How to evaluate $S=int_partial Dfrac1^2ds(y)$?A Particular Metric: $(mathbbR^2,d_2)$Linear transformation that does thisShow that $int_[0,1]^2g(y_1-y_2) Bbb1_y_1>y_2dy_1dy_2 = int_[0,1]g(m)(1-m), dm$Product metric spaces is again a metric spaceSum of two Dense setsHow do I show that for a multivariate function, Lipschitz continuity in each variable implies Lipschitz continuity for the whole function?Continuity of the inverse mapGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Partial derivative of function with respect to itselfLipschitz property of $f(x,y)= fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0.$
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How to evaluate $S=int_partial Dfrac1x-yds(y)$?
A Particular Metric: $(mathbbR^2,d_2)$Linear transformation that does thisShow that $int_[0,1]^2g(y_1-y_2) Bbb1_y_1>y_2dy_1dy_2 = int_[0,1]g(m)(1-m), dm$Product metric spaces is again a metric spaceSum of two Dense setsHow do I show that for a multivariate function, Lipschitz continuity in each variable implies Lipschitz continuity for the whole function?Continuity of the inverse mapGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Partial derivative of function with respect to itselfLipschitz property of $f(x,y)= fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0.$
$begingroup$
Let $D=B(0,r_0), x,y in mathbbR^3$ and $xin D$.
beginalign*
& S=int_partial Dfrac1x-yds(y)=int_partial Dfrac1(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2ds(y)
endalign*
I want to prove that $Sleq k$ ($k>0$ and $k$ only depends on $r_0$).
In case $x ne 0$, my idea is
beginsplit
S^+ &=int_y_1^2+y_2^2leq r_0^2frac1(x_1-y_1)^2+(x_2-y_2)^2+Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2\
&leq int_y_1^2+y_2^2leq r_0^2frac1Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2: text (assume x_3 ne 0)
endsplit
Let $y_1=rcosphi, y_2=rsinphi, 0leq phi leq 2pi$
beginsplit
S^+&=int_0^2pidphiint_0^r_0frac1Big(x_3-sqrtr_0^2-r^2Big)^2fracr_0sqrtr_0^2-r^2rdr\
&=2r_0pileft(frac1-x_3-frac1r_0-x_3right)\
endsplit
with $-r_0<x_3<r_0, x_3 ne 0$, I can't show that $left(frac1-x_3-frac1r_0-x_3right)$ is bounded. Please help me, thank you so much!
real-analysis singular-integrals
edited yesterday
folouer of kaklas
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
$endgroup$
add a comment |
$begingroup$
Let $D=B(0,r_0), x,y in mathbbR^3$ and $xin D$.
beginalign*
& S=int_partial Dfrac1x-yds(y)=int_partial Dfrac1(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2ds(y)
endalign*
I want to prove that $Sleq k$ ($k>0$ and $k$ only depends on $r_0$).
In case $x ne 0$, my idea is
beginsplit
S^+ &=int_y_1^2+y_2^2leq r_0^2frac1(x_1-y_1)^2+(x_2-y_2)^2+Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2\
&leq int_y_1^2+y_2^2leq r_0^2frac1Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2: text (assume x_3 ne 0)
endsplit
Let $y_1=rcosphi, y_2=rsinphi, 0leq phi leq 2pi$
beginsplit
S^+&=int_0^2pidphiint_0^r_0frac1Big(x_3-sqrtr_0^2-r^2Big)^2fracr_0sqrtr_0^2-r^2rdr\
&=2r_0pileft(frac1-x_3-frac1r_0-x_3right)\
endsplit
with $-r_0<x_3<r_0, x_3 ne 0$, I can't show that $left(frac1-x_3-frac1r_0-x_3right)$ is bounded. Please help me, thank you so much!
real-analysis singular-integrals
edited yesterday
folouer of kaklas
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
$endgroup$
1
$begingroup$
What has this to do withfunctional-analysis?
$endgroup$
– José Carlos Santos
yesterday
1
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday
add a comment |
$begingroup$
Let $D=B(0,r_0), x,y in mathbbR^3$ and $xin D$.
beginalign*
& S=int_partial Dfrac1x-yds(y)=int_partial Dfrac1(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2ds(y)
endalign*
I want to prove that $Sleq k$ ($k>0$ and $k$ only depends on $r_0$).
In case $x ne 0$, my idea is
beginsplit
S^+ &=int_y_1^2+y_2^2leq r_0^2frac1(x_1-y_1)^2+(x_2-y_2)^2+Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2\
&leq int_y_1^2+y_2^2leq r_0^2frac1Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2: text (assume x_3 ne 0)
endsplit
Let $y_1=rcosphi, y_2=rsinphi, 0leq phi leq 2pi$
beginsplit
S^+&=int_0^2pidphiint_0^r_0frac1Big(x_3-sqrtr_0^2-r^2Big)^2fracr_0sqrtr_0^2-r^2rdr\
&=2r_0pileft(frac1-x_3-frac1r_0-x_3right)\
endsplit
with $-r_0<x_3<r_0, x_3 ne 0$, I can't show that $left(frac1-x_3-frac1r_0-x_3right)$ is bounded. Please help me, thank you so much!
real-analysis singular-integrals
edited yesterday
folouer of kaklas
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
$endgroup$
Let $D=B(0,r_0), x,y in mathbbR^3$ and $xin D$.
beginalign*
& S=int_partial Dfrac1x-yds(y)=int_partial Dfrac1(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2ds(y)
endalign*
I want to prove that $Sleq k$ ($k>0$ and $k$ only depends on $r_0$).
