Mixing
Why Dirichlet form are interesting?
$begingroup$
I'm currentely studing the Dirichlet form and to be honest, I really don't see in what they are useful. I don't really get the point with them. I recall the definition :
Definition Let $(H,left<cdot ,cdot right>)$ a Hilbert space. Set $E=E^s+E^a$ a bilinear form defined on a dense subset $D$ of $B$ where $E^s$ is symmetric and $E^a$ antisymmetric. We Then $E$ is a Dirichlet form if
$E^s$ is positive definite on $D$
$(E^s+left<cdot ,cdot right>,D)$ is a Hilbert space,
$(E,D)$ is coercive, i.e. there is $K>0$ s.t. $$|E(x,y)+left<x,yright>|^2leq K |E^s(x,x)+left<x,xright>||E^s(y,y)+left<y,yright>|$$
for all $xin D$, we have $x^*=min(x^+,1)in D$ where $x^+=max{x,0}$ and $$E(x+x^*,x-x^*)geq 0quad text{and}quad E(x-x^*,x+x^*)geq 0.$$
Seeing this definition, what is the motivation behind ? Because as written, it looks a bit barbarous for me. I can accept the first point of the definition, but the 3 other assumption looks to arise from nowhere. Maybe someone knows a very good small introduction to get the point with these Dirichlet form ?
functional-analysis operator-theory
asked Jan 16 at 13:36
idmidm
8,58421445
$endgroup$
add a comment |
$begingroup$
I'm currentely studing the Dirichlet form and to be honest, I really don't see in what they are useful. I don't really get the point with them. I recall the definition :
Definition Let $(H,left<cdot ,cdot right>)$ a Hilbert space. Set $E=E^s+E^a$ a bilinear form defined on a dense subset $D$ of $B$ where $E^s$ is symmetric and $E^a$ antisymmetric. We Then $E$ is a Dirichlet form if
$E^s$ is positive definite on $D$
$(E^s+left<cdot ,cdot right>,D)$ is a Hilbert space,
$(E,D)$ is coercive, i.e. there is $K>0$ s.t. $$|E(x,y)+left<x,yright>|^2leq K |E^s(x,x)+left<x,xright>||E^s(y,y)+left<y,yright>|$$
for all $xin D$, we have $x^*=min(x^+,1)in D$ where $x^+=max{x,0}$ and $$E(x+x^*,x-x^*)geq 0quad text{and}quad E(x-x^*,x+x^*)geq 0.$$
Seeing this definition, what is the motivation behind ? Because as written, it looks a bit barbarous for me. I can accept the first point of the definition, but the 3 other assumption looks to arise from nowhere. Maybe someone knows a very good small introduction to get the point with these Dirichlet form ?
functional-analysis operator-theory
asked Jan 16 at 13:36
idmidm
8,58421445
$endgroup$
$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57
add a comment |
$begingroup$
I'm currentely studing the Dirichlet form and to be honest, I really don't see in what they are useful. I don't really get the point with them. I recall the definition :
Definition Let $(H,left<cdot ,cdot right>)$ a Hilbert space. Set $E=E^s+E^a$ a bilinear form defined on a dense subset $D$ of $B$ where $E^s$ is symmetric and $E^a$ antisymmetric. We Then $E$ is a Dirichlet form if
$E^s$ is positive definite on $D$
$(E^s+left<cdot ,cdot right>,D)$ is a Hilbert space,
$(E,D)$ is coercive, i.e. there is $K>0$ s.t. $$|E(x,y)+left<x,yright>|^2leq K |E^s(x,x)+left<x,xright>||E^s(y,y)+left<y,yright>|$$
for all $xin D$, we have $x^*=min(x^+,1)in D$ where $x^+=max{x,0}$ and $$E(x+x^*,x-x^*)geq 0quad text{and}quad E(x-x^*,x+x^*)geq 0.$$
Seeing this definition, what is the motivation behind ? Because as written, it looks a bit barbarous for me. I can accept the first point of the definition, but the 3 other assumption looks to arise from nowhere. Maybe someone knows a very good small introduction to get the point with these Dirichlet form ?
