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Let $(X, \mu)$ be a finite measure space and let $A$ be a non-negative self-adjoint operator which generates a contraction semigroup $e^{tA}$ on $L^2(X, \mu)$. If additionally, we have that $e^{tA}$ is bounded on $L^\infty(X, \mu)$, is it true that $e^{tA}$ is bounded on $L^p(X, \mu)$, where $1 \leq p \leq \infty$?

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  • $\begingroup$ Did you mean to write $e^{-tA}$? $\endgroup$ – Nate Eldredge May 18 '15 at 21:03
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By interpolation $e^{tA}$ will be bounded on $L^p(X,\mu)$ for $2 \leq p \leq \infty$. Then the dual of $e^{tA}$ will be bounded on $L^p(X,\mu)$ for $1 < p \leq 2$; but since $A$ is self-adjoint this is just $e^{tA}$ itself. (I think there are no problems with this argument but I'm not 100% sure)

I'm not sure if you can say anything about boundedness on $L^1$. However, I haven't used the assumptions of finite measure or contractivity, so maybe these can be used.

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