Adjoint operator

Question

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 $1/p+1/q=1$.

Answer 1

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 $\|A\|_{L^p\to L^p} \geq \|A^*\|_{L^q\to L^q} - \epsilon$.