Show that the generator of a strongly continuous contraction semigroup on $L^2$ is nonpositive definite

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 $$\langle f,T(t)f\rangle_{L^2(\mu)}+\left\|f\right\|_{L^2(\mu)}^2\le 0.$$

• In your first displayed equation, where did you get the negative on the right side? $\langle f,T(t)f\rangle+\|f\|^2 = \langle f,T(t)f+f\rangle$ is correct. Mar 19 '19 at 4:47

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 $$A \le 0$$.
• Do we need the self-adjointness of $T(t)$ (or symmetry of $(\mathcal D(A),A)$) at all? As you noted, by contractivity, $\left\|T(t)\right\|_{L^2(\mu)}$ is nonincreasing in $t$. So, $$0\ge\lim_{s\to t}\left\|T(s)\right\|_{L^2(\mu)}^2=2\langle AT(t)f,T(t)f\rangle_{L^2(\mu)}$$ for all $f\in\mathcal D(A)$ and $t\ge0$; simply by definition of $\mathcal D(A)$ and the chain rule (note that we're dealing with a real Hilbert space). So, $0\ge\langle Af,f\rangle_{L^2(\mu)}$ and I don't where we needed the self-adjointness. Mar 19 '19 at 8:59