functional-analysis operator-theory
asked Jan 16 at 13:36
idmidm
8,58421445
$endgroup$
I'm currentely studing the Dirichlet form and to be honest, I really don't see in what they are useful. I don't really get the point with them. I recall the definition :
Definition Let $(H,left<cdot ,cdot right>)$ a Hilbert space. Set $E=E^s+E^a$ a bilinear form defined on a dense subset $D$ of $B$ where $E^s$ is symmetric and $E^a$ antisymmetric. We Then $E$ is a Dirichlet form if
$E^s$ is positive definite on $D$
$(E^s+left<cdot ,cdot right>,D)$ is a Hilbert space,
$(E,D)$ is coercive, i.e. there is $K>0$ s.t. $$|E(x,y)+left<x,yright>|^2leq K |E^s(x,x)+left<x,xright>||E^s(y,y)+left<y,yright>|$$
for all $xin D$, we have $x^*=min(x^+,1)in D$ where $x^+=max{x,0}$ and $$E(x+x^*,x-x^*)geq 0quad text{and}quad E(x-x^*,x+x^*)geq 0.$$
Seeing this definition, what is the motivation behind ? Because as written, it looks a bit barbarous for me. I can accept the first point of the definition, but the 3 other assumption looks to arise from nowhere. Maybe someone knows a very good small introduction to get the point with these Dirichlet form ?
functional-analysis operator-theory
functional-analysis operator-theory
asked Jan 16 at 13:36
idmidm
8,58421445
asked Jan 16 at 13:36
idmidm
8,58421445
edited Jan 16 at 13:43
idm
asked Jan 16 at 13:36
idmidm
8,58421445
asked Jan 16 at 13:36
idmidm
8,58421445
asked Jan 16 at 13:36
idmidm
8,58421445
8,58421445
$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57
add a comment |
$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57
$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57
add a comment |
1 Answer
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$begingroup$
1) Don't get caught up in the technical details too much. I will focus on symmetric Dirichlet forms because I think the situation is somewhat more transparent in this case (the coercivity condition is automatically satisfied for symmetric forms).
Condition 2 is there to ensure the existence of a generator of the form. More precisely, there is a bijective correspondence between positive self-adjoint operators and symmetric bilinear forms satisfying the first two conditions assigning to the form $E$ the operator
$$
D(L)={uin D(E)midexists vin H,forall win D(E)colon E(u,w)=langle v,wrangle},,Lu=v.
$$
One particularly important example is the Dirichlet energy
$$
D(E)=W^{1,2}(mathbb{R^n}),,E(u,v)=int_{mathbb{R^n}}nabla ucdotnabla v,dx.
$$
The corresponding operator is $L=-Delta$ with domain $D(L)=W^{2,2}(mathbb{R}^n)$.
For non-symmetric forms one needs the coercivity condition to extend this correspondence (with a wider class of operators).
2) The fourth condition is the most important one. In the symmetric case it can be reformulated as $E(C(u),C(u))leq E(u,u)$ for $Cin C^1(mathbb{R})$ with $C(0)=0$ and $|C'|leq 1$. As you see, this is satisfied for the Dirichlet energy above.
There is the following intuition behind: $E(u)$ is supposed to measure the oscillation of $u$ (if you know a little quantum mechanics, think of $u$ as wave function and of the oscillation of $u$ as measure of the energy of a paricle in state $u$). The fourth condition then says that the oscillation decreases if one damps the function $u$.
This property has several interesting consequences. The first (and easiest to prove) is that the semigroup generated by the generator of a Dirichlet form is Markovian. Markovian semigroups play a role in many places in mathematics, most notably in connection with Markov processes, which brings me to the second consequence.
Fukushima (later extended to the non-symmetric case by Ma and Röckner) showed that there is bijective correspondence between so-called regular Dirichlet forms and a class of symmetric Markov processes. This allows one to define Markov processes on non-smooth spaces like fractals or limit spaces of Riemannian manifolds.
Finally, there is another nice characterization of Dirichlet forms. As you see, the Dirichlet energy is of the form $E(u,v)=langle nabla u,nabla vrangle$ and $nabla$ satisfies the Leibniz rule. Cipriani and Sauvageot showed that every Dirichlet form can be written in the form $E(u,v)=langle partial u,partial vrangle_{mathcal{H}}$, where $partial$ is a derivation with values in the Hilbert space $mathcal{H}$. One can use this characterization for example to define the gradient of a function on metric spaces where you don't have a good geometric notion of tangent space.
