Does there exist an analytic function $f$ such satisfying the following three conditions? 
Does there exist an analytic function $f:\{z\in \mathbb C:|z|<1\}\to \{z\in \mathbb C:|z|<1\} $ such that, $f(0)=1/2$  , $f(1/2)=1/3$ , $f(1/3)=1/4$ ?

I tried through the Schwarz-Pick lemma & I found that Schwarz-Pick lemma is satisfied for pairwise two conditions. So I can't say that there does not exists such a function. Also I can not create a function satisfying the above three conditions.
How ,  I can test it whether the function exists or not ?
If exists give an example of such a function & please show how you create it ?
Update :Aug 20,2015
From Nivanlinna Pick Interpolation (suggested by João Ramos) I got the pick matrix as
$$\left[\begin{matrix}3/4&5/6&7/8\\5/6&32/27&11/10\\7/8&11/10&135/128\end{matrix}\right].$$
I found that one eigen value is negative.
So, the matrix is NOT non-positive definite. So we can conclude that such an analytic function does not exist satisfying the given three conditions.

But finding the matrix whether it is non-positive definite or NOT (in this cases) is too much laborious. So I want another approach to solve the problem...

 A: It's a Theorem from Pick that the interpolation problem 
$$ f(z_j) = w_j $$
Has a solution when $f$ is analytic on the unit disk and Bounded by one if and only if 
$$ Q(t_1,...,t_n) =\sum_{l,j} \frac{1-w_j\bar{w_l}}{1-z_j\bar{z_l}} t_j \bar{t_l} $$
Is a nonnegative quadratic form. Try to use this, as at least all the points are real. 
Alternatively, you can try to use functions of the form $\prod \frac{z-v_i}{1-\bar{v_i}z}$, as we generally know how to deal with Möbius Transformations.
A: To determine whether the Pick matrix is positive semi-definite, you don't have to find the eigenvalues.
The matrix is hermitean, so it's enough to check whether all principal minors are non-negative. I doubt there is a quicker method (in general).
A: (This is not a completely independent approach; in the deep it is same as Schwarz-Pick.)
You can remove the constraints one by one by the following trick. Suppose that we need to determine whether there is a function $f(z):D\to D$ ($D$ is the unit disk) satisfying some constraints $f(a_k)=b_k$ ($k=1,2\ldots,n$). 
(i) Of course, if $|b_k|>1$ for any $k$, then $f_1$ does not exist. If $|b_k|=1$ for some $k$ then $f$ must be constant.
Otherwise take a composition with $\dfrac{z-b_1}{1-\bar{b_1}z}$, so let $f_1(z)=\dfrac{f(z)-b_1}{1-\bar{b_1}f(z)}$. The function $f_1(z)$ must satisfy $f_1(a_k)=\dfrac{b_k-b_1}{1-\bar{b_1}b_k}$. Conversely, if such an $f_1$ exists, we can take $f(z)=\dfrac{f_1(z)+b_1}{1+\bar{b_1}f_1(z)}$. So $f$ exists if and only if $f_1$ exists.
(ii) The function $f_1$ has a root at $a_1$, so we can divide it by the Blascke-factor $\dfrac{z-a_1}{1-\bar{a_1}z}$ and define $f_2(z) = f_1(z) \cdot \dfrac{1-\bar{a_1}z}{z-a_1}$. The singularity at $a_1$ has been removed, so $f_2$ is analytic. Since the Blaske-factor has (close to) unit length at (close to) the boundary, $f_2$ also is a $D\to D$ function or a unit constant. Again, we can reverse this step, from $f_2$ we can construct $f_1$ by multiplying by $\dfrac{z-a_1}{1-\bar{a_1}z}$. Hence, $f_1$ exists if and only if $f_2$ exists.
Notice that we do not have any constraint on $f_2(a_1)$, so we have only $n-1$ constraints on $f_2$.
Repeating the steps (i) and (ii), we either find a constraint where the value must be greater than (or equal) $1$ -- that proves that $f$ does not exist or specifies the unit constant --, or we can remove all constraints, so at the end we just need the existence of a single $D\to D$ function without any constraint, which constructs all solutions of the question.

Concerning the concrete question, 
\begin{align*}
&
\text{there exists $f(z)$ with $f(0)=\tfrac12$, $f(\tfrac12)=\tfrac13$ and $f(\tfrac13)=\tfrac14$}
& \\
\Longleftrightarrow\quad &
\text{there exists $f_1(z)$ with $f_1(0)=0$, $f_1(\tfrac12)=\tfrac{-1}5$ and $f_1(\tfrac13)=\tfrac{-2}7$}
& \qquad (f_1(z)=\tfrac{f(z)-\frac12}{1-\frac12f(z)}) \\
\Longleftrightarrow\quad &
\text{there exists $f_2(z)$ with $f_2(\tfrac12)=-\tfrac{-2}5$ and $f_2(\tfrac13)=\tfrac{-6}7$}
& \qquad (f_2(z)=f_1(z) \cdot \tfrac1z) \\
\Longleftrightarrow\quad &
\text{there exists $f_3(z)$ with $f_3(\tfrac12)=0$ and $f_3(\tfrac13)=\tfrac{-16}{23}$}
& \qquad (f_3(z)=\tfrac{f_2(z)+\frac25}{1+\frac25f_2(z)}) \\
\Longleftrightarrow\quad &
\text{there exists $f_4(z)$ with $f_4(\tfrac13)=\tfrac{80}{23}$}
& \qquad (f_4(z)=f_3(z) \cdot \tfrac{1-\frac12z}{z-\frac12}) \\
\Longleftrightarrow\quad & \text{false} & 
\end{align*}
