In the text by Douglas, Banach Algebra techniques in Operator Theory, pg. 25, 1.47 Hardy spaces was introduced.
If $\mathbb {T}$ denotes the unit circle in the complex plane, let $\chi_n $ denote the function defined on $\mathbb {T}$ by $\chi_n (z) = z^n$. The set $$H^{\infty}=\{\psi \in L^{\infty}(\mathbb {T}): \frac {1}{2\pi} \int_0^{2\pi} \psi \chi_n dt=0\text { for n=1,2,3,..., } \} $$ is a closed subspace of $L^{\infty}(\mathbb {T})$. Moreover in this case, $H^{\infty}=\{\psi \in L^{\infty}(\mathbb {T}): \frac {1}{2\pi} \int_0^{2\pi} \psi \chi_n dt=0\text { for n=1,2,3,..., } \} $ is the null space of the $w^*$-continuous function $$\hat\chi_n (\psi)=\frac {1}{2\pi} \int_0^{2\pi} \psi \chi_n dt $$ and hence is $w^*$-closed.
My questions are:
I am very new to $w^*$-continuity, how can one show that the function $\hat\chi_n$ is $w^*$-continuous.
What is the relevance of the last statement. Can one directly use the last argument to conclude that the set $H^\infty$ is a closed subspace of $L^\infty$?. That is does $w^*$-closeness imply closeness.
Let $X$ be a Banach space and denote by $X'$ its (topological) dual. For $x\in X$ define the linear functional $F_x : X'\to\mathbb K$ by $F_x(x') := x'(x)$. The weak star topology on $X'$ is the smallest topology on $X'$ such each $F_x$ is continuous.
Here, you have $X = L^1(\mathbb T)$ and hence $X' = L^\infty(\mathbb T)$ in the sense that the map $L^\infty\to X'$, defined by $y\mapsto\left(x\in L^1\mapsto\int_{\mathbb T}xy\,dt\right)$ is an isometric isomorphism. Hence, for $x\in L^1$ you have $F_x(y) = \int xy\,dt$, $y\in L^\infty$. If you put $x_n = e^{int}$, your functional $\hat\chi_n$ exactly equals $F_{x_n}$ and is thus weak star-continuous.