Integral inequality of two variables function











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Suppose $u(x,y)$ is continuous in $D={(x,y)| 0le x le 1, 0le y le 1}$, $frac{partial u}{partial x}, frac{partial u}{partial y}, frac{partial^2 u}{partial x partial y}$ is absolutely integrable, prove



$$
sup_{(x,y) in D} |u| le iint_D |u| dxdy + iint_D left( left| frac{partial u}{partial x} right|+left|frac{partial u}{partial y}right| right) dxdy + iint_D left|frac{partial^2 u}{partial x partial y}right| dxdy.
$$



I have tried the Taylor formular,



$$
sup |u|=|u(xi_x,xi_y)+u(x_0,y_0)-u(xi_x,xi_y)|=left| u(xi_x,xi_y)+frac{partial u}{partial x}(xi_x,xi_y)(x_0-xi_x)+frac{partial u}{partial y}(xi_x,xi_y)(y_0-xi_y)+frac12 frac{partial^2 u}{partial x partial y}(eta_x,eta_y)[(x_0-xi_x)^2+(y_0-xi_y)^2] right|
$$
where $u(xi_x,xi_y)=iint_D u dxdy$.



but there seems to be some matters of detail.



Thanks for your help.






calculus analysis inequality integral-inequality







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  • I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
    – Eric Yau
    Nov 27 at 8:53












  • Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
    – gcfsjfcus
    Nov 27 at 11:16

















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Suppose $u(x,y)$ is continuous in $D={(x,y)| 0le x le 1, 0le y le 1}$, $frac{partial u}{partial x}, frac{partial u}{partial y}, frac{partial^2 u}{partial x partial y}$ is absolutely integrable, prove



$$
sup_{(x,y) in D} |u| le iint_D |u| dxdy + iint_D left( left| frac{partial u}{partial x} right|+left|frac{partial u}{partial y}right| right) dxdy + iint_D left|frac{partial^2 u}{partial x partial y}right| dxdy.
$$



I have tried the Taylor formular,



$$
sup |u|=|u(xi_x,xi_y)+u(x_0,y_0)-u(xi_x,xi_y)|=left| u(xi_x,xi_y)+frac{partial u}{partial x}(xi_x,xi_y)(x_0-xi_x)+frac{partial u}{partial y}(xi_x,xi_y)(y_0-xi_y)+frac12 frac{partial^2 u}{partial x partial y}(eta_x,eta_y)[(x_0-xi_x)^2+(y_0-xi_y)^2] right|
$$
where $u(xi_x,xi_y)=iint_D u dxdy$.



but there seems to be some matters of detail.



Thanks for your help.






calculus analysis inequality integral-inequality







share|cite|improve this question
























  • I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
    – Eric Yau
    Nov 27 at 8:53












  • Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
    – gcfsjfcus
    Nov 27 at 11:16















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose $u(x,y)$ is continuous in $D={(x,y)| 0le x le 1, 0le y le 1}$, $frac{partial u}{partial x}, frac{partial u}{partial y}, frac{partial^2 u}{partial x partial y}$ is absolutely integrable, prove



$$
sup_{(x,y) in D} |u| le iint_D |u| dxdy + iint_D left( left| frac{partial u}{partial x} right|+left|frac{partial u}{partial y}right| right) dxdy + iint_D left|frac{partial^2 u}{partial x partial y}right| dxdy.
$$



I have tried the Taylor formular,



$$
sup |u|=|u(xi_x,xi_y)+u(x_0,y_0)-u(xi_x,xi_y)|=left| u(xi_x,xi_y)+frac{partial u}{partial x}(xi_x,xi_y)(x_0-xi_x)+frac{partial u}{partial y}(xi_x,xi_y)(y_0-xi_y)+frac12 frac{partial^2 u}{partial x partial y}(eta_x,eta_y)[(x_0-xi_x)^2+(y_0-xi_y)^2] right|
$$
where $u(xi_x,xi_y)=iint_D u dxdy$.



but there seems to be some matters of detail.



Thanks for your help.






calculus analysis inequality integral-inequality







share|cite|improve this question















Suppose $u(x,y)$ is continuous in $D={(x,y)| 0le x le 1, 0le y le 1}$, $frac{partial u}{partial x}, frac{partial u}{partial y}, frac{partial^2 u}{partial x partial y}$ is absolutely integrable, prove



$$
sup_{(x,y) in D} |u| le iint_D |u| dxdy + iint_D left( left| frac{partial u}{partial x} right|+left|frac{partial u}{partial y}right| right) dxdy + iint_D left|frac{partial^2 u}{partial x partial y}right| dxdy.
$$



I have tried the Taylor formular,



$$
sup |u|=|u(xi_x,xi_y)+u(x_0,y_0)-u(xi_x,xi_y)|=left| u(xi_x,xi_y)+frac{partial u}{partial x}(xi_x,xi_y)(x_0-xi_x)+frac{partial u}{partial y}(xi_x,xi_y)(y_0-xi_y)+frac12 frac{partial^2 u}{partial x partial y}(eta_x,eta_y)[(x_0-xi_x)^2+(y_0-xi_y)^2] right|
$$
where $u(xi_x,xi_y)=iint_D u dxdy$.



but there seems to be some matters of detail.



Thanks for your help.






calculus analysis inequality integral-inequality




calculus analysis inequality integral-inequality






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share|cite|improve this question













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edited Nov 27 at 11:19

























asked Nov 27 at 6:07









gcfsjfcus

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596












  • I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
    – Eric Yau
    Nov 27 at 8:53












  • Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
    – gcfsjfcus
    Nov 27 at 11:16




















  • I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
    – Eric Yau
    Nov 27 at 8:53












  • Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
    – gcfsjfcus
    Nov 27 at 11:16


















I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
– Eric Yau
Nov 27 at 8:53






I think you are close to the answer. Just note that $max{|x_0-xi_x|,|y_0-xi_y|,frac{1}{2}|(x_0-xi_x)^2+(y_0-x_i_y)^2|}|le1$ when these variables are all in $D$.
– Eric Yau
Nov 27 at 8:53














Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
– gcfsjfcus
Nov 27 at 11:16






Thanks, but how to prove $iint_D (|frac{partial u}{partial x}(xi_x,xi_y)|+|frac{partial u}{partial y}(xi_x,xi_y)|) dxdy le iint_D (|frac{partial u}{partial x}|+|frac{partial u}{partial y}|) dxdy$ and $iint_D |frac{partial^2 u}{partial x partial y}(eta_x,eta_y|) dxdy le iint_D |frac{partial^2 u}{partial x partial y}dxdy$ ?
– gcfsjfcus
Nov 27 at 11:16

















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