Consider the PDE $$\partial_t u = L u$$ where $L = \Delta + \nabla V \cdot \nabla $ is a self-adjoint operator.

I read that if $L$ has a spectral gap $\lambda > 0$ then "[convergence of the initial condition to the stationary distribution $u_s(x) = e^{-V(x)}$] easily follows by elementary spectral analysis, or by noting that the existence of a spectral gap of size $\lambda$ for $L$ is equivalent to the statement that $e^{-V}$ satisfies a Poincaré inequality with constant $\lambda$." I.e. $$\int |\nabla f(x)|^2 e^{-V(x)} dx \geq \lambda \int f^2(x) e^{-V(x)}dx$$ for all $f$ $L^2$-integrable wrt the measure $e^{-V(x)}dx$ and such that $\int f e^{-V(x)}dx = 0$. I'm not sure why the latter is required.

Why is L having a spectral gap equivalent to $e^{-V}$ satisfying a Poincare inequality? And under what conditions on $L$ is this true?

  • $\begingroup$ Can you write the exact statement of such a Poincaré inequality? $\endgroup$ Jul 2, 2021 at 13:50
  • $\begingroup$ @GiuseppeNegro yes, edited. $\endgroup$
    – 900edges
    Jul 2, 2021 at 13:55
  • 1
    $\begingroup$ The condition on $f$ is required to exclude constant functions, which ofc violate the inequaltiy. $\endgroup$
    – daw
    Jul 2, 2021 at 14:04

1 Answer 1


We are working in $L^{2}(\mathbb{R}^{d}, e^{-V})$ just to make things clear. This is a Hilbert space and $L$ is self-adjoint and non-negative. (Non-negative means that $\langle Lf, f \rangle \geq 0$ for all $f$ in the domain of $L$.) We say that $L$ has a spectral gap if $\sigma(L) \subseteq \{0\} \cup [c,\infty)$ for some $c > 0$. Notice that we know that $0 \in \sigma(L)$ as $\text{Ker}(L)$ contains constant functions.

There are two key technical points.

$L$ as a "Dirichlet form": For each $f$ in the domain of $L$, we have \begin{equation*} \langle Lf, f \rangle = \int_{\mathbb{R}^{d}} \|Df\|^{2} e^{-V} \, dx. \end{equation*} (Here and henceforth $\langle \cdot,\cdot \rangle$ is the inner product in $L^{2}(\mathbb{R}^{d},e^{-V})$.)

Note one important implication of this: $Lf = 0$ if and only if $f$ is constant.

Condition on the spectrum: Let $\mathbf{1}$ be the constant function $\mathbf{1}(x) = 1$ and write $H = \langle \mathbf{1} \rangle^{\perp} \subseteq L^{2}(\mathbb{R}^{d};e^{-V})$. We know that $\sigma(L) \subseteq \{0\} \cup [c,\infty)$ if and only if \begin{equation*} \langle Lf, f \rangle \geq c \|f\|^{2} \quad \text{if} \, \, f \in H. \end{equation*} The reason this holds is $\text{Ker}(L) = \langle \mathbf{1} \rangle$ so we can "mod out" the kernel --- it is only necessary to show that $L \restriction_{H}$ is strictly positive.

From the two facts above, $L$ has a spectral gap if and only if $\langle Lf, f \rangle \geq c \|f\|^{2}$ for all $f \in H$, but then this is the same as showing that \begin{equation*} \int_{\mathbb{R}^{d}} \|Df\|^{2} e^{-V} \, dx = \langle Lf, f \rangle \geq c \|f\|^{2} = c \int_{\mathbb{R}^{d}} f^{2} e^{-V} \, dx. \end{equation*} In other words, $L$ has a spectral gap if and only if a Poincare inequality holds (in $H$).

  • $\begingroup$ Thanks for your answer! I think I follow the argument but I'm not sure about the first fact: I can see how this is true for $L= \Delta$, but not for my defined $L = \Delta + \nabla V \cdot \nabla$. And in your second fact, does $\langle \mathbf{1}\rangle ^{\perp}$ denote the set of $L^2$ functions orthogonal to the constant? $\endgroup$
    – 900edges
    Jul 3, 2021 at 17:40
  • $\begingroup$ With some googling, I mostly see poincare inequality refer to $||u|| \leq c||Lu||$ for $L = \Delta$; does it more generally hold for any $L$ with a spectral gap? Does $L$ need to be self-adjoint or nonnegative? $\endgroup$
    – 900edges
    Jul 3, 2021 at 17:49
  • 1
    $\begingroup$ The expression for $\langle Lf, f \rangle$ follows by integration by parts --- just like when $L = \Delta$. ($e^{-V}$ appears in $\langle \cdot, \cdot \rangle$ as we are interested in $L^{2}(\mathbb{R}^{d},e^{-V})$, not the standard unweighted one.) Poincare inequality can mean different things to different people, but in my answer (as in your original post) it is $\int_{\mathbb{R}^{d}} \|Df\|^{2} e^{-V} \, dx \geq c \int_{\mathbb{R}^{d}} f^{2} e^{-V} \, dx$. $\endgroup$
    – user711689
    Jul 3, 2021 at 19:56
  • $\begingroup$ Sorry for re-opening this thread. But I do not see why $\sigma(L)\subseteq\{0\}\cup [c,\infty)$ if and only if $\langle Lf,f\rangle\geq c\Vert f\Vert^2$ if $f\in H$. Could you give an explanation, please? $\endgroup$
    – selector
    Apr 29, 2023 at 16:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .