Why $f(x + x^3, y + y^3) \in L^1(\mathbb{R}^2)$, when $f(x, y) \in L^2(\mathbb{R}^2)$?

Question

How show that $f(x + x^3, y + y^3) \in L^1(\mathbb{R}^2)$, when $f(x, y) \in L^2(\mathbb{R}^2)$?

Can someone help me?

Thank you!

Answer 1

The statement that $f(x, y) \in L^2(\mathbb{R}^2)$ is the same as the statement that $(f(x + x^3, y + y^3))^2(1 + 3x^2)(1 + 3y^2)$ is in $L^1(\mathbb{R}^2)$, which can be seen after a change of variables to $(u,v) = (x + x^3, y + y^3)$ for the latter. Inspired thus, we write $$\int_{\mathbb{R}^2}|f(x + x^3, y + y^3)| = $$ $$\int_{\mathbb{R}^2}\bigg(\big|f(x + x^3, y + y^3)\big|\sqrt{(1 + 3x^2)(1 + 3y^2)}\bigg){1 \over \sqrt{(1 + 3x^2)(1 + 3y^2)} }\,dx\,dy$$ By Cauchy-Schwarz this is at most the square root of $$\int_{\mathbb{R}^2}(f(x + x^3, y + y^3))^2(1 + 3x^2)(1 + 3y^2)\,dx\,dy\int_{\mathbb{R}^2}{1 \over (1 + 3x^2)(1 + 3y^2) }\,dx\,dy$$ The first integral is finite as described above, and the second one can directly be computed to something finite. So the original integral is finite.