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I would like to know whether there exists an analytic function $f$ on the unit disk such that $$\sup_{|z|<1}|f(z)|(1-|z|^2)<\infty$$ and for every $|a|=1$, $$\limsup_{z\to a}|f(z)(1-z^2)|=\infty.$$

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Yes. Let $f$ be univalent, with image in the open right half-plane. Then $f^2$ is also univalent. Applying the standard growth estimate for univalent functions to $f^2$, we conclude that $f$ satisfies your growth condition.

To arrange that $f$ satisfies the second condition, notice that we only need $\limsup|f(z)|=\infty$ at every point of the unit circle (Then your condition easily follows. For the point $1$, you first take a sequence of boundary points tending to $1$, then a sequence tending to each point of this sequence where $f\to\infty$, then a diagonal sequence.).

To ensure that $\limsup|f(z)|=\infty$ at every point, one can use geometric construction of the image. Begin with some disc in the right half-plane. Then dig narrow channels from the circumference to infinity, and continue digging channels from dense sets of boundary points... Or look at the pictures in Collingwood Lohwater, The theory of cluster sets. Or look at the pictures of unbounded normality regions in transcendental dynamics, for example,; according to a theorem of Baker, all these regions have the property that infinity belongs to the impression of every prime end, which exactly means that your mapping function has the required property.

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To illustrate Alex's answer, the following old picture of mine shows a domain of the type described: Let f be the Riemann map from the unit circle to the blue domain at the bottom left of the picture. Then this map has the desired property. The union of the two white lines attached to the domain is actually precisely the cluster set of the function f at a certain point on the unit circle. – Lasse Rempe-Gillen Nov 23 '12 at 14:50
Exciting picture! Which function was used to generate it? – Alexandre Eremenko Nov 23 '12 at 14:53
It's an exponential map $z\mapsto e^z+a$, where the parameter $a$ is chosen to have an attracting periodic orbit of period $4$. – Lasse Rempe-Gillen Nov 23 '12 at 14:59
Alex - I think the theorem in question is due to Baker and Weinreich, and assumes that the domain is periodic but not a Baker domain? – Lasse Rempe-Gillen Nov 23 '12 at 15:09

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