To wrap it up, Dirichlet forms are related to a lot of interesting mathematical objects at the intersection of analysis, geometry and probability, and, what is nore, they often provide a technically easier approach.
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
$endgroup$
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
add a comment |
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$begingroup$
1) Don't get caught up in the technical details too much. I will focus on symmetric Dirichlet forms because I think the situation is somewhat more transparent in this case (the coercivity condition is automatically satisfied for symmetric forms).
Condition 2 is there to ensure the existence of a generator of the form. More precisely, there is a bijective correspondence between positive self-adjoint operators and symmetric bilinear forms satisfying the first two conditions assigning to the form $E$ the operator
$$
D(L)={uin D(E)midexists vin H,forall win D(E)colon E(u,w)=langle v,wrangle},,Lu=v.
$$
One particularly important example is the Dirichlet energy
$$
D(E)=W^{1,2}(mathbb{R^n}),,E(u,v)=int_{mathbb{R^n}}nabla ucdotnabla v,dx.
$$
The corresponding operator is $L=-Delta$ with domain $D(L)=W^{2,2}(mathbb{R}^n)$.
For non-symmetric forms one needs the coercivity condition to extend this correspondence (with a wider class of operators).
2) The fourth condition is the most important one. In the symmetric case it can be reformulated as $E(C(u),C(u))leq E(u,u)$ for $Cin C^1(mathbb{R})$ with $C(0)=0$ and $|C'|leq 1$. As you see, this is satisfied for the Dirichlet energy above.
There is the following intuition behind: $E(u)$ is supposed to measure the oscillation of $u$ (if you know a little quantum mechanics, think of $u$ as wave function and of the oscillation of $u$ as measure of the energy of a paricle in state $u$). The fourth condition then says that the oscillation decreases if one damps the function $u$.
This property has several interesting consequences. The first (and easiest to prove) is that the semigroup generated by the generator of a Dirichlet form is Markovian. Markovian semigroups play a role in many places in mathematics, most notably in connection with Markov processes, which brings me to the second consequence.
Fukushima (later extended to the non-symmetric case by Ma and Röckner) showed that there is bijective correspondence between so-called regular Dirichlet forms and a class of symmetric Markov processes. This allows one to define Markov processes on non-smooth spaces like fractals or limit spaces of Riemannian manifolds.
Finally, there is another nice characterization of Dirichlet forms. As you see, the Dirichlet energy is of the form $E(u,v)=langle nabla u,nabla vrangle$ and $nabla$ satisfies the Leibniz rule. Cipriani and Sauvageot showed that every Dirichlet form can be written in the form $E(u,v)=langle partial u,partial vrangle_{mathcal{H}}$, where $partial$ is a derivation with values in the Hilbert space $mathcal{H}$. One can use this characterization for example to define the gradient of a function on metric spaces where you don't have a good geometric notion of tangent space.
To wrap it up, Dirichlet forms are related to a lot of interesting mathematical objects at the intersection of analysis, geometry and probability, and, what is nore, they often provide a technically easier approach.
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
$endgroup$
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
add a comment |
$begingroup$
1) Don't get caught up in the technical details too much. I will focus on symmetric Dirichlet forms because I think the situation is somewhat more transparent in this case (the coercivity condition is automatically satisfied for symmetric forms).
Condition 2 is there to ensure the existence of a generator of the form. More precisely, there is a bijective correspondence between positive self-adjoint operators and symmetric bilinear forms satisfying the first two conditions assigning to the form $E$ the operator
$$
D(L)={uin D(E)midexists vin H,forall win D(E)colon E(u,w)=langle v,wrangle},,Lu=v.
$$
One particularly important example is the Dirichlet energy
$$
D(E)=W^{1,2}(mathbb{R^n}),,E(u,v)=int_{mathbb{R^n}}nabla ucdotnabla v,dx.
$$
The corresponding operator is $L=-Delta$ with domain $D(L)=W^{2,2}(mathbb{R}^n)$.
For non-symmetric forms one needs the coercivity condition to extend this correspondence (with a wider class of operators).
