Stitching two analytic functions? Let $f$ be an analytic function on the open unit disc and let $g$ be an analytic function on the complement of its closure.
Further assume that the two functions have a the same continuous limit on the common boundary of their domain.
Is it possible to build an entire function that settles with $f$ and $g$.
The question is a it vague I'm afraid...
 A: By the assumption

Further assume that the two functions have a the same continuous limit on the common boundary of their domain

the function
$$h(z) = \begin{cases} f(z) &, \lvert z\rvert \leqslant 1 \\ g(z) &, \lvert z\rvert \geqslant 1\end{cases}$$
is well-defined, and continuous on all of $\mathbb{C}$. Furthermore, it is holomorphic on $\mathbb{C}\setminus \partial \mathbb{D}$.
For a point $z_0 \in \partial\mathbb{D}$, consider the disk $U := D_{1/2}(z_0)$ and use Morera's theorem to find that $h$ is holomorphic on $U$: Let $\Delta \subset U$ be a closed triangle, and $A = \Delta \cap \overline{\mathbb{D}}$, $B = \Delta \setminus \mathbb{D}$. Then
$$\int_{\partial\Delta} h(z)\,dz = \int_{\partial A} h(z)\,dz + \int_{\partial B} h(z)\,dz$$
since the integrals over $\Delta\cap \partial\mathbb{D}$ cancel due to the opposite orientation this arc has in $\partial A$ and $\partial B$ [or $\Delta \cap \partial\mathbb{D}$ consists of one or two isolated points or is empty, in which cases that part of the integrals is $0$ trivially]. For $\varepsilon > 0$ let $A_\varepsilon = \Delta \cap \{ z : \lvert z\rvert \leqslant 1-\varepsilon\}$ and $B_\varepsilon = \Delta \cap \{ z : \lvert z\rvert \geqslant 1+\varepsilon\}$. By Cauchy's integral theorem, we have
$$\int_{\partial A_\varepsilon} h(z)\,dz = 0 = \int_{\partial B_\varepsilon} h(z)\,dz.$$
By the uniform continuity of $h$ in $U$, we have
$$\int_{\partial A} h(z)\,dz = \lim_{\varepsilon \searrow 0} \int_{\partial A_\varepsilon} h(z)\,dz = 0$$
and
$$\int_{\partial B} h(z)\,dz = \lim_{\varepsilon \searrow 0} \int_{\partial B_\varepsilon} h(z)\,dz = 0,$$
whence Morera's theorem tells us that $h$ is holomorphic on $U$.
Hence $h$ is entire.
A: Joining $f$ and $g_1$ would give you an entire function, which by the identity principle would have to agree with any other joining of $f$ with $g_2$ say, implying $g_1=g_2$. Thus if it is possible for $f$, it can only be done with one specific $g$.
This implies that it can be done if and only if $f$ is the restriction of some entire function to the unit disk $\mathbb{D}$, and $g$ is $f$ restricted to $\mathbb{C}\backslash\mathbb{D}$.
EDIT: I forgot to take into account behaviour on the boundary $\partial\mathbb{D}$. However as in Daniel's answer this follows from Morea's theorem and continuity at the boundary of $f$ and $g$.
A: Since both functions are continuous along the boundary (in particular, having no poles), they can be analytically continued. For example, let $p$ be a point on the unit circle, and let $U$ be a neighborhood of $p$ small enough so that both $f(z), g(z)$ can be analytically continued.
Then $f(z) - g(z) = 0$ along the arc of the unit circle contained in $U$, so $f(z) - g(z) = 0$ everywhere on $U$. In particular, the analytic continuation of $g(z)$ into $U$ just agrees with $f(z)$. This holds for arbitrary $p$.
So, yes -- the "join" of the two functions will be entire.
EDIT: This answer is not correct (see comments).
