Adjoint of the operator $(Mf)(x, \omega)=e^{3x/2}f(e^x\omega)$

Question

Let $r>1$ and $R=\ln{r}$ and $M: L^{2}(B(0, r)\setminus B(0, 1))\to L^{2}((0,R)\times S^{2})$ be the operator $(Mf)(x, \omega)=e^{3x/2}f(e^x\omega)$. What is its adjoint?

My attempt: Let $f\in L^{2}(B(0, r)\setminus B(0, 1))$ and $g\in L^{2}((0,R)\times S^{2})$, $$(Mf, g)_{L^{2}((0,R)\times S^{2})}=\int_{(0, R)\times S^2}e^{3x/2}f(e^x\omega)g(x, \omega) dx d\omega.$$

By substituting $u=e^{x}$ we get

$$\int_{(1,r)\times S^2} u^{1/2}f(u\omega)g(u, \omega) du d\omega.$$

Let $v=u\omega$, then

$$(Mf, g)_{L^{2}((0,R)\times S^{2}}=\int_{B(0, r)\setminus B(0, 1)} f(v) |v|^{1/2}g(v)=(f, M^{*}g)_{L^{2}(B(0, r)\setminus B(0, 1))}.$$

So that $(M^{*}g)(v)=|v|^{1/2}g(v)$.

Is this correct?

Answer 1

There are several mistakes. First, after substituting $u = e^x$, one should get

$$ \int_{(1,r)\times S^2} u^{1/2}f(u\omega)g( \ln u, \omega) du d\omega.$$

Next, when you change to polar coordinate, remember that $$ dV:= dx^1 dx^2 dx^3 = r^2 dr d\omega.$$ Since $|\omega|=1$, indeed $|v| = u=r$, thus

$$ \int_{(1,r)\times S^2} u^{1/2}f(u\omega)g( \ln u, \omega) du d\omega = \int_{B(0,r)\setminus B(0,1)} f(v) G(v) dV, $$ where

$$ G(v) = r^{-3/2} g (\ln u, \omega)= |v|^{-3/2} g(\ln |v|, v/|v|). $$

Thus $M^*$ is given by $(M^* g)(v) = |v|^{-3/2} g(\ln |v|, v/|v|)$.