Checking whether an operator is unitary

Question

I am given an operator $B:L^{2}([0, 1]) \mapsto L^{2}([0, 1])$ which can be expressed as following $$Bf(x) = \sqrt{3}xf(x^3).$$

I would like to check whether $B$ is unitary. Thus I need to find $B^*$. That's my attempt \begin{align} \langle Bf, g \rangle &= \int \limits_{0}^{1} \sqrt{3}xf(x^3) g(x) \, dx = \\ &= \int \limits_{0}^{1} f(t) \frac{\sqrt{3}}{3} t^{-1/3}g(t^{1/3}) \, dt. \end{align}

This means that $BB^* \neq I$. But according to the answers it should be unitary. What have I done wrong?

Answer 1

$B^*g (t)= \frac1{\sqrt 3} t^{-1/3} g(t^{1/3}) $, so $$ BB^*g(x) = B(B^*g)(x) = \sqrt 3 x( B^*g)(x^3)= \sqrt 3 x \left( \frac1{\sqrt3}(x^3)^{-1/3}g((x^3)^{1/3})\right) = x \frac1x g(x) = g(x)$$

Answer 2

You need to verify unitary $\langle Bf, Bg \rangle=\langle f, g \rangle$ not self adjoint $\langle Bf, g \rangle=\langle f, Bg \rangle$ property.

$\displaystyle \langle Bf, Bg \rangle=\int_0^1 3x^2f(x^3)g(x^3)\mathop{\rm{d}\!}x=\int_0^1 f(t)g(t)\mathop{\rm{d}\!}t$

Since $t=x^3$ gives $\mathop{\rm{d}\!}t=3x^2\mathop{\rm{d}\!}x$.