Showing that the function $f(x,y)=xsin y+ycos x$ is Lipschitz
Multi tool use
$begingroup$
I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.
According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.
real-analysis lipschitz-functions
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
$endgroup$
add a comment |
$begingroup$
I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.
According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.
real-analysis lipschitz-functions
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
$endgroup$
3
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52
add a comment |
$begingroup$
I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.
According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.
real-analysis lipschitz-functions
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
$endgroup$
I wanted to show that $f(x,y)=xsin y+ycos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$.
According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|le L|y2-y1|$$
I was able show that $x^2+y^2$ in the rectangle $|x|le a$, $|y|le b$ satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=xsin y+ycos x $.
real-analysis lipschitz-functions
real-analysis lipschitz-functions
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
edited Apr 5 '15 at 2:44
user147263
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
asked Apr 4 '15 at 5:44
James AnyanwuJames Anyanwu
61
61
3
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52
add a comment |
3
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52
3
3
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52
add a comment |
1 Answer
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It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
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add a comment |
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$begingroup$
It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
$endgroup$
add a comment |
$begingroup$
It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
$endgroup$
It seems that you want to prove that $f$ is Lipschitz with respect to $y$. When $f$ is viewed as a global function $f:>{mathbb R}^2to{mathbb R}$ the statement is wrong, because
$$left|{fbigl(x,{piover 2}bigr)-f(x,0)over {piover2}}right|geq |x|-{piover2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $fin C^1({mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${partial f(x,y)overpartial y}= xcos y+cos x$$
it follows that $bigl|{partial f(x,y)overpartial y}bigr|leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$left|{partial f(x,y)overpartial y}right|leq Mqquadforall (x,y)in W .$$
By means of the MVT we then conclude that
$$bigl|f(x,y_1)-f(x,y_2)bigr|leq M>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)in W$.
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
answered Apr 4 '15 at 15:22
Christian BlatterChristian Blatter
172k7113326
172k7113326
add a comment |
add a comment |
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3
$begingroup$
Do you want Lipschitz in $y$ or Lipschitz in both variables?
$endgroup$
– Jose27
Apr 4 '15 at 5:51
$begingroup$
according to my lecturer. the Lipschitz condition \f(x,y1)-f(x,y2)\<=Ly2-y1. I was able show that x^2+y^2 in the rectangle x<=a y<=b sastisfy the Lipschitz condition, with my L=2b. but I had problem showing f(x,y)=xsiny+ycosx
$endgroup$
– James Anyanwu
Apr 4 '15 at 6:52