Hardy Hilbert space: Multiplication by $H^\infty$ function If $H^2(\mathbb{D})$ is space of all analytic functions whose series of taylor coefficient is absolute square summable and $\phi$ is analytic in $\mathbb{D}$,then $\phi H^2 \subset H^2$  if and only if $\phi$ is in $H^\infty$.
Not able to solve it through elementary arguments. can't think of what theorem of functional analysis can i invoke. any hints ?
 A: Assume first that $\phi\in H^\infty$. Then one can prove that $\phi H^2\subseteq H^2$ by using the fact that if $f$ is a holomorphic function on $\mathbb D$, then 
$$\Vert f\Vert_{H^2}^2=\sup_{0\leq r<1}\int_0^{2\pi} \vert f(re^{it})\vert^2\,\frac{dt}{2\pi}\, \cdot$$
(So $f$ is in $H^2$ if and only if the right-hand side is finite.) Indeed, from this formula one can see at once that $\phi f\in H^2$ if $f\in H^2$, with $\Vert \phi f\Vert_{H^2}\leq \Vert\phi\Vert_\infty\,\Vert f\Vert_{H^2}$.
Conversely, assume that $\phi H^2\subseteq H^2$. Then, by the closed graph theorem, the operator $f\mapsto \phi f$ is bounded on $H^2$. Let us call denote this operator by $M_\phi$. 
Now, for any $\lambda\in\mathbb D$, let us denote by $k_\lambda\in H^2$ the so-called reproducing kernel at $\lambda$. By definition, $k_\lambda$ is the unique function in $H^2$ such that 
$$ \langle f,k_{\lambda}\rangle_{H^2}=f(\lambda)\qquad\hbox{for every $f\in H^2$}\, .$$
That such a function does exist follows from the Riesz representation theorem for linear functionals on Hilbert spaces, because the map $f\mapsto f(\lambda)$ is a continuous linear functional on $H^2$. (In fact $k_\lambda(z)=\frac{1}{1-\bar\lambda z}$; this follows from Cauchy's formula.)
By the very definition of $k_\lambda$, it follows rather easily that $k_\lambda$ is an eigenvector of the adjoint operator $M_\phi^*$; more precisely :
$$M_\phi^*k_\lambda=\phi(\lambda)\, k_\lambda\, .$$
To see this, one just has to check that $\langle M_\phi^*k_\lambda, f\rangle_{H^2}=\phi(\lambda)\,\langle k_\lambda ,f\rangle$ for every $f\in H^2$, which does follow immediately from the definition of $k_\lambda$.
From the above identity (and since $k_\lambda\neq 0$), we see that $\Vert M_\phi^*\Vert\geq \vert\phi (\lambda)\vert$ for every $\lambda\in\mathbb D$. So $\phi$ is bounded i.e. $\phi\in H^\infty$, with $\Vert\phi\Vert_\infty \leq \Vert M_\phi^*\Vert=\Vert M_\phi\Vert$.
(Altogether, we have in fact $\Vert \phi\Vert_\infty=\Vert M_\phi\Vert$ by the first part of the proof.)
