I'm trying to show that if $f$ and $1/f$ are in $H^2$ then $f$ is an outer function. I think I somewhat have a proof (if it is correct) but it doesn't feel like it is the easiest one.
We know that if $f$ is in $H^2$ and non-zero and $B$ is the Blanschke product of its zeros, then there exists a constant $K$ of modulus $1$ and a singular measure $\mu$ on $S^1$ such that $f(z)$ has the form:
$$f(z) = K B(z) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$
where the first exponential is the singular inner function part. So by uniqueness of this form (up to constants I believe?) and the fact that $1/f$ is in $H^2$ we can write: $$\frac{1}{f(z)} = \frac{1}{KB(z)} \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \, d\mu(\theta) \right ) \text{exp} \left (-\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$
So, we see that $B(z) = 1$. Further we can see that the singular inner function-part has modulus $>1$ a.e. in the open unit disk if $\mu \neq 0$, and so the limit $r \to 1^-$ is not $1$ because of the maximum modulus principle (maybe this is flawed since $f$ is not analytic on the boundary). So we are left with:
$$f(z) = K \text{exp} \left (\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i\theta} + z}{e^{i\theta} - z} \log |\tilde{f}(e^{i \theta}) | \, d\theta \right )$$
Which is an outer function. Are there any flaws in this? If not, is there a less "thick" way to show this (I mean, not invoke all the theorems about the form these functions must have and so on) since I'm not really satisfied with this?
Edit: Oh, a nice corollary of this result I found is that if $f$ in $H^\infty$ and $\|f\|_\infty < 1$, then $1 + f$ is outer. (By noting that if $f$ is in $H^\infty$, then $f$ in $H^2$.)