $$