Here $H^\infty:=\left\{f\in L^\infty(\Bbb{T}):\ \frac{1}{2\pi}\int\limits_0^{2\pi} f\chi_{-n}\ dt=0\ \forall n<0\right\}$ i.e. all those $L^\infty$ functions whose negative fourier coefficients are $0$.
I know that $H^\infty+C(\Bbb{T})$ is closed subalgebra of $L^\infty(\Bbb{T})$. I claim that $H^\infty+C(\Bbb{T})$ is not $\ast$ closed. Let's take $\psi(z)=\exp\left(\frac{z+1}{z-1}\right)$, this is an inner function in $H^\infty$ which is not continuous. I claim that $\overline{\psi(z)}=\exp\left(\frac{1+z}{1-z}\right)$ is not in $H^\infty+C(\Bbb{T})$. I can see that $\overline{\psi}$ is not in $H^\infty$ nor $C(\Bbb{T})$. But I'm unable to show that it cannot be written as a sum of such functions.
Another approach I thought is- A function $\phi\in H^\infty+C(\Bbb{T})$ is invertible in $H^\infty+C(\Bbb{T})$ if and only if there exist $\epsilon,\delta>0$ such that $|\hat{\phi}(re^{it})|\ge \epsilon\ \forall r\in (1-\delta,1)\ \forall t$ (Here $\hat{\phi}$ denotes the Gelfand transform)
If $H^\infty+C(\Bbb{T})$ is a $C^*$-subalgebra of $L^\infty$, then for any $\phi\in H^\infty+C(\Bbb{T})$, $\sigma_{H^\infty+C(\Bbb{T})}(\phi)=\sigma_{L^\infty}(\phi)$. I want to find a $\phi\in H^\infty+C(\Bbb{T})$ such that $\phi$ is invertible in $L^\infty$ but not in $H^\infty+C(\Bbb{T})$ i.e. the above sufficient condition is not satisfied.
Can anyone help with any idea to prove the statement? Thanks for your help in advance.