Total variation on a creased surface
$begingroup$
Suppose $S$ is a smooth surface without boundary and $f:Stomathbb R$ is continuous. Then, we can define the total variation of $f$ as:
$$
mathrm{TV}[f]:=sup_{|phi|_inftyleq1} int_S f(x)nablacdotphi(x),dA(x).
$$
Here the variable $phi$ is a smooth vector field whose norm is pointwise bounded by $1$. When $f$ is differentiable, this expression is the same as $int_S |nabla f(x)|_2,dA(x)$.
Now, suppose $overline S=cup_{i=1}^k overline S_i$ is the union of $k$ smooth surfaces that join along potentially sharp crease curves. It still makes sense to define a function $f$ to be continuous on $overline S$, and $f$ can even be differentiable away from $cup_i partial S_i$.
Is there a definition of total variation of a function $f$ on the creased surface $overline S$ that agrees with the formula above? In particular, we no longer can use $nablacdotphi$ on the creases since it's not clear what it means to have divergence (I think).
Ideally I'd like the formula to still work when the $overline S$ is smooth without creases, and if we take a limiting sequence of smoothed-out versions of $overline S$ (whatever that means) and apply the smooth formula, then we get our general formula in the limit.
Furthermore, if $f$ is smooth on each $overline S_i$ but potentially discontinuous along the edges between the patches, I'm hoping for a formula similar to Gauss-Bonnet with integrals along the creases:
$$
mathrm{TV}_{overline S}[f] = sum_i int_{overline S_i} |nabla f(x)|,dA(x) + sum_i int_{partial overline S_i} [mathrm{something}],dell.
$$
This seems possible via integration by parts.
Any pointers to papers or formulations would be much appreciated!
multivariable-calculus smooth-manifolds total-variation
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
$endgroup$
add a comment |
$begingroup$
Suppose $S$ is a smooth surface without boundary and $f:Stomathbb R$ is continuous. Then, we can define the total variation of $f$ as:
$$
mathrm{TV}[f]:=sup_{|phi|_inftyleq1} int_S f(x)nablacdotphi(x),dA(x).
$$
Here the variable $phi$ is a smooth vector field whose norm is pointwise bounded by $1$. When $f$ is differentiable, this expression is the same as $int_S |nabla f(x)|_2,dA(x)$.
Now, suppose $overline S=cup_{i=1}^k overline S_i$ is the union of $k$ smooth surfaces that join along potentially sharp crease curves. It still makes sense to define a function $f$ to be continuous on $overline S$, and $f$ can even be differentiable away from $cup_i partial S_i$.
Is there a definition of total variation of a function $f$ on the creased surface $overline S$ that agrees with the formula above? In particular, we no longer can use $nablacdotphi$ on the creases since it's not clear what it means to have divergence (I think).
Ideally I'd like the formula to still work when the $overline S$ is smooth without creases, and if we take a limiting sequence of smoothed-out versions of $overline S$ (whatever that means) and apply the smooth formula, then we get our general formula in the limit.
Furthermore, if $f$ is smooth on each $overline S_i$ but potentially discontinuous along the edges between the patches, I'm hoping for a formula similar to Gauss-Bonnet with integrals along the creases:
$$
mathrm{TV}_{overline S}[f] = sum_i int_{overline S_i} |nabla f(x)|,dA(x) + sum_i int_{partial overline S_i} [mathrm{something}],dell.
$$
This seems possible via integration by parts.
Any pointers to papers or formulations would be much appreciated!
multivariable-calculus smooth-manifolds total-variation
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
$endgroup$
$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18
add a comment |
$begingroup$
Suppose $S$ is a smooth surface without boundary and $f:Stomathbb R$ is continuous. Then, we can define the total variation of $f$ as:
$$
mathrm{TV}[f]:=sup_{|phi|_inftyleq1} int_S f(x)nablacdotphi(x),dA(x).
$$
Here the variable $phi$ is a smooth vector field whose norm is pointwise bounded by $1$. When $f$ is differentiable, this expression is the same as $int_S |nabla f(x)|_2,dA(x)$.
Now, suppose $overline S=cup_{i=1}^k overline S_i$ is the union of $k$ smooth surfaces that join along potentially sharp crease curves. It still makes sense to define a function $f$ to be continuous on $overline S$, and $f$ can even be differentiable away from $cup_i partial S_i$.
Is there a definition of total variation of a function $f$ on the creased surface $overline S$ that agrees with the formula above? In particular, we no longer can use $nablacdotphi$ on the creases since it's not clear what it means to have divergence (I think).
