Exercise about convolution of functions












2














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




















  • 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
















2














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




















  • 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














2












2








2


3





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















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 at 18:28

























asked Dec 6 at 16:29









mathlife

469




469








  • 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














  • 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








3




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






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






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




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






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




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




This reminds me of the Biot-Savart law.
– Giuseppe Negro
Dec 9 at 17:11










1 Answer
1






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0














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

















  • 1




    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 $p>1.$)
    – zhw.
    Dec 15 at 19:29













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0














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

















  • 1




    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 $p>1.$)
    – zhw.
    Dec 15 at 19:29


















0














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

















  • 1




    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 $p>1.$)
    – zhw.
    Dec 15 at 19:29
















0












0








0






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












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 at 19:19









Federico

4,369512




4,369512








  • 1




    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 $p>1.$)
    – zhw.
    Dec 15 at 19:29
















  • 1




    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 $p>1.$)
    – zhw.
    Dec 15 at 19:29










1




1




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 $p>1.$)
– zhw.
Dec 15 at 19:29






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 $p>1.$)
– zhw.
Dec 15 at 19:29




















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