# Exercise about convolution of functions

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 $$1 and $$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.

• This question is ill-posed. You use $(y, z)$ as arguments of $g$ but also in the convolution, it is like writing $f(x)=\int g(x)\, dx$, the $x$ is at the same time a true variable and a dummy one. What exactly do you mean by that convolution? Write it as an integral, please. – Giuseppe Negro Dec 9 '18 at 11:20
• I wan to say $f*g(x)=\int_{\mathbb{R^2}}f(x-z)g(z)dz$. In this exercise we have to see that the following function is in $L^\infty$: $\int_{\mathbb{R^2}}f((x_1-z_1,x_2-z_2))\frac{(-z_2,z_1)}{\sqrt{z_1^2+z_2^2}}dz_1dz_2$ – mathlife Dec 9 '18 at 11:53
• Yeah, please, edit your post. This edit will improve it and raise the odds that it gets a good answer. Make sure that the integral be dimensionally consistent; I see a vector there, is there a scalar product? Or is it a vector-valued integral? Edit the post and explain all these details, please. – Giuseppe Negro Dec 9 '18 at 11:58
• My book, where I found this excersice doesn't give more details about this. I suppose that it follows the definition of convolution in two dimensions – mathlife Dec 9 '18 at 12:20
• This reminds me of the Biot-Savart law. – Giuseppe Negro Dec 9 '18 at 17:11

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.
• Perhaps simpler: Let $U$ be the exterior of the unit disc, and set $f(y) = |y|^{-2}\chi_U(y).$ (This $f\in L^p$ for all $p>1.$) – zhw. Dec 15 '18 at 19:29