# Can an analytic function on the open unit disk blow up near the boundary? [duplicate]

Can we have a map $f$, holomorphic on the open unit disk, such that $|f(z)|\rightarrow \infty$ as $|z|\rightarrow 1$? I think not, (at least I can't think of any such map), but I'd like to be able to prove this.

## marked as duplicate by Daniel Fischer complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 26 '16 at 13:12

• Do you mean as $|z|\rightarrow 1$? 1/(z-1) blows up at one point, do you need it to blow up at all boundary points? – mathematician Nov 26 '16 at 8:20
Such a function cannot exist. Suppose such a function exists. Let $z_1,\ldots,z_k$ be the zeros of $f$ in the unit disc with corresponding multiplicites $m_1,\ldots,m_k$. Then $$g(z) = \frac{(z - z_1)^{m_1}\cdots(z - z_n)^{m_k}}{f(z)}$$ is an analytic function in the unit disc such that $\lim_{\vert z \vert \to 1}g(z) = 0$. This contradicts the maximum modulus principle.
• Note that we don't have $\lim\limits_{\lvert z\rvert \to 1} \lvert f(z)\rvert = +\infty$. That is not possible for holomorphic functions on the unit disk. Also, the denseness of dyadic rationals doesn't prove that the radial limit is $\infty$ at other points. – Daniel Fischer Nov 26 '16 at 11:26