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

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  • $\begingroup$ 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. $\endgroup$ – DisintegratingByParts Mar 19 at 4:47
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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 $A \le 0$.

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  • $\begingroup$ 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. $\endgroup$ – 0xbadf00d Mar 19 at 8:59

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