I have found this excercise in theory of convolution (I started it the last week). I have been thinking about it for two days but I don't get solve it:
Let be $1<p<2<q<\infty$ and $f:\mathbb{R^2}\rightarrow{\mathbb{R}}$ $f\in{L^p(\mathbb{R^2})}\textrm{ and }{L^q\mathbb{(R^2)}}$ prove that : $$g(y,z)=\displaystyle\frac{(-z,y)}{2\pi\sqrt{y^2+z^2}}*f\; \in{L^\infty(\mathbb{R^2})}$$ Where $*$ denotes covolution of two functions.
Edit
I want to show that the following function is in $L^\infty(\mathbb{R^2})$ $$\int_{\mathbb{R^2}}f(x_1-z_1,x_2-z_2)\frac{(-z_2,z_1)}{\sqrt{z_1^2+z_2^2}}dz_1dz_2$$
I would appreciate if someone help me. Thanks.
Take a positive function such that $f(x) \sim |x|^{-\alpha}$ near $0$ and $f(x) \sim |x|^{-\beta}$ near $\infty$ with $0<\alpha<\beta<2$.
Then $f\in L^p\cap L^q(\mathbb R^2)$ iff $\alpha<\frac2q<\frac2p<\beta$. In particular, $f\not\in L^1$, so the integral $$ \int_{\mathbb R^2} \frac{y_1}{|y|} f(x-y) \,dy $$ is not even defined in the Lebesgue sense.