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Michael Hardy
Michael Hardy
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Assume that $$\left<f,g\right>_R=\Re \int_U f(z) \overline{g(z)}dxdy, f,g \in L^2(U),$$$$\left<f,g\right>_R=\Re \int_U f(z) \overline{g(z)} \, dx \, dy, f,g \in L^2(U),$$ 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}$$\|A\|_{L^p\to L^p} = \|A^*\|_{L^q\to L^q}$, where $1/p+1/q=1$.

Assume that $$\left<f,g\right>_R=\Re \int_U f(z) \overline{g(z)}dxdy, f,g \in L^2(U),$$ 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 $1/p+1/q=1$.

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

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Lira
Lira
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Adjoint operator

Assume that $$\left<f,g\right>_R=\Re \int_U f(z) \overline{g(z)}dxdy, f,g \in L^2(U),$$ 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 $1/p+1/q=1$.