Assume that $$\left<f,g\right>\_R=\Re \int\_U f(z) \overline{g(z)} \, dx \, dy, f,g \in L^2(U),$$ where $U$ is the unit disk and assume that $A: L^2(U)\to L^2(U)$ is a real-linear operator. Assume also that $A^\*$ is its adoint, with respect to $\left<f,g\right>\_R$, that is $\left<Af,g\right>\_R= \left<f,A^\*g\right>\_R$. My question is, whether in that case we have $\|A\|\_{L^p\to L^p} = \|A^\*\|\_{L^q\to L^q}$, where $1/p+1/q=1$.



Adjoint operator
Yes, this is true, in the sense that if one side is finite then so is the other and they are equal. All you really need is to notice that $\|g\|\_q = \sup \{{\rm Re}\int fg: \|f\|\_p = 1\}$ (since you can multiply any $f$ by a scalar of modulus 1).


I'll show that $\|A^\*\|\_{L^q \to L^q} \leq \|A\|\_{L^p\to L^p}$; the reverse inequality follows by symmetry. Choose $f$ and $g$ with $\|f\|\_p = \|\bar{f}\|\_p= \|g\|\_q = \|\bar{g}\|\_q =1$ and $$\langle Af, g\rangle = \langle f, A^\*g\rangle = {\rm Re}\left(\int f\cdot \overline{A^\*g}\right) \geq \|A^\*\|\_{L^q\to L^q} - \epsilon.$$ (By truncating, we can assume $f$ and $g$ both lie in $L^2(U)$.) That is, $\int Af\cdot\bar{g} \geq \|A^\*\|\_{L^q\to L^q} - \epsilon$, which shows that $\|A\|\_{L^p\to L^p} \geq \|A^\*\|\_{L^q\to L^q} - \epsilon$.