In case $x ne 0$, my idea is
beginsplit
S^+ &=int_y_1^2+y_2^2leq r_0^2frac1(x_1-y_1)^2+(x_2-y_2)^2+Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2\
&leq int_y_1^2+y_2^2leq r_0^2frac1Big(x_3-sqrtr_0^2-y_1^2-y_2^2Big)^2fracr_0sqrtr_0^2-y_1^2-y_2^2dy_1dy_2: text (assume x_3 ne 0)
endsplit
Let $y_1=rcosphi, y_2=rsinphi, 0leq phi leq 2pi$
beginsplit
S^+&=int_0^2pidphiint_0^r_0frac1Big(x_3-sqrtr_0^2-r^2Big)^2fracr_0sqrtr_0^2-r^2rdr\
&=2r_0pileft(frac1-x_3-frac1r_0-x_3right)\
endsplit
with $-r_0<x_3<r_0, x_3 ne 0$, I can't show that $left(frac1-x_3-frac1r_0-x_3right)$ is bounded. Please help me, thank you so much!
real-analysis singular-integrals
real-analysis singular-integrals
edited yesterday
folouer of kaklas
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
edited yesterday
folouer of kaklas
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
edited yesterday
folouer of kaklas
340110
edited yesterday
folouer of kaklas
340110
edited yesterday
folouer of kaklas
340110
340110
asked yesterday
Chloe.SannonChloe.Sannon
1208
asked yesterday
Chloe.SannonChloe.Sannon
1208
asked yesterday
Chloe.SannonChloe.Sannon
1208
1208
1
$begingroup$
What has this to do withfunctional-analysis?
$endgroup$
– José Carlos Santos
yesterday
1
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday
add a comment |
1
$begingroup$
What has this to do withfunctional-analysis?
$endgroup$
– José Carlos Santos
yesterday
1
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday
1
1
$begingroup$
What has this to do with
functional-analysis?$endgroup$
– José Carlos Santos
yesterday
$begingroup$
What has this to do with
functional-analysis?$endgroup$
– José Carlos Santos
yesterday
1
1
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You may use spherical coordinates and the cosine-theorem:
$$
frac1x-y = frac1
$$
where $theta_x,y$ is the angle between $x$ and $y$.
Putting it into the integral you have:
$$
intlimits_fracdsx-y =
intlimits_frac r_0^2sin(theta) dphi dthetax=2pi r_0^2intlimits_-1^1frac dcos(theta)x
$$
where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $theta_x,y$ simply becomes the coordinate $theta$.
answered yesterday
denklodenklo
4457
$endgroup$
add a comment |
$begingroup$
Clearly,
$$
F(y)=int_fracds^2
$$
is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.
Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=sqrtr^2-x^2$ around the $x-$axis, then
$$
F(y)=2piint_-r_0^r_0 f(x)sqrt1+big(f'(x)big)^2g(x),dx
=cdots=2pi r_0int_-r_0^r_0 g(x),dx
$$
where
$$
g(x)=frac1big(x,sqrtr_0^2-x^2,0big)-(=frac1(x-=frac1r_0^2-2x
$$
and hence
$$
F(y)=2pi r_0int_-r_0^r_0fracdxy=
left.-fraclog(r_0^2+yright|_-r_0^r_0
=fraclogleft(fracyr^2_0-2r_0right)y=
frac
logleft(
fracr_0+y
right)
$$
Clearly,
$$
lim_yF(y)=infty
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may use spherical coordinates and the cosine-theorem:
$$
frac1x-y = frac1
$$
where $theta_x,y$ is the angle between $x$ and $y$.
Putting it into the integral you have:
$$
intlimits_fracdsx-y =
intlimits_frac r_0^2sin(theta) dphi dthetax=2pi r_0^2intlimits_-1^1frac dcos(theta)x
$$
where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $theta_x,y$ simply becomes the coordinate $theta$.
answered yesterday
denklodenklo
4457
$endgroup$
add a comment |
$begingroup$
You may use spherical coordinates and the cosine-theorem:
$$
frac1x-y = frac1
$$
where $theta_x,y$ is the angle between $x$ and $y$.
Putting it into the integral you have:
$$
intlimits_fracdsx-y =
intlimits_frac r_0^2sin(theta) dphi dthetax=2pi r_0^2intlimits_-1^1frac dcos(theta)x
$$
where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $theta_x,y$ simply becomes the coordinate $theta$.
answered yesterday
denklodenklo
4457
$endgroup$
add a comment |
$begingroup$
You may use spherical coordinates and the cosine-theorem:
$$
frac1x-y = frac1
$$
where $theta_x,y$ is the angle between $x$ and $y$.
