Is this function in $L^2(\mathbb{R}^6)$?

Question

I have to prove that the following function in $L^2(\mathbb{R}^6)$ $$F(x,y)=\frac{f(x)}{x^2+y^2+\frac{2}{m+1}x\cdot y+\lambda}$$ with $f\in H^{\frac{1}{2}}(\mathbb{R}^3)$, $x,y\in\mathbb{R}^3$ and $\lambda>0$, $m>0$. I've tried to use the Schwartz inequality, noting that $H^{\frac{1}{2}}(\mathbb{R}^3)\subset L^2(\mathbb{R}^3) $; so I have to find a good estimation for the denominator. I've worked in this way: $$ x^2+y^2+\frac{2}{m+1}x\cdot y+\lambda\geq\frac{m}{m+1}x^2+\frac{m}{m+1}y^2+\lambda $$ where I've used $2x\cdot y\geq -x^2-y^2$. Is this reasoning right?

Answer 1

$F(x,y)\le \dfrac{f(x)}{Cx^2+Cy^2+\lambda}$, say $\|f\|\le M$. $$\int_{\mathbb{R}^6}\dfrac{f^2(x)}{(Cx^2+Cy^2+\lambda)^2}dx_1 dx_2 dx_3 dy_1dy_2dy_3\le\int_{\mathbb{R}^6}\dfrac{f^2(x)}{(Cy^2+\lambda)^2}dx_1 dx_2 dx_3 dy_1dy_2dy_3\le M^2\int_{\mathbb{R}^3}\dfrac{1}{(Cy^2+\lambda)^2}dy_1dy_2dy_3 = \int_0^{\infty}\int_0^{2\pi} \dfrac{1}{(Cr^2+\lambda)^2}r^2\sin(\theta)drd\theta = C_1\int_0^{\infty} \dfrac{1}{(Cr^2+\lambda)^2}r^2dr$$

When $r$ is small, we take $\dfrac{1}{Cr^2+\lambda}\sim\dfrac{1}{\lambda}$

When $r$ is large, we take $\dfrac{1}{Cr^2+\lambda}\sim\dfrac{1}{Cr^2}$.

Thus

$$C_1\int_0^{\infty} \dfrac{1}{(Cr^2+\lambda)^2}r^2dr = \left(\int_0^{1} +\int_1^{\infty}\right)\dfrac{1}{(Cr^2+\lambda)^2}r^2dr\le\int_0^1 \dfrac{1}{\lambda^2}r^2dr+\int_1^{\infty}\dfrac{1}{C^2r^4}r^2dr$$

Which is finite.