Nonnegative selfadjoint operators generating strongly continuous semigroups Let $L$ be a nonnegative selfadjoint operator on $L^2(X, \mu)$ for $(X, \mu)$ an arbitrary measure space. (If we want to be concrete we can take euclidean space with Lebesgue measure, and take $L$ as a Fourier multiplier by a nonnegative function.)
Is it always true that $e^{-tL}$ is always a strongly continuous, contractive, positive semigroup? I can find a reference for this when $L$ is a diffusion operator but not in general. My intuition tells me yes - the spectrum of $L$ is real and contained in $[0, \infty)$ so the spectrum of $e^{-tL}$ is also real and contained in the image of $(0, 1]$, which is why I suspect $e^{-tL}$ is also contractive and positive. (If $L1 = 0$ as well then $e^{-tL}$ should also be Markov in that $e^{-tL}1 = 1$.)
If not, are there any counterexamples of operators $L$ which are nonnegative and selfadjoint but do not generate a strongly continuous contractive positive semigroup?
 A: There are two notions of positive operators on $L^2$. The first one is that (bounded) $T$ is positive if and only if $\langle f,Tf\rangle\geq 0$ for all $f\in L^2$. In this sense $e^{-tL}$ is indeed positive by the spectral theorem (but note that the spectrum of $e^{-tL}$ is not necessarily contained in $(0,1]$, but in the closed interval $[0,1]$).
The second notion is that a (bounded) operator is positive if $Tf\geq 0$ for all $f\in L^2$ with $f\geq 0$. I prefer to call these operators positivity-preserving. Your mention of Markov semigroups makes me suspect that you're after this notion of positivity. In this case, the answer is no. A nonnegative self-adjoint generates a strongly continuous semigroup of positivity-preserving operators if and only if $f\in D(L^{1/2})$ implies $f_\pm\in D(L^{1/2})$ and $\langle L^{1/2}f_+,L^{1/2}f_-\rangle\leq 0$. For example, $-\Delta$ generates a semigroup of positivity-preserving operators, while $\Delta^2$ does not.
A: Generation of strongly continuous contraction semigroups on a Hilbert space (or, more generally, a Banach space) is the content of the Hille-Yosida theorem.  When $L$ is self-adjoint, the theorem simplifies.
All we need to check is that $L$ is monotone, i.e. for all $v \in D(L)\subset H$, we have $\langle Lv,v\rangle\geq 0$, which holds true from the non-negativity assumption on $L$. Since self-adjoint operators are densely defined and closed and $L^*=L$ is also monotone we have that $L$ is maximal monotone (see the first Definition of Chapter 7 of Brezis' book http://www.math.utoronto.ca/almut/Brezis.pdf and then Remark 8). Then, Theorem 7.7 in Chapter 7 of Brezis tells us that we get a contraction semigroup which is strongly continuous.
