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Suppose we have the space $L^p(R,R^n)$ where $1 \leq p < \infty$ (i.e the space of functions that take values in $R^n$ and are $L^p$ integrable) and suppose $T_m: L^p(R,R^n) \to L^p(R,R^n) $ is a linear operator given by $T_mf(x)=m(x)f(x)$ where $m(x)$ is a an $n \times n$ matrix with coefficients that are functions of $x$. It is claimed that $T_m$ is bounded if and only if $m$ is essentially bounded and $||T_m||=||m||_\infty$. My question is what does it mean for $m$ to be essentially bounded and what is $||m||_\infty$.

I understand this in one dimension, i.e $m$ is essentially bounded means that the essential supremum of $m$ which is the infimum of the set $ \{M: m(x) \leq M \mbox{ almost everywhere } \}$ and the $L_\infty$ norm is precisely the infimum of this set.

What would the $L_\infty$ norm be in the matrix valued case?

Thank you.

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Let $\cal M_n$ the space of $n\times n$ matrices with, say, real entries. As it's a finite dimensional vector space, all the norms on it will be equivalent. If $m(x)$ is a matrix for all $x$, we can define $\lVert m\rVert=N(\widetilde m)$, where $\widetilde m$ is the matrix whose entries are $\widetilde m_{ij}=\lVert m_{ij}\rVert_{\infty}$.

Boundedness of $T_m$ won't depend on the choice of $N$.

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    Yes, that's what I mean.2012-11-07