Does there exist a holomorphic function with the following property?

Does there exist a holomorphic function $f$ defined over $D = \{ z : |z| < 1 \}$ such that $|f| \rightarrow \infty$ when $|z| \rightarrow 1$?

My approach:

If such an $f$ exists, then for a given $\varepsilon > 0$, there exists an $M > 0$ such that $|f(z)|>M$ if $1-\varepsilon < |z| < 1$

Since $f$ has no zeros in that set, I can consider $g = 1/f$, I would like to apply Maximum Modulus Principle and get a contradiction, but actually I am not sure I can.

Solution 1:

Since $f$ has no zeros in that set, I can consider $g=1/f$, I would like to apply Maximum modulus principle and get a contradiction, but actually I am not sure I can.

You need a small modification. We assume such an $f$ existed. Since $\lim\limits_{\lvert z\rvert \to 1} \lvert f(z)\rvert = \infty$, there is an $\varepsilon > 0$ with $\lvert f(z)\rvert \geqslant 1$ for $\lvert z\rvert > 1-\varepsilon$. Then $f$ can have only finitely many zeros in the compact set $B = \{ z : \lvert z\rvert \leqslant 1 - \varepsilon\}$, so let $P$ be a polynomial with the same zeros as $f$ in $B$, including multiplicity, and no other zeros. Since polynomials are bounded on the closed unit disk,

$$h(z) = \frac{f(z)}{P(z)}$$

would also have the property $\lvert h(z)\rvert \to \infty$ as $\lvert z\rvert \to 1$, and $h$ would have no zeros. Then consider $1/h$ and apply the maximum modulus principle. Since $\lim\limits_{\lvert z\rvert \to 1} \frac{1}{h(z)} = 0$, it would follow that $1/h \equiv 0$, but that contradicts the $\mathbb{C}$-valuedness of $h$.