Bounded operator Hardy space Let $T_f g = f \cdot g$ where $f, g, f \cdot g$ are in $H^2(\mathbb{D})$ (where $H^2$ is the Hardy space on the open unit disk). Now $T_f$ is a bounded operator.
I want to show this by showing that $f \in H^\infty$. So I try to write $f = G_1 h_1$ and $g = G_2 h_2$ where $G_i$ are outer functions and $h_i$ inner functions. So, what I need to do is if $G_1 G_2$ is in $H^2$ for all $G_2$ outer, then $G_1$ is in $H^\infty$. Does someone have a hint how I could obtain this?
 A: This isn't the same approach you had in mind, but you can show that $T_f$ is bounded using the closed graph theorem and the fact that evaluation at a point in the open disk is bounded on $H^2$.  You can then show that $f$ is in $H^\infty$ by showing that the complex conjugates of elements of its image on the disk are eigenvalues for the adjoint of $T_f$.
Here is an elaboration on the last sentence.  For each $w\in\mathbb{D}$, define $k_w:\mathbb{D}\to\mathbb{C}$ by $k_w(z)=\frac{1}{1-\overline{w}z}=\sum_{k=0}^\infty \overline{w}^k z^k$.  Each $k_w$ is in $H^\infty$ and thus in $H^2$.  Using the second expression for $k_w$ and the characterization of the inner product on $H^2$ in terms of the $\ell^2$ sequences of Maclaurin coefficients, notice that $\langle g,k_w\rangle=g(w)$ for all $w\in\mathbb{D}$ and all $g\in H^2$.  It then follows that for all $w$ and $z$ in $\mathbb{D}$, 
$$(T_f^*k_w)(z)=\langle T_f^* k_w,k_z\rangle=\overline{\langle T_f k_z,k_w\rangle}=\overline{f(w)k_z(w)}$$ $$=\overline{f(w)}\overline{\langle k_z,k_w\rangle}=\overline{f(w)}\langle k_w,k_z\rangle=\overline{f(w)}k_w(z).$$  Since $z$ was arbitrary, this shows that $T_f^*k_w=\overline{f(w)}k_w$, so $\overline{f(w)}$ is an eigenvalue for $T_f^*$ with eigenvector $k_w$.  Thus, $\|f\|_\infty\leq \|T_f^*\|=\|T_f\|<\infty$.
This is a standard fact about reproducing kernel Hilbert spaces, and only the particular form of the function $k_w$ is special to the Hardy space. The way I have presented this, it might seem that $k_w$ was summoned by magic, but in fact one could rediscover them without too much work.  The important point is that there exist elements of $H^2$ whose corresponding inner product functionals are point evaluations.  These exist by Riesz's lemma using continuity of the point evaluations, which can be shown by other means.  You don't need to know what these elements are for the argument to carry through.  However, if you did want to discover them, then "working backwards" and considering Maclaurin series would lead you to the second expression for $k_w$ given above.
