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Oct 5, 2020 at 11:15 comment added Francesco Nobili Okay, I was making first sure to identify your question. I did not report a proof in the answer, but links to read this fact. It is all in the references I gave, e.g. (a)-(b) in Theorem 4.16 of arxiv.org/abs/1106.2090.
Oct 5, 2020 at 11:12 comment added MaoWao Of course both of these facts can be proven, but the question was about how to prove this (or, more precisely, give a reference for these facts).
Oct 5, 2020 at 11:11 comment added Francesco Nobili For $p=\infty$, there is a 'weak maximum principle' to be shown, namely $$ f \le C, \quad a.e. \quad \Rightarrow \quad h_t(f) \le C, \quad a.e.,\forall t>0.$$ Again, this can be shown to be true to cover the case of $L^\infty$
Oct 5, 2020 at 10:50 comment added Francesco Nobili One can prove that, for $u \colon \mathbb{R}\rightarrow [0,\infty]$ convex and l.s.c. with $u(0)=0$, the mapping $t\mapsto \int u(h_t(f))\, dVol_g$ is nonincreasing. Apply this to $u(\cdot)=|\cdot|^p$, for any $p \in [1,\infty)$ to get $$\|h_t(f)\|_{L^p}\le \|f\|_{L^p}, \qquad \forall f \in L^2\cap L^p, t\ge 0.$$ The case $p=\infty$ is to be treated differently. Is this what you asked?
Oct 5, 2020 at 10:29 comment added MaoWao I fail to see how this proves contractivity on $L^p$ for $p\neq 2$. For that you should use the "Markovianity" of the Dirichlet energy as indicated in Mateusz's answer.
Oct 5, 2020 at 10:15 history edited Francesco Nobili CC BY-SA 4.0
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Oct 2, 2020 at 17:09 history edited Francesco Nobili CC BY-SA 4.0
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Oct 2, 2020 at 14:14 history edited Francesco Nobili CC BY-SA 4.0
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Oct 2, 2020 at 13:50 history edited Francesco Nobili CC BY-SA 4.0
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Oct 2, 2020 at 13:43 history answered Francesco Nobili CC BY-SA 4.0