Does there exist a function which is holomorphic on the open unit disc but goes to infinity on the boundary?

Question

Does there exist a function $f(z)$ holomorphic on the disc $|z| < 1$ but such that $\lim_{|z| \rightarrow 1} |f(z)| = \infty$?

I think the answer is no because $1/f(z)$ would go to $0$ as $|z| \to 1$.

Answer 1

Correct. Such a function would have finitely many zeros in the unit disk. Dividing by the product of $z - r_j$ where $r_j$ are the zeros (listed by multiplicity), you would get an analytic function $g$ on the disk which has no zeros there but still has $|g| \to \infty$ as $|z| \to 1-$. And then $1/g$ would violate the maximum modulus principle.

Answer 2

[Edit: The commenters kindly pointed out my earlier example didn't work.]

[Edit 2: Martin's right, this doesn't work because the function below must have zeros arbitrarily close to the boundary of the disk, so its modulus doesn't go to infinity.]

I hate to disagree with Robert, but if I understood the question correctly, this is an example of just such a function: $$ f(z)=\sum_{n=0}^\infty z^{n!} $$ This series converges absolutely on compact sets inside the unit disk $D$, and thus defines a holomorphic function $f : D \to \mathbb C$: If $K \subset D$ is a compact set, we can find a disk $D(0, r)$ centered at $0$ of radius $r < 1$ that contains $K$. On $K$, we then have $$ |f(z)| \leq \sum_{n=0}^{\infty} r^{n!} \leq \sum_{n=0}^{\infty} r^{n} = \frac{1}{1-r} < +\infty. $$ By exhausting the unit disk with disks $D(0,r)$ and defining $f$ on each one, we can define $f$ on the whole unit disk.

Now, if $\zeta$ is any unit root ($k$-th, say), we can consider what happens when we approach $\zeta$ along the line $t \mapsto t \zeta$, where $0 < t < 1$. The $n$-th term in the series is then simply $t^{n!}$ when $n \geq k$, so $$ f(t\zeta) = \text{polynomial of degree $k-1$ in $t$}\ + \sum_{n=k}^{\infty} t^{n!}, $$ which goes to infinity as $t \to 1$ (the series becomes $1 + 1 + 1 \cdots$).

Finally, the roots of unity are dense on the boundary of the unit disk, so $f$ is a holomorphic function on the unit disk that blows up as it approaches any point on its boundary.

Robert's argument showing that such a function must be constant doesn't work here: We can indeed divide out the zeros of $f$ to get a function $g$ (a bit tricky since there must be infinitely many of them, but it can be done) and consider $1/g$. The maximum modulus principle then implies that $g$ must be constant, but we can then only conclude that $f$ was equal to a constant multiple of what we divided out. The same actually happens for a function with finitely many zeros, but there we can conclude that our original function must have been a polynomial, which have no problems at the boundary of the unit disk.

In general, any domain (i.e., open set) in $\mathbb C$ is a domain of holomorphy, meaning that there are holomorphic functions on it that cannot be extended to any larger domain. We can construct functions like the above on any connected open set in $\mathbb C$ by using interpolation theorems.