Example of continuous function that is analytic on the interior but cannot be analytically continued? I am looking for an example of a function $f$ that is 1) continuous on the closed unit disk, 2) analytic in the interior and 3) cannot be extended analytically to any larger set. A concrete example would be the best but just a proof that some exist would also be nice. (In fact I am not sure they do.)
I know of examples of analytic functions that cannot be extended from the unit disk. Take a lacuanary power series for example with radius of convergence 1. But I am not sure if any of them define a continuous function on the closed unit disk. 
 A: One can invoke Carathéodory's theorem. 

If $U$ is a simply connected open
  subset of the complex plane with a
  Jordan curve as boundary then the
  Riemann map $f : U \to \mathbb D$
  extends continuosly to the boundary
  and the extension is a homeomorphism
  $\partial U \to S^1$ at the boundary.

To obtain the sougth function, it suffices to consider a simply connected open set $U\subset \mathbb C$ having  a nowhere analytic  Jordan curve as boundary and take the inverse of the Riemann map of $U$.  
A: I found some neat stuff in Remmert's Classical topics in complex function theory.
Fabry's gap theorem gives a way to construct many examples including some already mentioned.  Stated for the unit disk, it says:

If $m_1,m_2,\ldots$ is a sequence of positive integers such that $\displaystyle{\lim_{n\to\infty}}\frac{m_n}{n}=\infty$ and if $\displaystyle{f(z)=\sum_{n=1}^\infty a_nz^{m_n}}$ has radius of convergence 1, then the unit disk is the domain of holomorphy of $f$.

For example, if $p_n$ is the $n^{th}$ prime, then $$f(z)=\sum_{n=1}^\infty \frac{z^{p_n}}{n^2}$$ converges uniformly on the closed disk and is therefore continuous.  It is not analytically extendable to any larger set because it satisfies the hypotheses of Fabry's theorem.  

An interesting result that yields many such functions in a nonconstructive way is a theorem of Fatou-Hurwitz-Pólya:

If $\displaystyle{f(z)=\sum_{n=0}^\infty a_n z^n}$ has radius of convergence 1, then the set of functions $$f_\epsilon(z)=\sum_{n=0}^\infty \epsilon_na_nz^n$$ for $\epsilon_n\in\{\pm1\}$ whose domain of holomorphy is the unit disk has cardinality $2^{\aleph_0}$.

Hausdorff showed further that if $\displaystyle{\lim_{n\to\infty} |a_n|^{1/n}}$ exists (and equals 1) then the set of such functions whose domain of holomorphy is not the unit disk is at most countable.  This applies in particular to the function $\displaystyle{f(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}}$, which therefore yields examples by changing the signs of the coefficients in all but countably many ways.

One more, this time an explicit example from Remmert: The series $$f(z)=1+2z+\sum_{n=1}^\infty\frac{z^{2^n}}{2^{n^2}}$$ is one-to-one and has real derivatives of all orders on the closed disk, and has the open disk as domain of holomorphy.  
Reference: Remmert's Classical topics in complex function theory, pages 252-258.  (Fatou-Hurwitz-Pólya is stated on a page without preview.)
A: I suggest this function: $$f(z)=\sum_{n=1}^\infty \frac{z^{n!}}{n^2}.$$ It converges uniformly on the closed unit disk, and the derivatives blow up as you approach any root of unity radially.
A: Let $f(z) = \sum z^n/n^2$, which is continuous  and bounded on the closed unit disc but not analytic near $1$. Then consider 
$$\sum f(z^n)/n^2.$$
This should have a singularity at every root of unity; and should be analytic in the interior because it is uniformly convergent.
A: Here is a very concrete example:
$
g(z) = \sum_{n=0}^{\infty}\frac{z^{2^n + 1}}{2^n + 1}.
$
The power series converges uniformly to a continuous function on the closed unit disk. Differentiating we obtain $g'(z) = f(z)$ with
$
f(z) = \sum_{n=0}^{\infty}z^{2^n}.
$
This is the standard example of a function with a natural boundary. Clearly $f(x) \rightarrow +\infty$ as $x \rightarrow 1^{-}$ on the real axis. The functional equation 
$
f(z) = z + f(z^2)
$
shows that $f(x) \rightarrow + \infty$ as $x \rightarrow (-1)^{+}$ on the real axis, then 
$|f(z)| \rightarrow \infty$ as $z$ tends radially to ${\pm}i$, and so on, so that $|f(z)|$ tends to $\infty$ as $z$ tends radially to any root of unity of order $2^m$. Hence $f(z)$ has a dense set of singularities on the unit circle, and so does $g(z)$, thus $g(z)$ has the unit circle as natural boundary.
