{"id":65,"date":"2010-06-15T23:48:32","date_gmt":"2010-06-15T21:48:32","guid":{"rendered":"http:\/\/djalil.chafai.net\/blog\/?p=65"},"modified":"2019-05-09T14:07:53","modified_gmt":"2019-05-09T12:07:53","slug":"curvatures-of-potentials","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2010\/06\/15\/curvatures-of-potentials\/","title":{"rendered":"Curvatures of potentials"},"content":{"rendered":"

\"Plot<\/a><\/p>\n

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:<\/p>\n

    \n
  1. \\( {\\lim_{\\left|x\\right|\\rightarrow\\infty}V(x)=\\infty} \\) and \\( {\\inf_{\\mathbb{R}^d}(\\left|\\nabla V\\right|^2-\\Delta V)>-\\infty} \\)<\/li>\n
  2. 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} \\).<\/li>\n<\/ol>\n

    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,<\/p>\n

    \\[ \\begin{cases} dX_t&=\\sqrt{2}\\,dB_t-\\nabla V(X_t)\\,dt\\\\ X_0&=x \\end{cases} \\]<\/p>\n

    see e.g. the book of Royer<\/a>. 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<\/a> and based on the Brenier theorem and the Monge-Amp\u00e8re 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<\/p>\n

    \\[ L:=\\Delta-\\nabla V\\cdot\\nabla. \\]<\/p>\n

    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<\/a>, 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<\/p>\n

    \\[ 2\\Gamma(f,g):=L(fg)-fLg-gLf \\]<\/p>\n

    and<\/p>\n

    \\[ 2\\Gamma_{\\!\\!2}(f,g):=L\\Gamma(f,g)-\\Gamma(f,Lg)-\\Gamma(g,Lf) \\]<\/p>\n

    for every \\( {f,g\\in\\mathcal{A}} \\). Some algebraic computations reveal that <\/a><\/p>\n

    \\[ \\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) \\]<\/p>\n

    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} \\))<\/p>\n

    \\[ \\mathcal{E}(f,g):=\\int\\!\\Gamma(f,g)\\,d\\mu=-\\int\\!fLg\\,d\\mu=-\\int\\!gLf\\,d\\mu \\]<\/p>\n

    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}} \\), <\/a><\/p>\n

    \\[ \\int\\!\\Gamma_{\\!\\!2}(f,f)\\,d\\mu = -\\int\\!\\Gamma(f,Lf)\\,d\\mu = \\int\\!(Lf)^2\\,d\\mu\\geq0. \\ \\ \\ \\ \\ (2) \\]<\/p>\n

    By combining (2)<\/a> and (1)<\/a>, we obtain <\/a><\/p>\n

    \\[ \\int\\!\\Vert\\nabla^2f\\Vert_{HS}^2\\,d\\mu +\\int\\!\\nabla f\\cdot\\nabla^2V\\nabla f\\,d\\mu \\geq 0. \\ \\ \\ \\ \\ (3) \\]<\/p>\n

    All this is well known, see e.g. the fifth chapter of the ABC book<\/a>. 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)<\/a> gives<\/p>\n

    \\[ \\int\\!\\nabla^2 V\\,d\\mu \\succcurlyeq 0 \\quad\\text{i.e.}\\quad \\min\\mathrm{spec}\\left(\\int\\!\\nabla^2V\\,d\\mu\\right)\\geq0. \\]<\/p>\n

    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:<\/p>\n

    Question 1<\/b> <\/a> Do we have \\( {\\int\\!\\underline{\\lambda}\\,d\\mu\\geq0} \\) if \\( {d>1} \\)?<\/em><\/p><\/blockquote>\n

    Example 1 (Gaussian and log-concave cases)<\/b> 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<\/a> 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} \\).<\/em><\/p><\/blockquote>\n

    Example 2 (Gradient type ground state)<\/b> For every \\( {x\\in\\mathbb{R}^d} \\), let us pick an eigenvector \\( {u(x)} \\) associated to the eigenvalue \\( {\\underline{\\lambda}(x)} \\), namely<\/em><\/p>\n

    \\[ u(x)\\in\\arg\\min_{\\left\\Vert v\\right\\Vert_2=1}(v\\cdot(\\nabla^2 V(x))v). \\]<\/em><\/p>\n

    If \\( {\\mathrm{curl}(u)=0} \\) on \\( {\\mathbb{R}^d} \\) i.e. \\( {u=\\nabla f} \\) on \\( {\\mathbb{R}^d} \\) for some \\( {f} \\) (Poincar\u00e9 lemma for differential forms on contractible manifolds<\/a>), then, from (3)<\/a>,<\/em><\/p>\n

    \\[ \\int\\!\\underline{\\lambda}\\,d\\mu \\geq -n\\int\\!\\left\\Vert\\nabla u\\right\\Vert_2^2\\,d\\mu. \\]<\/em><\/p>\n

    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<\/em><\/p>\n

    \\[ S:x\\in\\mathbb{R}^d\\mapsto S(x)\\in\\mathcal{S}_d(\\mathbb{R}) \\]<\/em><\/p>\n

    is of the form \\( {\\nabla^2 V=S} \\) for some potential \\( {V} \\). The answer is given by the Saint-Venant<\/a> compatibility conditions, which constitute a matrix version of the Poincar\u00e9 lemma.<\/em><\/p><\/blockquote>\n

    Example 3 (Radial potentials)<\/b> 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<\/em><\/p>\n

    \\[ \\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. \\]<\/em><\/p>\n

    Consequently, if \\( {\\varphi} \\) is convex then \\( {\\underline{\\lambda}(x)=2\\varphi'(|x|^2)} \\), and thus, for \\( {d\\geq2} \\),<\/em><\/p>\n

    \\[ \\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 \\]<\/em><\/p>\n

    and therefore<\/em><\/p>\n

    \\[ \\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. \\]<\/em><\/p>\n

    The answer to Question 1<\/a> is thus positive in this case. The answer is probably negative in general when \\( {\\varphi} \\) is not convex.<\/em><\/p><\/blockquote>\n

    Example 4 (Tensor potentials)<\/b> 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<\/em><\/p>\n

    \\[ \\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)). \\]<\/em><\/p>\n

    Therefore, \\( {\\underline{\\lambda}(x)=\\min_{1\\leq i\\leq d}W_i''(x)} \\), and consequently<\/em><\/p>\n

    \\[ \\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. \\]<\/em><\/p>\n

    If we assume that \\( {W_1=\\cdots=W_d=W} \\) then we have the symmetric integral <\/a><\/em><\/p>\n

    \\[ \\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) \\]<\/em><\/p>\n

    Let us further specialize to \\( {d=2} \\) and \\( {W(u)=\\alpha u^4- u^2} \\) with \\( {\\alpha>0} \\) (double well):<\/p>\n

    \\[ \\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. \\]<\/p>\n

    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<\/a> seems to be negative in this case.<\/p><\/blockquote>\n

    Note:<\/b> this post is motivated by a question asked by Rapha\u00ebl Roux<\/a>. The answer to Question 1<\/a> 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)<\/a> 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.<\/p>\n","protected":false},"excerpt":{"rendered":"

    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}…<\/p>\n

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