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.

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  • 3
    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 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 at 11:53
  • 2
    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 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 at 12:20
  • This reminds me of the Biot-Savart law. – Giuseppe Negro Dec 9 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. 12 hours ago

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