Is the function Lipschitz continuous on $ overlineB_1(0) $?
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$$ f(x,y)=frac2x^2y+y^3sqrtx^2+y^2 $$ $$ I know that it is continuous on overlineB_1(0) . \ I think I have to show ||f(x_1,x_2)-f(y_1,y_2)||leq L||(x_1,x_2)-(y_1,y_2)|| , for
x_i,y_iin overlineB_1(0) ,\i=1,2 and Lin mathbbR_+ $$
lipschitz-functions
asked Sep 9 at 12:43
Matillo
114
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up vote
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down vote
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$$ f(x,y)=frac2x^2y+y^3sqrtx^2+y^2 $$ $$ I know that it is continuous on overlineB_1(0) . \ I think I have to show ||f(x_1,x_2)-f(y_1,y_2)||leq L||(x_1,x_2)-(y_1,y_2)|| , for
x_i,y_iin overlineB_1(0) ,\i=1,2 and Lin mathbbR_+ $$
lipschitz-functions
asked Sep 9 at 12:43
Matillo
114
What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$ f(x,y)=frac2x^2y+y^3sqrtx^2+y^2 $$ $$ I know that it is continuous on overlineB_1(0) . \ I think I have to show ||f(x_1,x_2)-f(y_1,y_2)||leq L||(x_1,x_2)-(y_1,y_2)|| , for
x_i,y_iin overlineB_1(0) ,\i=1,2 and Lin mathbbR_+ $$
lipschitz-functions
asked Sep 9 at 12:43
Matillo
114
$$ f(x,y)=frac2x^2y+y^3sqrtx^2+y^2 $$ $$ I know that it is continuous on overlineB_1(0) . \ I think I have to show ||f(x_1,x_2)-f(y_1,y_2)||leq L||(x_1,x_2)-(y_1,y_2)|| , for
x_i,y_iin overlineB_1(0) ,\i=1,2 and Lin mathbbR_+ $$
lipschitz-functions
lipschitz-functions
asked Sep 9 at 12:43
Matillo
114
asked Sep 9 at 12:43
Matillo
114
asked Sep 9 at 12:43
Matillo
114
asked Sep 9 at 12:43
Matillo
114
asked Sep 9 at 12:43
Matillo
114
114
What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37
add a comment |Â
What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37
What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37
add a comment |Â
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What have you tried so far? Note that $f(x,y) = ysqrtx^2+y^2$.
– amsmath
Sep 9 at 13:14
is it correct what I have to show ? $$ ysqrtx^2+y^2* sqrtx^2+y^2 =yx^2+y^3 . $$
– Matillo
Sep 9 at 13:37
Yes, that's correct by definition of Lipschitz continuity. Note that it's enough to show that $f$ is differentiable and that its derivative is bounded in the disk.
– amsmath
Sep 9 at 19:37