Exercise about convolution of functions












2












$begingroup$


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 $f:mathbb{R^2}rightarrow{mathbb{R}}$ $fin{L^p(mathbb{R^2})}textrm{ and }{L^qmathbb{(R^2)}}$ prove that :
$$g(y,z)=displaystylefrac{(-z,y)}{2pisqrt{y^2+z^2}}*f; in{L^infty(mathbb{R^2})}$$
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.






real-analysis functional-analysis convolution







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:20












  • $begingroup$
    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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
    $endgroup$
    – mathlife
    Dec 9 '18 at 11:53








  • 2




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:58












  • $begingroup$
    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
    $endgroup$
    – mathlife
    Dec 9 '18 at 12:20










  • $begingroup$
    This reminds me of the Biot-Savart law.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:11
















2












$begingroup$


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 $f:mathbb{R^2}rightarrow{mathbb{R}}$ $fin{L^p(mathbb{R^2})}textrm{ and }{L^qmathbb{(R^2)}}$ prove that :
$$g(y,z)=displaystylefrac{(-z,y)}{2pisqrt{y^2+z^2}}*f; in{L^infty(mathbb{R^2})}$$
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.






real-analysis functional-analysis convolution







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:20












  • $begingroup$
    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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
    $endgroup$
    – mathlife
    Dec 9 '18 at 11:53








  • 2




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:58












  • $begingroup$
    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
    $endgroup$
    – mathlife
    Dec 9 '18 at 12:20










  • $begingroup$
    This reminds me of the Biot-Savart law.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:11














2












2








2


3



$begingroup$


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 $f:mathbb{R^2}rightarrow{mathbb{R}}$ $fin{L^p(mathbb{R^2})}textrm{ and }{L^qmathbb{(R^2)}}$ prove that :
$$g(y,z)=displaystylefrac{(-z,y)}{2pisqrt{y^2+z^2}}*f; in{L^infty(mathbb{R^2})}$$
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.






real-analysis functional-analysis convolution







share|cite|improve this question











$endgroup$




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 $f:mathbb{R^2}rightarrow{mathbb{R}}$ $fin{L^p(mathbb{R^2})}textrm{ and }{L^qmathbb{(R^2)}}$ prove that :
$$g(y,z)=displaystylefrac{(-z,y)}{2pisqrt{y^2+z^2}}*f; in{L^infty(mathbb{R^2})}$$
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.






real-analysis functional-analysis convolution




real-analysis functional-analysis convolution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 18:28







mathlife

















asked Dec 6 '18 at 16:29









mathlifemathlife

629




629








  • 3




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:20












  • $begingroup$
    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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
    $endgroup$
    – mathlife
    Dec 9 '18 at 11:53








  • 2




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:58












  • $begingroup$
    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
    $endgroup$
    – mathlife
    Dec 9 '18 at 12:20










  • $begingroup$
    This reminds me of the Biot-Savart law.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:11














  • 3




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:20












  • $begingroup$
    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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
    $endgroup$
    – mathlife
    Dec 9 '18 at 11:53








  • 2




    $begingroup$
    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.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 11:58












  • $begingroup$
    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
    $endgroup$
    – mathlife
    Dec 9 '18 at 12:20










  • $begingroup$
    This reminds me of the Biot-Savart law.
    $endgroup$
    – Giuseppe Negro
    Dec 9 '18 at 17:11








3




3




$begingroup$
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.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 11:20






$begingroup$
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.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 11:20














$begingroup$
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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
$endgroup$
– mathlife
Dec 9 '18 at 11:53






$begingroup$
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 $: $int_{mathbb{R^2}}f((x_1-z_1,x_2-z_2))frac{(-z_2,z_1)}{sqrt{z_1^2+z_2^2}}dz_1dz_2$
$endgroup$
– mathlife
Dec 9 '18 at 11:53






2




2




$begingroup$
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.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 11:58






$begingroup$
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.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 11:58














$begingroup$
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
$endgroup$
– mathlife
Dec 9 '18 at 12:20




$begingroup$
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
$endgroup$
– mathlife
Dec 9 '18 at 12:20












$begingroup$
This reminds me of the Biot-Savart law.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:11




$begingroup$
This reminds me of the Biot-Savart law.
$endgroup$
– Giuseppe Negro
Dec 9 '18 at 17:11










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $fin L^pcap L^q(mathbb R^2)$ iff $alpha<frac2q<frac2p<beta$. In particular, $fnotin L^1$, so the integral
$$
int_{mathbb R^2} frac{y_1}{|y|} f(x-y) ,dy
$$
is not even defined in the Lebesgue sense.






share|cite|improve this answer









$endgroup$









  • 1




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













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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $fin L^pcap L^q(mathbb R^2)$ iff $alpha<frac2q<frac2p<beta$. In particular, $fnotin L^1$, so the integral
$$
int_{mathbb R^2} frac{y_1}{|y|} f(x-y) ,dy
$$
is not even defined in the Lebesgue sense.






share|cite|improve this answer









$endgroup$









  • 1




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


















1












$begingroup$

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 $fin L^pcap L^q(mathbb R^2)$ iff $alpha<frac2q<frac2p<beta$. In particular, $fnotin L^1$, so the integral
$$
int_{mathbb R^2} frac{y_1}{|y|} f(x-y) ,dy
$$
is not even defined in the Lebesgue sense.






share|cite|improve this answer









$endgroup$









  • 1




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
















1












1








1





$begingroup$

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 $fin L^pcap L^q(mathbb R^2)$ iff $alpha<frac2q<frac2p<beta$. In particular, $fnotin L^1$, so the integral
$$
int_{mathbb R^2} frac{y_1}{|y|} f(x-y) ,dy
$$
is not even defined in the Lebesgue sense.






share|cite|improve this answer









$endgroup$



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 $fin L^pcap L^q(mathbb R^2)$ iff $alpha<frac2q<frac2p<beta$. In particular, $fnotin L^1$, so the integral
$$
int_{mathbb R^2} frac{y_1}{|y|} f(x-y) ,dy
$$
is not even defined in the Lebesgue sense.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 19:19









FedericoFederico

4,954514




4,954514








  • 1




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
















  • 1




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










1




1




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






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




















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