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Dec 16, 2023 at 16:30 comment added Jochen Glueck Let $\mathcal{L}$ denote the generator of the heat semigroup on $L^1(\mathbb{R})$. The spectrum of $\mathcal{L}^2$ is $[0,\infty)$, so $\mathcal{L}^2$ does not generate a $C_0$-semigroup.
Dec 16, 2023 at 14:06 comment added matilda @JochenGlueck- Could you give me a reference for the fact, or perhaps an example? I apologize if this question is too elementary; I don't work in this field.
Dec 16, 2023 at 11:04 comment added Jochen Glueck I'm having difficulties to understand the question. In general, $\mathcal{L}^2$ does not generate a semigroup, as can be easily seen be considering the spectrum. Apart from this, could you please explain what you mean by "contractive to a higher order"?
Dec 16, 2023 at 2:32 history asked matilda CC BY-SA 4.0