Ideally I'd like the formula to still work when the $overline S$ is smooth without creases, and if we take a limiting sequence of smoothed-out versions of $overline S$ (whatever that means) and apply the smooth formula, then we get our general formula in the limit.
Furthermore, if $f$ is smooth on each $overline S_i$ but potentially discontinuous along the edges between the patches, I'm hoping for a formula similar to Gauss-Bonnet with integrals along the creases:
$$
mathrm{TV}_{overline S}[f] = sum_i int_{overline S_i} |nabla f(x)|,dA(x) + sum_i int_{partial overline S_i} [mathrm{something}],dell.
$$
This seems possible via integration by parts.
Any pointers to papers or formulations would be much appreciated!
multivariable-calculus smooth-manifolds total-variation
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
$endgroup$
Suppose $S$ is a smooth surface without boundary and $f:Stomathbb R$ is continuous. Then, we can define the total variation of $f$ as:
$$
mathrm{TV}[f]:=sup_{|phi|_inftyleq1} int_S f(x)nablacdotphi(x),dA(x).
$$
Here the variable $phi$ is a smooth vector field whose norm is pointwise bounded by $1$. When $f$ is differentiable, this expression is the same as $int_S |nabla f(x)|_2,dA(x)$.
Now, suppose $overline S=cup_{i=1}^k overline S_i$ is the union of $k$ smooth surfaces that join along potentially sharp crease curves. It still makes sense to define a function $f$ to be continuous on $overline S$, and $f$ can even be differentiable away from $cup_i partial S_i$.
Is there a definition of total variation of a function $f$ on the creased surface $overline S$ that agrees with the formula above? In particular, we no longer can use $nablacdotphi$ on the creases since it's not clear what it means to have divergence (I think).
Ideally I'd like the formula to still work when the $overline S$ is smooth without creases, and if we take a limiting sequence of smoothed-out versions of $overline S$ (whatever that means) and apply the smooth formula, then we get our general formula in the limit.
Furthermore, if $f$ is smooth on each $overline S_i$ but potentially discontinuous along the edges between the patches, I'm hoping for a formula similar to Gauss-Bonnet with integrals along the creases:
$$
mathrm{TV}_{overline S}[f] = sum_i int_{overline S_i} |nabla f(x)|,dA(x) + sum_i int_{partial overline S_i} [mathrm{something}],dell.
$$
This seems possible via integration by parts.
Any pointers to papers or formulations would be much appreciated!
multivariable-calculus smooth-manifolds total-variation
multivariable-calculus smooth-manifolds total-variation
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
asked Dec 7 '18 at 23:00
Justin SolomonJustin Solomon
200113
200113
$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18
add a comment |
$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18
$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18
add a comment |
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$begingroup$
Purely based on the intuition of what total variation measures, this should be ${rm TV}_{bar S}[f] = sum_i {rm TV}_{bar S_i}[f] + frac12 sum_i int_{partialbar S_i} |f^+(x) - f^-(x)|,mathrm dell$ where $f^+$ and $f^-$ are the values at either side of the crease, right? I put in a $frac12$ to account for double-counting the creases. // I presume you're looking for a rigorous proof, though. Have you tried connecting the patches via smooth strips of width $h$, then taking the limit as $hto0$?
$endgroup$
– Rahul
Dec 9 '18 at 5:35
$begingroup$
This certainly seems like a reasonable intuition! I wasn't sure the proper proof technique here (it's outside of my usual math comfort area! :-) ) --- is there an example of a proof you have in mind that uses this smoothing procedure for a measure like TV?
$endgroup$
– Justin Solomon
Dec 9 '18 at 14:54
$begingroup$
Sorry, no, I'm not very familiar with this area either. It was just the first thing that came to mind as something to try. (Though I notice now that you say $f$ is continuous on $bar S$, in which case the second term of my suggestion drops out and the total variation would simply be the sum of the TV over all the individual patches.) By the way, by "surface" do you mean a 2D surface embedded in $mathbb R^3$? an abstract differentiable manifold? or something else?
$endgroup$
– Rahul
Dec 10 '18 at 4:36
$begingroup$
For our purposes a 2D surface in R3 is good enough, although I'm betting everything carries through to the Riemannian submanifold case. After some discussion with my students, we think your formula is correct except without the 1/2 factor in front of the second term (to sanity check the 1/2, remember TV of an indicator function is perimeter). But we're still lacking a careful argument :-) --- to come soon hopefully!
$endgroup$
– Justin Solomon
Dec 11 '18 at 21:18