Boundary function of product of $H^\infty$ and $H^2$ function

Question

If $f\in H^2$ and $\phi \in H^\infty$ is it true that $\widetilde{\phi f} = \tilde{\phi}\tilde{f}$? It is is easy to see that $ \widetilde{z^nf} = \tilde{z^n}\tilde{f} $ and so for all polynomials $p$ it is true $ \widetilde{pf} = \tilde{p}\tilde{f} $ How can this be carried forward to $H^\infty$ functions if at all?

Answer 1

Boundary function $\tilde f$ of an $H^p$ function $f$ is (almost everywhere) just the ordinary limit of $f$ in radial direction (normal to the boundary, for general smooth domains). Since the limit of a product is the product of limits, the claim follows.

(Remark: the product of an $H^p$ function and an $H^\infty$ function is in $H^p$, since multiplication by a bounded function contributes at most a constant factor to the integrals hat determine the $H^p$ norm.)