Does the Riesz-Thorin interpolation theorem hold for real-linear operators?
The standard proof uses complex linearity, but maybe another proof avoids the assumption of complex-linearity?
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$\begingroup$ What is the problem. You can complexify a real-linear operator. Perhaps the non-obvious (though true) fact is that the complexification does not change the $L^p$-norm. $\endgroup$– Denis SerreCommented Jul 15, 2020 at 15:15
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1$\begingroup$ The norm of the complexification can change if the operator maps $L^p$ to $L^q$ with $q<p$. When such a situation occurs one needs a constant (2 suffices) in the real Riesz-Thorin, I guess. $\endgroup$– Giorgio MetafuneCommented Jul 15, 2020 at 15:39
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$\begingroup$ @Denis Serre. Assume that $Tf(z)= \int_U (A(z,w) f(z)+B(z,w) \overline{f(z)})dudv, w=u+iv, z=x+i y$, $U$ is the unit disk. This operator is real-linear, but the function $f:U\to U$ need not be real. So what is the complexification of this operator. $A$ and $B$ are certain kernels? $\endgroup$– LiraCommented Jul 18, 2020 at 14:50
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$\begingroup$ @Lira. You should not focus on the fact that $f$ is a complex-valued function. Your operator is real-linear, so just write it in terms of a pair of unctions $(g,h)$ (real and imaginary parts). Then you may complexify :). $\endgroup$– Denis SerreCommented Jul 18, 2020 at 20:18
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