# Spectral gap and Poincaré inequality

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?

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

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 $$H$$).

• 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? Jul 3, 2021 at 17:40
• 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? Jul 3, 2021 at 17:49
• 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 $\int_{\mathbb{R}^{d}} \|Df\|^{2} e^{-V} \, dx \geq c \int_{\mathbb{R}^{d}} f^{2} e^{-V} \, dx$.
– user711689
Jul 3, 2021 at 19:56
• 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? Apr 29, 2023 at 16:04