Let $F:\mathbb R^{2}\to \mathbb C$ be a function. Suppose $F\in L^{p,1}(\mathbb R \times \mathbb R); (1<p< \infty).$
Define $G:\mathbb R^{2}\to \mathbb C$ as follows: $$G(x,y):=F(y,x)$$
My Question is: Given $F\in L^{p,1}(\mathbb R^{2}); (1<p<\infty).$ Is it true that $G\in L^{p,1}(\mathbb R^{2}),$ that is, $\int_{\mathbb R} \left(\int_{\mathbb R}|G(x,y)|^{p}dx\right)^{1/p} dy < \infty$ ? If not, what is a counter example ?
Thanks,
No, the condition $\int_{\mathbb R} \left(\int_{\mathbb R}|F(x,y)|^{p}dx\right)^{1/p} dy < \infty$ is sensitive to the interchange of coordinates. For example, let $p=2$ and $$ F(x,y) = \begin{cases}y^{-1/2},\quad & 0\le x\le y\le 1 \\ 0 &\text{otherwise}\end{cases} $$ Then for $y\in [0,1]$ we have $$ \int_{\mathbb R}|F(x,y)|^{2}dx = y^{-1} \int_0^y \,dx = 1 $$ and subsequent integration over $y$ presents no problem.
However, after flipping the coordinates $G(x,y)=F(y,x)$ trying to integrate $|G|^2$ with respect to $x$ gives divergent integral for all $y\in [0,1]$.