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Is $int V^2e^-Vdtextvol0$?
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Suppose we have a measure $mu(dx)= e^-V(x)dtextvol$ on a Riemannian manifold M with $Vin C^2(M)$, $int e^-Vdtextvol = 1$ and $textRic+textHess(V)>0$.
Is it then true that $int V^2e^-Vdtextvol<infty$?
This holds on $mathbbR^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.
differential-geometry riemannian-geometry
edited Jul 24 at 4:09
John Ma
37.5k93669
asked Jul 23 at 21:55
Corram
885
add a comment |Â
up vote
8
down vote
favorite
Suppose we have a measure $mu(dx)= e^-V(x)dtextvol$ on a Riemannian manifold M with $Vin C^2(M)$, $int e^-Vdtextvol = 1$ and $textRic+textHess(V)>0$.
Is it then true that $int V^2e^-Vdtextvol<infty$?
This holds on $mathbbR^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.
differential-geometry riemannian-geometry
edited Jul 24 at 4:09
John Ma
37.5k93669
asked Jul 23 at 21:55
Corram
885
May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Suppose we have a measure $mu(dx)= e^-V(x)dtextvol$ on a Riemannian manifold M with $Vin C^2(M)$, $int e^-Vdtextvol = 1$ and $textRic+textHess(V)>0$.
Is it then true that $int V^2e^-Vdtextvol<infty$?
This holds on $mathbbR^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.
differential-geometry riemannian-geometry
edited Jul 24 at 4:09
John Ma
37.5k93669
asked Jul 23 at 21:55
Corram
885
Suppose we have a measure $mu(dx)= e^-V(x)dtextvol$ on a Riemannian manifold M with $Vin C^2(M)$, $int e^-Vdtextvol = 1$ and $textRic+textHess(V)>0$.
Is it then true that $int V^2e^-Vdtextvol<infty$?
This holds on $mathbbR^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.
differential-geometry riemannian-geometry
edited Jul 24 at 4:09
John Ma
37.5k93669
asked Jul 23 at 21:55
Corram
885
edited Jul 24 at 4:09
John Ma
37.5k93669
edited Jul 24 at 4:09
John Ma
37.5k93669
edited Jul 24 at 4:09
John Ma
37.5k93669
37.5k93669
asked Jul 23 at 21:55
Corram
885
asked Jul 23 at 21:55
Corram
885
asked Jul 23 at 21:55
Corram
885
885
May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41
add a comment |Â
May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41
May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41
add a comment |Â
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May I know how this is true for $mathbb R^n$?
– John Ma
Jul 24 at 4:09
My attempt on $mathbbR^n$: $V$ is coercive and thus bounded from below. Let $k_0:= lfloor inf Vrfloor>-infty$ and, for any $kinmathbbZ$, let $A_k := leftkleq V<k+1right$. \ $V$ grows at least linearly at infinity so that there is some $M>>0$ with $textvol(A_k)leq Mk^n-1$ for all $k$. Thus, beginalign* int V^2e^-V d x leq sum_k=k_0^infty (k+1)^2e^-ktextvol(A_k) leq M sum_k=k_0^infty (k+1)^2e^-kk^n-1 < infty ,endalign* where the last sum is finite by the integral test for convergence.
– Corram
Jul 24 at 8:41