Radial limits of bounded outer functions Let $f$ be  a non-invertible bounded outer function on the unit disk. Does $f$ has radial limit $0$ somewhere? Note that  such a property holds for singular inner functions.
 A: The answer is negative: $f$ may have a non-zero radial limit everywhere.

Part 1. Let us start with definitions and notation. A holomorphic function $f$ in the unit disk $D$ is an outer function if and only if
$$ -\log |f(r e^{i t})| = \int_0^{2\pi} \phi(s) K_r(t - s) ds $$
for some integrable function $\phi$ on $[0, 2\pi)$, where $K_r(t)$ is the Poisson kernel:
$$ K_r(t) = \frac{1}{2 \pi} \frac{1 - r^2}{1 + r^2 - 2 r \cos t} \, . $$
Furthermore, every integrable $\phi : [0, 2\pi) \to \mathbb R$ corresponds in the above sense to an outer function (a unique one up to multiplication by a constant with modulus $1$). For convenience, let us extend $\phi$ to all of $\mathbb R$ in such a way that $\phi$ is $2\pi$-periodic.
If $\phi$ is continuous at $t$ (or, more generally, if $t$ is a Lebesgue point of $\phi$) and $f$ corresponds to $\phi$, then
$$ \lim_{r \to 1^-} (-\log |f(r e^{it})|) = \phi(t) . $$
Due to symmetry of the Poisson kernel, we also have
$$ \lim_{r \to 1^-} (-\log |f(r e^{it})|) = \frac{\phi(t^+) + \phi(t^-)}{2} $$
whenever $\phi$ has one-sided limits $\phi(t^+)$ and $\phi(t^-)$ at $t$.
(To clarify the notation: $e^{-\phi}$ is the radial boundary limit of $|f|$ almost everywhere. Thus, $\phi \geqslant 0$ a.e. if and only if $|f|$ is bounded by $1$; more generally, $\phi \geqslant -\log C$ if and only if $|f|$ is bounded by $C$. For $p > 1$, $f$ is in the complex Hardy space $H^p$ if and only if $e^{-\phi}$ is in $L^p$.)

Part 2. We are now ready to construct a counter-example. Let
$$ \phi(t) = \begin{cases} 2^n & \text{if } t \in [\tfrac{1}{2^n}, \tfrac{1}{2^n} + \tfrac{1}{8^n}] \text{ for some } n = 1, 2, \ldots , \\ 0 & \text{otherwise.} \end{cases} $$
Let $f$ be an outer function corresponding to $f$. We claim that $f$ is a counter-example: its radial boundary limit is everywhere non-zero, and $1/f$ is not a bounded outer function.
Indeed, the limit of $-\log |f(r e^{it})|$ is equal to a finite number $\tfrac12(\phi(t^+) + \phi(t^-))$ whenever $t \in (0, 2 \pi)$, and so the radial boundary limit of $|f|$ at $e^{i t}$ is strictly positive (the radial limit of $f$ may fail to exist, though; see Part 3 below for more details). Thus, it remains to see what happens for $t = 0$. Clearly,
$$ -\log |f(r)| = \int_0^{2\pi} \phi(s) K_r(-s) ds = \int_0^{2\pi} \phi(s) K_r(s) ds . $$
Note that if $r \geqslant \tfrac14$ and $s \in (0, \pi)$, we have
$$ \begin{aligned} 0 \leqslant K_r(s) & = \frac{1}{2\pi} \, \frac{(1 + r) (1 - r)}{(1 - r)^2 + 4 r \sin^2 (s/2)} \\ & \leqslant \frac{(1 + r) (1 - r)}{2 \times (1 - r) \times 2 \sqrt{r} \sin(s/2)} \\ & = \frac{1 + r}{4 \sqrt{r} \sin(s/2)} \leqslant \frac{1}{\sin(s/2)} \leqslant \frac{\pi}{s} \, . \end{aligned} $$
Furthermore,
$$ \begin{aligned} \int_0^{2\pi} \phi(s) \, \frac{\pi}{s} \, ds & = \sum_{n = 1}^\infty \int_{1/2^n}^{1/2^n+1/8^n} 2^n \times \frac{\pi}{s} \, ds \\ & \leqslant \sum_{n = 1}^\infty \int_{1/2^n}^{1/2^n+1/8^n} 2^n \times 2^n \pi \, ds \\ & = \sum_{n = 1}^\infty \frac{1}{8^n} \times 4^n \pi = \frac{\pi}{3} < \infty . \end{aligned} $$
Thus, by the dominated convergence theorem,
$$ \begin{aligned} \lim_{r \to 1^-} (-\log |f(r)|) & = \lim_{r \to 1^-} \int_0^{2\pi} \phi(s) K_r(s) ds \\ & = \int_0^{2\pi} \phi(s) \lim_{r \to 1^-} K_r(s) ds = 0 . \end{aligned} $$
In other words, the radial boundary limit of $|f|$ at $0$ is $1$.
We have thus proved that at no boundary point the radial boundary limit of $f$ is equal to zero. On the other hand, the radial boundary limit of $-\log |1/f| = \log |f|$ at $e^{it}$ is equal to $\phi(t)$ for almost all $t$, and since $\phi$ is essentially unbounded, $1/f$ is not bounded in the unit disk. To be specific: the radial boundary limit of $|1/f|$ at $t = \tfrac{1}{2^n} + \tfrac{1}{2} \tfrac{1}{8^n}$ is equal to $e^{\phi(t)} = e^{2^n}$. In other words, $f$ is not invertible in the class of bounded outer functions.

Part 3. The above $f$ fails to have a radial boundary limits at $t = \tfrac{1}{2^n}$ and $t = \tfrac{1}{2^n} + \tfrac{1}{8^n}$. If we require $f$ to have a radial boundary limit everywhere, then we need to modify $\phi$ appropriately: we choose an auxiliary smooth "bump" function $\psi$ equal to $0$ outside of $(0, 1)$, equal to $1$ in $[\tfrac13, \tfrac23]$, and taking values in $[0, 1]$, and we set
$$ \phi(t) = \begin{cases} 2^n \psi(8^n (t - \tfrac{1}{2^n})) & \text{if } t \in [\tfrac{1}{2^n}, \tfrac{1}{2^n} + \tfrac{1}{8^n}] \text{ for some } n = 1, 2, \ldots , \\ 0 & \text{otherwise.} \end{cases} $$
Now $\phi$ is smooth except at $t = 0$, and hence $f$ is continuous in $\overline{D} \setminus \{1\}$. The same calculation as in Part 2 shows that $-\log |f| = -\Re \log f$ again has a radial boundary limit $0$ at $1$. A similar calculation with the conjugate Poisson kernel shows that also $-\Im \log f$ has a radial boundary limit at $1$, and so $f$ has a (non-zero) radial boundary limit at $1$.
A: It is worth to mention that there exists a special class of outer functions where the answer is positive: let $u$ be a non-constant inner function and $|\alpha|=1$. Then $f:=u-\alpha$ is an outer function which has radial limit $0$ somewhere. This is a direct consequence of a  theorem of Hoessjer-Frostman (see Noshiro's book, "Cluster sets" Theorem 6).
