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?
 A: 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 $H$).
