By Fatou's theorem every bounded holomorphic function on the unit disk has non-tangential limits almost everywhere on the unit circle $\mathbb T$. Is there an explicit example of a bounded continuous function on the disk without non-tangential limits but with radial limits everywhere?
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3$\begingroup$ Wouldn't it suffice to have something oscillating faster and faster as one approaches the boundary, such as $f(re^{i\theta}) = \sin ( (1 - r)^{-1})$ ? I admit that I have not thought through how one might show this has no non-tangential limits, so this is merely a suggestion rather than an answer $\endgroup$– Yemon ChoiCommented Oct 18, 2019 at 20:33
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$\begingroup$ @YemonChoi: yes, that works. $\endgroup$– Nik WeaverCommented Oct 18, 2019 at 21:01
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$\begingroup$ @Yemon Choi Sure. Thanks. Actually I wanted to have some of these functions without non-tangential limits, but with radial limits everywhere. $\endgroup$– rayCommented Oct 19, 2019 at 11:21
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1$\begingroup$ @ray OK, that is certainly more subtle. Perhaps you'd like to edit your question to impose this extra restriction? $\endgroup$– Yemon ChoiCommented Oct 19, 2019 at 15:38
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1$\begingroup$ @Yemon Choi sure; just added this restriction. $\endgroup$– rayCommented Oct 19, 2019 at 16:08
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Consider the function $$ f(z)=\exp\left(-\frac{1}{(1-z)^{2}}\right), $$ which is a classical example of a function analytic in the unit disk $\mathbb{D}$, with radial limits everywhere on the unit circle $\mathbb{T}$, and with no non-tangential limit at 1 when the angle of approach with respect to the real axis is larger than $\pi/4$. With $1-z=re^{i\theta}$, its real part is $$ \operatorname{Re}(f)(z)=e^{-r^{-2}\cos(2\theta)}\cos(r^{-2}\sin(2\theta)), $$ whose limits on $\mathbb{T}$ have the same properties as above. Then, the real continuous function $$\arctan(\operatorname{Re}(f))$$ has the same properties as $f$, but it is also bounded.
