sub-Markovianity and extensions of $L^2$ semigroup contractions to $L^p$

Question

I just read that a contraction semigroup $(T_t)_{t\geq 0}$ on $L^2(X,m;\mathbb{R})$ space can be extended to a contraction semigroup on $L^p$ for any $p\geq 2$ provided that it satisfies the sub-Markovian property, i.e., $0\leq f\leq 1 \rightarrow 0\leq T_tf\leq 1$. I understand that this is easy if one can associate a transition function to $T_t$ but is it possible to show this if we do not know that we can do that? I would appreciate all suggestions!

Answer 1

By assumption, $(T_t)_{t \geq 0}$ is a contraction on $L^2(X,m)$. If we can show that $(T_t)_{t \geq 0}$ extends to a contraction on $L^{\infty}(X,m)$, then it follows from the Riesz-Thorin interpolation theorem that $(T_t)_{t \geq 0}$ extends to a contraction on $L^p(X,m)$ for any $p \geq 2$. The following proof assumes that $(X,m)$ is a $\sigma$-finite measure space.

First, we show that $T_t$ is positive, i.e. $$f \in L^2(X,m), f \geq 0 \implies T_t f \geq 0. \tag{1}$$ If $f \in L^2(X,m) \cap L^{\infty}(X,m)$, this follows directly from the sub-Markov property. For general $f \in L^2(X,m)$, $f \geq 0$, we set $f_n := f \wedge n$. Then $f_n \in L^2(X,m) \cap L^{\infty}(X,m)$ and $f_n \uparrow f$. Note that $T_t f_n$ is increasing in $n$. Moreover, as $$\|T_t f_n-T_t f\|_{L^2} \leq \|f_n-f\|_{L^2} \to 0$$ we can choose a subsequence $(T_t f_{n(k)})_k$ such that $T_t f_{n(k)} \to T_t f$ almost surely as $k \to \infty$. By the monotonicity, $T_t f_n \uparrow T_t f$ almost surely. In particular, $$T_t f = \lim_{n \to \infty} T_t f_n \geq 0$$ as $T_t f_n \geq 0$.

Note that $(1)$ implies

$$f,g \in L^2(X,m), f \leq g \implies T_t f \leq T_t g, \tag{2}$$

i.e. $T_t$ is monotone.

Now let $f \in L^{\infty}(X,m)$, $f \geq 0$. Choose a sequence $(f_n)_{n \in \mathbb{N}} \subseteq L^2(X,m) \cap L^{\infty}(X,m)$ such that $f_n \uparrow f$ (e.g. $f_n := 1_{A_n} f$ where $m(A_n) <\infty$ and $A_n \uparrow X$). By $(2)$, $T_t f_n$ is increasing (in $n$). Therefore, $$T_t f := \sup_{n \in \mathbb{N}} T_t f_n$$ exists and is well-defined. (Well-defined means that $T_t f$ does not depend on the approximating sequence $(f_n)_{n \in \mathbb{N}}$; this follows again from the monotonicity.) In particular, by the sub-Markov property,

$$|T_t f| = T_t f \leq \sup_{n \in \mathbb{N}} T_t f_n \leq \sup_{n \in \mathbb{N}} f_n = f,$$

i.e. $T_t$ is a contraction on $L^{\infty}_+(X,m)$. For general $f \in L^{\infty}(X,m)$, we extend $T_t$ by linearity $$T_t f := T_t(f^+)- T_t(f^-)$$ where $f^+$ ($f^-$) denotes the positive (negative) part of $f$. Then

$$|T_t f| \leq \max\{T_t(f^+),T_t(f^-)\} \leq \max\{f^+,f^-\} \leq |f|$$

as $T_t(f^+) \geq 0$ and $T_t(f^-) \geq 0$. This shows that the so-defined operator is an $L^{\infty}(X,m)$-contraction.