2) The fourth condition is the most important one. In the symmetric case it can be reformulated as $E(C(u),C(u))leq E(u,u)$ for $Cin C^1(mathbb{R})$ with $C(0)=0$ and $|C'|leq 1$. As you see, this is satisfied for the Dirichlet energy above.
There is the following intuition behind: $E(u)$ is supposed to measure the oscillation of $u$ (if you know a little quantum mechanics, think of $u$ as wave function and of the oscillation of $u$ as measure of the energy of a paricle in state $u$). The fourth condition then says that the oscillation decreases if one damps the function $u$.
This property has several interesting consequences. The first (and easiest to prove) is that the semigroup generated by the generator of a Dirichlet form is Markovian. Markovian semigroups play a role in many places in mathematics, most notably in connection with Markov processes, which brings me to the second consequence.
Fukushima (later extended to the non-symmetric case by Ma and Röckner) showed that there is bijective correspondence between so-called regular Dirichlet forms and a class of symmetric Markov processes. This allows one to define Markov processes on non-smooth spaces like fractals or limit spaces of Riemannian manifolds.
Finally, there is another nice characterization of Dirichlet forms. As you see, the Dirichlet energy is of the form $E(u,v)=langle nabla u,nabla vrangle$ and $nabla$ satisfies the Leibniz rule. Cipriani and Sauvageot showed that every Dirichlet form can be written in the form $E(u,v)=langle partial u,partial vrangle_{mathcal{H}}$, where $partial$ is a derivation with values in the Hilbert space $mathcal{H}$. One can use this characterization for example to define the gradient of a function on metric spaces where you don't have a good geometric notion of tangent space.
To wrap it up, Dirichlet forms are related to a lot of interesting mathematical objects at the intersection of analysis, geometry and probability, and, what is nore, they often provide a technically easier approach.
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
$endgroup$
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
add a comment |
$begingroup$
1) Don't get caught up in the technical details too much. I will focus on symmetric Dirichlet forms because I think the situation is somewhat more transparent in this case (the coercivity condition is automatically satisfied for symmetric forms).
Condition 2 is there to ensure the existence of a generator of the form. More precisely, there is a bijective correspondence between positive self-adjoint operators and symmetric bilinear forms satisfying the first two conditions assigning to the form $E$ the operator
$$
D(L)={uin D(E)midexists vin H,forall win D(E)colon E(u,w)=langle v,wrangle},,Lu=v.
$$
One particularly important example is the Dirichlet energy
$$
D(E)=W^{1,2}(mathbb{R^n}),,E(u,v)=int_{mathbb{R^n}}nabla ucdotnabla v,dx.
$$
The corresponding operator is $L=-Delta$ with domain $D(L)=W^{2,2}(mathbb{R}^n)$.
For non-symmetric forms one needs the coercivity condition to extend this correspondence (with a wider class of operators).
2) The fourth condition is the most important one. In the symmetric case it can be reformulated as $E(C(u),C(u))leq E(u,u)$ for $Cin C^1(mathbb{R})$ with $C(0)=0$ and $|C'|leq 1$. As you see, this is satisfied for the Dirichlet energy above.
There is the following intuition behind: $E(u)$ is supposed to measure the oscillation of $u$ (if you know a little quantum mechanics, think of $u$ as wave function and of the oscillation of $u$ as measure of the energy of a paricle in state $u$). The fourth condition then says that the oscillation decreases if one damps the function $u$.
This property has several interesting consequences. The first (and easiest to prove) is that the semigroup generated by the generator of a Dirichlet form is Markovian. Markovian semigroups play a role in many places in mathematics, most notably in connection with Markov processes, which brings me to the second consequence.
Fukushima (later extended to the non-symmetric case by Ma and Röckner) showed that there is bijective correspondence between so-called regular Dirichlet forms and a class of symmetric Markov processes. This allows one to define Markov processes on non-smooth spaces like fractals or limit spaces of Riemannian manifolds.
Finally, there is another nice characterization of Dirichlet forms. As you see, the Dirichlet energy is of the form $E(u,v)=langle nabla u,nabla vrangle$ and $nabla$ satisfies the Leibniz rule. Cipriani and Sauvageot showed that every Dirichlet form can be written in the form $E(u,v)=langle partial u,partial vrangle_{mathcal{H}}$, where $partial$ is a derivation with values in the Hilbert space $mathcal{H}$. One can use this characterization for example to define the gradient of a function on metric spaces where you don't have a good geometric notion of tangent space.
