Let \( {V:\mathbb{R}^d\rightarrow\mathbb{R}} \) be a smooth “potential”. We do not assume that \( {e^{-V}} \) is Lebesgue integrable for the moment. Let \( {\mu} \) be the positive Borel measure on \( {\mathbb{R}^d} \) with Lebesgue density function \( {e^{-V}} \). Let us assume additionally that one the following properties holds:
- \( {\lim_{\left|x\right|\rightarrow\infty}V(x)=\infty} \) and \( {\inf_{\mathbb{R}^d}(\left|\nabla V\right|^2-\Delta V)>-\infty} \)
- there exists \( {a,b\in\mathbb{R}} \) such that \( {x\cdot \nabla V(x) \geq -a\left|x\right|^2-b} \) for all \( {x\in\mathbb{R}^d} \).
These conditions ensure the non explosion of the Langevin-Kolmogorov Markov diffusion process \( {(X_t)_{t\geq0}} \) on \( {\mathbb{R}^d} \) solving the stochastic differential equation, driven by a standard Brownian Motion,
\[ \begin{cases} dX_t&=\sqrt{2}\,dB_t-\nabla V(X_t)\,dt\\ X_0&=x \end{cases} \]
see e.g. the book of Royer. It can be shown without difficulty that these conditions on the potential are satisfied if for instance there exists a constant \( {\kappa\in\mathbb{R}} \) such that \( {\nabla^2 V(x) \succcurlyeq \kappa I_d} \) for every \( {x\in\mathbb{R}^d} \). Here “\( {\succcurlyeq} \)” stands for the Loewner partial order on quadratic forms (Hermitian matrices). If \( {\kappa>0} \) then \( {e^{-V}} \) is uniformly log-concave and is Lebesgue integrable. By the way, a nice result due to Caffarelli and based on the Brenier theorem and the Monge-Ampère equation states that \( {\mu} \) is then the image of the standard Gaussian law by a \( {\kappa} \) Lipschitz map. However, \( {e^{-V}} \) can be Lebesgue integrable beyond this log-concavity sufficient condition (e.g. multiple wells potentials). The infinitesimal generator of \( {(X_t)_{t\geq0}} \) is the second order differential operator
\[ L:=\Delta-\nabla V\cdot\nabla. \]
The Markov semigroup \( {(P_t)_{t\geq0}=(e^{tL})_{t\geq0}} \) is given by \( {P_t(f):=\mathbb{E}(f(X_t)|X_0=\cdot)} \). Following e.g. the Definition 2.4.2 in the ABC book, we assume that there exists a nice algebra of functions \( {\mathcal{A}} \) for our computations. We define the functional quadratic forms \( {\Gamma} \) and \( {\Gamma_{\!\!2}} \) by
\[ 2\Gamma(f,g):=L(fg)-fLg-gLf \]
and
\[ 2\Gamma_{\!\!2}(f,g):=L\Gamma(f,g)-\Gamma(f,Lg)-\Gamma(g,Lf) \]
for every \( {f,g\in\mathcal{A}} \). Some algebraic computations reveal that
\[ \Gamma(f,f)=\left|\nabla f\right|^2 \quad\text{and}\quad \Gamma_{\!\!2}(f,f)=\Vert\nabla^2f\Vert_{HS}^2+\nabla f\cdot\nabla^2V\nabla f. \ \ \ \ \ (1) \]
Here \( {\left\Vert A\right\Vert_{HS}^2:=\mathrm{Tr}(AA^*)=\sum_{i,j=1}^d|A_{i,j}|^2} \) is the Hilbert-Schmidt norm of the \( {d\times d} \) matrix \( {A} \). We have the integration by parts formula (\( {\mu} \) is symmetric invariant for \( {L} \))
\[ \mathcal{E}(f,g):=\int\!\Gamma(f,g)\,d\mu=-\int\!fLg\,d\mu=-\int\!gLf\,d\mu \]
for all \( {f,g\in\mathcal{A}} \). The functional quadratic form \( {\mathcal{E}} \) is the Dirichlet form. We have, using the definition of \( {\Gamma_{\!\!2}} \) and the integration by parts formula, for every \( {f\in\mathcal{A}} \),
\[ \int\!\Gamma_{\!\!2}(f,f)\,d\mu = -\int\!\Gamma(f,Lf)\,d\mu = \int\!(Lf)^2\,d\mu\geq0. \ \ \ \ \ (2) \]
By combining (2) and (1), we obtain
\[ \int\!\Vert\nabla^2f\Vert_{HS}^2\,d\mu +\int\!\nabla f\cdot\nabla^2V\nabla f\,d\mu \geq 0. \ \ \ \ \ (3) \]
All this is well known, see e.g. the fifth chapter of the ABC book. From now on, we assume that \( {\mu(\mathbb{R}^d)<\infty} \). By adding a constant to \( {V} \), we may further assume that \( {\mu} \) is a probability measure. Since \( {\mu} \) is tight, the approximation of linear functions by elements of \( {\mathcal{A}} \) in (3) gives
\[ \int\!\nabla^2 V\,d\mu \succcurlyeq 0 \quad\text{i.e.}\quad \min\mathrm{spec}\left(\int\!\nabla^2V\,d\mu\right)\geq0. \]
If we set \( {\underline{\lambda}(x):=\min\mathrm{spec}(\nabla^2 V(x))} \) then \( {\displaystyle{\int\!\underline{\lambda}\,d\mu\geq0}} \) when \( {d=1} \). We may ask:
Question 1 Do we have \( {\int\!\underline{\lambda}\,d\mu\geq0} \) if \( {d>1} \)?
