# 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.

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.)

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.