Putting it into the integral you have:
$$
intlimits_fracdsx-y =
intlimits_frac r_0^2sin(theta) dphi dthetax=2pi r_0^2intlimits_-1^1frac dcos(theta)x
$$
where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $theta_x,y$ simply becomes the coordinate $theta$.
answered yesterday
denklodenklo
4457
$endgroup$
You may use spherical coordinates and the cosine-theorem:
$$
frac1x-y = frac1
$$
where $theta_x,y$ is the angle between $x$ and $y$.
Putting it into the integral you have:
$$
intlimits_fracdsx-y =
intlimits_frac r_0^2sin(theta) dphi dthetax=2pi r_0^2intlimits_-1^1frac dcos(theta)x
$$
where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $theta_x,y$ simply becomes the coordinate $theta$.
answered yesterday
denklodenklo
4457
answered yesterday
denklodenklo
4457
answered yesterday
denklodenklo
4457
answered yesterday
denklodenklo
4457
4457
add a comment |
add a comment |
$begingroup$
Clearly,
$$
F(y)=int_fracds^2
$$
is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.
Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=sqrtr^2-x^2$ around the $x-$axis, then
$$
F(y)=2piint_-r_0^r_0 f(x)sqrt1+big(f'(x)big)^2g(x),dx
=cdots=2pi r_0int_-r_0^r_0 g(x),dx
$$
where
$$
g(x)=frac1big(x,sqrtr_0^2-x^2,0big)-(=frac1(x-=frac1r_0^2-2x
$$
and hence
$$
F(y)=2pi r_0int_-r_0^r_0fracdxy=
left.-fraclog(r_0^2+yright|_-r_0^r_0
=fraclogleft(fracyr^2_0-2r_0right)y=
frac
logleft(
fracr_0+y
right)
$$
Clearly,
$$
lim_yF(y)=infty
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
add a comment |
$begingroup$
Clearly,
$$
F(y)=int_fracds^2
$$
is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.
Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=sqrtr^2-x^2$ around the $x-$axis, then
$$
F(y)=2piint_-r_0^r_0 f(x)sqrt1+big(f'(x)big)^2g(x),dx
=cdots=2pi r_0int_-r_0^r_0 g(x),dx
$$
where
$$
g(x)=frac1big(x,sqrtr_0^2-x^2,0big)-(=frac1(x-=frac1r_0^2-2x
$$
and hence
$$
F(y)=2pi r_0int_-r_0^r_0fracdxy=
left.-fraclog(r_0^2+yright|_-r_0^r_0
=fraclogleft(fracyr^2_0-2r_0right)y=
frac
logleft(
fracr_0+y
right)
$$
Clearly,
$$
lim_yF(y)=infty
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
add a comment |
$begingroup$
Clearly,
$$
F(y)=int_fracds^2
$$
is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.
Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=sqrtr^2-x^2$ around the $x-$axis, then
$$
F(y)=2piint_-r_0^r_0 f(x)sqrt1+big(f'(x)big)^2g(x),dx
=cdots=2pi r_0int_-r_0^r_0 g(x),dx
$$
where
$$
g(x)=frac1big(x,sqrtr_0^2-x^2,0big)-(=frac1(x-=frac1r_0^2-2x
$$
and hence
$$
F(y)=2pi r_0int_-r_0^r_0fracdxy=
left.-fraclog(r_0^2+yright|_-r_0^r_0
=fraclogleft(fracyr^2_0-2r_0right)y=
frac
logleft(
fracr_0+y
right)
$$
Clearly,
$$
lim_yF(y)=infty
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
Clearly,
$$
F(y)=int_fracds^2
$$
is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.
Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=sqrtr^2-x^2$ around the $x-$axis, then
$$
F(y)=2piint_-r_0^r_0 f(x)sqrt1+big(f'(x)big)^2g(x),dx
=cdots=2pi r_0int_-r_0^r_0 g(x),dx
$$
where
$$
g(x)=frac1big(x,sqrtr_0^2-x^2,0big)-(=frac1(x-=frac1r_0^2-2x
$$
and hence
$$
F(y)=2pi r_0int_-r_0^r_0fracdxy=
left.-fraclog(r_0^2+yright|_-r_0^r_0
=fraclogleft(fracyr^2_0-2r_0right)y=
frac
logleft(
fracr_0+y
right)
$$
Clearly,
$$
lim_yF(y)=infty
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
63.6k1385165
add a comment |
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1
$begingroup$
What has this to do with
functional-analysis?$endgroup$
– José Carlos Santos
yesterday
1
$begingroup$
It's not true... for instance, when $|x-y|to 0$, it explose.
$endgroup$
– user657324
yesterday
$begingroup$
@user657324 since $xin D$ and $y in partial D$ there exists $delta>0$ with $|x-y|>delta$
$endgroup$
– Chloe.Sannon
yesterday
$begingroup$
Yes, but $delta $ depend on $x$, so you can't upper bound $S$ uniformly in $x$. @Chloe.Sannon
$endgroup$
– user657324
yesterday