The Multiplier Algebra of the Hardy Hilbert Space Let $H$ be a Hilbert space of analytic functions.
Define the multiplier algebra of H in the following manner:
$M(H)=\{ \phi \in H  :  \phi h \in H, \forall h \in H \}$
It is mentioned in countless places that the multiplier algebra for
$H^2=\{f \text{ analytic on the disk}, f= \sum a_nz^n : \sum |a_n|^2 < \infty \}$
is $H^\infty$, the space of bounded analytic functions on the unit disk.
It is obvious that $H^\infty \subset M(H^2)$.
My question is how does one prove the other inclusion?
 A: Let us assume that you are working with a reproducing kernel Hilbert space (otherwise multiplier algebras don't make all that much sense, and $H^2$ is certainly a RKHS). This means that the space $H$ consists of functions on some space $X$ and point evaluations (i.e. $H \ni h \mapsto h(x)$) are continuous functionals for all $x \in X$.
If $\phi \in M(H)$, and we assume that the constant function $1 \in H$, then $\phi\cdot 1 \in H$, so at least $\phi$ is holomorphic (if the functions in $H$ are, which is the case for $H^2$).
Secondly, you can prove that each $\phi \in M(H)$ defines a bounded linear mapping $M_\phi: H \to H$ by $M_\phi(h) = \phi h$. Boundedness follows from the closed-graph theorem and continuity of point evaluations.
Then, take $x \in X$ and let $k_x \in H$ represent point evaluation at $x$, i.e.:
$$
h(x) = \langle h, k_x \rangle
$$
for all $h \in H$. (Such a $k_x$ exists by Riesz' representation theorem). 
Next, look at the adjoint $M_\phi^*$ acting on $k_x$. Let $h \in H$ be arbitrary. Then
$$
\langle h, M_\phi^* k_x \rangle =
\langle M_\phi h, k_x \rangle =
\langle \phi h, k_x \rangle = \phi(x)h(x) = 
\langle h, \overline{\phi(x)} k_x \rangle
$$
but this is true for all $h$, so $M_\phi^* k_x = \overline{\phi(x)} k_x$, i.e. $k_x$ is an eigenvector of $M_\phi^*$ with eigenvalue $\overline{\phi(x)}$.
In particular $|\phi(x)| \le \| M_\phi^* \|$ for all $x \in X$ which shows that $\phi$ must be bounded on $X$ (since $M_\phi$ is a bounded operator, and consequently so is $M_\phi^*$).
