Can I show this using a contraction semigroup property? Let $H$ be a (real) Hilbert space, $L$ be an unbounded operator on $H$ with its domain $D(L)$ and $(e^{-tL})_{t\ge 0}$ be a contraction semigroup on $H$.
Then, the following holds from a semigroup property?
$$(Lh_0, h_0) \le \kappa (h_0, h_0) \Longrightarrow (Le^{-tL}h_0, e^{-tL}h_0) \le \kappa (e^{-tL}h_0, e^{-tL}h_0), \quad \text{ for }\,\forall h_0 \in D(L)$$
where $(\,\,,\,)$ is an inner product on $H$.
I'd appreciate it if you'd give me any advice.

If the operator $L$ generates a contraction semigroup on $H$, then
$$(L h_0, h_0) \le 0 \le \kappa \|h_0\|^2, \quad \forall h_0 \in D(L),$$
where $\kappa$ is any nonnegative real number. It suffices to apply the above inequality for $h_0'=e^{-tL}h_0$ where $h_0 \in D(L)$.
