Extending a bounded holomorphic function past its boundary Suppose I have a bounded holomorphic function on the unit disc, centred at the origin. Can I always extend this beyond the origin to say a disc of radius $1 + \epsilon$ for some $\epsilon > 0$? My guess is that this should be possible since we can take a Taylor expansion about 0, but I can't show that there are not going to be poles/ a limit point of poles at the boundary.
 A: No, this is in general not possible. Take e.g.,
$$
f(z) = \sum_{n=1}^\infty \frac{z^{n!}}{n!}
$$
This series converges absolutely and uniformly for $|z| \le 1$, with 
$$
|f(z)| \le \sum_{n=1}^\infty \frac{1}{n!} = e-1
$$
but the derivative
$$
f'(z) = \sum_{n=1}^\infty z^{n!-1}
$$
does not extend holomorphically to any domain strictly containing the open unit disk. (The radial limit at any root of unity is infinite.)
A: As Lukas Geyer already pointed out, the answer is generally no. Let $f\in L^{1}(\mathbb{T})$, and consider the function $\tilde{f}$ defined on the open unit disk $D$ by
$$\tilde{f}(z)=\dfrac{1}{2\pi i}\int_{\partial D}\dfrac{f(w)}{w-z}\mathrm{d}w,\qquad z\in D$$
For any compact set $K$ contained $D$,
$$\sup_{{z\in K}\atop{w\in\partial D}}\dfrac{1}{\left|w-z\right|}\leq M<\infty,$$
where $M$ may depend on the set $K$. By dominated convergence, $\tilde{f}$ is continuous is on $D$. For any triangle $T$ contained in $D$,
$$\int_{T}\left(\int_{\partial D}\left|\dfrac{f(w)}{w-z}\right|\left|\mathrm{d}w\right|\right)\left|\mathrm{d}z\right|<\infty,$$
whence by Fubini's theorem, we may change the order of integration to obtain
$$\int_{T}\tilde{f}(z)\mathrm{d}z=\dfrac{1}{2\pi i}\int_{\partial D}f(w)\left(\int_{T}\dfrac{1}{w-z}\mathrm{d}z\right)\mathrm{d}w=0,$$
since the inner integral vanishes because $w$ is not contained in the region enclosed by $T$. Morera's theorem then tells us that $\tilde{f}$ is analytic on the unit disk $D$.
Now suppose $f$ to be a real-valued bounded measurable function, and suppose $\tilde{f}$ has an analytic continuation $g$ on some larger open disk $D(0,1+\epsilon)$. Then by the Poisson integral formula,
$$\text{Re}(g)(z)=\text{Re}(\tilde{f})(z)=f\ast P_{r}(\theta),\qquad\forall z=re^{i\theta}\in D$$
Since $\left\{P_{r}\right\}_{0\leq r<1}$ forms an approximate identity and $f\in L^{1}$, there is a subsequence $\left\{r_{k}\right\}$ such that $P_{r_{k}}\ast f\rightarrow f$ a.e (actually, $P_{r}\ast f\rightarrow f$ a.e., but we don't need this). We conclude that
$$\text{Re}(g)(e^{i\theta})=\lim_{k\rightarrow\infty}f\ast P_{r_{k}}(\theta)=f(e^{i\theta}), \qquad \text{a.e.} \ \theta\in [0,2\pi)$$
In other words, $f$ is almost everywhere equal to a continuous function on the unit circle. If $f$ is appropriately discontinuous, for example $f(z)=\chi_{[0,\pi)}(\arg(z))$, where we take the principal branch, then we arrive at a contradiction.
