Let $(E,\mathcal E,\mu)$ be a finite measure space, $(T(t))_{t\ge0}$ be a strongly continuous contraction semigroup on $L^2(\mu)$ and $(\mathcal D(A,A)$ denote the generator of $(T(t))_{t\ge0}$. Assume $T(t)$ is self-adjoint for all $t>0$.
Are we able to conclude that $(\mathcal D(A,A)$ is nonpositive definite?
Let $f\in\mathcal D(A)$. We need to show that $\langle f,Af\rangle_{L^2(\mu)}\le0$. By definition, $$\frac12\left(\langle f,T(t)f\rangle_{L^2(\mu)}+\left\|f\right\|_{L^2(\mu)}^2\right)=\left\langle f,\frac{T(t)f-f}t\right\rangle_{L^2(\mu)}\xrightarrow{t\to0+}\langle f,Af\rangle_{L^2(\mu)}\tag1$$ Now, by contractivity $$\left\|T(t)f\right\|_{L^2(\mu)}\le\left\|f\right\|_{L^2(\mu)}\tag2.$$ However, in light of $(1)$ it seems like we need to show $$\langle f,T(t)f\rangle_{L^2(\mu)}+\left\|f\right\|_{L^2(\mu)}^2\le 0.$$
Best Answer
Because $T(t)$ is contractive, then $\|T(t)f\|^2$ is a non-increasing function of $t$ for each fixed $f$. Consequently, for all $f\in\mathcal{D}(A)$, $$ 0 \ge \left.\frac{d}{dt}\|T(t)f\|^2 \right|_{t=0} = \langle Af,f\rangle+\langle f,Af\rangle = 2\Re\langle Af,f\rangle. $$ Assuming that $A$ is selfadjoint gives $A \le 0$.