an isomorphism from $L^\infty(\mathbb{T})$ to $L^\infty([-1,1],\frac{2}{\pi } \sqrt{1-t^2}\mathrm{d}t)$ $\mathbb{T}$ is the boundary of unit ball.Consier $\phi:[-1,1]\rightarrow\mathbb{T},\phi(t)=e^{2i(\arcsin t+t\sqrt{1-t^2})},t\in[-1,1]$. It is easy to check that $L^2(\mathbb{T})\ni f\mapsto f\circ\phi\in L^2([-1,1],\displaystyle\frac{2}{\pi } \sqrt{1-t^2}\mathrm{d}t)$ is a unitary which implements an isomorphism from $L^\infty(\mathbb{T})$ to $L^\infty([-1,1],\displaystyle\frac{2}{\pi } \sqrt{1-t^2}\mathrm{d}t)$ (Why?). Is there anyone who would like to answer? Thanks.
 A: So take $f\in L^2(\mathbb T)$. To check that $\phi$ implements an isometry,
\begin{align}
\|f\circ\phi\|_2^2&=\frac2\pi\int_{-1}^1 |f(\phi(t))|^2\,\sqrt{1-t^2}\,dt
=\frac2\pi\int_{-1}^1 |f(e^{2i(\arcsin t+t\sqrt{1-t^2})})|^2\,\sqrt{1-t^2}\,dt\\ \ \\ &\ \ \ \ \ \ \textit{(setting $u={2arcsint+2t\sqrt{1-t^2}}$, $du=4\sqrt{1-t^2}\,dt$)}\\ \ \\
&=\frac1{2\pi}\,\int_{-\pi}^{\pi}|f(e^{iu})|^2\,du=\|f\|_2^2.
\end{align}
As $\phi$ is bijective, $f\longmapsto f\circ\phi$ is onto and so it is a unitary (surjective isometry). 
As for the isomorphism at the $L^\infty$ level, one can either consider $f\longmapsto f\circ\phi$ directly (since $L^\infty(\mathbb T)\subset L^2(\mathbb T)$), or consider $L^\infty(\mathbb T)$ as the multiplication operators acting on $L^2(\mathbb T)$ (and similarly for $[-1,1]$) and the implement the isomorphism by conjugation with the above unitary. The two constructions yield the same isomorphism $L^\infty(\mathbb T)\simeq L^\infty([-1,1],\displaystyle\frac{2}{\pi } \sqrt{1-t^2}\mathrm{d}t)$.