To wrap it up, Dirichlet forms are related to a lot of interesting mathematical objects at the intersection of analysis, geometry and probability, and, what is nore, they often provide a technically easier approach.
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
$endgroup$
1) Don't get caught up in the technical details too much. I will focus on symmetric Dirichlet forms because I think the situation is somewhat more transparent in this case (the coercivity condition is automatically satisfied for symmetric forms).
Condition 2 is there to ensure the existence of a generator of the form. More precisely, there is a bijective correspondence between positive self-adjoint operators and symmetric bilinear forms satisfying the first two conditions assigning to the form $E$ the operator
$$
D(L)={uin D(E)midexists vin H,forall win D(E)colon E(u,w)=langle v,wrangle},,Lu=v.
$$
One particularly important example is the Dirichlet energy
$$
D(E)=W^{1,2}(mathbb{R^n}),,E(u,v)=int_{mathbb{R^n}}nabla ucdotnabla v,dx.
$$
The corresponding operator is $L=-Delta$ with domain $D(L)=W^{2,2}(mathbb{R}^n)$.
For non-symmetric forms one needs the coercivity condition to extend this correspondence (with a wider class of operators).
2) The fourth condition is the most important one. In the symmetric case it can be reformulated as $E(C(u),C(u))leq E(u,u)$ for $Cin C^1(mathbb{R})$ with $C(0)=0$ and $|C'|leq 1$. As you see, this is satisfied for the Dirichlet energy above.
There is the following intuition behind: $E(u)$ is supposed to measure the oscillation of $u$ (if you know a little quantum mechanics, think of $u$ as wave function and of the oscillation of $u$ as measure of the energy of a paricle in state $u$). The fourth condition then says that the oscillation decreases if one damps the function $u$.
This property has several interesting consequences. The first (and easiest to prove) is that the semigroup generated by the generator of a Dirichlet form is Markovian. Markovian semigroups play a role in many places in mathematics, most notably in connection with Markov processes, which brings me to the second consequence.
Fukushima (later extended to the non-symmetric case by Ma and Röckner) showed that there is bijective correspondence between so-called regular Dirichlet forms and a class of symmetric Markov processes. This allows one to define Markov processes on non-smooth spaces like fractals or limit spaces of Riemannian manifolds.
Finally, there is another nice characterization of Dirichlet forms. As you see, the Dirichlet energy is of the form $E(u,v)=langle nabla u,nabla vrangle$ and $nabla$ satisfies the Leibniz rule. Cipriani and Sauvageot showed that every Dirichlet form can be written in the form $E(u,v)=langle partial u,partial vrangle_{mathcal{H}}$, where $partial$ is a derivation with values in the Hilbert space $mathcal{H}$. One can use this characterization for example to define the gradient of a function on metric spaces where you don't have a good geometric notion of tangent space.
To wrap it up, Dirichlet forms are related to a lot of interesting mathematical objects at the intersection of analysis, geometry and probability, and, what is nore, they often provide a technically easier approach.
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
answered Jan 16 at 15:08
MaoWaoMaoWao
3,318617
3,318617
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
add a comment |
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
$begingroup$
Thank you for your very nice answer. Just a questions : What do you mean by "generator of the form" ? In what $L=-Delta $ is relevant ?
$endgroup$
– idm
Jan 16 at 16:12
add a comment |
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$begingroup$
I' don't know much about Dirichlet forms specifically, but that looks suspiciously like a retroactive definition: a definition that's been written precisely so that some theorem or other is true about the things that it's defining.
$endgroup$
– user3482749
Jan 16 at 13:41
$begingroup$
There is a typo in the fourth bullet point. It should be $x^ast=min(x^+,1)$.
$endgroup$
– MaoWao
Jan 16 at 13:41
$begingroup$
@MaoWao: thank you :)
$endgroup$
– idm
Jan 16 at 13:43
$begingroup$
One more quibble: Dirichlet forms are not defined on arbitrary Hilbert spaces, but on $L^2$ spaces (otherwise, the $1$ in the fourth condition would not make sense).
$endgroup$
– MaoWao
Jan 16 at 14:57