Example 1 (Gaussian and log-concave cases) If \( {V(x)} \) is a positive semidefinite quadratic form then \( {\mu} \) is up to normalization a Gaussian law, \( {\nabla^2V\equiv\rho I_d} \) with \( {\rho>0} \) (we assumed that \( {\mu} \) is a finite measure), and \( {\underline{\lambda}\equiv\rho} \). The answer to Question 1 is thus positive here. More generally, the answer is obviously positive if \( {V} \) is more convex than the Gaussian, i.e. \( {\nabla^2 V\succcurlyeq \rho I_d} \) on \( {\mathbb{R}^d} \).
Example 2 (Gradient type ground state) For every \( {x\in\mathbb{R}^d} \), let us pick an eigenvector \( {u(x)} \) associated to the eigenvalue \( {\underline{\lambda}(x)} \), namely
\[ u(x)\in\arg\min_{\left\Vert v\right\Vert_2=1}(v\cdot(\nabla^2 V(x))v). \]
If \( {\mathrm{curl}(u)=0} \) on \( {\mathbb{R}^d} \) i.e. \( {u=\nabla f} \) on \( {\mathbb{R}^d} \) for some \( {f} \) (Poincaré lemma for differential forms on contractible manifolds), then, from (3),
\[ \int\!\underline{\lambda}\,d\mu \geq -n\int\!\left\Vert\nabla u\right\Vert_2^2\,d\mu. \]
One may ask if the right hand side may be zero for a non quadratic potential \( {V} \). By the way, one can ask under which condition a given field of symmetric matrices
\[ S:x\in\mathbb{R}^d\mapsto S(x)\in\mathcal{S}_d(\mathbb{R}) \]
is of the form \( {\nabla^2 V=S} \) for some potential \( {V} \). The answer is given by the Saint-Venant compatibility conditions, which constitute a matrix version of the Poincaré lemma.
Example 3 (Radial potentials) Suppose that the potential is radial, i.e. takes the form \( {V(x)=\varphi(|x|^2)} \) where \( {\varphi:\mathbb{R}_+\rightarrow\mathbb{R}} \) is a smooth function. We have
\[ \nabla V(x)=2\varphi'(|x|^2)x \quad\text{and}\quad \nabla^2 V(x)=2\varphi'(|x|^2)I_d+4\varphi”(|x|^2)xx^\top. \]
Consequently, if \( {\varphi} \) is convex then \( {\underline{\lambda}(x)=2\varphi'(|x|^2)} \), and thus, for \( {d\geq2} \),
\[ \omega_d^{-1}\!\int\!\underline{\lambda}\,d\mu =2\int_0^\infty\!\varphi'(r^2)e^{-\varphi(r^2)}\,r^{d-1}dr =\int_0^\infty\!(-e^{-\varphi(r^2)})’\,r^{d-2}dr \]
and therefore
\[ \omega_d^{-1}\!\int\!\underline{\lambda}\,d\mu =\left[-r^{d-2}e^{-\varphi(r^2)}\right]_0^\infty +(d-2)\int_0^\infty\!e^{-\varphi(r^2)}\,r^{d-3}dr\geq0. \]
The answer to Question 1 is thus positive in this case. The answer is probably negative in general when \( {\varphi} \) is not convex.
Example 4 (Tensor potentials) Consider the case where \( {\mu} \) is a tensor product i.e. \( {V(x)=W_1(x_1)+\cdots+W_d(x_d)} \) where all \( {W_i:\mathbb{R}\rightarrow\mathbb{R}} \) are smooth. We have
\[ \nabla V(x)=(W’_1(x_1),\ldots,W’_d(x_d))^\top \quad\text{and}\quad \nabla^2V(x)=\mathrm{Diag}(W_1”(x_1),\ldots,W_d”(x_d)). \]
Therefore, \( {\underline{\lambda}(x)=\min_{1\leq i\leq d}W_i”(x)} \), and consequently
\[ \int\!\underline{\lambda}\,d\mu =\int\!\min_{1\leq i\leq d}W”_i(x_i)e^{-W_1(x_1)-\cdots-W_n(x_d)}\,dx. \]
If we assume that \( {W_1=\cdots=W_d=W} \) then we have the symmetric integral
\[ \int\!\underline{\lambda}\,d\mu =\int\!\min_{1\leq i\leq d}W”(x_i)e^{-W(x_1)-\cdots-W(x_d)}\,dx_1\cdots dx_d. \ \ \ \ \ (4) \]
Let us further specialize to \( {d=2} \) and \( {W(u)=\alpha u^4- u^2} \) with \( {\alpha>0} \) (double well):
\[ \int\!\underline{\lambda}\,d\mu =2\int_{\mathbb{R}^2}\!(6\alpha\min(x^2,y^2)-1) e^{-\alpha (x^4+y^4)+x^2+y^2}\,dxdy. \]
Now a numerical quadrature using the int2d function of the Scilab 5.2.1 software package suggests that this integral is negative for e.g. \( {\alpha=1/4} \). The answer to Question 1 seems to be negative in this case.
Note: this post is motivated by a question asked by Raphaël Roux. The answer to Question 1 seems to be sometimes positive, and negative in general. One may then ask for more sufficient conditions on \( {V} \) ensuring a positive answer. More recently, Roux came with an interesting further remark: the right hand side of (4) writes \( {\mathbb{E}(\min_{1\leq i\leq d}W”(X_i))} \) where \( {X_1,\ldots,X_d} \) are i.i.d. of density \( {e^{-W}} \), and therefore, if \( {W”} \) is bounded with a negative minimum, then \( {\min_{1\leq i\leq d}W”(X_i)} \) converges in probability to a negative quantity as \( {d} \) goes to infinity.
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