Adjoint of the following operator:$$B: L_2[0,1]\to L_2[0,1]; \\ B(x(t))=e^tx(\sqrt t)$$
This is what I have tried: Using the knowledge that
$$<Bx,y^*>=<x,B^*y^*>$$
I have that : $$<Bx,y^*> = \int_{0}^{1}(Bx)(t)y^*(t)dt=\int_{0}^{1}e^tx(\sqrt t)y^*(t)dt=\int_{0}^{1}x(\sqrt t)e^ty^*(t)dt$$ and also: $$<x,B^*y^*> = \int_{0}^{1}x(t)(B^*y^*)(t)dt$$
This square root is messing this up, and I do not know how to finish. If in both integrals it was $x(t)$ this would be a peice of cake. But looking at this as it is, I don't know what to do!